diff --git "a/IMO/segmented/en-IMO2019SL.jsonl" "b/IMO/segmented/en-IMO2019SL.jsonl" deleted file mode 100644--- "a/IMO/segmented/en-IMO2019SL.jsonl" +++ /dev/null @@ -1,136 +0,0 @@ -{"year": "2019", "tier": "T0", "problem_label": "A1", "problem_type": "Algebra", "exam": "IMO", "problem": "Let $\\mathbb{Z}$ be the set of integers. Determine all functions $f: \\mathbb{Z} \\rightarrow \\mathbb{Z}$ such that, for all integers $a$ and $b$, $$ f(2 a)+2 f(b)=f(f(a+b)) $$ (South Africa)", "solution": "Substituting $a=0, b=n+1$ gives $f(f(n+1))=f(0)+2 f(n+1)$. Substituting $a=1, b=n$ gives $f(f(n+1))=f(2)+2 f(n)$. In particular, $f(0)+2 f(n+1)=f(2)+2 f(n)$, and so $f(n+1)-f(n)=\\frac{1}{2}(f(2)-f(0))$. Thus $f(n+1)-f(n)$ must be constant. Since $f$ is defined only on $\\mathbb{Z}$, this tells us that $f$ must be a linear function; write $f(n)=M n+K$ for arbitrary constants $M$ and $K$, and we need only determine which choices of $M$ and $K$ work. Now, (1) becomes $$ 2 M a+K+2(M b+K)=M(M(a+b)+K)+K $$ which we may rearrange to form $$ (M-2)(M(a+b)+K)=0 $$ Thus, either $M=2$, or $M(a+b)+K=0$ for all values of $a+b$. In particular, the only possible solutions are $f(n)=0$ and $f(n)=2 n+K$ for any constant $K \\in \\mathbb{Z}$, and these are easily seen to work.", "metadata": {"resource_path": "IMO/segmented/en-IMO2019SL.jsonl"}} -{"year": "2019", "tier": "T0", "problem_label": "A1", "problem_type": "Algebra", "exam": "IMO", "problem": "Let $\\mathbb{Z}$ be the set of integers. Determine all functions $f: \\mathbb{Z} \\rightarrow \\mathbb{Z}$ such that, for all integers $a$ and $b$, $$ f(2 a)+2 f(b)=f(f(a+b)) $$ (South Africa)", "solution": "Let $K=f(0)$. First, put $a=0$ in (1); this gives $$ f(f(b))=2 f(b)+K $$ for all $b \\in \\mathbb{Z}$. Now put $b=0$ in (1); this gives $$ f(2 a)+2 K=f(f(a))=2 f(a)+K $$ where the second equality follows from (2). Consequently, $$ f(2 a)=2 f(a)-K $$ for all $a \\in \\mathbb{Z}$. Substituting (2) and (3) into (1), we obtain $$ \\begin{aligned} f(2 a)+2 f(b) & =f(f(a+b)) \\\\ 2 f(a)-K+2 f(b) & =2 f(a+b)+K \\\\ f(a)+f(b) & =f(a+b)+K \\end{aligned} $$ Thus, if we set $g(n)=f(n)-K$ we see that $g$ satisfies the Cauchy equation $g(a+b)=$ $g(a)+g(b)$. The solution to the Cauchy equation over $\\mathbb{Z}$ is well-known; indeed, it may be proven by an easy induction that $g(n)=M n$ for each $n \\in \\mathbb{Z}$, where $M=g(1)$ is a constant. Therefore, $f(n)=M n+K$, and we may proceed as in Solution 1 . Comment 1. Instead of deriving (3) by substituting $b=0$ into (1), we could instead have observed that the right hand side of (1) is symmetric in $a$ and $b$, and thus $$ f(2 a)+2 f(b)=f(2 b)+2 f(a) $$ Thus, $f(2 a)-2 f(a)=f(2 b)-2 f(b)$ for any $a, b \\in \\mathbb{Z}$, and in particular $f(2 a)-2 f(a)$ is constant. Setting $a=0$ shows that this constant is equal to $-K$, and so we obtain (3). Comment 2. Some solutions initially prove that $f(f(n))$ is linear (sometimes via proving that $f(f(n))-3 K$ satisfies the Cauchy equation). However, one can immediately prove that $f$ is linear by substituting something of the form $f(f(n))=M^{\\prime} n+K^{\\prime}$ into (2).", "metadata": {"resource_path": "IMO/segmented/en-IMO2019SL.jsonl"}} -{"year": "2019", "tier": "T0", "problem_label": "A2", "problem_type": "Algebra", "exam": "IMO", "problem": "Let $u_{1}, u_{2}, \\ldots, u_{2019}$ be real numbers satisfying $$ u_{1}+u_{2}+\\cdots+u_{2019}=0 \\quad \\text { and } \\quad u_{1}^{2}+u_{2}^{2}+\\cdots+u_{2019}^{2}=1 . $$ Let $a=\\min \\left(u_{1}, u_{2}, \\ldots, u_{2019}\\right)$ and $b=\\max \\left(u_{1}, u_{2}, \\ldots, u_{2019}\\right)$. Prove that $$ a b \\leqslant-\\frac{1}{2019} $$ (Germany)", "solution": "Notice first that $b>0$ and $a<0$. Indeed, since $\\sum_{i=1}^{2019} u_{i}^{2}=1$, the variables $u_{i}$ cannot be all zero, and, since $\\sum_{i=1}^{2019} u_{i}=0$, the nonzero elements cannot be all positive or all negative. Let $P=\\left\\{i: u_{i}>0\\right\\}$ and $N=\\left\\{i: u_{i} \\leqslant 0\\right\\}$ be the indices of positive and nonpositive elements in the sequence, and let $p=|P|$ and $n=|N|$ be the sizes of these sets; then $p+n=2019$. By the condition $\\sum_{i=1}^{2019} u_{i}=0$ we have $0=\\sum_{i=1}^{2019} u_{i}=\\sum_{i \\in P} u_{i}-\\sum_{i \\in N}\\left|u_{i}\\right|$, so $$ \\sum_{i \\in P} u_{i}=\\sum_{i \\in N}\\left|u_{i}\\right| . $$ After this preparation, estimate the sum of squares of the positive and nonpositive elements as follows: $$ \\begin{aligned} & \\sum_{i \\in P} u_{i}^{2} \\leqslant \\sum_{i \\in P} b u_{i}=b \\sum_{i \\in P} u_{i}=b \\sum_{i \\in N}\\left|u_{i}\\right| \\leqslant b \\sum_{i \\in N}|a|=-n a b ; \\\\ & \\sum_{i \\in N} u_{i}^{2} \\leqslant \\sum_{i \\in N}|a| \\cdot\\left|u_{i}\\right|=|a| \\sum_{i \\in N}\\left|u_{i}\\right|=|a| \\sum_{i \\in P} u_{i} \\leqslant|a| \\sum_{i \\in P} b=-p a b . \\end{aligned} $$ The sum of these estimates is $$ 1=\\sum_{i=1}^{2019} u_{i}^{2}=\\sum_{i \\in P} u_{i}^{2}+\\sum_{i \\in N} u_{i}^{2} \\leqslant-(p+n) a b=-2019 a b ; $$ that proves $a b \\leqslant \\frac{-1}{2019}$. Comment 1. After observing $\\sum_{i \\in P} u_{i}^{2} \\leqslant b \\sum_{i \\in P} u_{i}$ and $\\sum_{i \\in N} u_{i}^{2} \\leqslant|a| \\sum_{i \\in P}\\left|u_{i}\\right|$, instead of $(2,3)$ an alternative continuation is $$ |a b| \\geqslant \\frac{\\sum_{i \\in P} u_{i}^{2}}{\\sum_{i \\in P} u_{i}} \\cdot \\frac{\\sum_{i \\in N} u_{i}^{2}}{\\sum_{i \\in N}\\left|u_{i}\\right|}=\\frac{\\sum_{i \\in P} u_{i}^{2}}{\\left(\\sum_{i \\in P} u_{i}\\right)^{2}} \\sum_{i \\in N} u_{i}^{2} \\geqslant \\frac{1}{p} \\sum_{i \\in N} u_{i}^{2} $$ (by the AM-QM or the Cauchy-Schwarz inequality) and similarly $|a b| \\geqslant \\frac{1}{n} \\sum_{i \\in P} u_{i}^{2}$.", "metadata": {"resource_path": "IMO/segmented/en-IMO2019SL.jsonl"}} -{"year": "2019", "tier": "T0", "problem_label": "A2", "problem_type": "Algebra", "exam": "IMO", "problem": "Let $u_{1}, u_{2}, \\ldots, u_{2019}$ be real numbers satisfying $$ u_{1}+u_{2}+\\cdots+u_{2019}=0 \\quad \\text { and } \\quad u_{1}^{2}+u_{2}^{2}+\\cdots+u_{2019}^{2}=1 . $$ Let $a=\\min \\left(u_{1}, u_{2}, \\ldots, u_{2019}\\right)$ and $b=\\max \\left(u_{1}, u_{2}, \\ldots, u_{2019}\\right)$. Prove that $$ a b \\leqslant-\\frac{1}{2019} $$ (Germany)", "solution": "As in the previous solution we conclude that $a<0$ and $b>0$. For every index $i$, the number $u_{i}$ is a convex combination of $a$ and $b$, so $$ u_{i}=x_{i} a+y_{i} b \\quad \\text { with some weights } 0 \\leqslant x_{i}, y_{i} \\leqslant 1, \\text { with } x_{i}+y_{i}=1 \\text {. } $$ Let $X=\\sum_{i=1}^{2019} x_{i}$ and $Y=\\sum_{i=1}^{2019} y_{i}$. From $0=\\sum_{i=1}^{2019} u_{i}=\\sum_{i=1}^{2019}\\left(x_{i} a+y_{i} b\\right)=-|a| X+b Y$, we get $$ |a| X=b Y $$ From $\\sum_{i=1}^{2019}\\left(x_{i}+y_{i}\\right)=2019$ we have $$ X+Y=2019 $$ The system of linear equations $(4,5)$ has a unique solution: $$ X=\\frac{2019 b}{|a|+b}, \\quad Y=\\frac{2019|a|}{|a|+b} $$ Now apply the following estimate to every $u_{i}^{2}$ in their sum: $$ u_{i}^{2}=x_{i}^{2} a^{2}+2 x_{i} y_{i} a b+y_{i}^{2} b^{2} \\leqslant x_{i} a^{2}+y_{i} b^{2} $$ we obtain that $$ 1=\\sum_{i=1}^{2019} u_{i}^{2} \\leqslant \\sum_{i=1}^{2019}\\left(x_{i} a^{2}+y_{i} b^{2}\\right)=X a^{2}+Y b^{2}=\\frac{2019 b}{|a|+b}|a|^{2}+\\frac{2019|a|}{|a|+b} b^{2}=2019|a| b=-2019 a b . $$ Hence, $a b \\leqslant \\frac{-1}{2019}$. Comment 2. The idea behind Solution 2 is the following thought. Suppose we fix $a<0$ and $b>0$, fix $\\sum u_{i}=0$ and vary the $u_{i}$ to achieve the maximum value of $\\sum u_{i}^{2}$. Considering varying any two of the $u_{i}$ while preserving their sum: the maximum value of $\\sum u_{i}^{2}$ is achieved when those two are as far apart as possible, so all but at most one of the $u_{i}$ are equal to $a$ or $b$. Considering a weighted version of the problem, we see the maximum (with fractional numbers of $u_{i}$ having each value) is achieved when $\\frac{2019 b}{|a|+b}$ of them are $a$ and $\\frac{2019|a|}{|a|+b}$ are $b$. In fact, this happens in the solution: the number $u_{i}$ is replaced by $x_{i}$ copies of $a$ and $y_{i}$ copies of $b$.", "metadata": {"resource_path": "IMO/segmented/en-IMO2019SL.jsonl"}} -{"year": "2019", "tier": "T0", "problem_label": "A3", "problem_type": "Algebra", "exam": "IMO", "problem": "Let $n \\geqslant 3$ be a positive integer and let $\\left(a_{1}, a_{2}, \\ldots, a_{n}\\right)$ be a strictly increasing sequence of $n$ positive real numbers with sum equal to 2 . Let $X$ be a subset of $\\{1,2, \\ldots, n\\}$ such that the value of $$ \\left|1-\\sum_{i \\in X} a_{i}\\right| $$ is minimised. Prove that there exists a strictly increasing sequence of $n$ positive real numbers $\\left(b_{1}, b_{2}, \\ldots, b_{n}\\right)$ with sum equal to 2 such that $$ \\sum_{i \\in X} b_{i}=1 $$ (New Zealand)", "solution": "Without loss of generality, assume $\\sum_{i \\in X} a_{i} \\leqslant 1$, and we may assume strict inequality as otherwise $b_{i}=a_{i}$ works. Also, $X$ clearly cannot be empty. If $n \\in X$, add $\\Delta$ to $a_{n}$, producing a sequence of $c_{i}$ with $\\sum_{i \\in X} c_{i}=\\sum_{i \\in X^{c}} c_{i}$, and then scale as described above to make the sum equal to 2 . Otherwise, there is some $k$ with $k \\in X$ and $k+1 \\in X^{c}$. Let $\\delta=a_{k+1}-a_{k}$. - If $\\delta>\\Delta$, add $\\Delta$ to $a_{k}$ and then scale. - If $\\delta<\\Delta$, then considering $X \\cup\\{k+1\\} \\backslash\\{k\\}$ contradicts $X$ being $\\left(a_{i}\\right)$-minimising. - If $\\delta=\\Delta$, choose any $j \\neq k, k+1$ (possible since $n \\geqslant 3$ ), and any $\\epsilon$ less than the least of $a_{1}$ and all the differences $a_{i+1}-a_{i}$. If $j \\in X$ then add $\\Delta-\\epsilon$ to $a_{k}$ and $\\epsilon$ to $a_{j}$, then scale; otherwise, add $\\Delta$ to $a_{k}$ and $\\epsilon / 2$ to $a_{k+1}$, and subtract $\\epsilon / 2$ from $a_{j}$, then scale.", "metadata": {"resource_path": "IMO/segmented/en-IMO2019SL.jsonl"}} -{"year": "2019", "tier": "T0", "problem_label": "A3", "problem_type": "Algebra", "exam": "IMO", "problem": "Let $n \\geqslant 3$ be a positive integer and let $\\left(a_{1}, a_{2}, \\ldots, a_{n}\\right)$ be a strictly increasing sequence of $n$ positive real numbers with sum equal to 2 . Let $X$ be a subset of $\\{1,2, \\ldots, n\\}$ such that the value of $$ \\left|1-\\sum_{i \\in X} a_{i}\\right| $$ is minimised. Prove that there exists a strictly increasing sequence of $n$ positive real numbers $\\left(b_{1}, b_{2}, \\ldots, b_{n}\\right)$ with sum equal to 2 such that $$ \\sum_{i \\in X} b_{i}=1 $$ (New Zealand)", "solution": "This is similar to Suppose there exists $1 \\leqslant j \\leqslant n-1$ such that $j \\in X$ but $j+1 \\in X^{c}$. Then $a_{j+1}-a_{j} \\geqslant \\Delta$, because otherwise considering $X \\cup\\{j+1\\} \\backslash\\{j\\}$ contradicts $X$ being $\\left(a_{i}\\right)$-minimising. If $a_{j+1}-a_{j}>\\Delta$, put $$ b_{i}= \\begin{cases}a_{j}+\\Delta / 2, & \\text { if } i=j \\\\ a_{j+1}-\\Delta / 2, & \\text { if } i=j+1 \\\\ a_{i}, & \\text { otherwise }\\end{cases} $$ If $a_{j+1}-a_{j}=\\Delta$, choose any $\\epsilon$ less than the least of $\\Delta / 2, a_{1}$ and all the differences $a_{i+1}-a_{i}$. If $|X| \\geqslant 2$, choose $k \\in X$ with $k \\neq j$, and put $$ b_{i}= \\begin{cases}a_{j}+\\Delta / 2-\\epsilon, & \\text { if } i=j \\\\ a_{j+1}-\\Delta / 2, & \\text { if } i=j+1 \\\\ a_{k}+\\epsilon, & \\text { if } i=k \\\\ a_{i}, & \\text { otherwise }\\end{cases} $$ Otherwise, $\\left|X^{c}\\right| \\geqslant 2$, so choose $k \\in X^{c}$ with $k \\neq j+1$, and put $$ b_{i}= \\begin{cases}a_{j}+\\Delta / 2, & \\text { if } i=j \\\\ a_{j+1}-\\Delta / 2+\\epsilon, & \\text { if } i=j+1 \\\\ a_{k}-\\epsilon, & \\text { if } i=k \\\\ a_{i}, & \\text { otherwise }\\end{cases} $$ If there is no $1 \\leqslant j \\leqslant n$ such that $j \\in X$ but $j+1 \\in X^{c}$, there must be some $1\\Delta$, as otherwise considering $X \\cup\\{1\\}$ contradicts $X$ being $\\left(a_{i}\\right)$-minimising. Now put $$ b_{i}= \\begin{cases}a_{1}-\\Delta / 2, & \\text { if } i=1 \\\\ a_{n}+\\Delta / 2, & \\text { if } i=n \\\\ a_{i}, & \\text { otherwise }\\end{cases} $$", "metadata": {"resource_path": "IMO/segmented/en-IMO2019SL.jsonl"}} -{"year": "2019", "tier": "T0", "problem_label": "A3", "problem_type": "Algebra", "exam": "IMO", "problem": "Let $n \\geqslant 3$ be a positive integer and let $\\left(a_{1}, a_{2}, \\ldots, a_{n}\\right)$ be a strictly increasing sequence of $n$ positive real numbers with sum equal to 2 . Let $X$ be a subset of $\\{1,2, \\ldots, n\\}$ such that the value of $$ \\left|1-\\sum_{i \\in X} a_{i}\\right| $$ is minimised. Prove that there exists a strictly increasing sequence of $n$ positive real numbers $\\left(b_{1}, b_{2}, \\ldots, b_{n}\\right)$ with sum equal to 2 such that $$ \\sum_{i \\in X} b_{i}=1 $$ (New Zealand)", "solution": "Without loss of generality, assume $\\sum_{i \\in X} a_{i} \\leqslant 1$, so $\\Delta \\geqslant 0$. If $\\Delta=0$ we can take $b_{i}=a_{i}$, so now assume that $\\Delta>0$. Suppose that there is some $k \\leqslant n$ such that $|X \\cap[k, n]|>\\left|X^{c} \\cap[k, n]\\right|$. If we choose the largest such $k$ then $|X \\cap[k, n]|-\\left|X^{c} \\cap[k, n]\\right|=1$. We can now find the required sequence $\\left(b_{i}\\right)$ by starting with $c_{i}=a_{i}$ for $i\\frac{n-k+1}{2}$ and $|X \\cap[\\ell, n]|<\\frac{n-\\ell+1}{2}$. We now construct our sequence $\\left(b_{i}\\right)$ using this claim. Let $k$ and $\\ell$ be the greatest values satisfying the claim, and without loss of generality suppose $k=n$ and $\\ell\\sum_{i \\in Y} a_{i}>1 $$ contradicting $X$ being $\\left(a_{i}\\right)$-minimising. Otherwise, we always have equality, meaning that $X=Y$. But now consider $Z=Y \\cup\\{n-1\\} \\backslash\\{n\\}$. Since $n \\geqslant 3$, we have $$ \\sum_{i \\in Y} a_{i}>\\sum_{i \\in Z} a_{i}>\\sum_{i \\in Y^{c}} a_{i}=2-\\sum_{i \\in Y} a_{i} $$ and so $Z$ contradicts $X$ being $\\left(a_{i}\\right)$-minimising.", "metadata": {"resource_path": "IMO/segmented/en-IMO2019SL.jsonl"}} -{"year": "2019", "tier": "T0", "problem_label": "A4", "problem_type": "Algebra", "exam": "IMO", "problem": "Let $n \\geqslant 2$ be a positive integer and $a_{1}, a_{2}, \\ldots, a_{n}$ be real numbers such that $$ a_{1}+a_{2}+\\cdots+a_{n}=0 $$ Define the set $A$ by $$ A=\\left\\{(i, j)\\left|1 \\leqslant i0 $$ Partition the indices into sets $P, Q, R$, and $S$ such that $$ \\begin{aligned} P & =\\left\\{i \\mid a_{i} \\leqslant-1\\right\\} & R & =\\left\\{i \\mid 01$ ). Therefore, $$ t_{+}+t_{-} \\leqslant \\frac{p^{2}+s^{2}}{2}+p q+r s+p r+p s+q s=\\frac{(p+q+r+s)^{2}}{2}-\\frac{(q+r)^{2}}{2}=-\\frac{(q+r)^{2}}{2} \\leqslant 0 $$ If $A$ is not empty and $p=s=0$, then there must exist $i \\in Q, j \\in R$ with $\\left|a_{i}-a_{j}\\right|>1$, and hence the earlier equality conditions cannot both occur. Comment. The RHS of the original inequality cannot be replaced with any constant $c<0$ (independent of $n$ ). Indeed, take $$ a_{1}=-\\frac{n}{n+2}, a_{2}=\\cdots=a_{n-1}=\\frac{1}{n+2}, a_{n}=\\frac{2}{n+2} . $$ Then $\\sum_{(i, j) \\in A} a_{i} a_{j}=-\\frac{2 n}{(n+2)^{2}}$, which converges to zero as $n \\rightarrow \\infty$. This page is intentionally left blank", "metadata": {"resource_path": "IMO/segmented/en-IMO2019SL.jsonl"}} -{"year": "2019", "tier": "T0", "problem_label": "A5", "problem_type": "Algebra", "exam": "IMO", "problem": "Let $x_{1}, x_{2}, \\ldots, x_{n}$ be different real numbers. Prove that $$ \\sum_{1 \\leqslant i \\leqslant n} \\prod_{j \\neq i} \\frac{1-x_{i} x_{j}}{x_{i}-x_{j}}= \\begin{cases}0, & \\text { if } n \\text { is even } \\\\ 1, & \\text { if } n \\text { is odd }\\end{cases} $$", "solution": null, "metadata": {"resource_path": "IMO/segmented/en-IMO2019SL.jsonl"}} -{"year": "2019", "tier": "T0", "problem_label": "A6", "problem_type": "Algebra", "exam": "IMO", "problem": "A polynomial $P(x, y, z)$ in three variables with real coefficients satisfies the identities $$ P(x, y, z)=P(x, y, x y-z)=P(x, z x-y, z)=P(y z-x, y, z) . $$ Prove that there exists a polynomial $F(t)$ in one variable such that $$ P(x, y, z)=F\\left(x^{2}+y^{2}+z^{2}-x y z\\right) . $$", "solution": "In the first two steps, we deal with any polynomial $P(x, y, z)$ satisfying $P(x, y, z)=$ $P(x, y, x y-z)$. Call such a polynomial weakly symmetric, and call a polynomial satisfying the full conditions in the problem symmetric. Step 1. We start with the description of weakly symmetric polynomials. We claim that they are exactly the polynomials in $x, y$, and $z(x y-z)$. Clearly, all such polynomials are weakly symmetric. For the converse statement, consider $P_{1}(x, y, z):=P\\left(x, y, z+\\frac{1}{2} x y\\right)$, which satisfies $P_{1}(x, y, z)=P_{1}(x, y,-z)$ and is therefore a polynomial in $x, y$, and $z^{2}$. This means that $P$ is a polynomial in $x, y$, and $\\left(z-\\frac{1}{2} x y\\right)^{2}=-z(x y-z)+\\frac{1}{4} x^{2} y^{2}$, and therefore a polynomial in $x, y$, and $z(x y-z)$. Step 2. Suppose that $P$ is weakly symmetric. Consider the monomials in $P(x, y, z)$ of highest total degree. Our aim is to show that in each such monomial $\\mu x^{a} y^{b} z^{c}$ we have $a, b \\geqslant c$. Consider the expansion $$ P(x, y, z)=\\sum_{i, j, k} \\mu_{i j k} x^{i} y^{j}(z(x y-z))^{k} $$ The maximal total degree of a summand in (1.1) is $m=\\max _{i, j, k: \\mu_{i j k} \\neq 0}(i+j+3 k)$. Now, for any $i, j, k$ satisfying $i+j+3 k=m$ the summand $\\mu_{i, j, k} x^{i} y^{j}(z(x y-z))^{k}$ has leading term of the form $\\mu x^{i+k} y^{j+k} z^{k}$. No other nonzero summand in (1.1) may have a term of this form in its expansion, hence this term does not cancel in the whole sum. Therefore, $\\operatorname{deg} P=m$, and the leading component of $P$ is exactly $$ \\sum_{i+j+3 k=m} \\mu_{i, j, k} x^{i+k} y^{j+k} z^{k} $$ and each summand in this sum satisfies the condition claimed above. Step 3. We now prove the problem statement by induction on $m=\\operatorname{deg} P$. For $m=0$ the claim is trivial. Consider now a symmetric polynomial $P$ with $\\operatorname{deg} P>0$. By Step 2, each of its monomials $\\mu x^{a} y^{b} z^{c}$ of the highest total degree satisfies $a, b \\geqslant c$. Applying other weak symmetries, we obtain $a, c \\geqslant b$ and $b, c \\geqslant a$; therefore, $P$ has a unique leading monomial of the form $\\mu(x y z)^{c}$. The polynomial $P_{0}(x, y, z)=P(x, y, z)-\\mu\\left(x y z-x^{2}-y^{2}-z^{2}\\right)^{c}$ has smaller total degree. Since $P_{0}$ is symmetric, it is representable as a polynomial function of $x y z-x^{2}-y^{2}-z^{2}$. Then $P$ is also of this form, completing the inductive step. Comment. We could alternatively carry out Step 1 by an induction on $n=\\operatorname{deg}_{z} P$, in a manner similar to Step 3. If $n=0$, the statement holds. Assume that $n>0$ and check the leading component of $P$ with respect to $z$ : $$ P(x, y, z)=Q_{n}(x, y) z^{n}+R(x, y, z) $$ where $\\operatorname{deg}_{z} Rv(x)$ and $v(x) \\geqslant v(y)$. Hence this $f$ satisfies the functional equation and 0 is an $f$-rare integer. Comment 3. In fact, if $v$ is an $f$-rare integer for an $f$ satisfying the functional equation, then its fibre $X_{v}=\\{v\\}$ must be a singleton. We may assume without loss of generality that $v=0$. We've already seen in Solution 1 that 0 is either the greatest or least element of $X_{0}$; replacing $f$ with the function $x \\mapsto-f(-x)$ if necessary, we may assume that 0 is the least element of $X_{0}$. We write $b$ for the largest element of $X_{0}$, supposing for contradiction that $b>0$, and write $N=(2 b)$ !. It now follows from (*) that we have $$ f(f(N b)+b)=f(f(0)+b)=f(b)=0 $$ from which we see that $f(N b)+b \\in X_{0} \\subseteq[0, b]$. It follows that $f(N b) \\in[-b, 0)$, since by construction $N b \\notin X_{v}$. Now it follows that $(f(N b)-0) \\cdot(f(N b)-b)$ is a divisor of $N$, so from ( $\\dagger$ ) we see that $f(N b)=f(0)=0$. This yields the desired contradiction.", "metadata": {"resource_path": "IMO/segmented/en-IMO2019SL.jsonl"}} -{"year": "2019", "tier": "T0", "problem_label": "C1", "problem_type": "Combinatorics", "exam": "IMO", "problem": "The infinite sequence $a_{0}, a_{1}, a_{2}, \\ldots$ of (not necessarily different) integers has the following properties: $0 \\leqslant a_{i} \\leqslant i$ for all integers $i \\geqslant 0$, and $$ \\binom{k}{a_{0}}+\\binom{k}{a_{1}}+\\cdots+\\binom{k}{a_{k}}=2^{k} $$ for all integers $k \\geqslant 0$. Prove that all integers $N \\geqslant 0$ occur in the sequence (that is, for all $N \\geqslant 0$, there exists $i \\geqslant 0$ with $a_{i}=N$ ). ## (Netherlands)", "solution": "We prove by induction on $k$ that every initial segment of the sequence, $a_{0}, a_{1}, \\ldots, a_{k}$, consists of the following elements (counted with multiplicity, and not necessarily in order), for some $\\ell \\geqslant 0$ with $2 \\ell \\leqslant k+1$ : $$ 0,1, \\ldots, \\ell-1, \\quad 0,1, \\ldots, k-\\ell $$ For $k=0$ we have $a_{0}=0$, which is of this form. Now suppose that for $k=m$ the elements $a_{0}, a_{1}, \\ldots, a_{m}$ are $0,0,1,1,2,2, \\ldots, \\ell-1, \\ell-1, \\ell, \\ell+1, \\ldots, m-\\ell-1, m-\\ell$ for some $\\ell$ with $0 \\leqslant 2 \\ell \\leqslant m+1$. It is given that $$ \\binom{m+1}{a_{0}}+\\binom{m+1}{a_{1}}+\\cdots+\\binom{m+1}{a_{m}}+\\binom{m+1}{a_{m+1}}=2^{m+1} $$ which becomes $$ \\begin{aligned} \\left(\\binom{m+1}{0}+\\binom{m+1}{1}\\right. & \\left.+\\cdots+\\binom{m+1}{\\ell-1}\\right) \\\\ & +\\left(\\binom{m+1}{0}+\\binom{m+1}{1}+\\cdots+\\binom{m+1}{m-\\ell}\\right)+\\binom{m+1}{a_{m+1}}=2^{m+1} \\end{aligned} $$ or, using $\\binom{m+1}{i}=\\binom{m+1}{m+1-i}$, that $$ \\begin{aligned} \\left(\\binom{m+1}{0}+\\binom{m+1}{1}\\right. & \\left.+\\cdots+\\binom{m+1}{\\ell-1}\\right) \\\\ & +\\left(\\binom{m+1}{m+1}+\\binom{m+1}{m}+\\cdots+\\binom{m+1}{\\ell+1}\\right)+\\binom{m+1}{a_{m+1}}=2^{m+1} \\end{aligned} $$ On the other hand, it is well known that $$ \\binom{m+1}{0}+\\binom{m+1}{1}+\\cdots+\\binom{m+1}{m+1}=2^{m+1} $$ and so, by subtracting, we get $$ \\binom{m+1}{a_{m+1}}=\\binom{m+1}{\\ell} $$ From this, using the fact that the binomial coefficients $\\binom{m+1}{i}$ are increasing for $i \\leqslant \\frac{m+1}{2}$ and decreasing for $i \\geqslant \\frac{m+1}{2}$, we conclude that either $a_{m+1}=\\ell$ or $a_{m+1}=m+1-\\ell$. In either case, $a_{0}, a_{1}, \\ldots, a_{m+1}$ is again of the claimed form, which concludes the induction. As a result of this description, any integer $N \\geqslant 0$ appears as a term of the sequence $a_{i}$ for some $0 \\leqslant i \\leqslant 2 N$.", "metadata": {"resource_path": "IMO/segmented/en-IMO2019SL.jsonl"}} -{"year": "2019", "tier": "T0", "problem_label": "C2", "problem_type": "Combinatorics", "exam": "IMO", "problem": "You are given a set of $n$ blocks, each weighing at least 1 ; their total weight is $2 n$. Prove that for every real number $r$ with $0 \\leqslant r \\leqslant 2 n-2$ you can choose a subset of the blocks whose total weight is at least $r$ but at most $r+2$. (Thailand)", "solution": "We prove the following more general statement by induction on $n$. Claim. Suppose that you have $n$ blocks, each of weight at least 1 , and of total weight $s \\leqslant 2 n$. Then for every $r$ with $-2 \\leqslant r \\leqslant s$, you can choose some of the blocks whose total weight is at least $r$ but at most $r+2$. Proof. The base case $n=1$ is trivial. To prove the inductive step, let $x$ be the largest block weight. Clearly, $x \\geqslant s / n$, so $s-x \\leqslant \\frac{n-1}{n} s \\leqslant 2(n-1)$. Hence, if we exclude a block of weight $x$, we can apply the inductive hypothesis to show the claim holds (for this smaller set) for any $-2 \\leqslant r \\leqslant s-x$. Adding the excluded block to each of those combinations, we see that the claim also holds when $x-2 \\leqslant r \\leqslant s$. So if $x-2 \\leqslant s-x$, then we have covered the whole interval $[-2, s]$. But each block weight is at least 1 , so we have $x-2 \\leqslant(s-(n-1))-2=s-(2 n-(n-1)) \\leqslant s-(s-(n-1)) \\leqslant s-x$, as desired. Comment. Instead of inducting on sets of blocks with total weight $s \\leqslant 2 n$, we could instead prove the result only for $s=2 n$. We would then need to modify the inductive step to scale up the block weights before applying the induction hypothesis.", "metadata": {"resource_path": "IMO/segmented/en-IMO2019SL.jsonl"}} -{"year": "2019", "tier": "T0", "problem_label": "C2", "problem_type": "Combinatorics", "exam": "IMO", "problem": "You are given a set of $n$ blocks, each weighing at least 1 ; their total weight is $2 n$. Prove that for every real number $r$ with $0 \\leqslant r \\leqslant 2 n-2$ you can choose a subset of the blocks whose total weight is at least $r$ but at most $r+2$. (Thailand)", "solution": "Let $x_{1}, \\ldots, x_{n}$ be the weights of the blocks in weakly increasing order. Consider the set $S$ of sums of the form $\\sum_{j \\in J} x_{j}$ for a subset $J \\subseteq\\{1,2, \\ldots, n\\}$. We want to prove that the mesh of $S$ - i.e. the largest distance between two adjacent elements - is at most 2. For $0 \\leqslant k \\leqslant n$, let $S_{k}$ denote the set of sums of the form $\\sum_{i \\in J} x_{i}$ for a subset $J \\subseteq\\{1,2, \\ldots, k\\}$. We will show by induction on $k$ that the mesh of $S_{k}$ is at most 2 . The base case $k=0$ is trivial (as $S_{0}=\\{0\\}$ ). For $k>0$ we have $$ S_{k}=S_{k-1} \\cup\\left(x_{k}+S_{k-1}\\right) $$ (where $\\left(x_{k}+S_{k-1}\\right)$ denotes $\\left\\{x_{k}+s: s \\in S_{k-1}\\right\\}$ ), so it suffices to prove that $x_{k} \\leqslant \\sum_{j\\sum_{j(n+1-k)(k+1)+k-1 $$ This rearranges to $n>k(n+1-k)$, which is false for $1 \\leqslant k \\leqslant n$, giving the desired contradiction.", "metadata": {"resource_path": "IMO/segmented/en-IMO2019SL.jsonl"}} -{"year": "2019", "tier": "T0", "problem_label": "C3", "problem_type": "Combinatorics", "exam": "IMO", "problem": "Let $n$ be a positive integer. Harry has $n$ coins lined up on his desk, each showing heads or tails. He repeatedly does the following operation: if there are $k$ coins showing heads and $k>0$, then he flips the $k^{\\text {th }}$ coin over; otherwise he stops the process. (For example, the process starting with THT would be THT $\\rightarrow H H T \\rightarrow H T T \\rightarrow T T T$, which takes three steps.) Letting $C$ denote the initial configuration (a sequence of $n H$ 's and $T$ 's), write $\\ell(C)$ for the number of steps needed before all coins show $T$. Show that this number $\\ell(C)$ is finite, and determine its average value over all $2^{n}$ possible initial configurations $C$.", "solution": "We represent the problem using a directed graph $G_{n}$ whose vertices are the length- $n$ strings of $H$ 's and $T$ 's. The graph features an edge from each string to its successor (except for $T T \\cdots T T$, which has no successor). We will also write $\\bar{H}=T$ and $\\bar{T}=H$. The graph $G_{0}$ consists of a single vertex: the empty string. The main claim is that $G_{n}$ can be described explicitly in terms of $G_{n-1}$ : - We take two copies, $X$ and $Y$, of $G_{n-1}$. - In $X$, we take each string of $n-1$ coins and just append a $T$ to it. In symbols, we replace $s_{1} \\cdots s_{n-1}$ with $s_{1} \\cdots s_{n-1} T$. - In $Y$, we take each string of $n-1$ coins, flip every coin, reverse the order, and append an $H$ to it. In symbols, we replace $s_{1} \\cdots s_{n-1}$ with $\\bar{s}_{n-1} \\bar{s}_{n-2} \\cdots \\bar{s}_{1} H$. - Finally, we add one new edge from $Y$ to $X$, namely $H H \\cdots H H H \\rightarrow H H \\cdots H H T$. We depict $G_{4}$ below, in a way which indicates this recursive construction: ![](https://cdn.mathpix.com/cropped/2024_11_18_9ee8c2ac2c2f5002ff14g-034.jpg?height=515&width=1135&top_left_y=1850&top_left_x=466) We prove the claim inductively. Firstly, $X$ is correct as a subgraph of $G_{n}$, as the operation on coins is unchanged by an extra $T$ at the end: if $s_{1} \\cdots s_{n-1}$ is sent to $t_{1} \\cdots t_{n-1}$, then $s_{1} \\cdots s_{n-1} T$ is sent to $t_{1} \\cdots t_{n-1} T$. Next, $Y$ is also correct as a subgraph of $G_{n}$, as if $s_{1} \\cdots s_{n-1}$ has $k$ occurrences of $H$, then $\\bar{s}_{n-1} \\cdots \\bar{s}_{1} H$ has $(n-1-k)+1=n-k$ occurrences of $H$, and thus (provided that $k>0$ ), if $s_{1} \\cdots s_{n-1}$ is sent to $t_{1} \\cdots t_{n-1}$, then $\\bar{s}_{n-1} \\cdots \\bar{s}_{1} H$ is sent to $\\bar{t}_{n-1} \\cdots \\bar{t}_{1} H$. Finally, the one edge from $Y$ to $X$ is correct, as the operation does send $H H \\cdots H H H$ to HH $\\cdots$ HHT. To finish, note that the sequences in $X$ take an average of $E(n-1)$ steps to terminate, whereas the sequences in $Y$ take an average of $E(n-1)$ steps to reach $H H \\cdots H$ and then an additional $n$ steps to terminate. Therefore, we have $$ E(n)=\\frac{1}{2}(E(n-1)+(E(n-1)+n))=E(n-1)+\\frac{n}{2} $$ We have $E(0)=0$ from our description of $G_{0}$. Thus, by induction, we have $E(n)=\\frac{1}{2}(1+\\cdots+$ $n)=\\frac{1}{4} n(n+1)$, which in particular is finite.", "metadata": {"resource_path": "IMO/segmented/en-IMO2019SL.jsonl"}} -{"year": "2019", "tier": "T0", "problem_label": "C3", "problem_type": "Combinatorics", "exam": "IMO", "problem": "Let $n$ be a positive integer. Harry has $n$ coins lined up on his desk, each showing heads or tails. He repeatedly does the following operation: if there are $k$ coins showing heads and $k>0$, then he flips the $k^{\\text {th }}$ coin over; otherwise he stops the process. (For example, the process starting with THT would be THT $\\rightarrow H H T \\rightarrow H T T \\rightarrow T T T$, which takes three steps.) Letting $C$ denote the initial configuration (a sequence of $n H$ 's and $T$ 's), write $\\ell(C)$ for the number of steps needed before all coins show $T$. Show that this number $\\ell(C)$ is finite, and determine its average value over all $2^{n}$ possible initial configurations $C$.", "solution": "We consider what happens with configurations depending on the coins they start and end with. - If a configuration starts with $H$, the last $n-1$ coins follow the given rules, as if they were all the coins, until they are all $T$, then the first coin is turned over. - If a configuration ends with $T$, the last coin will never be turned over, and the first $n-1$ coins follow the given rules, as if they were all the coins. - If a configuration starts with $T$ and ends with $H$, the middle $n-2$ coins follow the given rules, as if they were all the coins, until they are all $T$. After that, there are $2 n-1$ more steps: first coins $1,2, \\ldots, n-1$ are turned over in that order, then coins $n, n-1, \\ldots, 1$ are turned over in that order. As this covers all configurations, and the number of steps is clearly finite for 0 or 1 coins, it follows by induction on $n$ that the number of steps is always finite. We define $E_{A B}(n)$, where $A$ and $B$ are each one of $H, T$ or $*$, to be the average number of steps over configurations of length $n$ restricted to those that start with $A$, if $A$ is not $*$, and that end with $B$, if $B$ is not * (so * represents \"either $H$ or $T$ \"). The above observations tell us that, for $n \\geqslant 2$ : - $E_{H *}(n)=E(n-1)+1$. - $E_{* T}(n)=E(n-1)$. - $E_{H T}(n)=E(n-2)+1$ (by using both the observations for $H *$ and for $\\left.* T\\right)$. - $E_{T H}(n)=E(n-2)+2 n-1$. Now $E_{H *}(n)=\\frac{1}{2}\\left(E_{H H}(n)+E_{H T}(n)\\right)$, so $E_{H H}(n)=2 E(n-1)-E(n-2)+1$. Similarly, $E_{T T}(n)=2 E(n-1)-E(n-2)-1$. So $$ E(n)=\\frac{1}{4}\\left(E_{H T}(n)+E_{H H}(n)+E_{T T}(n)+E_{T H}(n)\\right)=E(n-1)+\\frac{n}{2} $$ We have $E(0)=0$ and $E(1)=\\frac{1}{2}$, so by induction on $n$ we have $E(n)=\\frac{1}{4} n(n+1)$.", "metadata": {"resource_path": "IMO/segmented/en-IMO2019SL.jsonl"}} -{"year": "2019", "tier": "T0", "problem_label": "C3", "problem_type": "Combinatorics", "exam": "IMO", "problem": "Let $n$ be a positive integer. Harry has $n$ coins lined up on his desk, each showing heads or tails. He repeatedly does the following operation: if there are $k$ coins showing heads and $k>0$, then he flips the $k^{\\text {th }}$ coin over; otherwise he stops the process. (For example, the process starting with THT would be THT $\\rightarrow H H T \\rightarrow H T T \\rightarrow T T T$, which takes three steps.) Letting $C$ denote the initial configuration (a sequence of $n H$ 's and $T$ 's), write $\\ell(C)$ for the number of steps needed before all coins show $T$. Show that this number $\\ell(C)$ is finite, and determine its average value over all $2^{n}$ possible initial configurations $C$.", "solution": "Let $H_{i}$ be the number of heads in positions 1 to $i$ inclusive (so $H_{n}$ is the total number of heads), and let $I_{i}$ be 1 if the $i^{\\text {th }}$ coin is a head, 0 otherwise. Consider the function $$ t(i)=I_{i}+2\\left(\\min \\left\\{i, H_{n}\\right\\}-H_{i}\\right) $$ We claim that $t(i)$ is the total number of times coin $i$ is turned over (which implies that the process terminates). Certainly $t(i)=0$ when all coins are tails, and $t(i)$ is always a nonnegative integer, so it suffices to show that when the $k^{\\text {th }}$ coin is turned over (where $k=H_{n}$ ), $t(k)$ goes down by 1 and all the other $t(i)$ are unchanged. We show this by splitting into cases: - If $ik, \\min \\left\\{i, H_{n}\\right\\}=H_{n}$ both before and after the coin flip, and both $H_{n}$ and $H_{i}$ change by the same amount, so $t(i)$ is unchanged. - If $i=k$ and the coin is heads, $I_{i}$ goes down by 1 , as do both $\\min \\left\\{i, H_{n}\\right\\}=H_{n}$ and $H_{i}$; so $t(i)$ goes down by 1 . - If $i=k$ and the coin is tails, $I_{i}$ goes up by $1, \\min \\left\\{i, H_{n}\\right\\}=i$ is unchanged and $H_{i}$ goes up by 1 ; so $t(i)$ goes down by 1 . We now need to compute the average value of $$ \\sum_{i=1}^{n} t(i)=\\sum_{i=1}^{n} I_{i}+2 \\sum_{i=1}^{n} \\min \\left\\{i, H_{n}\\right\\}-2 \\sum_{i=1}^{n} H_{i} . $$ The average value of the first term is $\\frac{1}{2} n$, and that of the third term is $-\\frac{1}{2} n(n+1)$. To compute the second term, we sum over choices for the total number of heads, and then over the possible values of $i$, getting $$ 2^{1-n} \\sum_{j=0}^{n}\\binom{n}{j} \\sum_{i=1}^{n} \\min \\{i, j\\}=2^{1-n} \\sum_{j=0}^{n}\\binom{n}{j}\\left(n j-\\binom{j}{2}\\right) . $$ Now, in terms of trinomial coefficients, $$ \\sum_{j=0}^{n} j\\binom{n}{j}=\\sum_{j=1}^{n}\\binom{n}{n-j, j-1,1}=n \\sum_{j=0}^{n-1}\\binom{n-1}{j}=2^{n-1} n $$ and $$ \\sum_{j=0}^{n}\\binom{j}{2}\\binom{n}{j}=\\sum_{j=2}^{n}\\binom{n}{n-j, j-2,2}=\\binom{n}{2} \\sum_{j=0}^{n-2}\\binom{n-2}{j}=2^{n-2}\\binom{n}{2} $$ So the second term above is $$ 2^{1-n}\\left(2^{n-1} n^{2}-2^{n-2}\\binom{n}{2}\\right)=n^{2}-\\frac{n(n-1)}{4} $$ and the required average is $$ E(n)=\\frac{1}{2} n+n^{2}-\\frac{n(n-1)}{4}-\\frac{1}{2} n(n+1)=\\frac{n(n+1)}{4} . $$", "metadata": {"resource_path": "IMO/segmented/en-IMO2019SL.jsonl"}} -{"year": "2019", "tier": "T0", "problem_label": "C3", "problem_type": "Combinatorics", "exam": "IMO", "problem": "Let $n$ be a positive integer. Harry has $n$ coins lined up on his desk, each showing heads or tails. He repeatedly does the following operation: if there are $k$ coins showing heads and $k>0$, then he flips the $k^{\\text {th }}$ coin over; otherwise he stops the process. (For example, the process starting with THT would be THT $\\rightarrow H H T \\rightarrow H T T \\rightarrow T T T$, which takes three steps.) Letting $C$ denote the initial configuration (a sequence of $n H$ 's and $T$ 's), write $\\ell(C)$ for the number of steps needed before all coins show $T$. Show that this number $\\ell(C)$ is finite, and determine its average value over all $2^{n}$ possible initial configurations $C$.", "solution": "Harry has built a Turing machine to flip the coins for him. The machine is initially positioned at the $k^{\\text {th }}$ coin, where there are $k$ heads (and the position before the first coin is considered to be the $0^{\\text {th }}$ coin). The machine then moves according to the following rules, stopping when it reaches the position before the first coin: if the coin at its current position is $H$, it flips the coin and moves to the previous coin, while if the coin at its current position is $T$, it flips the coin and moves to the next position. Consider the maximal sequences of consecutive moves in the same direction. Suppose the machine has $a$ consecutive moves to the next coin, before a move to the previous coin. After those $a$ moves, the $a$ coins flipped in those moves are all heads, as is the coin the machine is now at, so at least the next $a+1$ moves will all be moves to the previous coin. Similarly, $a$ consecutive moves to the previous coin are followed by at least $a+1$ consecutive moves to the next coin. There cannot be more than $n$ consecutive moves in the same direction, so this proves that the process terminates (with a move from the first coin to the position before the first coin). Thus we have a (possibly empty) sequence $a_{1}<\\cdotsj$, meaning that, as we follow the moves backwards, each coin is always in the correct state when flipped to result in a move in the required direction. (Alternatively, since there are $2^{n}$ possible configurations of coins and $2^{n}$ possible such ascending sequences, the fact that the sequence of moves determines at most one configuration of coins, and thus that there is an injection from configurations of coins to such ascending sequences, is sufficient for it to be a bijection, without needing to show that coins are in the right state as we move backwards.)", "metadata": {"resource_path": "IMO/segmented/en-IMO2019SL.jsonl"}} -{"year": "2019", "tier": "T0", "problem_label": "C3", "problem_type": "Combinatorics", "exam": "IMO", "problem": "Let $n$ be a positive integer. Harry has $n$ coins lined up on his desk, each showing heads or tails. He repeatedly does the following operation: if there are $k$ coins showing heads and $k>0$, then he flips the $k^{\\text {th }}$ coin over; otherwise he stops the process. (For example, the process starting with THT would be THT $\\rightarrow H H T \\rightarrow H T T \\rightarrow T T T$, which takes three steps.) Letting $C$ denote the initial configuration (a sequence of $n H$ 's and $T$ 's), write $\\ell(C)$ for the number of steps needed before all coins show $T$. Show that this number $\\ell(C)$ is finite, and determine its average value over all $2^{n}$ possible initial configurations $C$.", "solution": "We explicitly describe what happens with an arbitrary sequence $C$ of $n$ coins. Suppose that $C$ contain $k$ heads at positions $1 \\leqslant c_{1}1$ be an integer. Suppose we are given $2 n$ points in a plane such that no three of them are collinear. The points are to be labelled $A_{1}, A_{2}, \\ldots, A_{2 n}$ in some order. We then consider the $2 n$ angles $\\angle A_{1} A_{2} A_{3}, \\angle A_{2} A_{3} A_{4}, \\ldots, \\angle A_{2 n-2} A_{2 n-1} A_{2 n}, \\angle A_{2 n-1} A_{2 n} A_{1}$, $\\angle A_{2 n} A_{1} A_{2}$. We measure each angle in the way that gives the smallest positive value (i.e. between $0^{\\circ}$ and $180^{\\circ}$ ). Prove that there exists an ordering of the given points such that the resulting $2 n$ angles can be separated into two groups with the sum of one group of angles equal to the sum of the other group.", "solution": "Let $\\ell$ be a line separating the points into two groups ( $L$ and $R$ ) with $n$ points in each. Label the points $A_{1}, A_{2}, \\ldots, A_{2 n}$ so that $L=\\left\\{A_{1}, A_{3}, \\ldots, A_{2 n-1}\\right\\}$. We claim that this labelling works. Take a line $s=A_{2 n} A_{1}$. (a) Rotate $s$ around $A_{1}$ until it passes through $A_{2}$; the rotation is performed in a direction such that $s$ is never parallel to $\\ell$. (b) Then rotate the new $s$ around $A_{2}$ until it passes through $A_{3}$ in a similar manner. (c) Perform $2 n-2$ more such steps, after which $s$ returns to its initial position. The total (directed) rotation angle $\\Theta$ of $s$ is clearly a multiple of $180^{\\circ}$. On the other hand, $s$ was never parallel to $\\ell$, which is possible only if $\\Theta=0$. Now it remains to partition all the $2 n$ angles into those where $s$ is rotated anticlockwise, and the others.", "metadata": {"resource_path": "IMO/segmented/en-IMO2019SL.jsonl"}} -{"year": "2019", "tier": "T0", "problem_label": "C6", "problem_type": "Combinatorics", "exam": "IMO", "problem": "Let $n>1$ be an integer. Suppose we are given $2 n$ points in a plane such that no three of them are collinear. The points are to be labelled $A_{1}, A_{2}, \\ldots, A_{2 n}$ in some order. We then consider the $2 n$ angles $\\angle A_{1} A_{2} A_{3}, \\angle A_{2} A_{3} A_{4}, \\ldots, \\angle A_{2 n-2} A_{2 n-1} A_{2 n}, \\angle A_{2 n-1} A_{2 n} A_{1}$, $\\angle A_{2 n} A_{1} A_{2}$. We measure each angle in the way that gives the smallest positive value (i.e. between $0^{\\circ}$ and $180^{\\circ}$ ). Prove that there exists an ordering of the given points such that the resulting $2 n$ angles can be separated into two groups with the sum of one group of angles equal to the sum of the other group.", "solution": "When tracing a cyclic path through the $A_{i}$ in order, with straight line segments between consecutive points, let $\\theta_{i}$ be the exterior angle at $A_{i}$, with a sign convention that it is positive if the path turns left and negative if the path turns right. Then $\\sum_{i=1}^{2 n} \\theta_{i}=360 k^{\\circ}$ for some integer $k$. Let $\\phi_{i}=\\angle A_{i-1} A_{i} A_{i+1}($ indices $\\bmod 2 n)$, defined as in the problem; thus $\\phi_{i}=180^{\\circ}-\\left|\\theta_{i}\\right|$. Let $L$ be the set of $i$ for which the path turns left at $A_{i}$ and let $R$ be the set for which it turns right. Then $S=\\sum_{i \\in L} \\phi_{i}-\\sum_{i \\in R} \\phi_{i}=(180(|L|-|R|)-360 k)^{\\circ}$, which is a multiple of $360^{\\circ}$ since the number of points is even. We will show that the points can be labelled such that $S=0$, in which case $L$ and $R$ satisfy the required condition of the problem. Note that the value of $S$ is defined for a slightly larger class of configurations: it is OK for two points to coincide, as long as they are not consecutive, and OK for three points to be collinear, as long as $A_{i}, A_{i+1}$ and $A_{i+2}$ do not appear on a line in that order. In what follows it will be convenient, although not strictly necessary, to consider such configurations. Consider how $S$ changes if a single one of the $A_{i}$ is moved along some straight-line path (not passing through any $A_{j}$ and not lying on any line $A_{j} A_{k}$, but possibly crossing such lines). Because $S$ is a multiple of $360^{\\circ}$, and the angles change continuously, $S$ can only change when a point moves between $R$ and $L$. Furthermore, if $\\phi_{j}=0$ when $A_{j}$ moves between $R$ and $L, S$ is unchanged; it only changes if $\\phi_{j}=180^{\\circ}$ when $A_{j}$ moves between those sets. For any starting choice of points, we will now construct a new configuration, with labels such that $S=0$, that can be perturbed into the original one without any $\\phi_{i}$ passing through $180^{\\circ}$, so that $S=0$ for the original configuration with those labels as well. Take some line such that there are $n$ points on each side of that line. The new configuration has $n$ copies of a single point on each side of the line, and a path that alternates between sides of the line; all angles are 0 , so this configuration has $S=0$. Perturbing the points into their original positions, while keeping each point on its side of the line, no angle $\\phi_{i}$ can pass through $180^{\\circ}$, because no straight line can go from one side of the line to the other and back. So the perturbation process leaves $S=0$. Comment. More complicated variants of this solution are also possible; for example, a path defined using four quadrants of the plane rather than just two half-planes.", "metadata": {"resource_path": "IMO/segmented/en-IMO2019SL.jsonl"}} -{"year": "2019", "tier": "T0", "problem_label": "C6", "problem_type": "Combinatorics", "exam": "IMO", "problem": "Let $n>1$ be an integer. Suppose we are given $2 n$ points in a plane such that no three of them are collinear. The points are to be labelled $A_{1}, A_{2}, \\ldots, A_{2 n}$ in some order. We then consider the $2 n$ angles $\\angle A_{1} A_{2} A_{3}, \\angle A_{2} A_{3} A_{4}, \\ldots, \\angle A_{2 n-2} A_{2 n-1} A_{2 n}, \\angle A_{2 n-1} A_{2 n} A_{1}$, $\\angle A_{2 n} A_{1} A_{2}$. We measure each angle in the way that gives the smallest positive value (i.e. between $0^{\\circ}$ and $180^{\\circ}$ ). Prove that there exists an ordering of the given points such that the resulting $2 n$ angles can be separated into two groups with the sum of one group of angles equal to the sum of the other group.", "solution": "First, let $\\ell$ be a line in the plane such that there are $n$ points on one side and the other $n$ points on the other side. For convenience, assume $\\ell$ is horizontal (otherwise, we can rotate the plane). Then we can use the terms \"above\", \"below\", \"left\" and \"right\" in the usual way. We denote the $n$ points above the line in an arbitrary order as $P_{1}, P_{2}, \\ldots, P_{n}$, and the $n$ points below the line as $Q_{1}, Q_{2}, \\ldots, Q_{n}$. If we connect $P_{i}$ and $Q_{j}$ with a line segment, the line segment will intersect with the line $\\ell$. Denote the intersection as $I_{i j}$. If $P_{i}$ is connected to $Q_{j}$ and $Q_{k}$, where $jk$, then the sign of $\\angle Q_{j} P_{i} Q_{k}$ is taken to be the same as for $\\angle Q_{k} P_{i} Q_{j}$. Similarly, we can define the sign of $\\angle P_{j} Q_{i} P_{k}$ with $j1$ be an integer. Suppose we are given $2 n$ points in a plane such that no three of them are collinear. The points are to be labelled $A_{1}, A_{2}, \\ldots, A_{2 n}$ in some order. We then consider the $2 n$ angles $\\angle A_{1} A_{2} A_{3}, \\angle A_{2} A_{3} A_{4}, \\ldots, \\angle A_{2 n-2} A_{2 n-1} A_{2 n}, \\angle A_{2 n-1} A_{2 n} A_{1}$, $\\angle A_{2 n} A_{1} A_{2}$. We measure each angle in the way that gives the smallest positive value (i.e. between $0^{\\circ}$ and $180^{\\circ}$ ). Prove that there exists an ordering of the given points such that the resulting $2 n$ angles can be separated into two groups with the sum of one group of angles equal to the sum of the other group.", "solution": "We shall think instead of the problem as asking us to assign a weight $\\pm 1$ to each angle, such that the weighted sum of all the angles is zero. Given an ordering $A_{1}, \\ldots, A_{2 n}$ of the points, we shall assign weights according to the following recipe: walk in order from point to point, and assign the left turns +1 and the right turns -1 . This is the same weighting as in Solution 3, and as in that solution, the weighted sum is a multiple of $360^{\\circ}$. We now aim to show the following: Lemma. Transposing any two consecutive points in the ordering changes the weighted sum by $\\pm 360^{\\circ}$ or 0 . Knowing that, we can conclude quickly: if the ordering $A_{1}, \\ldots, A_{2 n}$ has weighted angle sum $360 k^{\\circ}$, then the ordering $A_{2 n}, \\ldots, A_{1}$ has weighted angle sum $-360 k^{\\circ}$ (since the angles are the same, but left turns and right turns are exchanged). We can reverse the ordering of $A_{1}$, $\\ldots, A_{2 n}$ by a sequence of transpositions of consecutive points, and in doing so the weighted angle sum must become zero somewhere along the way. We now prove that lemma: Proof. Transposing two points amounts to taking a section $A_{k} A_{k+1} A_{k+2} A_{k+3}$ as depicted, reversing the central line segment $A_{k+1} A_{k+2}$, and replacing its two neighbours with the dotted lines. ![](https://cdn.mathpix.com/cropped/2024_11_18_9ee8c2ac2c2f5002ff14g-048.jpg?height=424&width=1357&top_left_y=1156&top_left_x=354) Figure 1: Transposing two consecutive vertices: before (left) and afterwards (right) In each triangle, we alter the sum by $\\pm 180^{\\circ}$. Indeed, using (anticlockwise) directed angles modulo $360^{\\circ}$, we either add or subtract all three angles of each triangle. Hence both triangles together alter the sum by $\\pm 180 \\pm 180^{\\circ}$, which is $\\pm 360^{\\circ}$ or 0 .", "metadata": {"resource_path": "IMO/segmented/en-IMO2019SL.jsonl"}} -{"year": "2019", "tier": "T0", "problem_label": "C7", "problem_type": "Combinatorics", "exam": "IMO", "problem": "There are 60 empty boxes $B_{1}, \\ldots, B_{60}$ in a row on a table and an unlimited supply of pebbles. Given a positive integer $n$, Alice and Bob play the following game. In the first round, Alice takes $n$ pebbles and distributes them into the 60 boxes as she wishes. Each subsequent round consists of two steps: (a) Bob chooses an integer $k$ with $1 \\leqslant k \\leqslant 59$ and splits the boxes into the two groups $B_{1}, \\ldots, B_{k}$ and $B_{k+1}, \\ldots, B_{60}$. (b) Alice picks one of these two groups, adds one pebble to each box in that group, and removes one pebble from each box in the other group. Bob wins if, at the end of any round, some box contains no pebbles. Find the smallest $n$ such that Alice can prevent Bob from winning. (Czech Republic)", "solution": null, "metadata": {"resource_path": "IMO/segmented/en-IMO2019SL.jsonl"}} -{"year": "2019", "tier": "T0", "problem_label": "C8", "problem_type": "Combinatorics", "exam": "IMO", "problem": "Alice has a map of Wonderland, a country consisting of $n \\geqslant 2$ towns. For every pair of towns, there is a narrow road going from one town to the other. One day, all the roads are declared to be \"one way\" only. Alice has no information on the direction of the roads, but the King of Hearts has offered to help her. She is allowed to ask him a number of questions. For each question in turn, Alice chooses a pair of towns and the King of Hearts tells her the direction of the road connecting those two towns. Alice wants to know whether there is at least one town in Wonderland with at most one outgoing road. Prove that she can always find out by asking at most $4 n$ questions. Comment. This problem could be posed with an explicit statement about points being awarded for weaker bounds $c n$ for some $c>4$, in the style of IMO 2014 Problem 6. (Thailand)", "solution": "We will show Alice needs to ask at most $4 n-7$ questions. Her strategy has the following phases. In what follows, $S$ is the set of towns that Alice, so far, does not know to have more than one outgoing road (so initially $|S|=n$ ). Phase 1. Alice chooses any two towns, say $A$ and $B$. Without loss of generality, suppose that the King of Hearts' answer is that the road goes from $A$ to $B$. At the end of this phase, Alice has asked 1 question. Phase 2. During this phase there is a single (variable) town $T$ that is known to have at least one incoming road but not yet known to have any outgoing roads. Initially, $T$ is $B$. Alice does the following $n-2$ times: she picks a town $X$ she has not asked about before, and asks the direction of the road between $T$ and $X$. If it is from $X$ to $T, T$ is unchanged; if it is from $T$ to $X, X$ becomes the new choice of town $T$, as the previous $T$ is now known to have an outgoing road. At the end of this phase, Alice has asked a total of $n-1$ questions. The final town $T$ is not yet known to have any outgoing roads, while every other town has exactly one outgoing road known. The undirected graph of roads whose directions are known is a tree. Phase 3. During this phase, Alice asks about the directions of all roads between $T$ and another town she has not previously asked about, stopping if she finds two outgoing roads from $T$. This phase involves at most $n-2$ questions. If she does not find two outgoing roads from $T$, she has answered her original question with at most $2 n-3 \\leqslant 4 n-7$ questions, so in what follows we suppose that she does find two outgoing roads, asking a total of $k$ questions in this phase, where $2 \\leqslant k \\leqslant n-2$ (and thus $n \\geqslant 4$ for what follows). For every question where the road goes towards $T$, the town at the other end is removed from $S$ (as it already had one outgoing road known), while the last question resulted in $T$ being removed from $S$. So at the end of this phase, $|S|=n-k+1$, while a total of $n+k-1$ questions have been asked. Furthermore, the undirected graph of roads within $S$ whose directions are known contains no cycles (as $T$ is no longer a member of $S$, all questions asked in this phase involved $T$ and the graph was a tree before this phase started). Every town in $S$ has exactly one outgoing road known (not necessarily to another town in $S$ ). Phase 4. During this phase, Alice repeatedly picks any pair of towns in $S$ for which she does not know the direction of the road between them. Because every town in $S$ has exactly one outgoing road known, this always results in the removal of one of those two towns from $S$. Because there are no cycles in the graph of roads of known direction within $S$, this can continue until there are at most 2 towns left in $S$. If it ends with $t$ towns left, $n-k+1-t$ questions were asked in this phase, so a total of $2 n-t$ questions have been asked. Phase 5. During this phase, Alice asks about all the roads from the remaining towns in $S$ that she has not previously asked about. She has definitely already asked about any road between those towns (if $t=2$ ). She must also have asked in one of the first two phases about at least one other road involving one of those towns (as those phases resulted in a tree with $n>2$ vertices). So she asks at most $t(n-t)-1$ questions in this phase. At the end of this phase, Alice knows whether any town has at most one outgoing road. If $t=1$, at most $3 n-3 \\leqslant 4 n-7$ questions were needed in total, while if $t=2$, at most $4 n-7$ questions were needed in total. Comment 1. The version of this problem originally submitted asked only for an upper bound of $5 n$, which is much simpler to prove. The Problem Selection Committee preferred a version with an asymptotically optimal constant. In the following comment, we will show that the constant is optimal. Comment 2. We will show that Alice cannot always find out by asking at most $4 n-3\\left(\\log _{2} n\\right)-$ 15 questions, if $n \\geqslant 8$. To show this, we suppose the King of Hearts is choosing the directions as he goes along, only picking the direction of a road when Alice asks about it for the first time. We provide a strategy for the King of Hearts that ensures that, after the given number of questions, the map is still consistent both with the existence of a town with at most one outgoing road, and with the nonexistence of such a town. His strategy has the following phases. When describing how the King of Hearts' answer to a question is determined below, we always assume he is being asked about a road for the first time (otherwise, he just repeats his previous answer for that road). This strategy is described throughout in graph-theoretic terms (vertices and edges rather than towns and roads). Phase 1. In this phase, we consider the undirected graph formed by edges whose directions are known. The phase terminates when there are exactly 8 connected components whose undirected graphs are trees. The following invariant is maintained: in a component with $k$ vertices whose undirected graph is a tree, every vertex has at most $\\left[\\log _{2} k\\right\\rfloor$ edges into it. - If the King of Hearts is asked about an edge between two vertices in the same component, or about an edge between two components at least one of which is not a tree, he chooses any direction for that edge arbitrarily. - If he is asked about an edge between a vertex in component $A$ that has $a$ vertices and is a tree and a vertex in component $B$ that has $b$ vertices and is a tree, suppose without loss of generality that $a \\geqslant b$. He then chooses the edge to go from $A$ to $B$. In this case, the new number of edges into any vertex is at most $\\max \\left\\{\\left\\lfloor\\log _{2} a\\right\\rfloor,\\left\\lfloor\\log _{2} b\\right\\rfloor+1\\right\\} \\leqslant\\left\\lfloor\\log _{2}(a+b)\\right\\rfloor$. In all cases, the invariant is preserved, and the number of tree components either remains unchanged or goes down by 1. Assuming Alice does not repeat questions, the process must eventually terminate with 8 tree components, and at least $n-8$ questions having been asked. Note that each tree component contains at least one vertex with no outgoing edges. Colour one such vertex in each tree component red. Phase 2. Let $V_{1}, V_{2}$ and $V_{3}$ be the three of the red vertices whose components are smallest (so their components together have at most $\\left\\lfloor\\frac{3}{8} n\\right\\rfloor$ vertices, with each component having at most $\\left\\lfloor\\frac{3}{8} n-2\\right\\rfloor$ vertices). Let sets $C_{1}, C_{2}, \\ldots$ be the connected components after removing the $V_{j}$. By construction, there are no edges with known direction between $C_{i}$ and $C_{j}$ for $i \\neq j$, and there are at least five such components. If at any point during this phase, the King of Hearts is asked about an edge within one of the $C_{i}$, he chooses an arbitrary direction. If he is asked about an edge between $C_{i}$ and $C_{j}$ for $i \\neq j$, he answers so that all edges go from $C_{i}$ to $C_{i+1}$ and $C_{i+2}$, with indices taken modulo the number of components, and chooses arbitrarily for other pairs. This ensures that all vertices other than the $V_{j}$ will have more than one outgoing edge. For edges involving one of the $V_{j}$ he answers as follows, so as to remain consistent for as long as possible with both possibilities for whether one of those vertices has at most one outgoing edge. Note that as they were red vertices, they have no outgoing edges at the start of this phase. For edges between two of the $V_{j}$, he answers that the edges go from $V_{1}$ to $V_{2}$, from $V_{2}$ to $V_{3}$ and from $V_{3}$ to $V_{1}$. For edges between $V_{j}$ and some other vertex, he always answers that the edge goes into $V_{j}$, except for the last such edge for which he is asked the question for any given $V_{j}$, for which he answers that the edge goes out of $V_{j}$. Thus, as long as at least one of the $V_{j}$ has not had the question answered for all the vertices that are not among the $V_{j}$, his answers are still compatible both with all vertices having more than one outgoing edge, and with that $V_{j}$ having only one outgoing edge. At the start of this phase, each of the $V_{j}$ has at most $\\left\\lfloor\\log _{2}\\left\\lfloor\\frac{3}{8} n-2\\right\\rfloor\\right\\rfloor<\\left(\\log _{2} n\\right)-1$ incoming edges. Thus, Alice cannot determine whether some vertex has only one outgoing edge within $3(n-$ $\\left.3-\\left(\\left(\\log _{2} n\\right)-1\\right)\\right)-1$ questions in this phase; that is, $4 n-3\\left(\\log _{2} n\\right)-15$ questions total. Comment 3. We can also improve the upper bound slightly, to $4 n-2\\left(\\log _{2} n\\right)+1$. (We do not know where the precise minimum number of questions lies between $4 n-3\\left(\\log _{2} n\\right)+O(1)$ and $4 n-2\\left(\\log _{2} n\\right)+$ $O(1)$.) Suppose $n \\geqslant 5$ (otherwise no questions are required at all). To do this, we replace Phases 1 and 2 of the given solution with a different strategy that also results in a spanning tree where one vertex $V$ is not known to have any outgoing edges, and all other vertices have exactly one outgoing edge known, but where there is more control over the numbers of incoming edges. In Phases 3 and 4 we then take more care about the order in which pairs of towns are chosen, to ensure that each of the remaining towns has already had a question asked about at least $\\log _{2} n+O(1)$ edges. Define trees $T_{m}$ with $2^{m}$ vertices, exactly one of which (the root) has no outgoing edges and the rest of which have exactly one outgoing edge, as follows: $T_{0}$ is a single vertex, while $T_{m}$ is constructed by joining the roots of two copies of $T_{m-1}$ with an edge in either direction. If $n=2^{m}$ we can readily ask $n-1$ questions, resulting in a tree $T_{m}$ for the edges with known direction: first ask about $2^{m-1}$ disjoint pairs of vertices, then about $2^{m-2}$ disjoint pairs of the roots of the resulting $T_{1}$ trees, and so on. For the general case, where $n$ is not a power of 2 , after $k$ stages of this process we have $\\left\\lfloor n / 2^{k}\\right\\rfloor$ trees, each of which is like $T_{k}$ but may have some extra vertices (but, however, a unique root). If there are an even number of trees, then ask about pairs of their roots. If there are an odd number (greater than 1) of trees, when a single $T_{k}$ is left over, ask about its root together with that of one of the $T_{k+1}$ trees. Say $m=\\left\\lfloor\\log _{2} n\\right\\rfloor$. The result of that process is a single $T_{m}$ tree, possibly with some extra vertices but still a unique root $V$. That root has at least $m$ incoming edges, and we may list vertices $V_{0}$, $\\ldots, V_{m-1}$ with edges to $V$, such that, for all $0 \\leqslant i1$ we have $D\\left(x_{i}, x_{j}\\right)>d$, since $$ \\left|x_{i}-x_{j}\\right|=\\left|x_{i+1}-x_{i}\\right|+\\left|x_{j}-x_{i+1}\\right| \\geqslant 2^{d}+2^{d}=2^{d+1} . $$ Now choose the minimal $i \\leqslant t$ and the maximal $j \\geqslant t+1$ such that $D\\left(x_{i}, x_{i+1}\\right)=$ $D\\left(x_{i+1}, x_{i+2}\\right)=\\cdots=D\\left(x_{j-1}, x_{j}\\right)=d$. Let $E$ be the set of all the $x_{s}$ with even indices $i \\leqslant s \\leqslant j, O$ be the set of those with odd indices $i \\leqslant s \\leqslant j$, and $R$ be the rest of the elements (so that $\\mathcal{S}$ is the disjoint union of $E, O$ and $R$ ). Set $\\mathcal{S}_{O}=R \\cup O$ and $\\mathcal{S}_{E}=R \\cup E$; we have $\\left|\\mathcal{S}_{O}\\right|<|\\mathcal{S}|$ and $\\left|\\mathcal{S}_{E}\\right|<|\\mathcal{S}|$, so $w\\left(\\mathcal{S}_{O}\\right), w\\left(\\mathcal{S}_{E}\\right) \\leqslant 1$ by the inductive hypothesis. Clearly, $r_{\\mathcal{S}_{O}}(x) \\leqslant r_{\\mathcal{S}}(x)$ and $r_{\\mathcal{S}_{E}}(x) \\leqslant r_{\\mathcal{S}}(x)$ for any $x \\in R$, and thus $$ \\begin{aligned} \\sum_{x \\in R} 2^{-r_{\\mathcal{S}}(x)} & =\\frac{1}{2} \\sum_{x \\in R}\\left(2^{-r_{\\mathcal{S}}(x)}+2^{-r_{\\mathcal{S}}(x)}\\right) \\\\ & \\leqslant \\frac{1}{2} \\sum_{x \\in R}\\left(2^{-r_{\\mathcal{S}_{O}}(x)}+2^{-r_{\\mathcal{S}_{E}}(x)}\\right) \\end{aligned} $$ On the other hand, for every $x \\in O$, there is no $y \\in \\mathcal{S}_{O}$ such that $D_{\\mathcal{S}_{O}}(x, y)=d$ (as all candidates from $\\mathcal{S}$ were in $E$ ). Hence, we have $r_{\\mathcal{S}_{O}}(x) \\leqslant r_{\\mathcal{S}}(x)-1$, and thus $$ \\sum_{x \\in O} 2^{-r \\mathcal{S}_{\\mathcal{S}}(x)} \\leqslant \\frac{1}{2} \\sum_{x \\in O} 2^{-r s_{O}(x)} $$ Similarly, for every $x \\in E$, we have $$ \\sum_{x \\in E} 2^{-r_{\\mathcal{S}}(x)} \\leqslant \\frac{1}{2} \\sum_{x \\in E} 2^{-r_{\\mathcal{S}_{E}}(x)} $$ We can then combine these to give $$ \\begin{aligned} w(S) & =\\sum_{x \\in R} 2^{-r_{\\mathcal{S}}(x)}+\\sum_{x \\in O} 2^{-r_{\\mathcal{S}}(x)}+\\sum_{x \\in E} 2^{-r_{\\mathcal{S}}(x)} \\\\ & \\leqslant \\frac{1}{2} \\sum_{x \\in R}\\left(2^{-r_{\\mathcal{S}_{O}}(x)}+2^{-r_{\\mathcal{S}_{E}}(x)}\\right)+\\frac{1}{2} \\sum_{x \\in O} 2^{-r_{\\mathcal{S}_{O}}(x)}+\\frac{1}{2} \\sum_{x \\in E} 2^{-r_{\\mathcal{S}_{E}}(x)} \\\\ & \\left.=\\frac{1}{2}\\left(\\sum_{x \\in \\mathcal{S}_{O}} 2^{-r_{S_{O}}(x)}+\\sum_{x \\in \\mathcal{S}_{E}} 2^{-r s_{S_{E}}(x)}\\right) \\quad \\text { (since } \\mathcal{S}_{O}=O \\cup R \\text { and } \\mathcal{S}_{E}=E \\cup R\\right) \\\\ & \\left.\\left.=\\frac{1}{2}\\left(w\\left(\\mathcal{S}_{O}\\right)+w\\left(\\mathcal{S}_{E}\\right)\\right)\\right) \\quad \\text { (by definition of } w(\\cdot)\\right) \\\\ & \\leqslant 1 \\quad \\text { (by the inductive hypothesis) } \\end{aligned} $$ which completes the induction. Comment 1. The sets $O$ and $E$ above are not the only ones we could have chosen. Indeed, we could instead have used the following definitions: Let $d$ be the maximal scale between two distinct elements of $\\mathcal{S}$; that is, $d=D\\left(x_{1}, x_{n}\\right)$. Let $O=\\left\\{x \\in \\mathcal{S}: D\\left(x, x_{n}\\right)=d\\right\\}$ (a 'left' part of the set) and let $E=\\left\\{x \\in \\mathcal{S}: D\\left(x_{1}, x\\right)=d\\right\\}$ (a 'right' part of the set). Note that these two sets are disjoint, and nonempty (since they contain $x_{1}$ and $x_{n}$ respectively). The rest of the proof is then the same as in Solution 1. Comment 2. Another possible set $\\mathcal{F}$ containing $2^{k}$ members could arise from considering a binary tree of height $k$, allocating a real number to each leaf, and trying to make the scale between the values of two leaves dependent only on the (graph) distance between them. The following construction makes this more precise. We build up sets $\\mathcal{F}_{k}$ recursively. Let $\\mathcal{F}_{0}=\\{0\\}$, and then let $\\mathcal{F}_{k+1}=\\mathcal{F}_{k} \\cup\\left\\{x+3 \\cdot 4^{k}: x \\in \\mathcal{F}_{k}\\right\\}$ (i.e. each half of $\\mathcal{F}_{k+1}$ is a copy of $F_{k}$ ). We have that $\\mathcal{F}_{k}$ is contained in the interval $\\left[0,4^{k+1}\\right.$ ), and so it follows by induction on $k$ that every member of $F_{k+1}$ has $k$ different scales in its own half of $F_{k+1}$ (by the inductive hypothesis), and only the single scale $2 k+1$ in the other half of $F_{k+1}$. Both of the constructions presented here have the property that every member of $\\mathcal{F}$ has exactly $k$ different scales in $\\mathcal{F}$. Indeed, it can be seen that this must hold (up to a slight perturbation) for any such maximal set. Suppose there were some element $x$ with only $k-1$ different scales in $\\mathcal{F}$ (and every other element had at most $k$ different scales). Then we take some positive real $\\epsilon$, and construct a new set $\\mathcal{F}^{\\prime}=\\{y: y \\in \\mathcal{F}, y \\leqslant x\\} \\cup\\{y+\\epsilon: y \\in \\mathcal{F}, y \\geqslant x\\}$. We have $\\left|\\mathcal{F}^{\\prime}\\right|=|\\mathcal{F}|+1$, and if $\\epsilon$ is sufficiently small then $\\mathcal{F}^{\\prime}$ will also satisfy the property that no member has more than $k$ different scales in $\\mathcal{F}^{\\prime}$. This observation might be used to motivate the idea of weighting members of an arbitrary set $\\mathcal{S}$ of reals according to how many different scales they have in $\\mathcal{S}$.", "metadata": {"resource_path": "IMO/segmented/en-IMO2019SL.jsonl"}} -{"year": "2019", "tier": "T0", "problem_label": "G1", "problem_type": "Geometry", "exam": "IMO", "problem": "Let $A B C$ be a triangle. Circle $\\Gamma$ passes through $A$, meets segments $A B$ and $A C$ again at points $D$ and $E$ respectively, and intersects segment $B C$ at $F$ and $G$ such that $F$ lies between $B$ and $G$. The tangent to circle $B D F$ at $F$ and the tangent to circle $C E G$ at $G$ meet at point $T$. Suppose that points $A$ and $T$ are distinct. Prove that line $A T$ is parallel to $B C$. (Nigeria)", "solution": "Notice that $\\angle T F B=\\angle F D A$ because $F T$ is tangent to circle $B D F$, and moreover $\\angle F D A=\\angle C G A$ because quadrilateral $A D F G$ is cyclic. Similarly, $\\angle T G B=\\angle G E C$ because $G T$ is tangent to circle $C E G$, and $\\angle G E C=\\angle C F A$. Hence, $$ \\angle T F B=\\angle C G A \\text { and } \\quad \\angle T G B=\\angle C F A $$ ![](https://cdn.mathpix.com/cropped/2024_11_18_9ee8c2ac2c2f5002ff14g-058.jpg?height=624&width=1326&top_left_y=930&top_left_x=365) Triangles $F G A$ and $G F T$ have a common side $F G$, and by (1) their angles at $F, G$ are the same. So, these triangles are congruent. So, their altitudes starting from $A$ and $T$, respectively, are equal and hence $A T$ is parallel to line $B F G C$. Comment. Alternatively, we can prove first that $T$ lies on $\\Gamma$. For example, this can be done by showing that $\\angle A F T=\\angle A G T$ using (1). Then the statement follows as $\\angle T A F=\\angle T G F=\\angle G F A$.", "metadata": {"resource_path": "IMO/segmented/en-IMO2019SL.jsonl"}} -{"year": "2019", "tier": "T0", "problem_label": "G2", "problem_type": "Geometry", "exam": "IMO", "problem": "Let $A B C$ be an acute-angled triangle and let $D, E$, and $F$ be the feet of altitudes from $A, B$, and $C$ to sides $B C, C A$, and $A B$, respectively. Denote by $\\omega_{B}$ and $\\omega_{C}$ the incircles of triangles $B D F$ and $C D E$, and let these circles be tangent to segments $D F$ and $D E$ at $M$ and $N$, respectively. Let line $M N$ meet circles $\\omega_{B}$ and $\\omega_{C}$ again at $P \\neq M$ and $Q \\neq N$, respectively. Prove that $M P=N Q$. (Vietnam)", "solution": "Denote the centres of $\\omega_{B}$ and $\\omega_{C}$ by $O_{B}$ and $O_{C}$, let their radii be $r_{B}$ and $r_{C}$, and let $B C$ be tangent to the two circles at $T$ and $U$, respectively. ![](https://cdn.mathpix.com/cropped/2024_11_18_9ee8c2ac2c2f5002ff14g-059.jpg?height=801&width=1014&top_left_y=659&top_left_x=521) From the cyclic quadrilaterals $A F D C$ and $A B D E$ we have $$ \\angle M D O_{B}=\\frac{1}{2} \\angle F D B=\\frac{1}{2} \\angle B A C=\\frac{1}{2} \\angle C D E=\\angle O_{C} D N, $$ so the right-angled triangles $D M O_{B}$ and $D N O_{C}$ are similar. The ratio of similarity between the two triangles is $$ \\frac{D N}{D M}=\\frac{O_{C} N}{O_{B} M}=\\frac{r_{C}}{r_{B}} . $$ Let $\\varphi=\\angle D M N$ and $\\psi=\\angle M N D$. The lines $F M$ and $E N$ are tangent to $\\omega_{B}$ and $\\omega_{C}$, respectively, so $$ \\angle M T P=\\angle F M P=\\angle D M N=\\varphi \\quad \\text { and } \\quad \\angle Q U N=\\angle Q N E=\\angle M N D=\\psi $$ (It is possible that $P$ or $Q$ coincides with $T$ or $U$, or lie inside triangles $D M T$ or $D U N$, respectively. To reduce case-sensitivity, we may use directed angles or simply ignore angles $M T P$ and $Q U N$.) In the circles $\\omega_{B}$ and $\\omega_{C}$ the lengths of chords $M P$ and $N Q$ are $$ M P=2 r_{B} \\cdot \\sin \\angle M T P=2 r_{B} \\cdot \\sin \\varphi \\quad \\text { and } \\quad N Q=2 r_{C} \\cdot \\sin \\angle Q U N=2 r_{C} \\cdot \\sin \\psi $$ By applying the sine rule to triangle $D N M$ we get $$ \\frac{D N}{D M}=\\frac{\\sin \\angle D M N}{\\sin \\angle M N D}=\\frac{\\sin \\varphi}{\\sin \\psi} $$ Finally, putting the above observations together, we get $$ \\frac{M P}{N Q}=\\frac{2 r_{B} \\sin \\varphi}{2 r_{C} \\sin \\psi}=\\frac{r_{B}}{r_{C}} \\cdot \\frac{\\sin \\varphi}{\\sin \\psi}=\\frac{D M}{D N} \\cdot \\frac{\\sin \\varphi}{\\sin \\psi}=\\frac{\\sin \\psi}{\\sin \\varphi} \\cdot \\frac{\\sin \\varphi}{\\sin \\psi}=1, $$ so $M P=N Q$ as required.", "metadata": {"resource_path": "IMO/segmented/en-IMO2019SL.jsonl"}} -{"year": "2019", "tier": "T0", "problem_label": "G3", "problem_type": "Geometry", "exam": "IMO", "problem": "In triangle $A B C$, let $A_{1}$ and $B_{1}$ be two points on sides $B C$ and $A C$, and let $P$ and $Q$ be two points on segments $A A_{1}$ and $B B_{1}$, respectively, so that line $P Q$ is parallel to $A B$. On ray $P B_{1}$, beyond $B_{1}$, let $P_{1}$ be a point so that $\\angle P P_{1} C=\\angle B A C$. Similarly, on ray $Q A_{1}$, beyond $A_{1}$, let $Q_{1}$ be a point so that $\\angle C Q_{1} Q=\\angle C B A$. Show that points $P, Q, P_{1}$, and $Q_{1}$ are concyclic. (Ukraine)", "solution": "Throughout the solution we use oriented angles. Let rays $A A_{1}$ and $B B_{1}$ intersect the circumcircle of $\\triangle A C B$ at $A_{2}$ and $B_{2}$, respectively. By $$ \\angle Q P A_{2}=\\angle B A A_{2}=\\angle B B_{2} A_{2}=\\angle Q B_{2} A_{2} $$ points $P, Q, A_{2}, B_{2}$ are concyclic; denote the circle passing through these points by $\\omega$. We shall prove that $P_{1}$ and $Q_{1}$ also lie on $\\omega$. ![](https://cdn.mathpix.com/cropped/2024_11_18_9ee8c2ac2c2f5002ff14g-060.jpg?height=615&width=858&top_left_y=909&top_left_x=602) By $$ \\angle C A_{2} A_{1}=\\angle C A_{2} A=\\angle C B A=\\angle C Q_{1} Q=\\angle C Q_{1} A_{1}, $$ points $C, Q_{1}, A_{2}, A_{1}$ are also concyclic. From that we get $$ \\angle Q Q_{1} A_{2}=\\angle A_{1} Q_{1} A_{2}=\\angle A_{1} C A_{2}=\\angle B C A_{2}=\\angle B A A_{2}=\\angle Q P A_{2} $$ so $Q_{1}$ lies on $\\omega$. It follows similarly that $P_{1}$ lies on $\\omega$.", "metadata": {"resource_path": "IMO/segmented/en-IMO2019SL.jsonl"}} -{"year": "2019", "tier": "T0", "problem_label": "G3", "problem_type": "Geometry", "exam": "IMO", "problem": "In triangle $A B C$, let $A_{1}$ and $B_{1}$ be two points on sides $B C$ and $A C$, and let $P$ and $Q$ be two points on segments $A A_{1}$ and $B B_{1}$, respectively, so that line $P Q$ is parallel to $A B$. On ray $P B_{1}$, beyond $B_{1}$, let $P_{1}$ be a point so that $\\angle P P_{1} C=\\angle B A C$. Similarly, on ray $Q A_{1}$, beyond $A_{1}$, let $Q_{1}$ be a point so that $\\angle C Q_{1} Q=\\angle C B A$. Show that points $P, Q, P_{1}$, and $Q_{1}$ are concyclic. (Ukraine)", "solution": "First consider the case when lines $P P_{1}$ and $Q Q_{1}$ intersect each other at some point $R$. Let line $P Q$ meet the sides $A C$ and $B C$ at $E$ and $F$, respectively. Then $$ \\angle P P_{1} C=\\angle B A C=\\angle P E C $$ so points $C, E, P, P_{1}$ lie on a circle; denote that circle by $\\omega_{P}$. It follows analogously that points $C, F, Q, Q_{1}$ lie on another circle; denote it by $\\omega_{Q}$. Let $A Q$ and $B P$ intersect at $T$. Applying Pappus' theorem to the lines $A A_{1} P$ and $B B_{1} Q$ provides that points $C=A B_{1} \\cap B A_{1}, R=A_{1} Q \\cap B_{1} P$ and $T=A Q \\cap B P$ are collinear. Let line $R C T$ meet $P Q$ and $A B$ at $S$ and $U$, respectively. From $A B \\| P Q$ we obtain $$ \\frac{S P}{S Q}=\\frac{U B}{U A}=\\frac{S F}{S E} $$ so $$ S P \\cdot S E=S Q \\cdot S F $$ ![](https://cdn.mathpix.com/cropped/2024_11_18_9ee8c2ac2c2f5002ff14g-061.jpg?height=792&width=1072&top_left_y=204&top_left_x=498) So, point $S$ has equal powers with respect to $\\omega_{P}$ and $\\omega_{Q}$, hence line $R C S$ is their radical axis; then $R$ also has equal powers to the circles, so $R P \\cdot R P_{1}=R Q \\cdot R Q_{1}$, proving that points $P, P_{1}, Q, Q_{1}$ are indeed concyclic. Now consider the case when $P P_{1}$ and $Q Q_{1}$ are parallel. Like in the previous case, let $A Q$ and $B P$ intersect at $T$. Applying Pappus' theorem again to the lines $A A_{1} P$ and $B B_{1} Q$, in this limit case it shows that line $C T$ is parallel to $P P_{1}$ and $Q Q_{1}$. Let line $C T$ meet $P Q$ and $A B$ at $S$ and $U$, as before. The same calculation as in the previous case shows that $S P \\cdot S E=S Q \\cdot S F$, so $S$ lies on the radical axis between $\\omega_{P}$ and $\\omega_{Q}$. ![](https://cdn.mathpix.com/cropped/2024_11_18_9ee8c2ac2c2f5002ff14g-061.jpg?height=800&width=1043&top_left_y=1413&top_left_x=512) Line $C S T$, that is the radical axis between $\\omega_{P}$ and $\\omega_{Q}$, is perpendicular to the line $\\ell$ of centres of $\\omega_{P}$ and $\\omega_{Q}$. Hence, the chords $P P_{1}$ and $Q Q_{1}$ are perpendicular to $\\ell$. So the quadrilateral $P P_{1} Q_{1} Q$ is an isosceles trapezium with symmetry axis $\\ell$, and hence is cyclic. Comment. There are several ways of solving the problem involving Pappus' theorem. For example, one may consider the points $K=P B_{1} \\cap B C$ and $L=Q A_{1} \\cap A C$. Applying Pappus' theorem to the lines $A A_{1} P$ and $Q B_{1} B$ we get that $K, L$, and $P Q \\cap A B$ are collinear, i.e. that $K L \\| A B$. Therefore, cyclicity of $P, Q, P_{1}$, and $Q_{1}$ is equivalent to that of $K, L, P_{1}$, and $Q_{1}$. The latter is easy after noticing that $C$ also lies on that circle. Indeed, e.g. $\\angle(L K, L C)=\\angle(A B, A C)=\\angle\\left(P_{1} K, P_{1} C\\right)$ shows that $K$ lies on circle $K L C$. This approach also has some possible degeneracy, as the points $K$ and $L$ may happen to be ideal.", "metadata": {"resource_path": "IMO/segmented/en-IMO2019SL.jsonl"}} -{"year": "2019", "tier": "T0", "problem_label": "G4", "problem_type": "Geometry", "exam": "IMO", "problem": "Let $P$ be a point inside triangle $A B C$. Let $A P$ meet $B C$ at $A_{1}$, let $B P$ meet $C A$ at $B_{1}$, and let $C P$ meet $A B$ at $C_{1}$. Let $A_{2}$ be the point such that $A_{1}$ is the midpoint of $P A_{2}$, let $B_{2}$ be the point such that $B_{1}$ is the midpoint of $P B_{2}$, and let $C_{2}$ be the point such that $C_{1}$ is the midpoint of $P C_{2}$. Prove that points $A_{2}, B_{2}$, and $C_{2}$ cannot all lie strictly inside the circumcircle of triangle $A B C$. (Australia)", "solution": "Since $$ \\angle A P B+\\angle B P C+\\angle C P A=2 \\pi=(\\pi-\\angle A C B)+(\\pi-\\angle B A C)+(\\pi-\\angle C B A), $$ at least one of the following inequalities holds: $$ \\angle A P B \\geqslant \\pi-\\angle A C B, \\quad \\angle B P C \\geqslant \\pi-\\angle B A C, \\quad \\angle C P A \\geqslant \\pi-\\angle C B A . $$ Without loss of generality, we assume that $\\angle B P C \\geqslant \\pi-\\angle B A C$. We have $\\angle B P C>\\angle B A C$ because $P$ is inside $\\triangle A B C$. So $\\angle B P C \\geqslant \\max (\\angle B A C, \\pi-\\angle B A C)$ and hence $$ \\sin \\angle B P C \\leqslant \\sin \\angle B A C . $$ Let the rays $A P, B P$, and $C P$ cross the circumcircle $\\Omega$ again at $A_{3}, B_{3}$, and $C_{3}$, respectively. We will prove that at least one of the ratios $\\frac{P B_{1}}{B_{1} B_{3}}$ and $\\frac{P C_{1}}{C_{1} C_{3}}$ is at least 1 , which yields that one of the points $B_{2}$ and $C_{2}$ does not lie strictly inside $\\Omega$. Because $A, B, C, B_{3}$ lie on a circle, the triangles $C B_{1} B_{3}$ and $B B_{1} A$ are similar, so $$ \\frac{C B_{1}}{B_{1} B_{3}}=\\frac{B B_{1}}{B_{1} A} $$ Applying the sine rule we obtain $$ \\frac{P B_{1}}{B_{1} B_{3}}=\\frac{P B_{1}}{C B_{1}} \\cdot \\frac{C B_{1}}{B_{1} B_{3}}=\\frac{P B_{1}}{C B_{1}} \\cdot \\frac{B B_{1}}{B_{1} A}=\\frac{\\sin \\angle A C P}{\\sin \\angle B P C} \\cdot \\frac{\\sin \\angle B A C}{\\sin \\angle P B A} . $$ Similarly, $$ \\frac{P C_{1}}{C_{1} C_{3}}=\\frac{\\sin \\angle P B A}{\\sin \\angle B P C} \\cdot \\frac{\\sin \\angle B A C}{\\sin \\angle A C P} $$ Multiplying these two equations we get $$ \\frac{P B_{1}}{B_{1} B_{3}} \\cdot \\frac{P C_{1}}{C_{1} C_{3}}=\\frac{\\sin ^{2} \\angle B A C}{\\sin ^{2} \\angle B P C} \\geqslant 1 $$ using (*), which yields the desired conclusion. Comment. It also cannot happen that all three points $A_{2}, B_{2}$, and $C_{2}$ lie strictly outside $\\Omega$. The same proof works almost literally, starting by assuming without loss of generality that $\\angle B P C \\leqslant \\pi-\\angle B A C$ and using $\\angle B P C>\\angle B A C$ to deduce that $\\sin \\angle B P C \\geqslant \\sin \\angle B A C$. It is possible for $A_{2}, B_{2}$, and $C_{2}$ all to lie on the circumcircle; from the above solution we may derive that this happens if and only if $P$ is the orthocentre of the triangle $A B C$, (which lies strictly inside $A B C$ if and only if $A B C$ is acute).", "metadata": {"resource_path": "IMO/segmented/en-IMO2019SL.jsonl"}} -{"year": "2019", "tier": "T0", "problem_label": "G4", "problem_type": "Geometry", "exam": "IMO", "problem": "Let $P$ be a point inside triangle $A B C$. Let $A P$ meet $B C$ at $A_{1}$, let $B P$ meet $C A$ at $B_{1}$, and let $C P$ meet $A B$ at $C_{1}$. Let $A_{2}$ be the point such that $A_{1}$ is the midpoint of $P A_{2}$, let $B_{2}$ be the point such that $B_{1}$ is the midpoint of $P B_{2}$, and let $C_{2}$ be the point such that $C_{1}$ is the midpoint of $P C_{2}$. Prove that points $A_{2}, B_{2}$, and $C_{2}$ cannot all lie strictly inside the circumcircle of triangle $A B C$. (Australia)", "solution": "Define points $A_{3}, B_{3}$, and $C_{3}$ as in Observe that $\\triangle P B C_{3} \\sim \\triangle P C B_{3}$. Let $X$ be the point on side $P B_{3}$ that corresponds to point $C_{1}$ on side $P C_{3}$ under this similarity. In other words, $X$ lies on segment $P B_{3}$ and satisfies $P X: X B_{3}=P C_{1}: C_{1} C_{3}$. It follows that $$ \\angle X C P=\\angle P B C_{1}=\\angle B_{3} B A=\\angle B_{3} C B_{1} . $$ Hence lines $C X$ and $C B_{1}$ are isogonal conjugates in $\\triangle P C B_{3}$. ![](https://cdn.mathpix.com/cropped/2024_11_18_9ee8c2ac2c2f5002ff14g-063.jpg?height=812&width=807&top_left_y=1207&top_left_x=533) Let $Y$ be the foot of the bisector of $\\angle B_{3} C P$ in $\\triangle P C B_{3}$. Since $P C_{1}0$ and $\\alpha+\\beta+\\gamma=1$. Then $$ A_{1}=\\frac{\\beta B+\\gamma C}{\\beta+\\gamma}=\\frac{1}{1-\\alpha} P-\\frac{\\alpha}{1-\\alpha} A, $$ so $$ A_{2}=2 A_{1}-P=\\frac{1+\\alpha}{1-\\alpha} P-\\frac{2 \\alpha}{1-\\alpha} A $$ Hence $$ \\left|A_{2}\\right|^{2}=\\left(\\frac{1+\\alpha}{1-\\alpha}\\right)^{2}|P|^{2}+\\left(\\frac{2 \\alpha}{1-\\alpha}\\right)^{2}|A|^{2}-\\frac{4 \\alpha(1+\\alpha)}{(1-\\alpha)^{2}} A \\cdot P $$ Using $|A|^{2}=1$ we obtain $$ \\frac{(1-\\alpha)^{2}}{2(1+\\alpha)}\\left|A_{2}\\right|^{2}=\\frac{1+\\alpha}{2}|P|^{2}+\\frac{2 \\alpha^{2}}{1+\\alpha}-2 \\alpha A \\cdot P $$ Likewise $$ \\frac{(1-\\beta)^{2}}{2(1+\\beta)}\\left|B_{2}\\right|^{2}=\\frac{1+\\beta}{2}|P|^{2}+\\frac{2 \\beta^{2}}{1+\\beta}-2 \\beta B \\cdot P $$ and $$ \\frac{(1-\\gamma)^{2}}{2(1+\\gamma)}\\left|C_{2}\\right|^{2}=\\frac{1+\\gamma}{2}|P|^{2}+\\frac{2 \\gamma^{2}}{1+\\gamma}-2 \\gamma C \\cdot P $$ Summing (1), (2) and (3) we obtain on the LHS the positive linear combination $$ \\text { LHS }=\\frac{(1-\\alpha)^{2}}{2(1+\\alpha)}\\left|A_{2}\\right|^{2}+\\frac{(1-\\beta)^{2}}{2(1+\\beta)}\\left|B_{2}\\right|^{2}+\\frac{(1-\\gamma)^{2}}{2(1+\\gamma)}\\left|C_{2}\\right|^{2} $$ and on the RHS the quantity $$ \\left(\\frac{1+\\alpha}{2}+\\frac{1+\\beta}{2}+\\frac{1+\\gamma}{2}\\right)|P|^{2}+\\left(\\frac{2 \\alpha^{2}}{1+\\alpha}+\\frac{2 \\beta^{2}}{1+\\beta}+\\frac{2 \\gamma^{2}}{1+\\gamma}\\right)-2(\\alpha A \\cdot P+\\beta B \\cdot P+\\gamma C \\cdot P) . $$ The first term is $2|P|^{2}$ and the last term is $-2 P \\cdot P$, so $$ \\begin{aligned} \\text { RHS } & =\\left(\\frac{2 \\alpha^{2}}{1+\\alpha}+\\frac{2 \\beta^{2}}{1+\\beta}+\\frac{2 \\gamma^{2}}{1+\\gamma}\\right) \\\\ & =\\frac{3 \\alpha-1}{2}+\\frac{(1-\\alpha)^{2}}{2(1+\\alpha)}+\\frac{3 \\beta-1}{2}+\\frac{(1-\\beta)^{2}}{2(1+\\beta)}+\\frac{3 \\gamma-1}{2}+\\frac{(1-\\gamma)^{2}}{2(1+\\gamma)} \\\\ & =\\frac{(1-\\alpha)^{2}}{2(1+\\alpha)}+\\frac{(1-\\beta)^{2}}{2(1+\\beta)}+\\frac{(1-\\gamma)^{2}}{2(1+\\gamma)} \\end{aligned} $$ Here we used the fact that $$ \\frac{3 \\alpha-1}{2}+\\frac{3 \\beta-1}{2}+\\frac{3 \\gamma-1}{2}=0 . $$ We have shown that a linear combination of $\\left|A_{1}\\right|^{2},\\left|B_{1}\\right|^{2}$, and $\\left|C_{1}\\right|^{2}$ with positive coefficients is equal to the sum of the coefficients. Therefore at least one of $\\left|A_{1}\\right|^{2},\\left|B_{1}\\right|^{2}$, and $\\left|C_{1}\\right|^{2}$ must be at least 1 , as required. Comment. This proof also works when $P$ is any point for which $\\alpha, \\beta, \\gamma>-1, \\alpha+\\beta+\\gamma=1$, and $\\alpha, \\beta, \\gamma \\neq 1$. (In any cases where $\\alpha=1$ or $\\beta=1$ or $\\gamma=1$, some points in the construction are not defined.) This page is intentionally left blank", "metadata": {"resource_path": "IMO/segmented/en-IMO2019SL.jsonl"}} -{"year": "2019", "tier": "T0", "problem_label": "G5", "problem_type": "Geometry", "exam": "IMO", "problem": "Let $A B C D E$ be a convex pentagon with $C D=D E$ and $\\angle E D C \\neq 2 \\cdot \\angle A D B$. Suppose that a point $P$ is located in the interior of the pentagon such that $A P=A E$ and $B P=B C$. Prove that $P$ lies on the diagonal $C E$ if and only if $\\operatorname{area}(B C D)+\\operatorname{area}(A D E)=$ $\\operatorname{area}(A B D)+\\operatorname{area}(A B P)$. (Hungary)", "solution": "Let $P^{\\prime}$ be the reflection of $P$ across line $A B$, and let $M$ and $N$ be the midpoints of $P^{\\prime} E$ and $P^{\\prime} C$ respectively. Convexity ensures that $P^{\\prime}$ is distinct from both $E$ and $C$, and hence from both $M$ and $N$. We claim that both the area condition and the collinearity condition in the problem are equivalent to the condition that the (possibly degenerate) right-angled triangles $A P^{\\prime} M$ and $B P^{\\prime} N$ are directly similar (equivalently, $A P^{\\prime} E$ and $B P^{\\prime} C$ are directly similar). ![](https://cdn.mathpix.com/cropped/2024_11_18_9ee8c2ac2c2f5002ff14g-066.jpg?height=535&width=549&top_left_y=806&top_left_x=756) For the equivalence with the collinearity condition, let $F$ denote the foot of the perpendicular from $P^{\\prime}$ to $A B$, so that $F$ is the midpoint of $P P^{\\prime}$. We have that $P$ lies on $C E$ if and only if $F$ lies on $M N$, which occurs if and only if we have the equality $\\angle A F M=\\angle B F N$ of signed angles modulo $\\pi$. By concyclicity of $A P^{\\prime} F M$ and $B F P^{\\prime} N$, this is equivalent to $\\angle A P^{\\prime} M=\\angle B P^{\\prime} N$, which occurs if and only if $A P^{\\prime} M$ and $B P^{\\prime} N$ are directly similar. ![](https://cdn.mathpix.com/cropped/2024_11_18_9ee8c2ac2c2f5002ff14g-066.jpg?height=306&width=441&top_left_y=1663&top_left_x=813) For the other equivalence with the area condition, we have the equality of signed areas $\\operatorname{area}(A B D)+\\operatorname{area}(A B P)=\\operatorname{area}\\left(A P^{\\prime} B D\\right)=\\operatorname{area}\\left(A P^{\\prime} D\\right)+\\operatorname{area}\\left(B D P^{\\prime}\\right)$. Using the identity $\\operatorname{area}(A D E)-\\operatorname{area}\\left(A P^{\\prime} D\\right)=\\operatorname{area}(A D E)+\\operatorname{area}\\left(A D P^{\\prime}\\right)=2$ area $(A D M)$, and similarly for $B$, we find that the area condition is equivalent to the equality $$ \\operatorname{area}(D A M)=\\operatorname{area}(D B N) . $$ Now note that $A$ and $B$ lie on the perpendicular bisectors of $P^{\\prime} E$ and $P^{\\prime} C$, respectively. If we write $G$ and $H$ for the feet of the perpendiculars from $D$ to these perpendicular bisectors respectively, then this area condition can be rewritten as $$ M A \\cdot G D=N B \\cdot H D $$ (In this condition, we interpret all lengths as signed lengths according to suitable conventions: for instance, we orient $P^{\\prime} E$ from $P^{\\prime}$ to $E$, orient the parallel line $D H$ in the same direction, and orient the perpendicular bisector of $P^{\\prime} E$ at an angle $\\pi / 2$ clockwise from the oriented segment $P^{\\prime} E$ - we adopt the analogous conventions at $B$.) ![](https://cdn.mathpix.com/cropped/2024_11_18_9ee8c2ac2c2f5002ff14g-067.jpg?height=529&width=575&top_left_y=204&top_left_x=758) To relate the signed lengths $G D$ and $H D$ to the triangles $A P^{\\prime} M$ and $B P^{\\prime} N$, we use the following calculation. Claim. Let $\\Gamma$ denote the circle centred on $D$ with both $E$ and $C$ on the circumference, and $h$ the power of $P^{\\prime}$ with respect to $\\Gamma$. Then we have the equality $$ G D \\cdot P^{\\prime} M=H D \\cdot P^{\\prime} N=\\frac{1}{4} h \\neq 0 . $$ Proof. Firstly, we have $h \\neq 0$, since otherwise $P^{\\prime}$ would lie on $\\Gamma$, and hence the internal angle bisectors of $\\angle E D P^{\\prime}$ and $\\angle P^{\\prime} D C$ would pass through $A$ and $B$ respectively. This would violate the angle inequality $\\angle E D C \\neq 2 \\cdot \\angle A D B$ given in the question. Next, let $E^{\\prime}$ denote the second point of intersection of $P^{\\prime} E$ with $\\Gamma$, and let $E^{\\prime \\prime}$ denote the point on $\\Gamma$ diametrically opposite $E^{\\prime}$, so that $E^{\\prime \\prime} E$ is perpendicular to $P^{\\prime} E$. The point $G$ lies on the perpendicular bisectors of the sides $P^{\\prime} E$ and $E E^{\\prime \\prime}$ of the right-angled triangle $P^{\\prime} E E^{\\prime \\prime}$; it follows that $G$ is the midpoint of $P^{\\prime} E^{\\prime \\prime}$. Since $D$ is the midpoint of $E^{\\prime} E^{\\prime \\prime}$, we have that $G D=\\frac{1}{2} P^{\\prime} E^{\\prime}$. Since $P^{\\prime} M=\\frac{1}{2} P^{\\prime} E$, we have $G D \\cdot P^{\\prime} M=\\frac{1}{4} P^{\\prime} E^{\\prime} \\cdot P^{\\prime} E=\\frac{1}{4} h$. The other equality $H D \\cdot P^{\\prime} N$ follows by exactly the same argument. ![](https://cdn.mathpix.com/cropped/2024_11_18_9ee8c2ac2c2f5002ff14g-067.jpg?height=727&width=672&top_left_y=1664&top_left_x=726) From this claim, we see that the area condition is equivalent to the equality $$ \\left(M A: P^{\\prime} M\\right)=\\left(N B: P^{\\prime} N\\right) $$ of ratios of signed lengths, which is equivalent to direct similarity of $A P^{\\prime} M$ and $B P^{\\prime} N$, as desired.", "metadata": {"resource_path": "IMO/segmented/en-IMO2019SL.jsonl"}} -{"year": "2019", "tier": "T0", "problem_label": "G5", "problem_type": "Geometry", "exam": "IMO", "problem": "Let $A B C D E$ be a convex pentagon with $C D=D E$ and $\\angle E D C \\neq 2 \\cdot \\angle A D B$. Suppose that a point $P$ is located in the interior of the pentagon such that $A P=A E$ and $B P=B C$. Prove that $P$ lies on the diagonal $C E$ if and only if $\\operatorname{area}(B C D)+\\operatorname{area}(A D E)=$ $\\operatorname{area}(A B D)+\\operatorname{area}(A B P)$. (Hungary)", "solution": "Along the perpendicular bisector of $C E$, define the linear function $$ f(X)=\\operatorname{area}(B C X)+\\operatorname{area}(A X E)-\\operatorname{area}(A B X)-\\operatorname{area}(A B P), $$ where, from now on, we always use signed areas. Thus, we want to show that $C, P, E$ are collinear if and only if $f(D)=0$. ![](https://cdn.mathpix.com/cropped/2024_11_18_9ee8c2ac2c2f5002ff14g-068.jpg?height=433&width=512&top_left_y=503&top_left_x=772) Let $P^{\\prime}$ be the reflection of $P$ across line $A B$. The point $P^{\\prime}$ does not lie on the line $C E$. To see this, we let $A^{\\prime \\prime}$ and $B^{\\prime \\prime}$ be the points obtained from $A$ and $B$ by dilating with scale factor 2 about $P^{\\prime}$, so that $P$ is the orthogonal projection of $P^{\\prime}$ onto $A^{\\prime \\prime} B^{\\prime \\prime}$. Since $A$ lies on the perpendicular bisector of $P^{\\prime} E$, the triangle $A^{\\prime \\prime} E P^{\\prime}$ is right-angled at $E$ (and $B^{\\prime \\prime} C P^{\\prime}$ similarly). If $P^{\\prime}$ were to lie on $C E$, then the lines $A^{\\prime \\prime} E$ and $B^{\\prime \\prime} C$ would be perpendicular to $C E$ and $A^{\\prime \\prime}$ and $B^{\\prime \\prime}$ would lie on the opposite side of $C E$ to $D$. It follows that the line $A^{\\prime \\prime} B^{\\prime \\prime}$ does not meet triangle $C D E$, and hence point $P$ does not lie inside $C D E$. But then $P$ must lie inside $A B C E$, and it is clear that such a point cannot reflect to a point $P^{\\prime}$ on $C E$. We thus let $O$ be the centre of the circle $C E P^{\\prime}$. The lines $A O$ and $B O$ are the perpendicular bisectors of $E P^{\\prime}$ and $C P^{\\prime}$, respectively, so $$ \\begin{aligned} \\operatorname{area}(B C O)+\\operatorname{area}(A O E) & =\\operatorname{area}\\left(O P^{\\prime} B\\right)+\\operatorname{area}\\left(P^{\\prime} O A\\right)=\\operatorname{area}\\left(P^{\\prime} B O A\\right) \\\\ & =\\operatorname{area}(A B O)+\\operatorname{area}\\left(B A P^{\\prime}\\right)=\\operatorname{area}(A B O)+\\operatorname{area}(A B P), \\end{aligned} $$ and hence $f(O)=0$. Notice that if point $O$ coincides with $D$ then points $A, B$ lie in angle domain $C D E$ and $\\angle E O C=2 \\cdot \\angle A O B$, which is not allowed. So, $O$ and $D$ must be distinct. Since $f$ is linear and vanishes at $O$, it follows that $f(D)=0$ if and only if $f$ is constant zero - we want to show this occurs if and only if $C, P, E$ are collinear. ![](https://cdn.mathpix.com/cropped/2024_11_18_9ee8c2ac2c2f5002ff14g-068.jpg?height=556&width=1254&top_left_y=1906&top_left_x=408) In the one direction, suppose firstly that $C, P, E$ are not collinear, and let $T$ be the centre of the circle $C E P$. The same calculation as above provides $$ \\operatorname{area}(B C T)+\\operatorname{area}(A T E)=\\operatorname{area}(P B T A)=\\operatorname{area}(A B T)-\\operatorname{area}(A B P) $$ $$ f(T)=-2 \\operatorname{area}(A B P) \\neq 0 $$ Hence, the linear function $f$ is nonconstant with its zero is at $O$, so that $f(D) \\neq 0$. In the other direction, suppose that the points $C, P, E$ are collinear. We will show that $f$ is constant zero by finding a second point (other than $O$ ) at which it vanishes. ![](https://cdn.mathpix.com/cropped/2024_11_18_9ee8c2ac2c2f5002ff14g-069.jpg?height=410&width=1098&top_left_y=366&top_left_x=479) Let $Q$ be the reflection of $P$ across the midpoint of $A B$, so $P A Q B$ is a parallelogram. It is easy to see that $Q$ is on the perpendicular bisector of $C E$; for instance if $A^{\\prime}$ and $B^{\\prime}$ are the points produced from $A$ and $B$ by dilating about $P$ with scale factor 2, then the projection of $Q$ to $C E$ is the midpoint of the projections of $A^{\\prime}$ and $B^{\\prime}$, which are $E$ and $C$ respectively. The triangles $B C Q$ and $A Q E$ are indirectly congruent, so $$ f(Q)=(\\operatorname{area}(B C Q)+\\operatorname{area}(A Q E))-(\\operatorname{area}(A B Q)-\\operatorname{area}(B A P))=0-0=0 $$ The points $O$ and $Q$ are distinct. To see this, consider the circle $\\omega$ centred on $Q$ with $P^{\\prime}$ on the circumference; since triangle $P P^{\\prime} Q$ is right-angled at $P^{\\prime}$, it follows that $P$ lies outside $\\omega$. On the other hand, $P$ lies between $C$ and $E$ on the line $C P E$. It follows that $C$ and $E$ cannot both lie on $\\omega$, so that $\\omega$ is not the circle $C E P^{\\prime}$ and $Q \\neq O$. Since $O$ and $Q$ are distinct zeroes of the linear function $f$, we have $f(D)=0$ as desired. Comment 1. The condition $\\angle E D C \\neq 2 \\cdot \\angle A D B$ cannot be omitted. If $D$ is the centre of circle $C E P^{\\prime}$, then the condition on triangle areas is satisfied automatically, without having $P$ on line $C E$. Comment 2. The \"only if\" part of this problem is easier than the \"if\" part. For example, in the second part of Solution 2, the triangles $E A Q$ and $Q B C$ are indirectly congruent, so the sum of their areas is 0 , and $D C Q E$ is a kite. Now one can easily see that $\\angle(A Q, D E)=\\angle(C D, C B)$ and $\\angle(B Q, D C)=\\angle(E D, E A)$, whence area $(B C D)=\\operatorname{area}(A Q D)+\\operatorname{area}(E Q A)$ and $\\operatorname{area}(A D E)=$ $\\operatorname{area}(B D Q)+\\operatorname{area}(B Q C)$, which yields the result. Comment 3. The origin of the problem is the following observation. Let $A B D H$ be a tetrahedron and consider the sphere $\\mathcal{S}$ that is tangent to the four face planes, internally to planes $A D H$ and $B D H$ and externally to $A B D$ and $A B H$ (or vice versa). It is known that the sphere $\\mathcal{S}$ exists if and only if area $(A D H)+\\operatorname{area}(B D H) \\neq \\operatorname{area}(A B H)+\\operatorname{area}(A B D)$; this relation comes from the usual formula for the volume of the tetrahedron. Let $T, T_{a}, T_{b}, T_{d}$ be the points of tangency between the sphere and the four planes, as shown in the picture. Rotate the triangle $A B H$ inward, the triangles $B D H$ and $A D H$ outward, into the triangles $A B P, B D C$ and $A D E$, respectively, in the plane $A B D$. Notice that the points $T_{d}, T_{a}, T_{b}$ are rotated to $T$, so we have $H T_{a}=H T_{b}=H T_{d}=P T=C T=E T$. Therefore, the point $T$ is the centre of the circle $C E P$. Hence, if the sphere exists then $C, E, P$ cannot be collinear. If the condition $\\angle E D C \\neq 2 \\cdot \\angle A D B$ is replaced by the constraint that the angles $\\angle E D A, \\angle A D B$ and $\\angle B D C$ satisfy the triangle inequality, it enables reconstructing the argument with the tetrahedron and the tangent sphere. ![](https://cdn.mathpix.com/cropped/2024_11_18_9ee8c2ac2c2f5002ff14g-070.jpg?height=1069&width=1306&top_left_y=825&top_left_x=381)", "metadata": {"resource_path": "IMO/segmented/en-IMO2019SL.jsonl"}} -{"year": "2019", "tier": "T0", "problem_label": "G6", "problem_type": "Geometry", "exam": "IMO", "problem": "Let $I$ be the incentre of acute-angled triangle $A B C$. Let the incircle meet $B C, C A$, and $A B$ at $D, E$, and $F$, respectively. Let line $E F$ intersect the circumcircle of the triangle at $P$ and $Q$, such that $F$ lies between $E$ and $P$. Prove that $\\angle D P A+\\angle A Q D=\\angle Q I P$. (Slovakia)", "solution": "Let $N$ and $M$ be the midpoints of the arcs $\\widehat{B C}$ of the circumcircle, containing and opposite vertex $A$, respectively. By $\\angle F A E=\\angle B A C=\\angle B N C$, the right-angled kites $A F I E$ and $N B M C$ are similar. Consider the spiral similarity $\\varphi$ (dilation in case of $A B=A C$ ) that moves $A F I E$ to $N B M C$. The directed angle in which $\\varphi$ changes directions is $\\angle(A F, N B)$, same as $\\angle(A P, N P)$ and $\\angle(A Q, N Q)$; so lines $A P$ and $A Q$ are mapped to lines $N P$ and $N Q$, respectively. Line $E F$ is mapped to $B C$; we can see that the intersection points $P=E F \\cap A P$ and $Q=E F \\cap A Q$ are mapped to points $B C \\cap N P$ and $B C \\cap N Q$, respectively. Denote these points by $P^{\\prime}$ and $Q^{\\prime}$, respectively. ![](https://cdn.mathpix.com/cropped/2024_11_18_9ee8c2ac2c2f5002ff14g-071.jpg?height=689&width=1512&top_left_y=883&top_left_x=272) Let $L$ be the midpoint of $B C$. We claim that points $P, Q, D, L$ are concyclic (if $D=L$ then line $B C$ is tangent to circle $P Q D$ ). Let $P Q$ and $B C$ meet at $Z$. By applying Menelaus' theorem to triangle $A B C$ and line $E F Z$, we have $$ \\frac{B D}{D C}=\\frac{B F}{F A} \\cdot \\frac{A E}{E C}=-\\frac{B Z}{Z C}, $$ so the pairs $B, C$ and $D, Z$ are harmonic. It is well-known that this implies $Z B \\cdot Z C=Z D \\cdot Z L$. (The inversion with pole $Z$ that swaps $B$ and $C$ sends $Z$ to infinity and $D$ to the midpoint of $B C$, because the cross-ratio is preserved.) Hence, $Z D \\cdot Z L=Z B \\cdot Z C=Z P \\cdot Z Q$ by the power of $Z$ with respect to the circumcircle; this proves our claim. By $\\angle M P P^{\\prime}=\\angle M Q Q^{\\prime}=\\angle M L P^{\\prime}=\\angle M L Q^{\\prime}=90^{\\circ}$, the quadrilaterals $M L P P^{\\prime}$ and $M L Q Q^{\\prime}$ are cyclic. Then the problem statement follows by $$ \\begin{aligned} \\angle D P A+\\angle A Q D & =360^{\\circ}-\\angle P A Q-\\angle Q D P=360^{\\circ}-\\angle P N Q-\\angle Q L P \\\\ & =\\angle L P N+\\angle N Q L=\\angle P^{\\prime} M L+\\angle L M Q^{\\prime}=\\angle P^{\\prime} M Q^{\\prime}=\\angle P I Q . \\end{aligned} $$", "metadata": {"resource_path": "IMO/segmented/en-IMO2019SL.jsonl"}} -{"year": "2019", "tier": "T0", "problem_label": "G6", "problem_type": "Geometry", "exam": "IMO", "problem": "Let $I$ be the incentre of acute-angled triangle $A B C$. Let the incircle meet $B C, C A$, and $A B$ at $D, E$, and $F$, respectively. Let line $E F$ intersect the circumcircle of the triangle at $P$ and $Q$, such that $F$ lies between $E$ and $P$. Prove that $\\angle D P A+\\angle A Q D=\\angle Q I P$. (Slovakia)", "solution": "Define the point $M$ and the same spiral similarity $\\varphi$ as in the previous solution. (The point $N$ is not necessary.) It is well-known that the centre of the spiral similarity that maps $F, E$ to $B, C$ is the Miquel point of the lines $F E, B C, B F$ and $C E$; that is, the second intersection of circles $A B C$ and $A E F$. Denote that point by $S$. By $\\varphi(F)=B$ and $\\varphi(E)=C$ the triangles $S B F$ and $S C E$ are similar, so we have $$ \\frac{S B}{S C}=\\frac{B F}{C E}=\\frac{B D}{C D} $$ By the converse of the angle bisector theorem, that indicates that line $S D$ bisects $\\angle B S C$ and hence passes through $M$. Let $K$ be the intersection point of lines $E F$ and $S I$. Notice that $\\varphi$ sends points $S, F, E, I$ to $S, B, C, M$, so $\\varphi(K)=\\varphi(F E \\cap S I)=B C \\cap S M=D$. By $\\varphi(I)=M$, we have $K D \\| I M$. ![](https://cdn.mathpix.com/cropped/2024_11_18_9ee8c2ac2c2f5002ff14g-072.jpg?height=669&width=706&top_left_y=839&top_left_x=675) We claim that triangles $S P I$ and $S D Q$ are similar, and so are triangles $S P D$ and $S I Q$. Let ray $S I$ meet the circumcircle again at $L$. Note that the segment $E F$ is perpendicular to the angle bisector $A M$. Then by $\\angle A M L=\\angle A S L=\\angle A S I=90^{\\circ}$, we have $M L \\| P Q$. Hence, $\\widetilde{P L}=\\widetilde{M Q}$ and therefore $\\angle P S L=\\angle M S Q=\\angle D S Q$. By $\\angle Q P S=\\angle Q M S$, the triangles $S P K$ and $S M Q$ are similar. Finally, $$ \\frac{S P}{S I}=\\frac{S P}{S K} \\cdot \\frac{S K}{S I}=\\frac{S M}{S Q} \\cdot \\frac{S D}{S M}=\\frac{S D}{S Q} $$ shows that triangles $S P I$ and $S D Q$ are similar. The second part of the claim can be proved analogously. Now the problem statement can be proved by $$ \\angle D P A+\\angle A Q D=\\angle D P S+\\angle S Q D=\\angle Q I S+\\angle S I P=\\angle Q I P . $$", "metadata": {"resource_path": "IMO/segmented/en-IMO2019SL.jsonl"}} -{"year": "2019", "tier": "T0", "problem_label": "G6", "problem_type": "Geometry", "exam": "IMO", "problem": "Let $I$ be the incentre of acute-angled triangle $A B C$. Let the incircle meet $B C, C A$, and $A B$ at $D, E$, and $F$, respectively. Let line $E F$ intersect the circumcircle of the triangle at $P$ and $Q$, such that $F$ lies between $E$ and $P$. Prove that $\\angle D P A+\\angle A Q D=\\angle Q I P$. (Slovakia)", "solution": "Denote the circumcircle of triangle $A B C$ by $\\Gamma$, and let rays $P D$ and $Q D$ meet $\\Gamma$ again at $V$ and $U$, respectively. We will show that $A U \\perp I P$ and $A V \\perp I Q$. Then the problem statement will follow as $$ \\angle D P A+\\angle A Q D=\\angle V U A+\\angle A V U=180^{\\circ}-\\angle U A V=\\angle Q I P . $$ Let $M$ be the midpoint of arc $\\widehat{B U V C}$ and let $N$ be the midpoint of arc $\\widehat{C A B}$; the lines $A I M$ and $A N$ being the internal and external bisectors of angle $B A C$, respectively, are perpendicular. Let the tangents drawn to $\\Gamma$ at $B$ and $C$ meet at $R$; let line $P Q$ meet $A U, A I, A V$ and $B C$ at $X, T, Y$ and $Z$, respectively. As in Solution 1, we observe that the pairs $B, C$ and $D, Z$ are harmonic. Projecting these points from $Q$ onto the circumcircle, we can see that $B, C$ and $U, P$ are also harmonic. Analogously, the pair $V, Q$ is harmonic with $B, C$. Consider the inversion about the circle with centre $R$, passing through $B$ and $C$. Points $B$ and $C$ are fixed points, so this inversion exchanges every point of $\\Gamma$ by its harmonic pair with respect to $B, C$. In particular, the inversion maps points $B, C, N, U, V$ to points $B, C, M, P, Q$, respectively. Combine the inversion with projecting $\\Gamma$ from $A$ to line $P Q$; the points $B, C, M, P, Q$ are projected to $F, E, T, P, Q$, respectively. ![](https://cdn.mathpix.com/cropped/2024_11_18_9ee8c2ac2c2f5002ff14g-073.jpg?height=1006&width=1384&top_left_y=1082&top_left_x=339) The combination of these two transformations is projective map from the lines $A B, A C$, $A N, A U, A V$ to $I F, I E, I T, I P, I Q$, respectively. On the other hand, we have $A B \\perp I F$, $A C \\perp I E$ and $A N \\perp A T$, so the corresponding lines in these two pencils are perpendicular. This proves $A U \\perp I P$ and $A V \\perp I Q$, and hence completes the solution.", "metadata": {"resource_path": "IMO/segmented/en-IMO2019SL.jsonl"}} -{"year": "2019", "tier": "T0", "problem_label": "G7", "problem_type": "Geometry", "exam": "IMO", "problem": "The incircle $\\omega$ of acute-angled scalene triangle $A B C$ has centre $I$ and meets sides $B C$, $C A$, and $A B$ at $D, E$, and $F$, respectively. The line through $D$ perpendicular to $E F$ meets $\\omega$ again at $R$. Line $A R$ meets $\\omega$ again at $P$. The circumcircles of triangles $P C E$ and $P B F$ meet again at $Q \\neq P$. Prove that lines $D I$ and $P Q$ meet on the external bisector of angle $B A C$. (India)", "solution": "Step 1. The external bisector of $\\angle B A C$ is the line through $A$ perpendicular to $I A$. Let $D I$ meet this line at $L$ and let $D I$ meet $\\omega$ at $K$. Let $N$ be the midpoint of $E F$, which lies on $I A$ and is the pole of line $A L$ with respect to $\\omega$. Since $A N \\cdot A I=A E^{2}=A R \\cdot A P$, the points $R$, $N, I$, and $P$ are concyclic. As $I R=I P$, the line $N I$ is the external bisector of $\\angle P N R$, so $P N$ meets $\\omega$ again at the point symmetric to $R$ with respect to $A N-i . e$ at $K$. Let $D N$ cross $\\omega$ again at $S$. Opposite sides of any quadrilateral inscribed in the circle $\\omega$ meet on the polar line of the intersection of the diagonals with respect to $\\omega$. Since $L$ lies on the polar line $A L$ of $N$ with respect to $\\omega$, the line $P S$ must pass through $L$. Thus it suffices to prove that the points $S, Q$, and $P$ are collinear. ![](https://cdn.mathpix.com/cropped/2024_11_18_9ee8c2ac2c2f5002ff14g-074.jpg?height=875&width=1350&top_left_y=1176&top_left_x=361) Step 2. Let $\\Gamma$ be the circumcircle of $\\triangle B I C$. Notice that $$ \\begin{aligned} & \\angle(B Q, Q C)=\\angle(B Q, Q P)+\\angle(P Q, Q C)=\\angle(B F, F P)+\\angle(P E, E C) \\\\ &=\\angle(E F, E P)+\\angle(F P, F E)=\\angle(F P, E P)=\\angle(D F, D E)=\\angle(B I, I C) \\end{aligned} $$ so $Q$ lies on $\\Gamma$. Let $Q P$ meet $\\Gamma$ again at $T$. It will now suffice to prove that $S, P$, and $T$ are collinear. Notice that $\\angle(B I, I T)=\\angle(B Q, Q T)=\\angle(B F, F P)=\\angle(F K, K P)$. Note $F D \\perp F K$ and $F D \\perp B I$ so $F K \\| B I$ and hence $I T$ is parallel to the line $K N P$. Since $D I=I K$, the line $I T$ crosses $D N$ at its midpoint $M$. Step 3. Let $F^{\\prime}$ and $E^{\\prime}$ be the midpoints of $D E$ and $D F$, respectively. Since $D E^{\\prime} \\cdot E^{\\prime} F=D E^{\\prime 2}=$ $B E^{\\prime} \\cdot E^{\\prime} I$, the point $E^{\\prime}$ lies on the radical axis of $\\omega$ and $\\Gamma$; the same holds for $F^{\\prime}$. Therefore, this radical axis is $E^{\\prime} F^{\\prime}$, and it passes through $M$. Thus $I M \\cdot M T=D M \\cdot M S$, so $S, I, D$, and $T$ are concyclic. This shows $\\angle(D S, S T)=\\angle(D I, I T)=\\angle(D K, K P)=\\angle(D S, S P)$, whence the points $S, P$, and $T$ are collinear, as desired. ![](https://cdn.mathpix.com/cropped/2024_11_18_9ee8c2ac2c2f5002ff14g-075.jpg?height=1089&width=992&top_left_y=198&top_left_x=532) Comment. Here is a longer alternative proof in step 1 that $P, S$, and $L$ are collinear, using a circular inversion instead of the fact that opposite sides of a quadrilateral inscribed in a circle $\\omega$ meet on the polar line with respect to $\\omega$ of the intersection of the diagonals. Let $G$ be the foot of the altitude from $N$ to the line $D I K L$. Observe that $N, G, K, S$ are concyclic (opposite right angles) so $$ \\angle D I P=2 \\angle D K P=\\angle G K N+\\angle D S P=\\angle G S N+\\angle N S P=\\angle G S P, $$ hence $I, G, S, P$ are concyclic. We have $I G \\cdot I L=I N \\cdot I A=r^{2}$ since $\\triangle I G N \\sim \\triangle I A L$. Inverting the circle $I G S P$ in circle $\\omega$, points $P$ and $S$ are fixed and $G$ is taken to $L$ so we find that $P, S$, and $L$ are collinear.", "metadata": {"resource_path": "IMO/segmented/en-IMO2019SL.jsonl"}} -{"year": "2019", "tier": "T0", "problem_label": "G7", "problem_type": "Geometry", "exam": "IMO", "problem": "The incircle $\\omega$ of acute-angled scalene triangle $A B C$ has centre $I$ and meets sides $B C$, $C A$, and $A B$ at $D, E$, and $F$, respectively. The line through $D$ perpendicular to $E F$ meets $\\omega$ again at $R$. Line $A R$ meets $\\omega$ again at $P$. The circumcircles of triangles $P C E$ and $P B F$ meet again at $Q \\neq P$. Prove that lines $D I$ and $P Q$ meet on the external bisector of angle $B A C$. (India)", "solution": "We start as in Step 1. Let $A R$ meet the circumcircle $\\Omega$ of $A B C$ again at $X$. The lines $A R$ and $A K$ are isogonal in the angle $B A C$; it is well known that in this case $X$ is the tangency point of $\\Omega$ with the $A$-mixtilinear circle. It is also well known that for this point $X$, the line $X I$ crosses $\\Omega$ again at the midpoint $M^{\\prime}$ of arc $B A C$. Step 2. Denote the circles $B F P$ and $C E P$ by $\\Omega_{B}$ and $\\Omega_{C}$, respectively. Let $\\Omega_{B}$ cross $A R$ and $E F$ again at $U$ and $Y$, respectively. We have $$ \\angle(U B, B F)=\\angle(U P, P F)=\\angle(R P, P F)=\\angle(R F, F A) $$ so $U B \\| R F$. ![](https://cdn.mathpix.com/cropped/2024_11_18_9ee8c2ac2c2f5002ff14g-076.jpg?height=1009&width=1474&top_left_y=883&top_left_x=248) Next, we show that the points $B, I, U$, and $X$ are concyclic. Since $$ \\angle(U B, U X)=\\angle(R F, R X)=\\angle(A F, A R)+\\angle(F R, F A)=\\angle\\left(M^{\\prime} B, M^{\\prime} X\\right)+\\angle(D R, D F) $$ it suffices to prove $\\angle(I B, I X)=\\angle\\left(M^{\\prime} B, M^{\\prime} X\\right)+\\angle(D R, D F)$, or $\\angle\\left(I B, M^{\\prime} B\\right)=\\angle(D R, D F)$. But both angles equal $\\angle(C I, C B)$, as desired. (This is where we used the fact that $M^{\\prime}$ is the midpoint of $\\operatorname{arc} B A C$ of $\\Omega$.) It follows now from circles $B U I X$ and $B P U F Y$ that $$ \\begin{aligned} \\angle(I U, U B)=\\angle(I X, B X)=\\angle\\left(M^{\\prime} X, B X\\right)= & \\frac{\\pi-\\angle A}{2} \\\\ & =\\angle(E F, A F)=\\angle(Y F, B F)=\\angle(Y U, B U) \\end{aligned} $$ so the points $Y, U$, and $I$ are collinear. Let $E F$ meet $B C$ at $W$. We have $$ \\angle(I Y, Y W)=\\angle(U Y, F Y)=\\angle(U B, F B)=\\angle(R F, A F)=\\angle(C I, C W) $$ so the points $W, Y, I$, and $C$ are concyclic. Similarly, if $V$ and $Z$ are the second meeting points of $\\Omega_{C}$ with $A R$ and $E F$, we get that the 4-tuples $(C, V, I, X)$ and $(B, I, Z, W)$ are both concyclic. Step 3. Let $Q^{\\prime}=C Y \\cap B Z$. We will show that $Q^{\\prime}=Q$. First of all, we have $$ \\begin{aligned} & \\angle\\left(Q^{\\prime} Y, Q^{\\prime} B\\right)=\\angle(C Y, Z B)=\\angle(C Y, Z Y)+\\angle(Z Y, B Z) \\\\ & =\\angle(C I, I W)+\\angle(I W, I B)=\\angle(C I, I B)=\\frac{\\pi-\\angle A}{2}=\\angle(F Y, F B), \\end{aligned} $$ so $Q^{\\prime} \\in \\Omega_{B}$. Similarly, $Q^{\\prime} \\in \\Omega_{C}$. Thus $Q^{\\prime} \\in \\Omega_{B} \\cap \\Omega_{C}=\\{P, Q\\}$ and it remains to prove that $Q^{\\prime} \\neq P$. If we had $Q^{\\prime}=P$, we would have $\\angle(P Y, P Z)=\\angle\\left(Q^{\\prime} Y, Q^{\\prime} Z\\right)=\\angle(I C, I B)$. This would imply $$ \\angle(P Y, Y F)+\\angle(E Z, Z P)=\\angle(P Y, P Z)=\\angle(I C, I B)=\\angle(P E, P F), $$ so circles $\\Omega_{B}$ and $\\Omega_{C}$ would be tangent at $P$. That is excluded in the problem conditions, so $Q^{\\prime}=Q$. ![](https://cdn.mathpix.com/cropped/2024_11_18_9ee8c2ac2c2f5002ff14g-077.jpg?height=966&width=1468&top_left_y=1070&top_left_x=251) Step 4. Now we are ready to show that $P, Q$, and $S$ are collinear. Notice that $A$ and $D$ are the poles of $E W$ and $D W$ with respect to $\\omega$, so $W$ is the pole of $A D$. Hence, $W I \\perp A D$. Since $C I \\perp D E$, this yields $\\angle(I C, W I)=\\angle(D E, D A)$. On the other hand, $D A$ is a symmedian in $\\triangle D E F$, so $\\angle(D E, D A)=\\angle(D N, D F)=\\angle(D S, D F)$. Therefore, $$ \\begin{aligned} \\angle(P S, P F)=\\angle(D S, D F)=\\angle(D E, D A)= & \\angle(I C, I W) \\\\ & =\\angle(Y C, Y W)=\\angle(Y Q, Y F)=\\angle(P Q, P F), \\end{aligned} $$ which yields the desired collinearity.", "metadata": {"resource_path": "IMO/segmented/en-IMO2019SL.jsonl"}} -{"year": "2019", "tier": "T0", "problem_label": "G8", "problem_type": "Geometry", "exam": "IMO", "problem": "Let $\\mathcal{L}$ be the set of all lines in the plane and let $f$ be a function that assigns to each line $\\ell \\in \\mathcal{L}$ a point $f(\\ell)$ on $\\ell$. Suppose that for any point $X$, and for any three lines $\\ell_{1}, \\ell_{2}, \\ell_{3}$ passing through $X$, the points $f\\left(\\ell_{1}\\right), f\\left(\\ell_{2}\\right), f\\left(\\ell_{3}\\right)$ and $X$ lie on a circle. Prove that there is a unique point $P$ such that $f(\\ell)=P$ for any line $\\ell$ passing through $P$. (Australia)", "solution": "We provide a complete characterisation of the functions satisfying the given condition. Write $\\angle\\left(\\ell_{1}, \\ell_{2}\\right)$ for the directed angle modulo $180^{\\circ}$ between the lines $\\ell_{1}$ and $\\ell_{2}$. Given a point $P$ and an angle $\\alpha \\in\\left(0,180^{\\circ}\\right)$, for each line $\\ell$, let $\\ell^{\\prime}$ be the line through $P$ satisfying $\\angle\\left(\\ell^{\\prime}, \\ell\\right)=\\alpha$, and let $h_{P, \\alpha}(\\ell)$ be the intersection point of $\\ell$ and $\\ell^{\\prime}$. We will prove that there is some pair $(P, \\alpha)$ such that $f$ and $h_{P, \\alpha}$ are the same function. Then $P$ is the unique point in the problem statement. Given an angle $\\alpha$ and a point $P$, let a line $\\ell$ be called $(P, \\alpha)$-good if $f(\\ell)=h_{P, \\alpha}(\\ell)$. Let a point $X \\neq P$ be called $(P, \\alpha)$-good if the circle $g(X)$ passes through $P$ and some point $Y \\neq P, X$ on $g(X)$ satisfies $\\angle(P Y, Y X)=\\alpha$. It follows from this definition that if $X$ is $(P, \\alpha)$ good then every point $Y \\neq P, X$ of $g(X)$ satisfies this angle condition, so $h_{P, \\alpha}(X Y)=Y$ for every $Y \\in g(X)$. Equivalently, $f(\\ell) \\in\\left\\{X, h_{P, \\alpha}(\\ell)\\right\\}$ for each line $\\ell$ passing through $X$. This shows the following lemma. Lemma 1. If $X$ is $(P, \\alpha)$-good and $\\ell$ is a line passing through $X$ then either $f(\\ell)=X$ or $\\ell$ is $(P, \\alpha)$-good. Lemma 2. If $X$ and $Y$ are different $(P, \\alpha)$-good points, then line $X Y$ is $(P, \\alpha)$-good. Proof. If $X Y$ is not $(P, \\alpha)$-good then by the previous Lemma, $f(X Y)=X$ and similarly $f(X Y)=Y$, but clearly this is impossible as $X \\neq Y$. Lemma 3. If $\\ell_{1}$ and $\\ell_{2}$ are different $(P, \\alpha)$-good lines which intersect at $X \\neq P$, then either $f\\left(\\ell_{1}\\right)=X$ or $f\\left(\\ell_{2}\\right)=X$ or $X$ is $(P, \\alpha)$-good. Proof. If $f\\left(\\ell_{1}\\right), f\\left(\\ell_{2}\\right) \\neq X$, then $g(X)$ is the circumcircle of $X, f\\left(\\ell_{1}\\right)$ and $f\\left(\\ell_{2}\\right)$. Since $\\ell_{1}$ and $\\ell_{2}$ are $(P, \\alpha)$-good lines, the angles $$ \\angle\\left(P f\\left(\\ell_{1}\\right), f\\left(\\ell_{1}\\right) X\\right)=\\angle\\left(P f\\left(\\ell_{2}\\right), f\\left(\\ell_{2}\\right) X\\right)=\\alpha $$ so $P$ lies on $g(X)$. Hence, $X$ is $(P, \\alpha)$-good. Lemma 4. If $\\ell_{1}, \\ell_{2}$ and $\\ell_{3}$ are different $(P, \\alpha)$ good lines which intersect at $X \\neq P$, then $X$ is $(P, \\alpha)$-good. Proof. This follows from the previous Lemma since at most one of the three lines $\\ell_{i}$ can satisfy $f\\left(\\ell_{i}\\right)=X$ as the three lines are all $(P, \\alpha)$-good. Lemma 5. If $A B C$ is a triangle such that $A, B, C, f(A B), f(A C)$ and $f(B C)$ are all different points, then there is some point $P$ and some angle $\\alpha$ such that $A, B$ and $C$ are $(P, \\alpha)$-good points and $A B, B C$ and $C A$ are $(P, \\alpha)-\\operatorname{good}$ lines. ![](https://cdn.mathpix.com/cropped/2024_11_18_9ee8c2ac2c2f5002ff14g-079.jpg?height=415&width=612&top_left_y=207&top_left_x=728) Proof. Let $D, E, F$ denote the points $f(B C), f(A C), f(A B)$, respectively. Then $g(A)$, $g(B)$ and $g(C)$ are the circumcircles of $A E F, B D F$ and $C D E$, respectively. Let $P \\neq F$ be the second intersection of circles $g(A)$ and $g(B)$ (or, if these circles are tangent at $F$, then $P=F$ ). By Miquel's theorem (or an easy angle chase), $g(C)$ also passes through $P$. Then by the cyclic quadrilaterals, the directed angles $$ \\angle(P D, D C)=\\angle(P F, F B)=\\angle(P E, E A)=\\alpha $$ for some angle $\\alpha$. Hence, lines $A B, B C$ and $C A$ are all $(P, \\alpha)$-good, so by Lemma $3, A, B$ and $C$ are $(P, \\alpha)$-good. (In the case where $P=D$, the line $P D$ in the equation above denotes the line which is tangent to $g(B)$ at $P=D$. Similar definitions are used for $P E$ and $P F$ in the cases where $P=E$ or $P=F$.) Consider the set $\\Omega$ of all points $(x, y)$ with integer coordinates $1 \\leqslant x, y \\leqslant 1000$, and consider the set $L_{\\Omega}$ of all horizontal, vertical and diagonal lines passing through at least one point in $\\Omega$. A simple counting argument shows that there are 5998 lines in $L_{\\Omega}$. For each line $\\ell$ in $L_{\\Omega}$ we colour the point $f(\\ell)$ red. Then there are at most 5998 red points. Now we partition the points in $\\Omega$ into 10000 ten by ten squares. Since there are at most 5998 red points, at least one of these squares $\\Omega_{10}$ contains no red points. Let $(m, n)$ be the bottom left point in $\\Omega_{10}$. Then the triangle with vertices $(m, n),(m+1, n)$ and $(m, n+1)$ satisfies the condition of Lemma 5 , so these three vertices are all $(P, \\alpha)$-good for some point $P$ and angle $\\alpha$, as are the lines joining them. From this point on, we will simply call a point or line good if it is $(P, \\alpha)$-good for this particular pair $(P, \\alpha)$. Now by Lemma 1, the line $x=m+1$ is good, as is the line $y=n+1$. Then Lemma 3 implies that $(m+1, n+1)$ is good. By applying these two lemmas repeatedly, we can prove that the line $x+y=m+n+2$ is good, then the points $(m, n+2)$ and $(m+2, n)$ then the lines $x=m+2$ and $y=n+2$, then the points $(m+2, n+1),(m+1, n+2)$ and $(m+2, n+2)$ and so on until we have prove that all points in $\\Omega_{10}$ are good. Now we will use this to prove that every point $S \\neq P$ is good. Since $g(S)$ is a circle, it passes through at most two points of $\\Omega_{10}$ on any vertical line, so at most 20 points in total. Moreover, any line $\\ell$ through $S$ intersects at most 10 points in $\\Omega_{10}$. Hence, there are at least eight lines $\\ell$ through $S$ which contain a point $Q$ in $\\Omega_{10}$ which is not on $g(S)$. Since $Q$ is not on $g(S)$, the point $f(\\ell) \\neq Q$. Hence, by Lemma 1 , the line $\\ell$ is good. Hence, at least eight good lines pass through $S$, so by Lemma 4, the point $S$ is good. Hence, every point $S \\neq P$ is good, so by Lemma 2 , every line is good. In particular, every line $\\ell$ passing through $P$ is good, and therefore satisfies $f(\\ell)=P$, as required.", "metadata": {"resource_path": "IMO/segmented/en-IMO2019SL.jsonl"}} -{"year": "2019", "tier": "T0", "problem_label": "G8", "problem_type": "Geometry", "exam": "IMO", "problem": "Let $\\mathcal{L}$ be the set of all lines in the plane and let $f$ be a function that assigns to each line $\\ell \\in \\mathcal{L}$ a point $f(\\ell)$ on $\\ell$. Suppose that for any point $X$, and for any three lines $\\ell_{1}, \\ell_{2}, \\ell_{3}$ passing through $X$, the points $f\\left(\\ell_{1}\\right), f\\left(\\ell_{2}\\right), f\\left(\\ell_{3}\\right)$ and $X$ lie on a circle. Prove that there is a unique point $P$ such that $f(\\ell)=P$ for any line $\\ell$ passing through $P$. (Australia)", "solution": "Note that for any distinct points $X, Y$, the circles $g(X)$ and $g(Y)$ meet on $X Y$ at the point $f(X Y) \\in g(X) \\cap g(Y) \\cap(X Y)$. We write $s(X, Y)$ for the second intersection point of circles $g(X)$ and $g(Y)$. Lemma 1. Suppose that $X, Y$ and $Z$ are not collinear, and that $f(X Y) \\notin\\{X, Y\\}$ and similarly for $Y Z$ and $Z X$. Then $s(X, Y)=s(Y, Z)=s(Z, X)$. Proof. The circles $g(X), g(Y)$ and $g(Z)$ through the vertices of triangle $X Y Z$ meet pairwise on the corresponding edges (produced). By Miquel's theorem, the second points of intersection of any two of the circles coincide. (See the diagram for Lemma 5 of Solution 1.) Now pick any line $\\ell$ and any six different points $Y_{1}, \\ldots, Y_{6}$ on $\\ell \\backslash\\{f(\\ell)\\}$. Pick a point $X$ not on $\\ell$ or any of the circles $g\\left(Y_{i}\\right)$. Reordering the indices if necessary, we may suppose that $Y_{1}, \\ldots, Y_{4}$ do not lie on $g(X)$, so that $f\\left(X Y_{i}\\right) \\notin\\left\\{X, Y_{i}\\right\\}$ for $1 \\leqslant i \\leqslant 4$. By applying the above lemma to triangles $X Y_{i} Y_{j}$ for $1 \\leqslant i0$, there is a point $P_{\\epsilon}$ with $g\\left(P_{\\epsilon}\\right)$ of radius at most $\\epsilon$. Then there is a point $P$ with the given property. Proof. Consider a sequence $\\epsilon_{i}=2^{-i}$ and corresponding points $P_{\\epsilon_{i}}$. Because the two circles $g\\left(P_{\\epsilon_{i}}\\right)$ and $g\\left(P_{\\epsilon_{j}}\\right)$ meet, the distance between $P_{\\epsilon_{i}}$ and $P_{\\epsilon_{j}}$ is at most $2^{1-i}+2^{1-j}$. As $\\sum_{i} \\epsilon_{i}$ converges, these points converge to some point $P$. For all $\\epsilon>0$, the point $P$ has distance at most $2 \\epsilon$ from $P_{\\epsilon}$, and all circles $g(X)$ pass through a point with distance at most $2 \\epsilon$ from $P_{\\epsilon}$, so distance at most $4 \\epsilon$ from $P$. A circle that passes distance at most $4 \\epsilon$ from $P$ for all $\\epsilon>0$ must pass through $P$, so by Lemma 1 the point $P$ has the given property. Lemma 3. Suppose no two of the circles $g(X)$ cut. Then there is a point $P$ with the given property. Proof. Consider a circle $g(X)$ with centre $Y$. The circle $g(Y)$ must meet $g(X)$ without cutting it, so has half the radius of $g(X)$. Repeating this argument, there are circles with arbitrarily small radius and the result follows by Lemma 2. Lemma 4. Suppose there are six different points $A, B_{1}, B_{2}, B_{3}, B_{4}, B_{5}$ such that no three are collinear, no four are concyclic, and all the circles $g\\left(B_{i}\\right)$ cut pairwise at $A$. Then there is a point $P$ with the given property. Proof. Consider some line $\\ell$ through $A$ that does not pass through any of the $B_{i}$ and is not tangent to any of the $g\\left(B_{i}\\right)$. Fix some direction along that line, and let $X_{\\epsilon}$ be the point on $\\ell$ that has distance $\\epsilon$ from $A$ in that direction. In what follows we consider only those $\\epsilon$ for which $X_{\\epsilon}$ does not lie on any $g\\left(B_{i}\\right)$ (this restriction excludes only finitely many possible values of $\\epsilon$ ). Consider the circle $g\\left(X_{\\epsilon}\\right)$. Because no four of the $B_{i}$ are concyclic, at most three of them lie on this circle, so at least two of them do not. There must be some sequence of $\\epsilon \\rightarrow 0$ such that it is the same two of the $B_{i}$ for all $\\epsilon$ in that sequence, so now restrict attention to that sequence, and suppose without loss of generality that $B_{1}$ and $B_{2}$ do not lie on $g\\left(X_{\\epsilon}\\right)$ for any $\\epsilon$ in that sequence. Then $f\\left(X_{\\epsilon} B_{1}\\right)$ is not $B_{1}$, so must be the other point of intersection of $X_{\\epsilon} B_{1}$ with $g\\left(B_{1}\\right)$, and the same applies with $B_{2}$. Now consider the three points $X_{\\epsilon}, f\\left(X_{\\epsilon} B_{1}\\right)$ and $f\\left(X_{\\epsilon} B_{2}\\right)$. As $\\epsilon \\rightarrow 0$, the angle at $X_{\\epsilon}$ tends to $\\angle B_{1} A B_{2}$ or $180^{\\circ}-\\angle B_{1} A B_{2}$, which is not 0 or $180^{\\circ}$ because no three of the points were collinear. All three distances between those points are bounded above by constant multiples of $\\epsilon$ (in fact, if the triangle is scaled by a factor of $1 / \\epsilon$, it tends to a fixed triangle). Thus the circumradius of those three points, which is the radius of $g\\left(X_{\\epsilon}\\right)$, is also bounded above by a constant multiple of $\\epsilon$, and so the result follows by Lemma 2 . Lemma 5. Suppose there are two points $A$ and $B$ such that $g(A)$ and $g(B)$ cut. Then there is a point $P$ with the given property. Proof. Suppose that $g(A)$ and $g(B)$ cut at $C$ and $D$. One of those points, without loss of generality $C$, must be $f(A B)$, and so lie on the line $A B$. We now consider two cases, according to whether $D$ also lies on that line.", "metadata": {"resource_path": "IMO/segmented/en-IMO2019SL.jsonl"}} -{"year": "2019", "tier": "T0", "problem_label": "G8", "problem_type": "Geometry", "exam": "IMO", "problem": "Let $\\mathcal{L}$ be the set of all lines in the plane and let $f$ be a function that assigns to each line $\\ell \\in \\mathcal{L}$ a point $f(\\ell)$ on $\\ell$. Suppose that for any point $X$, and for any three lines $\\ell_{1}, \\ell_{2}, \\ell_{3}$ passing through $X$, the points $f\\left(\\ell_{1}\\right), f\\left(\\ell_{2}\\right), f\\left(\\ell_{3}\\right)$ and $X$ lie on a circle. Prove that there is a unique point $P$ such that $f(\\ell)=P$ for any line $\\ell$ passing through $P$. (Australia)", "solution": "For any point $X$, denote by $t(X)$ the line tangent to $g(X)$ at $X$; notice that $f(t(X))=X$, so $f$ is surjective. Step 1: We find a point $P$ for which there are at least two different lines $p_{1}$ and $p_{2}$ such that $f\\left(p_{i}\\right)=P$. Choose any point $X$. If $X$ does not have this property, take any $Y \\in g(X) \\backslash\\{X\\}$; then $f(X Y)=Y$. If $Y$ does not have the property, $t(Y)=X Y$, and the circles $g(X)$ and $g(Y)$ meet again at some point $Z$. Then $f(X Z)=Z=f(Y Z)$, so $Z$ has the required property. We will show that $P$ is the desired point. From now on, we fix two different lines $p_{1}$ and $p_{2}$ with $f\\left(p_{1}\\right)=f\\left(p_{2}\\right)=P$. Assume for contradiction that $f(\\ell)=Q \\neq P$ for some line $\\ell$ through $P$. We fix $\\ell$, and note that $Q \\in g(P)$. Step 2: We prove that $P \\in g(Q)$. Take an arbitrary point $X \\in \\ell \\backslash\\{P, Q\\}$. Two cases are possible for the position of $t(X)$ in relation to the $p_{i}$; we will show that each case (and subcase) occurs for only finitely many positions of $X$, yielding a contradiction. Case 2.1: $t(X)$ is parallel to one of the $p_{i}$; say, to $p_{1}$. Let $t(X) \\operatorname{cross} p_{2}$ at $R$. Then $g(R)$ is the circle $(P R X)$, as $f(R P)=P$ and $f(R X)=X$. Let $R Q$ cross $g(R)$ again at $S$. Then $f(R Q) \\in\\{R, S\\} \\cap g(Q)$, so $g(Q)$ contains one of the points $R$ and $S$. If $R \\in g(Q)$, then $R$ is one of finitely many points in the intersection $g(Q) \\cap p_{2}$, and each of them corresponds to a unique position of $X$, since $R X$ is parallel to $p_{1}$. If $S \\in g(Q)$, then $\\angle(Q S, S P)=\\angle(R S, S P)=\\angle(R X, X P)=\\angle\\left(p_{1}, \\ell\\right)$, so $\\angle(Q S, S P)$ is constant for all such points $X$, and all points $S$ obtained in such a way lie on one circle $\\gamma$ passing through $P$ and $Q$. Since $g(Q)$ does not contain $P$, it is different from $\\gamma$, so there are only finitely many points $S$. Each of them uniquely determines $R$ and thus $X$. ![](https://cdn.mathpix.com/cropped/2024_11_18_9ee8c2ac2c2f5002ff14g-083.jpg?height=843&width=909&top_left_y=207&top_left_x=568) So, Case 2.1 can occur for only finitely many points $X$. Case 2.2: $t(X)$ crosses $p_{1}$ and $p_{2}$ at $R_{1}$ and $R_{2}$, respectively. Clearly, $R_{1} \\neq R_{2}$, as $t(X)$ is the tangent to $g(X)$ at $X$, and $g(X)$ meets $\\ell$ only at $X$ and $Q$. Notice that $g\\left(R_{i}\\right)$ is the circle $\\left(P X R_{i}\\right)$. Let $R_{i} Q$ meet $g\\left(R_{i}\\right)$ again at $S_{i}$; then $S_{i} \\neq Q$, as $g\\left(R_{i}\\right)$ meets $\\ell$ only at $P$ and $X$. Then $f\\left(R_{i} Q\\right) \\in\\left\\{R_{i}, S_{i}\\right\\}$, and we distinguish several subcases. ![](https://cdn.mathpix.com/cropped/2024_11_18_9ee8c2ac2c2f5002ff14g-083.jpg?height=824&width=866&top_left_y=1501&top_left_x=595) Subcase 2.2.1: $f\\left(R_{1} Q\\right)=S_{1}, f\\left(R_{2} Q\\right)=S_{2}$; so $S_{1}, S_{2} \\in g(Q)$. In this case we have $0=\\angle\\left(R_{1} X, X P\\right)+\\angle\\left(X P, R_{2} X\\right)=\\angle\\left(R_{1} S_{1}, S_{1} P\\right)+\\angle\\left(S_{2} P, S_{2} R_{2}\\right)=$ $\\angle\\left(Q S_{1}, S_{1} P\\right)+\\angle\\left(S_{2} P, S_{2} Q\\right)$, which shows $P \\in g(Q)$. Subcase 2.2.2: $f\\left(R_{1} Q\\right)=R_{1}, f\\left(R_{2} Q\\right)=R_{2}$; so $R_{1}, R_{2} \\in g(Q)$. This can happen for at most four positions of $X$ - namely, at the intersections of $\\ell$ with a line of the form $K_{1} K_{2}$, where $K_{i} \\in g(Q) \\cap p_{i}$. Subcase 2.2.3: $f\\left(R_{1} Q\\right)=S_{1}, f\\left(R_{2} Q\\right)=R_{2}$ (the case $f\\left(R_{1} Q\\right)=R_{1}, f\\left(R_{2} Q\\right)=S_{2}$ is similar). In this case, there are at most two possible positions for $R_{2}$ - namely, the meeting points of $g(Q)$ with $p_{2}$. Consider one of them. Let $X$ vary on $\\ell$. Then $R_{1}$ is the projection of $X$ to $p_{1}$ via $R_{2}, S_{1}$ is the projection of $R_{1}$ to $g(Q)$ via $Q$. Finally, $\\angle\\left(Q S_{1}, S_{1} X\\right)=\\angle\\left(R_{1} S_{1}, S_{1} X\\right)=$ $\\angle\\left(R_{1} P, P X\\right)=\\angle\\left(p_{1}, \\ell\\right) \\neq 0$, so $X$ is obtained by a fixed projective transform $g(Q) \\rightarrow \\ell$ from $S_{1}$. So, if there were three points $X$ satisfying the conditions of this subcase, the composition of the three projective transforms would be the identity. But, if we apply it to $X=Q$, we successively get some point $R_{1}^{\\prime}$, then $R_{2}$, and then some point different from $Q$, a contradiction. Thus Case 2.2 also occurs for only finitely many points $X$, as desired. Step 3: We show that $f(P Q)=P$, as desired. The argument is similar to that in Step 2, with the roles of $Q$ and $X$ swapped. Again, we show that there are only finitely many possible positions for a point $X \\in \\ell \\backslash\\{P, Q\\}$, which is absurd. Case 3.1: $t(Q)$ is parallel to one of the $p_{i}$; say, to $p_{1}$. Let $t(Q)$ cross $p_{2}$ at $R$; then $g(R)$ is the circle $(P R Q)$. Let $R X \\operatorname{cross} g(R)$ again at $S$. Then $f(R X) \\in\\{R, S\\} \\cap g(X)$, so $g(X)$ contains one of the points $R$ and $S$. ![](https://cdn.mathpix.com/cropped/2024_11_18_9ee8c2ac2c2f5002ff14g-084.jpg?height=729&width=909&top_left_y=1126&top_left_x=568) Subcase 3.1.1: $S=f(R X) \\in g(X)$. We have $\\angle(t(X), Q X)=\\angle(S X, S Q)=\\angle(S R, S Q)=\\angle(P R, P Q)=\\angle\\left(p_{2}, \\ell\\right)$. Hence $t(X) \\| p_{2}$. Now we recall Case 2.1: we let $t(X)$ cross $p_{1}$ at $R^{\\prime}$, so $g\\left(R^{\\prime}\\right)=\\left(P R^{\\prime} X\\right)$, and let $R^{\\prime} Q$ meet $g\\left(R^{\\prime}\\right)$ again at $S^{\\prime}$; notice that $S^{\\prime} \\neq Q$. Excluding one position of $X$, we may assume that $R^{\\prime} \\notin g(Q)$, so $R^{\\prime} \\neq f\\left(R^{\\prime} Q\\right)$. Therefore, $S^{\\prime}=f\\left(R^{\\prime} Q\\right) \\in g(Q)$. But then, as in Case 2.1, we get $\\angle(t(Q), P Q)=\\angle\\left(Q S^{\\prime}, S^{\\prime} P\\right)=\\angle\\left(R^{\\prime} X, X P\\right)=\\angle\\left(p_{2}, \\ell\\right)$. This means that $t(Q)$ is parallel to $p_{2}$, which is impossible. Subcase 3.1.2: $R=f(R X) \\in g(X)$. In this case, we have $\\angle(t(X), \\ell)=\\angle(R X, R Q)=\\angle\\left(R X, p_{1}\\right)$. Again, let $R^{\\prime}=t(X) \\cap p_{1}$; this point exists for all but at most one position of $X$. Then $g\\left(R^{\\prime}\\right)=\\left(R^{\\prime} X P\\right)$; let $R^{\\prime} Q$ meet $g\\left(R^{\\prime}\\right)$ again at $S^{\\prime}$. Due to $\\angle\\left(R^{\\prime} X, X R\\right)=\\angle(Q X, Q R)=\\angle\\left(\\ell, p_{1}\\right), R^{\\prime}$ determines $X$ in at most two ways, so for all but finitely many positions of $X$ we have $R^{\\prime} \\notin g(Q)$. Therefore, for those positions we have $S^{\\prime}=f\\left(R^{\\prime} Q\\right) \\in g(Q)$. But then $\\angle\\left(R X, p_{1}\\right)=\\angle\\left(R^{\\prime} X, X P\\right)=\\angle\\left(R^{\\prime} S^{\\prime}, S^{\\prime} P\\right)=$ $\\angle\\left(Q S^{\\prime}, S^{\\prime} P\\right)=\\angle(t(Q), Q P)$ is fixed, so this case can hold only for one specific position of $X$ as well. Thus, in Case 3.1, there are only finitely many possible positions of $X$, yielding a contradiction. Case 3.2: $t(Q)$ crosses $p_{1}$ and $p_{2}$ at $R_{1}$ and $R_{2}$, respectively. By Step $2, R_{1} \\neq R_{2}$. Notice that $g\\left(R_{i}\\right)$ is the circle $\\left(P Q R_{i}\\right)$. Let $R_{i} X$ meet $g\\left(R_{i}\\right)$ at $S_{i}$; then $S_{i} \\neq X$. Then $f\\left(R_{i} X\\right) \\in\\left\\{R_{i}, S_{i}\\right\\}$, and we distinguish several subcases. ![](https://cdn.mathpix.com/cropped/2024_11_18_9ee8c2ac2c2f5002ff14g-085.jpg?height=854&width=866&top_left_y=424&top_left_x=589) Subcase 3.2.1: $f\\left(R_{1} X\\right)=S_{1}$ and $f\\left(R_{2} X\\right)=S_{2}$, so $S_{1}, S_{2} \\in g(X)$. As in Subcase 2.2.1, we have $0=\\angle\\left(R_{1} Q, Q P\\right)+\\angle\\left(Q P, R_{2} Q\\right)=\\angle\\left(X S_{1}, S_{1} P\\right)+\\angle\\left(S_{2} P, S_{2} X\\right)$, which shows $P \\in g(X)$. But $X, Q \\in g(X)$ as well, so $g(X)$ meets $\\ell$ at three distinct points, which is absurd. Subcase 3.2.2: $f\\left(R_{1} X\\right)=R_{1}, f\\left(R_{2} X\\right)=R_{2}$, so $R_{1}, R_{2} \\in g(X)$. Now three distinct collinear points $R_{1}, R_{2}$, and $Q$ belong to $g(X)$, which is impossible. Subcase 3.2.3: $f\\left(R_{1} X\\right)=S_{1}, f\\left(R_{2} X\\right)=R_{2}$ (the case $f\\left(R_{1} X\\right)=R_{1}, f\\left(R_{2} X\\right)=S_{2}$ is similar). We have $\\angle\\left(X R_{2}, R_{2} Q\\right)=\\angle\\left(X S_{1}, S_{1} Q\\right)=\\angle\\left(R_{1} S_{1}, S_{1} Q\\right)=\\angle\\left(R_{1} P, P Q\\right)=\\angle\\left(p_{1}, \\ell\\right)$, so this case can occur for a unique position of $X$. Thus, in Case 3.2 , there is only a unique position of $X$, again yielding the required contradiction.", "metadata": {"resource_path": "IMO/segmented/en-IMO2019SL.jsonl"}} -{"year": "2019", "tier": "T0", "problem_label": "G8", "problem_type": "Geometry", "exam": "IMO", "problem": "Let $\\mathcal{L}$ be the set of all lines in the plane and let $f$ be a function that assigns to each line $\\ell \\in \\mathcal{L}$ a point $f(\\ell)$ on $\\ell$. Suppose that for any point $X$, and for any three lines $\\ell_{1}, \\ell_{2}, \\ell_{3}$ passing through $X$, the points $f\\left(\\ell_{1}\\right), f\\left(\\ell_{2}\\right), f\\left(\\ell_{3}\\right)$ and $X$ lie on a circle. Prove that there is a unique point $P$ such that $f(\\ell)=P$ for any line $\\ell$ passing through $P$. (Australia)", "solution": "We will get an upper bound on $n$ from the speed at which $v_{2}\\left(L_{n}\\right)$ grows. From $$ L_{n}=\\left(2^{n}-1\\right)\\left(2^{n}-2\\right) \\cdots\\left(2^{n}-2^{n-1}\\right)=2^{1+2+\\cdots+(n-1)}\\left(2^{n}-1\\right)\\left(2^{n-1}-1\\right) \\cdots\\left(2^{1}-1\\right) $$ we read $$ v_{2}\\left(L_{n}\\right)=1+2+\\cdots+(n-1)=\\frac{n(n-1)}{2} . $$ On the other hand, $v_{2}(m!)$ is expressed by the Legendre formula as $$ v_{2}(m!)=\\sum_{i=1}^{\\infty}\\left\\lfloor\\frac{m}{2^{i}}\\right\\rfloor . $$ As usual, by omitting the floor functions, $$ v_{2}(m!)<\\sum_{i=1}^{\\infty} \\frac{m}{2^{i}}=m $$ Thus, $L_{n}=m$ ! implies the inequality $$ \\frac{n(n-1)}{2}1.3 \\cdot 10^{12}$. For $n \\geqslant 7$ we prove (3) by the following inequalities: $$ \\begin{aligned} \\left(\\frac{n(n-1)}{2}\\right)! & =15!\\cdot 16 \\cdot 17 \\cdots \\frac{n(n-1)}{2}>2^{36} \\cdot 16^{\\frac{n(n-1)}{2}-15} \\\\ & =2^{2 n(n-1)-24}=2^{n^{2}} \\cdot 2^{n(n-2)-24}>2^{n^{2}} . \\end{aligned} $$ Putting together (2) and (3), for $n \\geqslant 6$ we get a contradiction, since $$ L_{n}<2^{n^{2}}<\\left(\\frac{n(n-1)}{2}\\right)!a^{3} . $$ So $3 a^{3} \\geqslant(a b c)^{2}>a^{3}$ and hence $3 a \\geqslant b^{2} c^{2}>a$. Now $b^{3}+c^{3}=a^{2}\\left(b^{2} c^{2}-a\\right) \\geqslant a^{2}$, and so $$ 18 b^{3} \\geqslant 9\\left(b^{3}+c^{3}\\right) \\geqslant 9 a^{2} \\geqslant b^{4} c^{4} \\geqslant b^{3} c^{5} $$ so $18 \\geqslant c^{5}$ which yields $c=1$. Now, note that we must have $a>b$, as otherwise we would have $2 b^{3}+1=b^{4}$ which has no positive integer solutions. So $$ a^{3}-b^{3} \\geqslant(b+1)^{3}-b^{3}>1 $$ and $$ 2 a^{3}>1+a^{3}+b^{3}>a^{3} $$ which implies $2 a^{3}>a^{2} b^{2}>a^{3}$ and so $2 a>b^{2}>a$. Therefore $$ 4\\left(1+b^{3}\\right)=4 a^{2}\\left(b^{2}-a\\right) \\geqslant 4 a^{2}>b^{4} $$ so $4>b^{3}(b-4)$; that is, $b \\leqslant 4$. Now, for each possible value of $b$ with $2 \\leqslant b \\leqslant 4$ we obtain a cubic equation for $a$ with constant coefficients. These are as follows: $$ \\begin{array}{rlrl} b=2: & & & a^{3}-4 a^{2}+9=0 \\\\ b=3: & & a^{3}-9 a^{2}+28=0 \\\\ b=4: & & a^{3}-16 a^{2}+65=0 . \\end{array} $$ The only case with an integer solution for $a$ with $b \\leqslant a$ is $b=2$, leading to $(a, b, c)=(3,2,1)$. Comment 1.1. Instead of writing down each cubic equation explicitly, we could have just observed that $a^{2} \\mid b^{3}+1$, and for each choice of $b$ checked each square factor of $b^{3}+1$ for $a^{2}$. We could also have observed that, with $c=1$, the relation $18 b^{3} \\geqslant b^{4} c^{4}$ becomes $b \\leqslant 18$, and we can simply check all possibilities for $b$ (instead of working to prove that $b \\leqslant 4$ ). This check becomes easier after using the factorisation $b^{3}+1=(b+1)\\left(b^{2}-b+1\\right)$ and observing that no prime besides 3 can divide both of the factors. Comment 1.2. Another approach to finish the problem after establishing that $c \\leqslant 1$ is to set $k=b^{2} c^{2}-a$, which is clearly an integer and must be positive as it is equal to $\\left(b^{3}+c^{3}\\right) / a^{2}$. Then we divide into cases based on whether $k=1$ or $k \\geqslant 2$; in the first case, we have $b^{3}+1=a^{2}=\\left(b^{2}-1\\right)^{2}$ whose only positive root is $b=2$, and in the second case we have $b^{2} \\leqslant 3 a$, and so $$ b^{4} \\leqslant(3 a)^{2} \\leqslant \\frac{9}{2}\\left(k a^{2}\\right)=\\frac{9}{2}\\left(b^{3}+1\\right), $$ which implies that $b \\leqslant 4$.", "metadata": {"resource_path": "IMO/segmented/en-IMO2019SL.jsonl"}} -{"year": "2019", "tier": "T0", "problem_label": "N2", "problem_type": "Number Theory", "exam": "IMO", "problem": "Find all triples $(a, b, c)$ of positive integers such that $a^{3}+b^{3}+c^{3}=(a b c)^{2}$. (Nigeria)", "solution": "Again, we will start by proving that $c=1$. Suppose otherwise that $c \\geqslant 2$. We have $a^{3}+b^{3}+c^{3} \\leqslant 3 a^{3}$, so $b^{2} c^{2} \\leqslant 3 a$. Since $c \\geqslant 2$, this tells us that $b \\leqslant \\sqrt{3 a / 4}$. As the right-hand side of the original equation is a multiple of $a^{2}$, we have $a^{2} \\leqslant 2 b^{3} \\leqslant 2(3 a / 4)^{3 / 2}$. In other words, $a \\leqslant \\frac{27}{16}<2$, which contradicts the assertion that $a \\geqslant c \\geqslant 2$. So there are no solutions in this case, and so we must have $c=1$. Now, the original equation becomes $a^{3}+b^{3}+1=a^{2} b^{2}$. Observe that $a \\geqslant 2$, since otherwise $a=b=1$ as $a \\geqslant b$. The right-hand side is a multiple of $a^{2}$, so the left-hand side must be as well. Thus, $b^{3}+1 \\geqslant$ $a^{2}$. Since $a \\geqslant b$, we also have $$ b^{2}=a+\\frac{b^{3}+1}{a^{2}} \\leqslant 2 a+\\frac{1}{a^{2}} $$ and so $b^{2} \\leqslant 2 a$ since $b^{2}$ is an integer. Thus $(2 a)^{3 / 2}+1 \\geqslant b^{3}+1 \\geqslant a^{2}$, from which we deduce $a \\leqslant 8$. Now, for each possible value of $a$ with $2 \\leqslant a \\leqslant 8$ we obtain a cubic equation for $b$ with constant coefficients. These are as follows: $$ \\begin{array}{ll} a=2: & \\\\ b^{3}-4 b^{2}+9=0 \\\\ a=3: & \\\\ b^{3}-9 b^{2}+28=0 \\\\ a=4: & \\\\ a=5: 16 b^{2}+65=0 \\\\ a=6: & \\\\ b^{3}-25 b^{2}+126=0 \\\\ a=7: & \\\\ b^{3}-36 b^{2}+217=0 \\\\ a=8: & \\\\ b^{3}-49 b^{2}+344=0 \\\\ b^{3}-64 b^{2}+513=0 . \\end{array} $$ The only case with an integer solution for $b$ with $a \\geqslant b$ is $a=3$, leading to $(a, b, c)=(3,2,1)$. Comment 2.1. As in Solution 1, instead of writing down each cubic equation explicitly, we could have just observed that $b^{2} \\mid a^{3}+1$, and for each choice of $a$ checked each square factor of $a^{3}+1$ for $b^{2}$. Comment 2.2. This solution does not require initially proving that $c=1$, in which case the bound would become $a \\leqslant 108$. The resulting cases could, in principle, be checked by a particularly industrious student.", "metadata": {"resource_path": "IMO/segmented/en-IMO2019SL.jsonl"}} -{"year": "2019", "tier": "T0", "problem_label": "N2", "problem_type": "Number Theory", "exam": "IMO", "problem": "Find all triples $(a, b, c)$ of positive integers such that $a^{3}+b^{3}+c^{3}=(a b c)^{2}$. (Nigeria)", "solution": "Set $k=\\left(b^{3}+c^{3}\\right) / a^{2} \\leqslant 2 a$, and rewrite the original equation as $a+k=(b c)^{2}$. Since $b^{3}$ and $c^{3}$ are positive integers, we have $(b c)^{3} \\geqslant b^{3}+c^{3}-1=k a^{2}-1$, so $$ a+k \\geqslant\\left(k a^{2}-1\\right)^{2 / 3} $$ As in Comment 1.2, $k$ is a positive integer; for each value of $k \\geqslant 1$, this gives us a polynomial inequality satisfied by $a$ : $$ k^{2} a^{4}-a^{3}-5 k a^{2}-3 k^{2} a-\\left(k^{3}-1\\right) \\leqslant 0 . $$ We now prove that $a \\leqslant 3$. Indeed, $$ 0 \\geqslant \\frac{k^{2} a^{4}-a^{3}-5 k a^{2}-3 k^{2} a-\\left(k^{3}-1\\right)}{k^{2}} \\geqslant a^{4}-a^{3}-5 a^{2}-3 a-k \\geqslant a^{4}-a^{3}-5 a^{2}-5 a $$ which fails when $a \\geqslant 4$. This leaves ten triples with $3 \\geqslant a \\geqslant b \\geqslant c \\geqslant 1$, which may be checked manually to give $(a, b, c)=(3,2,1)$.", "metadata": {"resource_path": "IMO/segmented/en-IMO2019SL.jsonl"}} -{"year": "2019", "tier": "T0", "problem_label": "N2", "problem_type": "Number Theory", "exam": "IMO", "problem": "Find all triples $(a, b, c)$ of positive integers such that $a^{3}+b^{3}+c^{3}=(a b c)^{2}$. (Nigeria)", "solution": "Again, observe that $b^{3}+c^{3}=a^{2}\\left(b^{2} c^{2}-a\\right)$, so $b \\leqslant a \\leqslant b^{2} c^{2}-1$. We consider the function $f(x)=x^{2}\\left(b^{2} c^{2}-x\\right)$. It can be seen that that on the interval $\\left[0, b^{2} c^{2}-1\\right]$ the function $f$ is increasing if $x<\\frac{2}{3} b^{2} c^{2}$ and decreasing if $x>\\frac{2}{3} b^{2} c^{2}$. Consequently, it must be the case that $$ b^{3}+c^{3}=f(a) \\geqslant \\min \\left(f(b), f\\left(b^{2} c^{2}-1\\right)\\right) $$ First, suppose that $b^{3}+c^{3} \\geqslant f\\left(b^{2} c^{2}-1\\right)$. This may be written $b^{3}+c^{3} \\geqslant\\left(b^{2} c^{2}-1\\right)^{2}$, and so $$ 2 b^{3} \\geqslant b^{3}+c^{3} \\geqslant\\left(b^{2} c^{2}-1\\right)^{2}>b^{4} c^{4}-2 b^{2} c^{2} \\geqslant b^{4} c^{4}-2 b^{3} c^{4} $$ Thus, $(b-2) c^{4}<2$, and the only solutions to this inequality have $(b, c)=(2,2)$ or $b \\leqslant 3$ and $c=1$. It is easy to verify that the only case giving a solution for $a \\geqslant b$ is $(a, b, c)=(3,2,1)$. Otherwise, suppose that $b^{3}+c^{3}=f(a) \\geqslant f(b)$. Then, we have $$ 2 b^{3} \\geqslant b^{3}+c^{3}=a^{2}\\left(b^{2} c^{2}-a\\right) \\geqslant b^{2}\\left(b^{2} c^{2}-b\\right) . $$ Consequently $b c^{2} \\leqslant 3$, with strict inequality in the case that $b \\neq c$. Hence $c=1$ and $b \\leqslant 2$. Both of these cases have been considered already, so we are done. Comment 4.1. Instead of considering which of $f(b)$ and $f\\left(b^{2} c^{2}-1\\right)$ is less than $f(a)$, we may also proceed by explicitly dividing into cases based on whether $a \\geqslant \\frac{2}{3} b^{2} c^{2}$ or $a<\\frac{2}{3} b^{2} c^{2}$. The first case may now be dealt with as follows. We have $b^{3} c^{3}+1 \\geqslant b^{3}+c^{3}$ as $b^{3}$ and $c^{3}$ are positive integers, so we have $$ b^{3} c^{3}+1 \\geqslant b^{3}+c^{3} \\geqslant a^{2} \\geqslant \\frac{4}{9} b^{4} c^{4} $$ This implies $b c \\leqslant 2$, and hence $c=1$ and $b \\leqslant 2$.", "metadata": {"resource_path": "IMO/segmented/en-IMO2019SL.jsonl"}} -{"year": "2019", "tier": "T0", "problem_label": "N3", "problem_type": "Number Theory", "exam": "IMO", "problem": "We say that a set $S$ of integers is rootiful if, for any positive integer $n$ and any $a_{0}, a_{1}, \\ldots, a_{n} \\in S$, all integer roots of the polynomial $a_{0}+a_{1} x+\\cdots+a_{n} x^{n}$ are also in $S$. Find all rootiful sets of integers that contain all numbers of the form $2^{a}-2^{b}$ for positive integers $a$ and $b$. (Czech Republic)", "solution": "The set $\\mathbb{Z}$ of all integers is clearly rootiful. We shall prove that any rootiful set $S$ containing all the numbers of the form $2^{a}-2^{b}$ for $a, b \\in \\mathbb{Z}_{>0}$ must be all of $\\mathbb{Z}$. First, note that $0=2^{1}-2^{1} \\in S$ and $2=2^{2}-2^{1} \\in S$. Now, $-1 \\in S$, since it is a root of $2 x+2$, and $1 \\in S$, since it is a root of $2 x^{2}-x-1$. Also, if $n \\in S$ then $-n$ is a root of $x+n$, so it suffices to prove that all positive integers must be in $S$. Now, we claim that any positive integer $n$ has a multiple in $S$. Indeed, suppose that $n=2^{\\alpha} \\cdot t$ for $\\alpha \\in \\mathbb{Z}_{\\geqslant 0}$ and $t$ odd. Then $t \\mid 2^{\\phi(t)}-1$, so $n \\mid 2^{\\alpha+\\phi(t)+1}-2^{\\alpha+1}$. Moreover, $2^{\\alpha+\\phi(t)+1}-2^{\\alpha+1} \\in S$, and so $S$ contains a multiple of every positive integer $n$. We will now prove by induction that all positive integers are in $S$. Suppose that $0,1, \\ldots, n-$ $1 \\in S$; furthermore, let $N$ be a multiple of $n$ in $S$. Consider the base- $n$ expansion of $N$, say $N=a_{k} n^{k}+a_{k-1} n^{k-1}+\\cdots+a_{1} n+a_{0}$. Since $0 \\leqslant a_{i}0}$. We show that, in fact, every integer $k$ with $|k|>2$ can be expressed as a root of a polynomial whose coefficients are of the form $2^{a}-2^{b}$. Observe that it suffices to consider the case where $k$ is positive, as if $k$ is a root of $a_{n} x^{n}+\\cdots+a_{1} x+a_{0}=0$, then $-k$ is a root of $(-1)^{n} a_{n} x^{n}+\\cdots-$ $a_{1} x+a_{0}=0$. Note that $$ \\left(2^{a_{n}}-2^{b_{n}}\\right) k^{n}+\\cdots+\\left(2^{a_{0}}-2^{b_{0}}\\right)=0 $$ is equivalent to $$ 2^{a_{n}} k^{n}+\\cdots+2^{a_{0}}=2^{b_{n}} k^{n}+\\cdots+2^{b_{0}} $$ Hence our aim is to show that two numbers of the form $2^{a_{n}} k^{n}+\\cdots+2^{a_{0}}$ are equal, for a fixed value of $n$. We consider such polynomials where every term $2^{a_{i}} k^{i}$ is at most $2 k^{n}$; in other words, where $2 \\leqslant 2^{a_{i}} \\leqslant 2 k^{n-i}$, or, equivalently, $1 \\leqslant a_{i} \\leqslant 1+(n-i) \\log _{2} k$. Therefore, there must be $1+\\left\\lfloor(n-i) \\log _{2} k\\right\\rfloor$ possible choices for $a_{i}$ satisfying these constraints. The number of possible polynomials is then $$ \\prod_{i=0}^{n}\\left(1+\\left\\lfloor(n-i) \\log _{2} k\\right\\rfloor\\right) \\geqslant \\prod_{i=0}^{n-1}(n-i) \\log _{2} k=n!\\left(\\log _{2} k\\right)^{n} $$ where the inequality holds as $1+\\lfloor x\\rfloor \\geqslant x$. As there are $(n+1)$ such terms in the polynomial, each at most $2 k^{n}$, such a polynomial must have value at most $2 k^{n}(n+1)$. However, for large $n$, we have $n!\\left(\\log _{2} k\\right)^{n}>2 k^{n}(n+1)$. Therefore there are more polynomials than possible values, so some two must be equal, as required.", "metadata": {"resource_path": "IMO/segmented/en-IMO2019SL.jsonl"}} -{"year": "2019", "tier": "T0", "problem_label": "N4", "problem_type": "Number Theory", "exam": "IMO", "problem": "Let $\\mathbb{Z}_{>0}$ be the set of positive integers. A positive integer constant $C$ is given. Find all functions $f: \\mathbb{Z}_{>0} \\rightarrow \\mathbb{Z}_{>0}$ such that, for all positive integers $a$ and $b$ satisfying $a+b>C$, $$ a+f(b) \\mid a^{2}+b f(a) $$ (Croatia)", "solution": "First, we show that $b \\mid f(b)^{2}$ for all $b$. To do this, we choose a large positive integer $n$ so that $n b-f(b) \\geqslant C$. Setting $a=n b-f(b)$ in (*) then shows that $$ n b \\mid(n b-f(b))^{2}+b f(n b-f(b)) $$ so that $b \\mid f(b)^{2}$ as claimed. Now in particular we have that $p \\mid f(p)$ for every prime $p$. If we write $f(p)=k(p) \\cdot p$, then the bound $f(p) \\leqslant f(1) \\cdot p$ (valid for $p$ sufficiently large) shows that some value $k$ of $k(p)$ must be attained for infinitely many $p$. We will show that $f(a)=k a$ for all positive integers $a$. To do this, we substitute $b=p$ in (*), where $p$ is any sufficiently large prime for which $k(p)=k$, obtaining $$ a+k p \\mid\\left(a^{2}+p f(a)\\right)-a(a+k p)=p f(a)-p k a . $$ For suitably large $p$ we have $\\operatorname{gcd}(a+k p, p)=1$, and hence we have $$ a+k p \\mid f(a)-k a $$ But the only way this can hold for arbitrarily large $p$ is if $f(a)-k a=0$. This concludes the proof. Comment. There are other ways to obtain the divisibility $p \\mid f(p)$ for primes $p$, which is all that is needed in this proof. For instance, if $f(p)$ were not divisible by $p$ then the arithmetic progression $p^{2}+b f(p)$ would attain prime values for infinitely many $b$ by Dirichlet's Theorem: hence, for these pairs p , b, we would have $p+f(b)=p^{2}+b f(p)$. Substituting $a \\mapsto b$ and $b \\mapsto p$ in $(*)$ then shows that $\\left(f(p)^{2}-p^{2}\\right)(p-1)$ is divisible by $b+f(p)$ and hence vanishes, which is impossible since $p \\nmid f(p)$ by assumption.", "metadata": {"resource_path": "IMO/segmented/en-IMO2019SL.jsonl"}} -{"year": "2019", "tier": "T0", "problem_label": "N4", "problem_type": "Number Theory", "exam": "IMO", "problem": "Let $\\mathbb{Z}_{>0}$ be the set of positive integers. A positive integer constant $C$ is given. Find all functions $f: \\mathbb{Z}_{>0} \\rightarrow \\mathbb{Z}_{>0}$ such that, for all positive integers $a$ and $b$ satisfying $a+b>C$, $$ a+f(b) \\mid a^{2}+b f(a) $$ (Croatia)", "solution": "First, we substitute $b=1$ in (*) and rearrange to find that $$ \\frac{f(a)+f(1)^{2}}{a+f(1)}=f(1)-a+\\frac{a^{2}+f(a)}{a+f(1)} $$ is a positive integer for sufficiently large $a$. Since $f(a) \\leqslant a f(1)$, for all sufficiently large $a$, it follows that $\\frac{f(a)+f(1)^{2}}{a+f(1)} \\leqslant f(1)$ also and hence there is a positive integer $k$ such that $\\frac{f(a)+f(1)^{2}}{a+f(1)}=k$ for infinitely many values of $a$. In other words, $$ f(a)=k a+f(1) \\cdot(k-f(1)) $$ for infinitely many $a$. Fixing an arbitrary choice of $a$ in (*), we have that $$ \\frac{a^{2}+b f(a)}{a+k b+f(1) \\cdot(k-f(1))} $$ is an integer for infinitely many $b$ (the same $b$ as above, maybe with finitely many exceptions). On the other hand, for $b$ taken sufficiently large, this quantity becomes arbitrarily close to $\\frac{f(a)}{k}$; this is only possible if $\\frac{f(a)}{k}$ is an integer and $$ \\frac{a^{2}+b f(a)}{a+k b+f(1) \\cdot(k-f(1))}=\\frac{f(a)}{k} $$ for infinitely many $b$. This rearranges to $$ \\frac{f(a)}{k} \\cdot(a+f(1) \\cdot(k-f(1)))=a^{2} $$ Hence $a^{2}$ is divisible by $a+f(1) \\cdot(k-f(1))$, and hence so is $f(1)^{2}(k-f(1))^{2}$. The only way this can occur for all $a$ is if $k=f(1)$, in which case $(* *)$ provides that $f(a)=k a$ for all $a$, as desired.", "metadata": {"resource_path": "IMO/segmented/en-IMO2019SL.jsonl"}} -{"year": "2019", "tier": "T0", "problem_label": "N4", "problem_type": "Number Theory", "exam": "IMO", "problem": "Let $\\mathbb{Z}_{>0}$ be the set of positive integers. A positive integer constant $C$ is given. Find all functions $f: \\mathbb{Z}_{>0} \\rightarrow \\mathbb{Z}_{>0}$ such that, for all positive integers $a$ and $b$ satisfying $a+b>C$, $$ a+f(b) \\mid a^{2}+b f(a) $$ (Croatia)", "solution": "Fix any two distinct positive integers $a$ and $b$. From (*) it follows that the two integers $$ \\left(a^{2}+c f(a)\\right) \\cdot(b+f(c)) \\text { and }\\left(b^{2}+c f(b)\\right) \\cdot(a+f(c)) $$ are both multiples of $(a+f(c)) \\cdot(b+f(c))$ for all sufficiently large $c$. Taking an appropriate linear combination to eliminate the $c f(c)$ term, we find after expanding out that the integer $$ \\left[a^{2} f(b)-b^{2} f(a)\\right] \\cdot f(c)+[(b-a) f(a) f(b)] \\cdot c+[a b(a f(b)-b f(a))] $$ is also a multiple of $(a+f(c)) \\cdot(b+f(c))$. But as $c$ varies, $(\\dagger)$ is bounded above by a positive multiple of $c$ while $(a+f(c)) \\cdot(b+f(c))$ is bounded below by a positive multiple of $c^{2}$. The only way that such a divisibility can hold is if in fact $$ \\left[a^{2} f(b)-b^{2} f(a)\\right] \\cdot f(c)+[(b-a) f(a) f(b)] \\cdot c+[a b(a f(b)-b f(a))]=0 $$ for sufficiently large $c$. Since the coefficient of $c$ in this linear relation is nonzero, it follows that there are constants $k, \\ell$ such that $f(c)=k c+\\ell$ for all sufficiently large $c$; the constants $k$ and $\\ell$ are necessarily integers. The value of $\\ell$ satisfies $$ \\left[a^{2} f(b)-b^{2} f(a)\\right] \\cdot \\ell+[a b(a f(b)-b f(a))]=0 $$ and hence $b \\mid \\ell a^{2} f(b)$ for all $a$ and $b$. Taking $b$ sufficiently large so that $f(b)=k b+\\ell$, we thus have that $b \\mid \\ell^{2} a^{2}$ for all sufficiently large $b$; this implies that $\\ell=0$. From ( $\\dagger \\dagger \\dagger$ ) it then follows that $\\frac{f(a)}{a}=\\frac{f(b)}{b}$ for all $a \\neq b$, so that there is a constant $k$ such that $f(a)=k a$ for all $a$ ( $k$ is equal to the constant defined earlier).", "metadata": {"resource_path": "IMO/segmented/en-IMO2019SL.jsonl"}} -{"year": "2019", "tier": "T0", "problem_label": "N4", "problem_type": "Number Theory", "exam": "IMO", "problem": "Let $\\mathbb{Z}_{>0}$ be the set of positive integers. A positive integer constant $C$ is given. Find all functions $f: \\mathbb{Z}_{>0} \\rightarrow \\mathbb{Z}_{>0}$ such that, for all positive integers $a$ and $b$ satisfying $a+b>C$, $$ a+f(b) \\mid a^{2}+b f(a) $$ (Croatia)", "solution": "Let $\\Gamma$ denote the set of all points $(a, f(a))$, so that $\\Gamma$ is an infinite subset of the upper-right quadrant of the plane. For a point $A=(a, f(a))$ in $\\Gamma$, we define a point $A^{\\prime}=\\left(-f(a),-f(a)^{2} / a\\right)$ in the lower-left quadrant of the plane, and let $\\Gamma^{\\prime}$ denote the set of all such points $A^{\\prime}$. ![](https://cdn.mathpix.com/cropped/2024_11_18_9ee8c2ac2c2f5002ff14g-095.jpg?height=552&width=652&top_left_y=449&top_left_x=702) Claim. For any point $A \\in \\Gamma$, the set $\\Gamma$ is contained in finitely many lines through the point $A^{\\prime}$. Proof. Let $A=(a, f(a))$. The functional equation (with $a$ and $b$ interchanged) can be rewritten as $b+f(a) \\mid a f(b)-b f(a)$, so that all but finitely many points in $\\Gamma$ are contained in one of the lines with equation $$ a y-f(a) x=m(x+f(a)) $$ for $m$ an integer. Geometrically, these are the lines through $A^{\\prime}=\\left(-f(a),-f(a)^{2} / a\\right)$ with gradient $\\frac{f(a)+m}{a}$. Since $\\Gamma$ is contained, with finitely many exceptions, in the region $0 \\leqslant y \\leqslant$ $f(1) \\cdot x$ and the point $A^{\\prime}$ lies strictly in the lower-left quadrant of the plane, there are only finitely many values of $m$ for which this line meets $\\Gamma$. This concludes the proof of the claim. Now consider any distinct points $A, B \\in \\Gamma$. It is clear that $A^{\\prime}$ and $B^{\\prime}$ are distinct. A line through $A^{\\prime}$ and a line through $B^{\\prime}$ only meet in more than one point if these two lines are equal to the line $A^{\\prime} B^{\\prime}$. It then follows from the above claim that the line $A^{\\prime} B^{\\prime}$ must contain all but finitely many points of $\\Gamma$. If $C$ is another point of $\\Gamma$, then the line $A^{\\prime} C^{\\prime}$ also passes through all but finitely many points of $\\Gamma$, which is only possible if $A^{\\prime} C^{\\prime}=A^{\\prime} B^{\\prime}$. We have thus seen that there is a line $\\ell$ passing through all points of $\\Gamma^{\\prime}$ and through all but finitely many points of $\\Gamma$. We claim that this line passes through the origin $O$ and passes through every point of $\\Gamma$. To see this, note that by construction $A, O, A^{\\prime}$ are collinear for every point $A \\in \\Gamma$. Since $\\ell=A A^{\\prime}$ for all but finitely many points $A \\in \\Gamma$, it thus follows that $O \\in \\ell$. Thus any $A \\in \\Gamma$ lies on the line $\\ell=A^{\\prime} O$. Since $\\Gamma$ is contained in a line through $O$, it follows that there is a real constant $k$ (the gradient of $\\ell$ ) such that $f(a)=k a$ for all $a$. The number $k$ is, of course, a positive integer. Comment. Without the $a+b>C$ condition, this problem is approachable by much more naive methods. For instance, using the given divisibility for $a, b \\in\\{1,2,3\\}$ one can prove by a somewhat tedious case-check that $f(2)=2 f(1)$ and $f(3)=3 f(1)$; this then forms the basis of an induction establishing that $f(n)=n f(1)$ for all $n$.", "metadata": {"resource_path": "IMO/segmented/en-IMO2019SL.jsonl"}} -{"year": "2019", "tier": "T0", "problem_label": "N5", "problem_type": "Number Theory", "exam": "IMO", "problem": "Let $a$ be a positive integer. We say that a positive integer $b$ is $a$-good if $\\binom{a n}{b}-1$ is divisible by $a n+1$ for all positive integers $n$ with $a n \\geqslant b$. Suppose $b$ is a positive integer such that $b$ is $a$-good, but $b+2$ is not $a$-good. Prove that $b+1$ is prime. (Netherlands)", "solution": "For $p$ a prime and $n$ a nonzero integer, we write $v_{p}(n)$ for the $p$-adic valuation of $n$ : the largest integer $t$ such that $p^{t} \\mid n$. We first show that $b$ is $a$-good if and only if $b$ is even, and $p \\mid a$ for all primes $p \\leqslant b$. To start with, the condition that $a n+1 \\left\\lvert\\,\\binom{ a n}{b}-1\\right.$ can be rewritten as saying that $$ \\frac{a n(a n-1) \\cdots(a n-b+1)}{b!} \\equiv 1 \\quad(\\bmod a n+1) $$ Suppose, on the one hand, there is a prime $p \\leqslant b$ with $p \\nmid a$. Take $t=v_{p}(b!)$. Then there exist positive integers $c$ such that $a c \\equiv 1\\left(\\bmod p^{t+1}\\right)$. If we take $c$ big enough, and then take $n=(p-1) c$, then $a n=a(p-1) c \\equiv p-1\\left(\\bmod p^{t+1}\\right)$ and $a n \\geqslant b$. Since $p \\leqslant b$, one of the terms of the numerator an $(a n-1) \\cdots(a n-b+1)$ is $a n-p+1$, which is divisible by $p^{t+1}$. Hence the $p$-adic valuation of the numerator is at least $t+1$, but that of the denominator is exactly $t$. This means that $p \\left\\lvert\\,\\binom{ a n}{b}\\right.$, so $p \\nmid\\binom{a n}{b}-1$. As $p \\mid a n+1$, we get that $a n+1 \\nmid\\binom{a n}{b}$, so $b$ is not $a$-good. On the other hand, if for all primes $p \\leqslant b$ we have $p \\mid a$, then every factor of $b$ ! is coprime to $a n+1$, and hence invertible modulo $a n+1$ : hence $b$ ! is also invertible modulo $a n+1$. Then equation (1) reduces to: $$ a n(a n-1) \\cdots(a n-b+1) \\equiv b!\\quad(\\bmod a n+1) $$ However, we can rewrite the left-hand side as follows: $$ a n(a n-1) \\cdots(a n-b+1) \\equiv(-1)(-2) \\cdots(-b) \\equiv(-1)^{b} b!\\quad(\\bmod a n+1) $$ Provided that $a n>1$, if $b$ is even we deduce $(-1)^{b} b!\\equiv b$ ! as needed. On the other hand, if $b$ is odd, and we take $a n+1>2(b!)$, then we will not have $(-1)^{b} b!\\equiv b$ !, so $b$ is not $a$-good. This completes the claim. To conclude from here, suppose that $b$ is $a$-good, but $b+2$ is not. Then $b$ is even, and $p \\mid a$ for all primes $p \\leqslant b$, but there is a prime $q \\leqslant b+2$ for which $q \\nmid a$ : so $q=b+1$ or $q=b+2$. We cannot have $q=b+2$, as that is even too, so we have $q=b+1$ : in other words, $b+1$ is prime.", "metadata": {"resource_path": "IMO/segmented/en-IMO2019SL.jsonl"}} -{"year": "2019", "tier": "T0", "problem_label": "N5", "problem_type": "Number Theory", "exam": "IMO", "problem": "Let $a$ be a positive integer. We say that a positive integer $b$ is $a$-good if $\\binom{a n}{b}-1$ is divisible by $a n+1$ for all positive integers $n$ with $a n \\geqslant b$. Suppose $b$ is a positive integer such that $b$ is $a$-good, but $b+2$ is not $a$-good. Prove that $b+1$ is prime. (Netherlands)", "solution": "We show only half of the claim of the previous solution: we show that if $b$ is $a$-good, then $p \\mid a$ for all primes $p \\leqslant b$. We do this with Lucas' theorem. Suppose that we have $p \\leqslant b$ with $p \\nmid a$. Then consider the expansion of $b$ in base $p$; there will be some digit (not the final digit) which is nonzero, because $p \\leqslant b$. Suppose it is the $p^{t}$ digit for $t \\geqslant 1$. Now, as $n$ varies over the integers, an +1 runs over all residue classes modulo $p^{t+1}$; in particular, there is a choice of $n$ (with $a n>b$ ) such that the $p^{0}$ digit of $a n$ is $p-1$ (so $p \\mid a n+1$ ) and the $p^{t}$ digit of $a n$ is 0 . Consequently, $p \\mid a n+1$ but $p \\left\\lvert\\,\\binom{ a n}{b}\\right.$ (by Lucas' theorem) so $p \\nmid\\binom{a n}{b}-1$. Thus $b$ is not $a$-good. Now we show directly that if $b$ is $a$-good but $b+2$ fails to be so, then there must be a prime dividing $a n+1$ for some $n$, which also divides $(b+1)(b+2)$. Indeed, the ratio between $\\binom{a n}{b+2}$ and $\\binom{a n}{b}$ is $(b+1)(b+2) /(a n-b)(a n-b-1)$. We know that there must be a choice of $a n+1$ such that the former binomial coefficient is 1 modulo $a n+1$ but the latter is not, which means that the given ratio must not be $1 \\bmod a n+1$. If $b+1$ and $b+2$ are both coprime to $a n+1$ then the ratio is 1 , so that must not be the case. In particular, as any prime less than $b$ divides $a$, it must be the case that either $b+1$ or $b+2$ is prime. However, we can observe that $b$ must be even by insisting that $a n+1$ is prime (which is possible by Dirichlet's theorem) and hence $\\binom{a n}{b} \\equiv(-1)^{b}=1$. Thus $b+2$ cannot be prime, so $b+1$ must be prime.", "metadata": {"resource_path": "IMO/segmented/en-IMO2019SL.jsonl"}} -{"year": "2019", "tier": "T0", "problem_label": "N6", "problem_type": "Number Theory", "exam": "IMO", "problem": "Let $H=\\left\\{\\lfloor i \\sqrt{2}\\rfloor: i \\in \\mathbb{Z}_{>0}\\right\\}=\\{1,2,4,5,7, \\ldots\\}$, and let $n$ be a positive integer. Prove that there exists a constant $C$ such that, if $A \\subset\\{1,2, \\ldots, n\\}$ satisfies $|A| \\geqslant C \\sqrt{n}$, then there exist $a, b \\in A$ such that $a-b \\in H$. (Brazil)", "solution": "First, observe that if $n$ is a positive integer, then $n \\in H$ exactly when $$ \\left\\{\\frac{n}{\\sqrt{2}}\\right\\}>1-\\frac{1}{\\sqrt{2}} . $$ To see why, observe that $n \\in H$ if and only if $00}$. In other words, $01$ since $a_{i}-a_{1} \\notin H$. Furthermore, we must have $\\left\\{a_{i} / \\sqrt{2}\\right\\}<\\left\\{a_{j} / \\sqrt{2}\\right\\}$ whenever $i1 / \\sqrt{2}>$ $1-1 / \\sqrt{2}$, contradicting (1). Now, we have a sequence $0=a_{1}\\frac{1}{2 d \\sqrt{2}} $$ To see why this is the case, let $h=\\lfloor d / \\sqrt{2}\\rfloor$, so $\\{d / \\sqrt{2}\\}=d / \\sqrt{2}-h$. Then $$ \\left\\{\\frac{d}{\\sqrt{2}}\\right\\}\\left(\\frac{d}{\\sqrt{2}}+h\\right)=\\frac{d^{2}-2 h^{2}}{2} \\geqslant \\frac{1}{2} $$ since the numerator is a positive integer. Because $d / \\sqrt{2}+h<2 d / \\sqrt{2}$, inequality (2) follows. Let $d_{i}=a_{i+1}-a_{i}$, for $1 \\leqslant i\\sum_{i}\\left\\{\\frac{d_{i}}{\\sqrt{2}}\\right\\}>\\frac{1}{2 \\sqrt{2}} \\sum_{i} \\frac{1}{d_{i}} \\geqslant \\frac{(k-1)^{2}}{2 \\sqrt{2}} \\frac{1}{\\sum_{i} d_{i}}>\\frac{(k-1)^{2}}{2 \\sqrt{2}} \\cdot \\frac{1}{n} . $$ Here, the first inequality holds because $\\left\\{a_{k} / \\sqrt{2}\\right\\}<1-1 / \\sqrt{2}$, the second follows from (2), the third follows from an easy application of the AM-HM inequality (or Cauchy-Schwarz), and the fourth follows from the fact that $\\sum_{i} d_{i}=a_{k}k-1 $$ which provides the required bound on $k$.", "metadata": {"resource_path": "IMO/segmented/en-IMO2019SL.jsonl"}} -{"year": "2019", "tier": "T0", "problem_label": "N6", "problem_type": "Number Theory", "exam": "IMO", "problem": "Let $H=\\left\\{\\lfloor i \\sqrt{2}\\rfloor: i \\in \\mathbb{Z}_{>0}\\right\\}=\\{1,2,4,5,7, \\ldots\\}$, and let $n$ be a positive integer. Prove that there exists a constant $C$ such that, if $A \\subset\\{1,2, \\ldots, n\\}$ satisfies $|A| \\geqslant C \\sqrt{n}$, then there exist $a, b \\in A$ such that $a-b \\in H$. (Brazil)", "solution": "Let $\\alpha=2+\\sqrt{2}$, so $(1 / \\alpha)+(1 / \\sqrt{2})=1$. Thus, $J=\\left\\{\\lfloor i \\alpha\\rfloor: i \\in \\mathbb{Z}_{>0}\\right\\}$ is the complementary Beatty sequence to $H$ (in other words, $H$ and $J$ are disjoint with $H \\cup J=\\mathbb{Z}_{>0}$ ). Write $A=\\left\\{a_{1}0}$. For any $j>i$, we have $a_{j}-a_{i}=\\left\\lfloor\\alpha b_{j}\\right\\rfloor-\\left\\lfloor\\alpha b_{i}\\right\\rfloor$. Because $a_{j}-a_{i} \\in J$, we also have $a_{j}-a_{i}=\\lfloor\\alpha t\\rfloor$ for some positive integer $t$. Thus, $\\lfloor\\alpha t\\rfloor=\\left\\lfloor\\alpha b_{j}\\right\\rfloor-\\left\\lfloor\\alpha b_{i}\\right\\rfloor$. The right hand side must equal either $\\left\\lfloor\\alpha\\left(b_{j}-b_{i}\\right)\\right\\rfloor$ or $\\left\\lfloor\\alpha\\left(b_{j}-b_{i}\\right)\\right\\rfloor-1$, the latter of which is not a member of $J$ as $\\alpha>2$. Therefore, $t=b_{j}-b_{i}$ and so we have $\\left\\lfloor\\alpha b_{j}\\right\\rfloor-\\left\\lfloor\\alpha b_{i}\\right\\rfloor=\\left\\lfloor\\alpha\\left(b_{j}-b_{i}\\right)\\right\\rfloor$. For $1 \\leqslant i1 /(2 d \\sqrt{2})$ for positive integers $d)$ proves that $1>\\left((k-1)^{2} /(2 \\sqrt{2})\\right)(\\alpha / n)$, which again rearranges to give $$ \\sqrt{2 \\sqrt{2}-2} \\cdot \\sqrt{n}>k-1 $$ Comment. The use of Beatty sequences in Solution 2 is essentially a way to bypass (1). Both Solutions 1 and 2 use the fact that $\\sqrt{2}<2$; the statement in the question would still be true if $\\sqrt{2}$ did not have this property (for instance, if it were replaced with $\\alpha$ ), but any argument along the lines of Solutions 1 or 2 would be more complicated.", "metadata": {"resource_path": "IMO/segmented/en-IMO2019SL.jsonl"}} -{"year": "2019", "tier": "T0", "problem_label": "N6", "problem_type": "Number Theory", "exam": "IMO", "problem": "Let $H=\\left\\{\\lfloor i \\sqrt{2}\\rfloor: i \\in \\mathbb{Z}_{>0}\\right\\}=\\{1,2,4,5,7, \\ldots\\}$, and let $n$ be a positive integer. Prove that there exists a constant $C$ such that, if $A \\subset\\{1,2, \\ldots, n\\}$ satisfies $|A| \\geqslant C \\sqrt{n}$, then there exist $a, b \\in A$ such that $a-b \\in H$. (Brazil)", "solution": "Again, define $J=\\mathbb{Z}_{>0} \\backslash H$, so all differences between elements of $A$ are in $J$. We start by making the following observation. Suppose we have a set $B \\subseteq\\{1,2, \\ldots, n\\}$ such that all of the differences between elements of $B$ are in $H$. Then $|A| \\cdot|B| \\leqslant 2 n$. To see why, observe that any two sums of the form $a+b$ with $a \\in A, b \\in B$ are different; otherwise, we would have $a_{1}+b_{1}=a_{2}+b_{2}$, and so $\\left|a_{1}-a_{2}\\right|=\\left|b_{2}-b_{1}\\right|$. However, then the left hand side is in $J$ whereas the right hand side is in $H$. Thus, $\\{a+b: a \\in A, b \\in B\\}$ is a set of size $|A| \\cdot|B|$ all of whose elements are no greater than $2 n$, yielding the claimed inequality. With this in mind, it suffices to construct a set $B$, all of whose differences are in $H$ and whose size is at least $C^{\\prime} \\sqrt{n}$ for some constant $C^{\\prime}>0$. To do so, we will use well-known facts about the negative Pell equation $X^{2}-2 Y^{2}=-1$; in particular, that there are infinitely many solutions and the values of $X$ are given by the recurrence $X_{1}=1, X_{2}=7$ and $X_{m}=6 X_{m-1}-X_{m-2}$. Therefore, we may choose $X$ to be a solution with $\\sqrt{n} / 6\\left(\\frac{X}{\\sqrt{2}}-Y\\right) \\geqslant \\frac{-3}{\\sqrt{2 n}}, $$ from which it follows that $\\{X / \\sqrt{2}\\}>1-(3 / \\sqrt{2 n})$. Combined with (1), this shows that all differences between elements of $B$ are in $H$. Comment. Some of the ideas behind Solution 3 may be used to prove that the constant $C=\\sqrt{2 \\sqrt{2}-2}$ (from Solutions 1 and 2) is optimal, in the sense that there are arbitrarily large values of $n$ and sets $A_{n} \\subseteq\\{1,2, \\ldots, n\\}$ of size roughly $C \\sqrt{n}$, all of whose differences are contained in $J$. To see why, choose $X$ to come from a sufficiently large solution to the Pell equation $X^{2}-2 Y^{2}=1$, so $\\{X / \\sqrt{2}\\} \\approx 1 /(2 X \\sqrt{2})$. In particular, $\\{X\\},\\{2 X\\}, \\ldots,\\{[2 X \\sqrt{2}(1-1 / \\sqrt{2})\\rfloor X\\}$ are all less than $1-1 / \\sqrt{2}$. Thus, by (1) any positive integer of the form $i X$ for $1 \\leqslant i \\leqslant\\lfloor 2 X \\sqrt{2}(1-1 / \\sqrt{2})\\rfloor$ lies in $J$. Set $n \\approx 2 X^{2} \\sqrt{2}(1-1 / \\sqrt{2})$. We now have a set $A=\\{i X: i \\leqslant\\lfloor 2 X \\sqrt{2}(1-1 / \\sqrt{2})\\rfloor\\}$ containing roughly $2 X \\sqrt{2}(1-1 / \\sqrt{2})$ elements less than or equal to $n$ such that all of the differences lie in $J$, and we can see that $|A| \\approx C \\sqrt{n}$ with $C=\\sqrt{2 \\sqrt{2}-2}$.", "metadata": {"resource_path": "IMO/segmented/en-IMO2019SL.jsonl"}} -{"year": "2019", "tier": "T0", "problem_label": "N6", "problem_type": "Number Theory", "exam": "IMO", "problem": "Let $H=\\left\\{\\lfloor i \\sqrt{2}\\rfloor: i \\in \\mathbb{Z}_{>0}\\right\\}=\\{1,2,4,5,7, \\ldots\\}$, and let $n$ be a positive integer. Prove that there exists a constant $C$ such that, if $A \\subset\\{1,2, \\ldots, n\\}$ satisfies $|A| \\geqslant C \\sqrt{n}$, then there exist $a, b \\in A$ such that $a-b \\in H$. (Brazil)", "solution": "As in Choose $Y$ to be a solution to the Pell-like equation $X^{2}-2 Y^{2}= \\pm 1$; such solutions are given by the recurrence $Y_{1}=1, Y_{2}=2$ and $Y_{m}=2 Y_{m-1}+Y_{m-2}$, and so we can choose $Y$ such that $n /(3 \\sqrt{2})\\sqrt{Y}$ elements; if that subsequence is decreasing, we may reflect (using (4) or (5)) to ensure that it is increasing. Call the subsequence $\\left\\{y_{1} \\sqrt{2}\\right\\},\\left\\{y_{2} \\sqrt{2}\\right\\}, \\ldots,\\left\\{y_{t} \\sqrt{2}\\right\\}$ for $t>\\sqrt{Y}$. Now, set $B=\\left\\{\\left\\lfloor y_{i} \\sqrt{2}\\right\\rfloor: 1 \\leqslant i \\leqslant t\\right\\}$. We have $\\left\\lfloor y_{j} \\sqrt{2}\\right\\rfloor-\\left\\lfloor y_{i} \\sqrt{2}\\right\\rfloor=\\left\\lfloor\\left(y_{j}-y_{i}\\right) \\sqrt{2}\\right\\rfloor$ for $i\\sqrt{Y}>\\sqrt{n} / \\sqrt{3 \\sqrt{2}}$, this is the required set. Comment. Any solution to this problem will need to use the fact that $\\sqrt{2}$ cannot be approximated well by rationals, either directly or implicitly (for example, by using facts about solutions to Pelllike equations). If $\\sqrt{2}$ were replaced by a value of $\\theta$ with very good rational approximations (from below), then an argument along the lines of Solution 3 would give long arithmetic progressions in $\\{\\lfloor i \\theta\\rfloor: 0 \\leqslant i0$ and infinitely many positive integers $n$ with the following property: there are infinitely many positive integers that cannot be expressed as the sum of fewer than $c n \\log (n)$ pairwise coprime $n^{\\text {th }}$ powers. (Canada)", "solution": "Suppose, for an integer $n$, that we can find another integer $N$ satisfying the following property: $n$ is divisible by $\\varphi\\left(p^{e}\\right)$ for every prime power $p^{e}$ exactly dividing $N$. This property ensures that all $n^{\\text {th }}$ powers are congruent to 0 or 1 modulo each such prime power $p^{e}$, and hence that any sum of $m$ pairwise coprime $n^{\\text {th }}$ powers is congruent to $m$ or $m-1$ modulo $p^{e}$, since at most one of the $n^{\\text {th }}$ powers is divisible by $p$. Thus, if $k$ denotes the number of distinct prime factors of $N$, we find by the Chinese Remainder Theorem at most $2^{k} m$ residue classes modulo $N$ which are sums of at most $m$ pairwise coprime $n^{\\text {th }}$ powers. In particular, if $N>2^{k} m$ then there are infinitely many positive integers not expressible as a sum of at most $m$ pairwise coprime $n^{\\text {th }}$ powers. It thus suffices to prove that there are arbitrarily large pairs $(n, N)$ of integers satisfying $(\\dagger)$ such that $$ N>c \\cdot 2^{k} n \\log (n) $$ for some positive constant $c$. We construct such pairs as follows. Fix a positive integer $t$ and choose (distinct) prime numbers $p \\mid 2^{2^{t-1}}+1$ and $q \\mid 2^{2^{t}}+1$; we set $N=p q$. It is well-known that $2^{t} \\mid p-1$ and $2^{t+1} \\mid q-1$, hence $$ n=\\frac{(p-1)(q-1)}{2^{t}} $$ is an integer and the pair $(n, N)$ satisfies $(\\dagger)$. Estimating the size of $N$ and $n$ is now straightforward. We have $$ \\log _{2}(n) \\leqslant 2^{t-1}+2^{t}-t<2^{t+1}<2 \\cdot \\frac{N}{n} $$ which rearranges to $$ N>\\frac{1}{8} \\cdot 2^{2} n \\log _{2}(n) $$ and so we are done if we choose $c<\\frac{1}{8 \\log (2)} \\approx 0.18$. Comment 1. The trick in the above solution was to find prime numbers $p$ and $q$ congruent to 1 modulo some $d=2^{t}$ and which are not too large. An alternative way to do this is via Linnik's Theorem, which says that there are absolute constants $b$ and $L>1$ such that for any coprime integers $a$ and $d$, there is a prime congruent to $a$ modulo $d$ and of size $\\leqslant b d^{L}$. If we choose some $d$ not divisible by 3 and choose two distinct primes $p, q \\leqslant b \\cdot(3 d)^{L}$ congruent to 1 modulo $d$ (and, say, distinct modulo 3 ), then we obtain a pair $(n, N)$ satisfying $(\\dagger)$ with $N=p q$ and $n=\\frac{(p-1)(q-1)}{d}$. A straightforward computation shows that $$ N>C n^{1+\\frac{1}{2 L-1}} $$ for some constant $C$, which is in particular larger than any $c \\cdot 2^{2} n \\log (n)$ for $p$ large. Thus, the statement of the problem is true for any constant $c$. More strongly, the statement of the problem is still true when $c n \\log (n)$ is replaced by $n^{1+\\delta}$ for a sufficiently small $\\delta>0$. Solution 2, obtaining better bounds. As in the preceding solution, we seek arbitrarily large pairs of integers $n$ and $N$ satisfying ( $\\dagger$ ) such that $N>c 2^{k} n \\log (n)$. This time, to construct such pairs, we fix an integer $x \\geqslant 4$, set $N$ to be the lowest common multiple of $1,2, \\ldots, 2 x$, and set $n$ to be twice the lowest common multiple of $1,2, \\ldots, x$. The pair $(n, N)$ does indeed satisfy the condition, since if $p^{e}$ is a prime power divisor of $N$ then $\\frac{\\varphi\\left(p^{e}\\right)}{2} \\leqslant x$ is a factor of $\\frac{n}{2}=\\operatorname{lcm}_{r \\leqslant x}(r)$. Now $2 N / n$ is the product of all primes having a power lying in the interval $(x, 2 x]$, and hence $2 N / n>x^{\\pi(2 x)-\\pi(x)}$. Thus for sufficiently large $x$ we have $$ \\log \\left(\\frac{2 N}{2^{\\pi(2 x)} n}\\right)>(\\pi(2 x)-\\pi(x)) \\log (x)-\\log (2) \\pi(2 x) \\sim x $$ using the Prime Number Theorem $\\pi(t) \\sim t / \\log (t)$. On the other hand, $n$ is a product of at most $\\pi(x)$ prime powers less than or equal to $x$, and so we have the upper bound $$ \\log (n) \\leqslant \\pi(x) \\log (x) \\sim x $$ again by the Prime Number Theorem. Combined with the above inequality, we find that for any $\\epsilon>0$, the inequality $$ \\log \\left(\\frac{N}{2^{\\pi(2 x)} n}\\right)>(1-\\epsilon) \\log (n) $$ holds for sufficiently large $x$. Rearranging this shows that $$ N>2^{\\pi(2 x)} n^{2-\\epsilon}>2^{\\pi(2 x)} n \\log (n) $$ for all sufficiently large $x$ and we are done. Comment 2. The stronger bound $N>2^{\\pi(2 x)} n^{2-\\epsilon}$ obtained in the above proof of course shows that infinitely many positive integers cannot be written as a sum of at most $n^{2-\\epsilon}$ pairwise coprime $n^{\\text {th }}$ powers. By refining the method in Solution 2, these bounds can be improved further to show that infinitely many positive integers cannot be written as a sum of at most $n^{\\alpha}$ pairwise coprime $n^{\\text {th }}$ powers for any positive $\\alpha>0$. To do this, one fixes a positive integer $d$, sets $N$ equal to the product of the primes at most $d x$ which are congruent to 1 modulo $d$, and $n=d \\mathrm{lcm}_{r \\leqslant x}(r)$. It follows as in Solution 2 that $(n, N)$ satisfies $(\\dagger)$. Now the Prime Number Theorem in arithmetic progressions provides the estimates $\\log (N) \\sim \\frac{d}{\\varphi(d)} x$, $\\log (n) \\sim x$ and $\\pi(d x) \\sim \\frac{d x}{\\log (x)}$ for any fixed $d$. Combining these provides a bound $$ N>2^{\\pi(d x)} n^{d / \\varphi(d)-\\epsilon} $$ for any positive $\\epsilon$, valid for $x$ sufficiently large. Since the ratio $\\frac{d}{\\varphi(d)}$ can be made arbitrarily large by a judicious choice of $d$, we obtain the $n^{\\alpha}$ bound claimed. Comment 3. While big results from analytic number theory such as the Prime Number Theorem or Linnik's Theorem certainly can be used in this problem, they do not seem to substantially simplify matters: all known solutions involve first reducing to condition ( $\\dagger$ ), and even then analytic results do not make it clear how to proceed. For this reason, we regard this problem as suitable for the IMO. Rather than simplifying the problem, what nonelementary results from analytic number theory allow one to achieve is a strengthening of the main bound, typically replacing the $n \\log (n)$ growth with a power $n^{1+\\delta}$. However, we believe that such stronger bounds are unlikely to be found by students in the exam. The strongest bound we know how to achieve using purely elementary methods is a bound of the form $N>2^{k} n \\log (n)^{M}$ for any positive integer $M$. This is achieved by a variant of the argument in Solution 1, choosing primes $p_{0}, \\ldots, p_{M}$ with $p_{i} \\mid 2^{2^{t+i-1}}+1$ and setting $N=\\prod_{i} p_{i}$ and $n=$ $2^{-t M} \\prod_{i}\\left(p_{i}-1\\right)$.", "metadata": {"resource_path": "IMO/segmented/en-IMO2019SL.jsonl"}} -{"year": "2019", "tier": "T0", "problem_label": "N8", "problem_type": "Number Theory", "exam": "IMO", "problem": "Let $a$ and $b$ be two positive integers. Prove that the integer $$ a^{2}+\\left\\lceil\\frac{4 a^{2}}{b}\\right\\rceil $$ is not a square. (Here $\\lceil z\\rceil$ denotes the least integer greater than or equal to $z$.)", "solution": "Arguing indirectly, assume that $$ a^{2}+\\left\\lceil\\frac{4 a^{2}}{b}\\right\\rceil=(a+k)^{2}, \\quad \\text { or } \\quad\\left\\lceil\\frac{(2 a)^{2}}{b}\\right\\rceil=(2 a+k) k $$ Clearly, $k \\geqslant 1$. In other words, the equation $$ \\left\\lceil\\frac{c^{2}}{b}\\right\\rceil=(c+k) k $$ has a positive integer solution $(c, k)$, with an even value of $c$. Choose a positive integer solution of (1) with minimal possible value of $k$, without regard to the parity of $c$. From $$ \\frac{c^{2}}{b}>\\left\\lceil\\frac{c^{2}}{b}\\right\\rceil-1=c k+k^{2}-1 \\geqslant c k $$ and $$ \\frac{(c-k)(c+k)}{b}<\\frac{c^{2}}{b} \\leqslant\\left\\lceil\\frac{c^{2}}{b}\\right\\rceil=(c+k) k $$ it can be seen that $c>b k>c-k$, so $$ c=k b+r \\quad \\text { with some } 00$ and $0a$, so $$ \\begin{aligned} & c^{2}-10$ and $a<0$. Indeed, since $\\sum_{i=1}^{2019} u_{i}^{2}=1$, the variables $u_{i}$ cannot be all zero, and, since $\\sum_{i=1}^{2019} u_{i}=0$, the nonzero elements cannot be all positive or all negative. Let $P=\\left\\{i: u_{i}>0\\right\\}$ and $N=\\left\\{i: u_{i} \\leqslant 0\\right\\}$ be the indices of positive and nonpositive elements in the sequence, and let $p=|P|$ and $n=|N|$ be the sizes of these sets; then $p+n=2019$. By the condition $\\sum_{i=1}^{2019} u_{i}=0$ we have $0=\\sum_{i=1}^{2019} u_{i}=\\sum_{i \\in P} u_{i}-\\sum_{i \\in N}\\left|u_{i}\\right|$, so $$ \\sum_{i \\in P} u_{i}=\\sum_{i \\in N}\\left|u_{i}\\right| . $$ After this preparation, estimate the sum of squares of the positive and nonpositive elements as follows: $$ \\begin{aligned} & \\sum_{i \\in P} u_{i}^{2} \\leqslant \\sum_{i \\in P} b u_{i}=b \\sum_{i \\in P} u_{i}=b \\sum_{i \\in N}\\left|u_{i}\\right| \\leqslant b \\sum_{i \\in N}|a|=-n a b ; \\\\ & \\sum_{i \\in N} u_{i}^{2} \\leqslant \\sum_{i \\in N}|a| \\cdot\\left|u_{i}\\right|=|a| \\sum_{i \\in N}\\left|u_{i}\\right|=|a| \\sum_{i \\in P} u_{i} \\leqslant|a| \\sum_{i \\in P} b=-p a b . \\end{aligned} $$ The sum of these estimates is $$ 1=\\sum_{i=1}^{2019} u_{i}^{2}=\\sum_{i \\in P} u_{i}^{2}+\\sum_{i \\in N} u_{i}^{2} \\leqslant-(p+n) a b=-2019 a b ; $$ that proves $a b \\leqslant \\frac{-1}{2019}$. Comment 1. After observing $\\sum_{i \\in P} u_{i}^{2} \\leqslant b \\sum_{i \\in P} u_{i}$ and $\\sum_{i \\in N} u_{i}^{2} \\leqslant|a| \\sum_{i \\in P}\\left|u_{i}\\right|$, instead of $(2,3)$ an alternative continuation is $$ |a b| \\geqslant \\frac{\\sum_{i \\in P} u_{i}^{2}}{\\sum_{i \\in P} u_{i}} \\cdot \\frac{\\sum_{i \\in N} u_{i}^{2}}{\\sum_{i \\in N}\\left|u_{i}\\right|}=\\frac{\\sum_{i \\in P} u_{i}^{2}}{\\left(\\sum_{i \\in P} u_{i}\\right)^{2}} \\sum_{i \\in N} u_{i}^{2} \\geqslant \\frac{1}{p} \\sum_{i \\in N} u_{i}^{2} $$ (by the AM-QM or the Cauchy-Schwarz inequality) and similarly $|a b| \\geqslant \\frac{1}{n} \\sum_{i \\in P} u_{i}^{2}$.", "metadata": {"resource_path": "IMO/segmented/en-IMO2019SL.jsonl"}} -{"year": "2019", "tier": "T0", "problem_label": "A2", "problem_type": "Algebra", "exam": "IMO", "problem": "Let $u_{1}, u_{2}, \\ldots, u_{2019}$ be real numbers satisfying $$ u_{1}+u_{2}+\\cdots+u_{2019}=0 \\quad \\text { and } \\quad u_{1}^{2}+u_{2}^{2}+\\cdots+u_{2019}^{2}=1 $$ Let $a=\\min \\left(u_{1}, u_{2}, \\ldots, u_{2019}\\right)$ and $b=\\max \\left(u_{1}, u_{2}, \\ldots, u_{2019}\\right)$. Prove that $$ a b \\leqslant-\\frac{1}{2019} $$ (Germany)", "solution": "As in the previous solution we conclude that $a<0$ and $b>0$. For every index $i$, the number $u_{i}$ is a convex combination of $a$ and $b$, so $$ u_{i}=x_{i} a+y_{i} b \\quad \\text { with some weights } 0 \\leqslant x_{i}, y_{i} \\leqslant 1, \\text { with } x_{i}+y_{i}=1 \\text {. } $$ Let $X=\\sum_{i=1}^{2019} x_{i}$ and $Y=\\sum_{i=1}^{2019} y_{i}$. From $0=\\sum_{i=1}^{2019} u_{i}=\\sum_{i=1}^{2019}\\left(x_{i} a+y_{i} b\\right)=-|a| X+b Y$, we get $$ |a| X=b Y $$ From $\\sum_{i=1}^{2019}\\left(x_{i}+y_{i}\\right)=2019$ we have $$ X+Y=2019 $$ The system of linear equations $(4,5)$ has a unique solution: $$ X=\\frac{2019 b}{|a|+b}, \\quad Y=\\frac{2019|a|}{|a|+b} $$ Now apply the following estimate to every $u_{i}^{2}$ in their sum: $$ u_{i}^{2}=x_{i}^{2} a^{2}+2 x_{i} y_{i} a b+y_{i}^{2} b^{2} \\leqslant x_{i} a^{2}+y_{i} b^{2} $$ we obtain that $$ 1=\\sum_{i=1}^{2019} u_{i}^{2} \\leqslant \\sum_{i=1}^{2019}\\left(x_{i} a^{2}+y_{i} b^{2}\\right)=X a^{2}+Y b^{2}=\\frac{2019 b}{|a|+b}|a|^{2}+\\frac{2019|a|}{|a|+b} b^{2}=2019|a| b=-2019 a b . $$ Hence, $a b \\leqslant \\frac{-1}{2019}$. Comment 2. The idea behind Solution 2 is the following thought. Suppose we fix $a<0$ and $b>0$, fix $\\sum u_{i}=0$ and vary the $u_{i}$ to achieve the maximum value of $\\sum u_{i}^{2}$. Considering varying any two of the $u_{i}$ while preserving their sum: the maximum value of $\\sum u_{i}^{2}$ is achieved when those two are as far apart as possible, so all but at most one of the $u_{i}$ are equal to $a$ or $b$. Considering a weighted version of the problem, we see the maximum (with fractional numbers of $u_{i}$ having each value) is achieved when $\\frac{2019 b}{|a|+b}$ of them are $a$ and $\\frac{2019|a|}{|a|+b}$ are $b$. In fact, this happens in the solution: the number $u_{i}$ is replaced by $x_{i}$ copies of $a$ and $y_{i}$ copies of $b$.", "metadata": {"resource_path": "IMO/segmented/en-IMO2019SL.jsonl"}} -{"year": "2019", "tier": "T0", "problem_label": "A3", "problem_type": "Algebra", "exam": "IMO", "problem": "Let $n \\geqslant 3$ be a positive integer and let $\\left(a_{1}, a_{2}, \\ldots, a_{n}\\right)$ be a strictly increasing sequence of $n$ positive real numbers with sum equal to 2 . Let $X$ be a subset of $\\{1,2, \\ldots, n\\}$ such that the value of $$ \\left|1-\\sum_{i \\in X} a_{i}\\right| $$ is minimised. Prove that there exists a strictly increasing sequence of $n$ positive real numbers $\\left(b_{1}, b_{2}, \\ldots, b_{n}\\right)$ with sum equal to 2 such that $$ \\sum_{i \\in X} b_{i}=1 $$ (New Zealand) Common remarks. In all solutions, we say an index set $X$ is $\\left(a_{i}\\right)$-minimising if it has the property in the problem for the given sequence $\\left(a_{i}\\right)$. Write $X^{c}$ for the complement of $X$, and $[a, b]$ for the interval of integers $k$ such that $a \\leqslant k \\leqslant b$. Note that $$ \\left|1-\\sum_{i \\in X} a_{i}\\right|=\\left|1-\\sum_{i \\in X^{c}} a_{i}\\right|, $$ so we may exchange $X$ and $X^{c}$ where convenient. Let $$ \\Delta=\\sum_{i \\in X^{c}} a_{i}-\\sum_{i \\in X} a_{i} $$ and note that $X$ is $\\left(a_{i}\\right)$-minimising if and only if it minimises $|\\Delta|$, and that $\\sum_{i \\in X} a_{i}=1$ if and only if $\\Delta=0$. In some solutions, a scaling process is used. If we have a strictly increasing sequence of positive real numbers $c_{i}$ (typically obtained by perturbing the $a_{i}$ in some way) such that $$ \\sum_{i \\in X} c_{i}=\\sum_{i \\in X^{c}} c_{i} $$ then we may put $b_{i}=2 c_{i} / \\sum_{j=1}^{n} c_{j}$. So it suffices to construct such a sequence without needing its sum to be 2 . The solutions below show various possible approaches to the problem. Solutions 1 and 2 perturb a few of the $a_{i}$ to form the $b_{i}$ (with scaling in the case of Solution 1, without scaling in the case of Solution 2). Solutions 3 and 4 look at properties of the index set $X$. Solution 3 then perturbs many of the $a_{i}$ to form the $b_{i}$, together with scaling. Rather than using such perturbations, Solution 4 constructs a sequence $\\left(b_{i}\\right)$ directly from the set $X$ with the required properties. Solution 4 can be used to give a complete description of sets $X$ that are $\\left(a_{i}\\right)$-minimising for some $\\left(a_{i}\\right)$.", "solution": "Without loss of generality, assume $\\sum_{i \\in X} a_{i} \\leqslant 1$, and we may assume strict inequality as otherwise $b_{i}=a_{i}$ works. Also, $X$ clearly cannot be empty. If $n \\in X$, add $\\Delta$ to $a_{n}$, producing a sequence of $c_{i}$ with $\\sum_{i \\in X} c_{i}=\\sum_{i \\in X^{c}} c_{i}$, and then scale as described above to make the sum equal to 2 . Otherwise, there is some $k$ with $k \\in X$ and $k+1 \\in X^{c}$. Let $\\delta=a_{k+1}-a_{k}$. - If $\\delta>\\Delta$, add $\\Delta$ to $a_{k}$ and then scale. - If $\\delta<\\Delta$, then considering $X \\cup\\{k+1\\} \\backslash\\{k\\}$ contradicts $X$ being $\\left(a_{i}\\right)$-minimising. - If $\\delta=\\Delta$, choose any $j \\neq k, k+1$ (possible since $n \\geqslant 3$ ), and any $\\epsilon$ less than the least of $a_{1}$ and all the differences $a_{i+1}-a_{i}$. If $j \\in X$ then add $\\Delta-\\epsilon$ to $a_{k}$ and $\\epsilon$ to $a_{j}$, then scale; otherwise, add $\\Delta$ to $a_{k}$ and $\\epsilon / 2$ to $a_{k+1}$, and subtract $\\epsilon / 2$ from $a_{j}$, then scale.", "metadata": {"resource_path": "IMO/segmented/en-IMO2019SL.jsonl"}} -{"year": "2019", "tier": "T0", "problem_label": "A3", "problem_type": "Algebra", "exam": "IMO", "problem": "Let $n \\geqslant 3$ be a positive integer and let $\\left(a_{1}, a_{2}, \\ldots, a_{n}\\right)$ be a strictly increasing sequence of $n$ positive real numbers with sum equal to 2 . Let $X$ be a subset of $\\{1,2, \\ldots, n\\}$ such that the value of $$ \\left|1-\\sum_{i \\in X} a_{i}\\right| $$ is minimised. Prove that there exists a strictly increasing sequence of $n$ positive real numbers $\\left(b_{1}, b_{2}, \\ldots, b_{n}\\right)$ with sum equal to 2 such that $$ \\sum_{i \\in X} b_{i}=1 $$ (New Zealand) Common remarks. In all solutions, we say an index set $X$ is $\\left(a_{i}\\right)$-minimising if it has the property in the problem for the given sequence $\\left(a_{i}\\right)$. Write $X^{c}$ for the complement of $X$, and $[a, b]$ for the interval of integers $k$ such that $a \\leqslant k \\leqslant b$. Note that $$ \\left|1-\\sum_{i \\in X} a_{i}\\right|=\\left|1-\\sum_{i \\in X^{c}} a_{i}\\right|, $$ so we may exchange $X$ and $X^{c}$ where convenient. Let $$ \\Delta=\\sum_{i \\in X^{c}} a_{i}-\\sum_{i \\in X} a_{i} $$ and note that $X$ is $\\left(a_{i}\\right)$-minimising if and only if it minimises $|\\Delta|$, and that $\\sum_{i \\in X} a_{i}=1$ if and only if $\\Delta=0$. In some solutions, a scaling process is used. If we have a strictly increasing sequence of positive real numbers $c_{i}$ (typically obtained by perturbing the $a_{i}$ in some way) such that $$ \\sum_{i \\in X} c_{i}=\\sum_{i \\in X^{c}} c_{i} $$ then we may put $b_{i}=2 c_{i} / \\sum_{j=1}^{n} c_{j}$. So it suffices to construct such a sequence without needing its sum to be 2 . The solutions below show various possible approaches to the problem. Solutions 1 and 2 perturb a few of the $a_{i}$ to form the $b_{i}$ (with scaling in the case of Solution 1, without scaling in the case of Solution 2). Solutions 3 and 4 look at properties of the index set $X$. Solution 3 then perturbs many of the $a_{i}$ to form the $b_{i}$, together with scaling. Rather than using such perturbations, Solution 4 constructs a sequence $\\left(b_{i}\\right)$ directly from the set $X$ with the required properties. Solution 4 can be used to give a complete description of sets $X$ that are $\\left(a_{i}\\right)$-minimising for some $\\left(a_{i}\\right)$.", "solution": "This is similar to Suppose there exists $1 \\leqslant j \\leqslant n-1$ such that $j \\in X$ but $j+1 \\in X^{c}$. Then $a_{j+1}-a_{j} \\geqslant \\Delta$, because otherwise considering $X \\cup\\{j+1\\} \\backslash\\{j\\}$ contradicts $X$ being $\\left(a_{i}\\right)$-minimising. If $a_{j+1}-a_{j}>\\Delta$, put $$ b_{i}= \\begin{cases}a_{j}+\\Delta / 2, & \\text { if } i=j \\\\ a_{j+1}-\\Delta / 2, & \\text { if } i=j+1 \\\\ a_{i}, & \\text { otherwise }\\end{cases} $$ If $a_{j+1}-a_{j}=\\Delta$, choose any $\\epsilon$ less than the least of $\\Delta / 2, a_{1}$ and all the differences $a_{i+1}-a_{i}$. If $|X| \\geqslant 2$, choose $k \\in X$ with $k \\neq j$, and put $$ b_{i}= \\begin{cases}a_{j}+\\Delta / 2-\\epsilon, & \\text { if } i=j \\\\ a_{j+1}-\\Delta / 2, & \\text { if } i=j+1 \\\\ a_{k}+\\epsilon, & \\text { if } i=k \\\\ a_{i}, & \\text { otherwise }\\end{cases} $$ Otherwise, $\\left|X^{c}\\right| \\geqslant 2$, so choose $k \\in X^{c}$ with $k \\neq j+1$, and put $$ b_{i}= \\begin{cases}a_{j}+\\Delta / 2, & \\text { if } i=j \\\\ a_{j+1}-\\Delta / 2+\\epsilon, & \\text { if } i=j+1 \\\\ a_{k}-\\epsilon, & \\text { if } i=k \\\\ a_{i}, & \\text { otherwise }\\end{cases} $$ If there is no $1 \\leqslant j \\leqslant n$ such that $j \\in X$ but $j+1 \\in X^{c}$, there must be some $1\\Delta$, as otherwise considering $X \\cup\\{1\\}$ contradicts $X$ being $\\left(a_{i}\\right)$-minimising. Now put $$ b_{i}= \\begin{cases}a_{1}-\\Delta / 2, & \\text { if } i=1 \\\\ a_{n}+\\Delta / 2, & \\text { if } i=n \\\\ a_{i}, & \\text { otherwise }\\end{cases} $$", "metadata": {"resource_path": "IMO/segmented/en-IMO2019SL.jsonl"}} -{"year": "2019", "tier": "T0", "problem_label": "A3", "problem_type": "Algebra", "exam": "IMO", "problem": "Let $n \\geqslant 3$ be a positive integer and let $\\left(a_{1}, a_{2}, \\ldots, a_{n}\\right)$ be a strictly increasing sequence of $n$ positive real numbers with sum equal to 2 . Let $X$ be a subset of $\\{1,2, \\ldots, n\\}$ such that the value of $$ \\left|1-\\sum_{i \\in X} a_{i}\\right| $$ is minimised. Prove that there exists a strictly increasing sequence of $n$ positive real numbers $\\left(b_{1}, b_{2}, \\ldots, b_{n}\\right)$ with sum equal to 2 such that $$ \\sum_{i \\in X} b_{i}=1 $$ (New Zealand) Common remarks. In all solutions, we say an index set $X$ is $\\left(a_{i}\\right)$-minimising if it has the property in the problem for the given sequence $\\left(a_{i}\\right)$. Write $X^{c}$ for the complement of $X$, and $[a, b]$ for the interval of integers $k$ such that $a \\leqslant k \\leqslant b$. Note that $$ \\left|1-\\sum_{i \\in X} a_{i}\\right|=\\left|1-\\sum_{i \\in X^{c}} a_{i}\\right|, $$ so we may exchange $X$ and $X^{c}$ where convenient. Let $$ \\Delta=\\sum_{i \\in X^{c}} a_{i}-\\sum_{i \\in X} a_{i} $$ and note that $X$ is $\\left(a_{i}\\right)$-minimising if and only if it minimises $|\\Delta|$, and that $\\sum_{i \\in X} a_{i}=1$ if and only if $\\Delta=0$. In some solutions, a scaling process is used. If we have a strictly increasing sequence of positive real numbers $c_{i}$ (typically obtained by perturbing the $a_{i}$ in some way) such that $$ \\sum_{i \\in X} c_{i}=\\sum_{i \\in X^{c}} c_{i} $$ then we may put $b_{i}=2 c_{i} / \\sum_{j=1}^{n} c_{j}$. So it suffices to construct such a sequence without needing its sum to be 2 . The solutions below show various possible approaches to the problem. Solutions 1 and 2 perturb a few of the $a_{i}$ to form the $b_{i}$ (with scaling in the case of Solution 1, without scaling in the case of Solution 2). Solutions 3 and 4 look at properties of the index set $X$. Solution 3 then perturbs many of the $a_{i}$ to form the $b_{i}$, together with scaling. Rather than using such perturbations, Solution 4 constructs a sequence $\\left(b_{i}\\right)$ directly from the set $X$ with the required properties. Solution 4 can be used to give a complete description of sets $X$ that are $\\left(a_{i}\\right)$-minimising for some $\\left(a_{i}\\right)$.", "solution": "Without loss of generality, assume $\\sum_{i \\in X} a_{i} \\leqslant 1$, so $\\Delta \\geqslant 0$. If $\\Delta=0$ we can take $b_{i}=a_{i}$, so now assume that $\\Delta>0$. Suppose that there is some $k \\leqslant n$ such that $|X \\cap[k, n]|>\\left|X^{c} \\cap[k, n]\\right|$. If we choose the largest such $k$ then $|X \\cap[k, n]|-\\left|X^{c} \\cap[k, n]\\right|=1$. We can now find the required sequence $\\left(b_{i}\\right)$ by starting with $c_{i}=a_{i}$ for $i\\frac{n-k+1}{2}$ and $|X \\cap[\\ell, n]|<\\frac{n-\\ell+1}{2}$. We now construct our sequence $\\left(b_{i}\\right)$ using this claim. Let $k$ and $\\ell$ be the greatest values satisfying the claim, and without loss of generality suppose $k=n$ and $\\ell\\sum_{i \\in Y} a_{i}>1 $$ contradicting $X$ being $\\left(a_{i}\\right)$-minimising. Otherwise, we always have equality, meaning that $X=Y$. But now consider $Z=Y \\cup\\{n-1\\} \\backslash\\{n\\}$. Since $n \\geqslant 3$, we have $$ \\sum_{i \\in Y} a_{i}>\\sum_{i \\in Z} a_{i}>\\sum_{i \\in Y^{c}} a_{i}=2-\\sum_{i \\in Y} a_{i} $$ and so $Z$ contradicts $X$ being $\\left(a_{i}\\right)$-minimising.", "metadata": {"resource_path": "IMO/segmented/en-IMO2019SL.jsonl"}} -{"year": "2019", "tier": "T0", "problem_label": "A5", "problem_type": "Algebra", "exam": "IMO", "problem": "Let $x_{1}, x_{2}, \\ldots, x_{n}$ be different real numbers. Prove that $$ \\sum_{1 \\leqslant i \\leqslant n} \\prod_{j \\neq i} \\frac{1-x_{i} x_{j}}{x_{i}-x_{j}}= \\begin{cases}0, & \\text { if } n \\text { is even } \\\\ 1, & \\text { if } n \\text { is odd }\\end{cases} $$ (Kazakhstan) Common remarks. Let $G\\left(x_{1}, x_{2}, \\ldots, x_{n}\\right)$ be the function of the $n$ variables $x_{1}, x_{2}, \\ldots, x_{n}$ on the LHS of the required identity. Solution 1 (Lagrange interpolation). Since both sides of the identity are rational functions, it suffices to prove it when all $x_{i} \\notin\\{ \\pm 1\\}$. Define $$ f(t)=\\prod_{i=1}^{n}\\left(1-x_{i} t\\right) $$ and note that $$ f\\left(x_{i}\\right)=\\left(1-x_{i}^{2}\\right) \\prod_{j \\neq i} 1-x_{i} x_{j} $$ Using the nodes $+1,-1, x_{1}, \\ldots, x_{n}$, the Lagrange interpolation formula gives us the following expression for $f$ : $$ \\sum_{i=1}^{n} f\\left(x_{i}\\right) \\frac{(x-1)(x+1)}{\\left(x_{i}-1\\right)\\left(x_{i}+1\\right)} \\prod_{j \\neq i} \\frac{x-x_{j}}{x_{i}-x_{j}}+f(1) \\frac{x+1}{1+1} \\prod_{1 \\leqslant i \\leqslant n} \\frac{x-x_{i}}{1-x_{i}}+f(-1) \\frac{x-1}{-1-1} \\prod_{1 \\leqslant i \\leqslant n} \\frac{x-x_{i}}{1-x_{i}} $$ The coefficient of $t^{n+1}$ in $f(t)$ is 0 , since $f$ has degree $n$. The coefficient of $t^{n+1}$ in the above expression of $f$ is $$ \\begin{aligned} 0 & =\\sum_{1 \\leqslant i \\leqslant n} \\frac{f\\left(x_{i}\\right)}{\\prod_{j \\neq i}\\left(x_{i}-x_{j}\\right) \\cdot\\left(x_{i}-1\\right)\\left(x_{i}+1\\right)}+\\frac{f(1)}{\\prod_{1 \\leqslant j \\leqslant n}\\left(1-x_{j}\\right) \\cdot(1+1)}+\\frac{f(-1)}{\\prod_{1 \\leqslant j \\leqslant n}\\left(-1-x_{j}\\right) \\cdot(-1-1)} \\\\ & =-G\\left(x_{1}, \\ldots, x_{n}\\right)+\\frac{1}{2}+\\frac{(-1)^{n+1}}{2} \\end{aligned} $$ Comment. The main difficulty is to think of including the two extra nodes $\\pm 1$ and evaluating the coefficient $t^{n+1}$ in $f$ when $n+1$ is higher than the degree of $f$. It is possible to solve the problem using Lagrange interpolation on the nodes $x_{1}, \\ldots, x_{n}$, but the definition of the polynomial being interpolated should depend on the parity of $n$. For $n$ even, consider the polynomial $$ P(x)=\\prod_{i}\\left(1-x x_{i}\\right)-\\prod_{i}\\left(x-x_{i}\\right) $$ Lagrange interpolation shows that $G$ is the coefficient of $x^{n-1}$ in the polynomial $P(x) /\\left(1-x^{2}\\right)$, i.e. 0 . For $n$ odd, consider the polynomial $$ P(x)=\\prod_{i}\\left(1-x x_{i}\\right)-x \\prod_{i}\\left(x-x_{i}\\right) $$ Now $G$ is the coefficient of $x^{n-1}$ in $P(x) /\\left(1-x^{2}\\right)$, which is 1 . Solution 2 (using symmetries). Observe that $G$ is symmetric in the variables $x_{1}, \\ldots, x_{n}$. Define $V=\\prod_{i0$. By Step 2, each of its monomials $\\mu x^{a} y^{b} z^{c}$ of the highest total degree satisfies $a, b \\geqslant c$. Applying other weak symmetries, we obtain $a, c \\geqslant b$ and $b, c \\geqslant a$; therefore, $P$ has a unique leading monomial of the form $\\mu(x y z)^{c}$. The polynomial $P_{0}(x, y, z)=P(x, y, z)-\\mu\\left(x y z-x^{2}-y^{2}-z^{2}\\right)^{c}$ has smaller total degree. Since $P_{0}$ is symmetric, it is representable as a polynomial function of $x y z-x^{2}-y^{2}-z^{2}$. Then $P$ is also of this form, completing the inductive step. Comment. We could alternatively carry out Step 1 by an induction on $n=\\operatorname{deg}_{z} P$, in a manner similar to Step 3. If $n=0$, the statement holds. Assume that $n>0$ and check the leading component of $P$ with respect to $z$ : $$ P(x, y, z)=Q_{n}(x, y) z^{n}+R(x, y, z) $$ where $\\operatorname{deg}_{z} Rv(x)$ and $v(x) \\geqslant v(y)$. Hence this $f$ satisfies the functional equation and 0 is an $f$-rare integer. Comment 3. In fact, if $v$ is an $f$-rare integer for an $f$ satisfying the functional equation, then its fibre $X_{v}=\\{v\\}$ must be a singleton. We may assume without loss of generality that $v=0$. We've already seen in Solution 1 that 0 is either the greatest or least element of $X_{0}$; replacing $f$ with the function $x \\mapsto-f(-x)$ if necessary, we may assume that 0 is the least element of $X_{0}$. We write $b$ for the largest element of $X_{0}$, supposing for contradiction that $b>0$, and write $N=(2 b)$ !. It now follows from (*) that we have $$ f(f(N b)+b)=f(f(0)+b)=f(b)=0 $$ from which we see that $f(N b)+b \\in X_{0} \\subseteq[0, b]$. It follows that $f(N b) \\in[-b, 0)$, since by construction $N b \\notin X_{v}$. Now it follows that $(f(N b)-0) \\cdot(f(N b)-b)$ is a divisor of $N$, so from ( $\\dagger$ ) we see that $f(N b)=f(0)=0$. This yields the desired contradiction.", "metadata": {"resource_path": "IMO/segmented/en-IMO2019SL.jsonl"}} -{"year": "2019", "tier": "T0", "problem_label": "C1", "problem_type": "Combinatorics", "exam": "IMO", "problem": "The infinite sequence $a_{0}, a_{1}, a_{2}, \\ldots$ of (not necessarily different) integers has the following properties: $0 \\leqslant a_{i} \\leqslant i$ for all integers $i \\geqslant 0$, and $$ \\binom{k}{a_{0}}+\\binom{k}{a_{1}}+\\cdots+\\binom{k}{a_{k}}=2^{k} $$ for all integers $k \\geqslant 0$. Prove that all integers $N \\geqslant 0$ occur in the sequence (that is, for all $N \\geqslant 0$, there exists $i \\geqslant 0$ with $\\left.a_{i}=N\\right)$. (Netherlands)", "solution": "We prove by induction on $k$ that every initial segment of the sequence, $a_{0}, a_{1}, \\ldots, a_{k}$, consists of the following elements (counted with multiplicity, and not necessarily in order), for some $\\ell \\geqslant 0$ with $2 \\ell \\leqslant k+1$ : $$ 0,1, \\ldots, \\ell-1, \\quad 0,1, \\ldots, k-\\ell $$ For $k=0$ we have $a_{0}=0$, which is of this form. Now suppose that for $k=m$ the elements $a_{0}, a_{1}, \\ldots, a_{m}$ are $0,0,1,1,2,2, \\ldots, \\ell-1, \\ell-1, \\ell, \\ell+1, \\ldots, m-\\ell-1, m-\\ell$ for some $\\ell$ with $0 \\leqslant 2 \\ell \\leqslant m+1$. It is given that $$ \\binom{m+1}{a_{0}}+\\binom{m+1}{a_{1}}+\\cdots+\\binom{m+1}{a_{m}}+\\binom{m+1}{a_{m+1}}=2^{m+1} $$ which becomes $$ \\begin{aligned} \\left(\\binom{m+1}{0}+\\binom{m+1}{1}\\right. & \\left.+\\cdots+\\binom{m+1}{\\ell-1}\\right) \\\\ & +\\left(\\binom{m+1}{0}+\\binom{m+1}{1}+\\cdots+\\binom{m+1}{m-\\ell}\\right)+\\binom{m+1}{a_{m+1}}=2^{m+1} \\end{aligned} $$ or, using $\\binom{m+1}{i}=\\binom{m+1}{m+1-i}$, that $$ \\begin{aligned} \\left(\\binom{m+1}{0}+\\binom{m+1}{1}\\right. & \\left.+\\cdots+\\binom{m+1}{\\ell-1}\\right) \\\\ & +\\left(\\binom{m+1}{m+1}+\\binom{m+1}{m}+\\cdots+\\binom{m+1}{\\ell+1}\\right)+\\binom{m+1}{a_{m+1}}=2^{m+1} \\end{aligned} $$ On the other hand, it is well known that $$ \\binom{m+1}{0}+\\binom{m+1}{1}+\\cdots+\\binom{m+1}{m+1}=2^{m+1} $$ and so, by subtracting, we get $$ \\binom{m+1}{a_{m+1}}=\\binom{m+1}{\\ell} $$ From this, using the fact that the binomial coefficients $\\binom{m+1}{i}$ are increasing for $i \\leqslant \\frac{m+1}{2}$ and decreasing for $i \\geqslant \\frac{m+1}{2}$, we conclude that either $a_{m+1}=\\ell$ or $a_{m+1}=m+1-\\ell$. In either case, $a_{0}, a_{1}, \\ldots, a_{m+1}$ is again of the claimed form, which concludes the induction. As a result of this description, any integer $N \\geqslant 0$ appears as a term of the sequence $a_{i}$ for some $0 \\leqslant i \\leqslant 2 N$.", "metadata": {"resource_path": "IMO/segmented/en-IMO2019SL.jsonl"}} -{"year": "2019", "tier": "T0", "problem_label": "C3", "problem_type": "Combinatorics", "exam": "IMO", "problem": "Let $n$ be a positive integer. Harry has $n$ coins lined up on his desk, each showing heads or tails. He repeatedly does the following operation: if there are $k$ coins showing heads and $k>0$, then he flips the $k^{\\text {th }}$ coin over; otherwise he stops the process. (For example, the process starting with THT would be THT $\\rightarrow H H T \\rightarrow H T T \\rightarrow T T T$, which takes three steps.) Letting $C$ denote the initial configuration (a sequence of $n H$ 's and $T$ 's), write $\\ell(C)$ for the number of steps needed before all coins show $T$. Show that this number $\\ell(C)$ is finite, and determine its average value over all $2^{n}$ possible initial configurations $C$. Answer: The average is $\\frac{1}{4} n(n+1)$. Common remarks. Throughout all these solutions, we let $E(n)$ denote the desired average value.", "solution": "We represent the problem using a directed graph $G_{n}$ whose vertices are the length- $n$ strings of $H$ 's and $T$ 's. The graph features an edge from each string to its successor (except for $T T \\cdots T T$, which has no successor). We will also write $\\bar{H}=T$ and $\\bar{T}=H$. The graph $G_{0}$ consists of a single vertex: the empty string. The main claim is that $G_{n}$ can be described explicitly in terms of $G_{n-1}$ : - We take two copies, $X$ and $Y$, of $G_{n-1}$. - In $X$, we take each string of $n-1$ coins and just append a $T$ to it. In symbols, we replace $s_{1} \\cdots s_{n-1}$ with $s_{1} \\cdots s_{n-1} T$. - In $Y$, we take each string of $n-1$ coins, flip every coin, reverse the order, and append an $H$ to it. In symbols, we replace $s_{1} \\cdots s_{n-1}$ with $\\bar{s}_{n-1} \\bar{s}_{n-2} \\cdots \\bar{s}_{1} H$. - Finally, we add one new edge from $Y$ to $X$, namely $H H \\cdots H H H \\rightarrow H H \\cdots H H T$. We depict $G_{4}$ below, in a way which indicates this recursive construction: ![](https://cdn.mathpix.com/cropped/2024_11_18_9ee8c2ac2c2f5002ff14g-034.jpg?height=515&width=1135&top_left_y=1850&top_left_x=466) We prove the claim inductively. Firstly, $X$ is correct as a subgraph of $G_{n}$, as the operation on coins is unchanged by an extra $T$ at the end: if $s_{1} \\cdots s_{n-1}$ is sent to $t_{1} \\cdots t_{n-1}$, then $s_{1} \\cdots s_{n-1} T$ is sent to $t_{1} \\cdots t_{n-1} T$. Next, $Y$ is also correct as a subgraph of $G_{n}$, as if $s_{1} \\cdots s_{n-1}$ has $k$ occurrences of $H$, then $\\bar{s}_{n-1} \\cdots \\bar{s}_{1} H$ has $(n-1-k)+1=n-k$ occurrences of $H$, and thus (provided that $k>0$ ), if $s_{1} \\cdots s_{n-1}$ is sent to $t_{1} \\cdots t_{n-1}$, then $\\bar{s}_{n-1} \\cdots \\bar{s}_{1} H$ is sent to $\\bar{t}_{n-1} \\cdots \\bar{t}_{1} H$. Finally, the one edge from $Y$ to $X$ is correct, as the operation does send $H H \\cdots H H H$ to HH $\\cdots$ HHT. To finish, note that the sequences in $X$ take an average of $E(n-1)$ steps to terminate, whereas the sequences in $Y$ take an average of $E(n-1)$ steps to reach $H H \\cdots H$ and then an additional $n$ steps to terminate. Therefore, we have $$ E(n)=\\frac{1}{2}(E(n-1)+(E(n-1)+n))=E(n-1)+\\frac{n}{2} $$ We have $E(0)=0$ from our description of $G_{0}$. Thus, by induction, we have $E(n)=\\frac{1}{2}(1+\\cdots+$ $n)=\\frac{1}{4} n(n+1)$, which in particular is finite.", "metadata": {"resource_path": "IMO/segmented/en-IMO2019SL.jsonl"}} -{"year": "2019", "tier": "T0", "problem_label": "C3", "problem_type": "Combinatorics", "exam": "IMO", "problem": "Let $n$ be a positive integer. Harry has $n$ coins lined up on his desk, each showing heads or tails. He repeatedly does the following operation: if there are $k$ coins showing heads and $k>0$, then he flips the $k^{\\text {th }}$ coin over; otherwise he stops the process. (For example, the process starting with THT would be THT $\\rightarrow H H T \\rightarrow H T T \\rightarrow T T T$, which takes three steps.) Letting $C$ denote the initial configuration (a sequence of $n H$ 's and $T$ 's), write $\\ell(C)$ for the number of steps needed before all coins show $T$. Show that this number $\\ell(C)$ is finite, and determine its average value over all $2^{n}$ possible initial configurations $C$. Answer: The average is $\\frac{1}{4} n(n+1)$. Common remarks. Throughout all these solutions, we let $E(n)$ denote the desired average value.", "solution": "We consider what happens with configurations depending on the coins they start and end with. - If a configuration starts with $H$, the last $n-1$ coins follow the given rules, as if they were all the coins, until they are all $T$, then the first coin is turned over. - If a configuration ends with $T$, the last coin will never be turned over, and the first $n-1$ coins follow the given rules, as if they were all the coins. - If a configuration starts with $T$ and ends with $H$, the middle $n-2$ coins follow the given rules, as if they were all the coins, until they are all $T$. After that, there are $2 n-1$ more steps: first coins $1,2, \\ldots, n-1$ are turned over in that order, then coins $n, n-1, \\ldots, 1$ are turned over in that order. As this covers all configurations, and the number of steps is clearly finite for 0 or 1 coins, it follows by induction on $n$ that the number of steps is always finite. We define $E_{A B}(n)$, where $A$ and $B$ are each one of $H, T$ or $*$, to be the average number of steps over configurations of length $n$ restricted to those that start with $A$, if $A$ is not $*$, and that end with $B$, if $B$ is not * (so * represents \"either $H$ or $T$ \"). The above observations tell us that, for $n \\geqslant 2$ : - $E_{H *}(n)=E(n-1)+1$. - $E_{* T}(n)=E(n-1)$. - $E_{H T}(n)=E(n-2)+1$ (by using both the observations for $H *$ and for $\\left.* T\\right)$. - $E_{T H}(n)=E(n-2)+2 n-1$. Now $E_{H *}(n)=\\frac{1}{2}\\left(E_{H H}(n)+E_{H T}(n)\\right)$, so $E_{H H}(n)=2 E(n-1)-E(n-2)+1$. Similarly, $E_{T T}(n)=2 E(n-1)-E(n-2)-1$. So $$ E(n)=\\frac{1}{4}\\left(E_{H T}(n)+E_{H H}(n)+E_{T T}(n)+E_{T H}(n)\\right)=E(n-1)+\\frac{n}{2} $$ We have $E(0)=0$ and $E(1)=\\frac{1}{2}$, so by induction on $n$ we have $E(n)=\\frac{1}{4} n(n+1)$.", "metadata": {"resource_path": "IMO/segmented/en-IMO2019SL.jsonl"}} -{"year": "2019", "tier": "T0", "problem_label": "C3", "problem_type": "Combinatorics", "exam": "IMO", "problem": "Let $n$ be a positive integer. Harry has $n$ coins lined up on his desk, each showing heads or tails. He repeatedly does the following operation: if there are $k$ coins showing heads and $k>0$, then he flips the $k^{\\text {th }}$ coin over; otherwise he stops the process. (For example, the process starting with THT would be THT $\\rightarrow H H T \\rightarrow H T T \\rightarrow T T T$, which takes three steps.) Letting $C$ denote the initial configuration (a sequence of $n H$ 's and $T$ 's), write $\\ell(C)$ for the number of steps needed before all coins show $T$. Show that this number $\\ell(C)$ is finite, and determine its average value over all $2^{n}$ possible initial configurations $C$. Answer: The average is $\\frac{1}{4} n(n+1)$. Common remarks. Throughout all these solutions, we let $E(n)$ denote the desired average value.", "solution": "Let $H_{i}$ be the number of heads in positions 1 to $i$ inclusive (so $H_{n}$ is the total number of heads), and let $I_{i}$ be 1 if the $i^{\\text {th }}$ coin is a head, 0 otherwise. Consider the function $$ t(i)=I_{i}+2\\left(\\min \\left\\{i, H_{n}\\right\\}-H_{i}\\right) $$ We claim that $t(i)$ is the total number of times coin $i$ is turned over (which implies that the process terminates). Certainly $t(i)=0$ when all coins are tails, and $t(i)$ is always a nonnegative integer, so it suffices to show that when the $k^{\\text {th }}$ coin is turned over (where $k=H_{n}$ ), $t(k)$ goes down by 1 and all the other $t(i)$ are unchanged. We show this by splitting into cases: - If $ik, \\min \\left\\{i, H_{n}\\right\\}=H_{n}$ both before and after the coin flip, and both $H_{n}$ and $H_{i}$ change by the same amount, so $t(i)$ is unchanged. - If $i=k$ and the coin is heads, $I_{i}$ goes down by 1 , as do both $\\min \\left\\{i, H_{n}\\right\\}=H_{n}$ and $H_{i}$; so $t(i)$ goes down by 1 . - If $i=k$ and the coin is tails, $I_{i}$ goes up by $1, \\min \\left\\{i, H_{n}\\right\\}=i$ is unchanged and $H_{i}$ goes up by 1 ; so $t(i)$ goes down by 1 . We now need to compute the average value of $$ \\sum_{i=1}^{n} t(i)=\\sum_{i=1}^{n} I_{i}+2 \\sum_{i=1}^{n} \\min \\left\\{i, H_{n}\\right\\}-2 \\sum_{i=1}^{n} H_{i} . $$ The average value of the first term is $\\frac{1}{2} n$, and that of the third term is $-\\frac{1}{2} n(n+1)$. To compute the second term, we sum over choices for the total number of heads, and then over the possible values of $i$, getting $$ 2^{1-n} \\sum_{j=0}^{n}\\binom{n}{j} \\sum_{i=1}^{n} \\min \\{i, j\\}=2^{1-n} \\sum_{j=0}^{n}\\binom{n}{j}\\left(n j-\\binom{j}{2}\\right) . $$ Now, in terms of trinomial coefficients, $$ \\sum_{j=0}^{n} j\\binom{n}{j}=\\sum_{j=1}^{n}\\binom{n}{n-j, j-1,1}=n \\sum_{j=0}^{n-1}\\binom{n-1}{j}=2^{n-1} n $$ and $$ \\sum_{j=0}^{n}\\binom{j}{2}\\binom{n}{j}=\\sum_{j=2}^{n}\\binom{n}{n-j, j-2,2}=\\binom{n}{2} \\sum_{j=0}^{n-2}\\binom{n-2}{j}=2^{n-2}\\binom{n}{2} $$ So the second term above is $$ 2^{1-n}\\left(2^{n-1} n^{2}-2^{n-2}\\binom{n}{2}\\right)=n^{2}-\\frac{n(n-1)}{4} $$ and the required average is $$ E(n)=\\frac{1}{2} n+n^{2}-\\frac{n(n-1)}{4}-\\frac{1}{2} n(n+1)=\\frac{n(n+1)}{4} . $$", "metadata": {"resource_path": "IMO/segmented/en-IMO2019SL.jsonl"}} -{"year": "2019", "tier": "T0", "problem_label": "C3", "problem_type": "Combinatorics", "exam": "IMO", "problem": "Let $n$ be a positive integer. Harry has $n$ coins lined up on his desk, each showing heads or tails. He repeatedly does the following operation: if there are $k$ coins showing heads and $k>0$, then he flips the $k^{\\text {th }}$ coin over; otherwise he stops the process. (For example, the process starting with THT would be THT $\\rightarrow H H T \\rightarrow H T T \\rightarrow T T T$, which takes three steps.) Letting $C$ denote the initial configuration (a sequence of $n H$ 's and $T$ 's), write $\\ell(C)$ for the number of steps needed before all coins show $T$. Show that this number $\\ell(C)$ is finite, and determine its average value over all $2^{n}$ possible initial configurations $C$. Answer: The average is $\\frac{1}{4} n(n+1)$. Common remarks. Throughout all these solutions, we let $E(n)$ denote the desired average value.", "solution": "Harry has built a Turing machine to flip the coins for him. The machine is initially positioned at the $k^{\\text {th }}$ coin, where there are $k$ heads (and the position before the first coin is considered to be the $0^{\\text {th }}$ coin). The machine then moves according to the following rules, stopping when it reaches the position before the first coin: if the coin at its current position is $H$, it flips the coin and moves to the previous coin, while if the coin at its current position is $T$, it flips the coin and moves to the next position. Consider the maximal sequences of consecutive moves in the same direction. Suppose the machine has $a$ consecutive moves to the next coin, before a move to the previous coin. After those $a$ moves, the $a$ coins flipped in those moves are all heads, as is the coin the machine is now at, so at least the next $a+1$ moves will all be moves to the previous coin. Similarly, $a$ consecutive moves to the previous coin are followed by at least $a+1$ consecutive moves to the next coin. There cannot be more than $n$ consecutive moves in the same direction, so this proves that the process terminates (with a move from the first coin to the position before the first coin). Thus we have a (possibly empty) sequence $a_{1}<\\cdotsj$, meaning that, as we follow the moves backwards, each coin is always in the correct state when flipped to result in a move in the required direction. (Alternatively, since there are $2^{n}$ possible configurations of coins and $2^{n}$ possible such ascending sequences, the fact that the sequence of moves determines at most one configuration of coins, and thus that there is an injection from configurations of coins to such ascending sequences, is sufficient for it to be a bijection, without needing to show that coins are in the right state as we move backwards.)", "metadata": {"resource_path": "IMO/segmented/en-IMO2019SL.jsonl"}} -{"year": "2019", "tier": "T0", "problem_label": "C3", "problem_type": "Combinatorics", "exam": "IMO", "problem": "Let $n$ be a positive integer. Harry has $n$ coins lined up on his desk, each showing heads or tails. He repeatedly does the following operation: if there are $k$ coins showing heads and $k>0$, then he flips the $k^{\\text {th }}$ coin over; otherwise he stops the process. (For example, the process starting with THT would be THT $\\rightarrow H H T \\rightarrow H T T \\rightarrow T T T$, which takes three steps.) Letting $C$ denote the initial configuration (a sequence of $n H$ 's and $T$ 's), write $\\ell(C)$ for the number of steps needed before all coins show $T$. Show that this number $\\ell(C)$ is finite, and determine its average value over all $2^{n}$ possible initial configurations $C$. Answer: The average is $\\frac{1}{4} n(n+1)$. Common remarks. Throughout all these solutions, we let $E(n)$ denote the desired average value.", "solution": "We explicitly describe what happens with an arbitrary sequence $C$ of $n$ coins. Suppose that $C$ contain $k$ heads at positions $1 \\leqslant c_{1}1$ be an integer. Suppose we are given $2 n$ points in a plane such that no three of them are collinear. The points are to be labelled $A_{1}, A_{2}, \\ldots, A_{2 n}$ in some order. We then consider the $2 n$ angles $\\angle A_{1} A_{2} A_{3}, \\angle A_{2} A_{3} A_{4}, \\ldots, \\angle A_{2 n-2} A_{2 n-1} A_{2 n}, \\angle A_{2 n-1} A_{2 n} A_{1}$, $\\angle A_{2 n} A_{1} A_{2}$. We measure each angle in the way that gives the smallest positive value (i.e. between $0^{\\circ}$ and $180^{\\circ}$ ). Prove that there exists an ordering of the given points such that the resulting $2 n$ angles can be separated into two groups with the sum of one group of angles equal to the sum of the other group. Comment. The first three solutions all use the same construction involving a line separating the points into groups of $n$ points each, but give different proofs that this construction works. Although Solution 1 is very short, the Problem Selection Committee does not believe any of the solutions is easy to find and thus rates this as a problem of medium difficulty.", "solution": "Let $\\ell$ be a line separating the points into two groups ( $L$ and $R$ ) with $n$ points in each. Label the points $A_{1}, A_{2}, \\ldots, A_{2 n}$ so that $L=\\left\\{A_{1}, A_{3}, \\ldots, A_{2 n-1}\\right\\}$. We claim that this labelling works. Take a line $s=A_{2 n} A_{1}$. (a) Rotate $s$ around $A_{1}$ until it passes through $A_{2}$; the rotation is performed in a direction such that $s$ is never parallel to $\\ell$. (b) Then rotate the new $s$ around $A_{2}$ until it passes through $A_{3}$ in a similar manner. (c) Perform $2 n-2$ more such steps, after which $s$ returns to its initial position. The total (directed) rotation angle $\\Theta$ of $s$ is clearly a multiple of $180^{\\circ}$. On the other hand, $s$ was never parallel to $\\ell$, which is possible only if $\\Theta=0$. Now it remains to partition all the $2 n$ angles into those where $s$ is rotated anticlockwise, and the others.", "metadata": {"resource_path": "IMO/segmented/en-IMO2019SL.jsonl"}} -{"year": "2019", "tier": "T0", "problem_label": "C6", "problem_type": "Combinatorics", "exam": "IMO", "problem": "Let $n>1$ be an integer. Suppose we are given $2 n$ points in a plane such that no three of them are collinear. The points are to be labelled $A_{1}, A_{2}, \\ldots, A_{2 n}$ in some order. We then consider the $2 n$ angles $\\angle A_{1} A_{2} A_{3}, \\angle A_{2} A_{3} A_{4}, \\ldots, \\angle A_{2 n-2} A_{2 n-1} A_{2 n}, \\angle A_{2 n-1} A_{2 n} A_{1}$, $\\angle A_{2 n} A_{1} A_{2}$. We measure each angle in the way that gives the smallest positive value (i.e. between $0^{\\circ}$ and $180^{\\circ}$ ). Prove that there exists an ordering of the given points such that the resulting $2 n$ angles can be separated into two groups with the sum of one group of angles equal to the sum of the other group. Comment. The first three solutions all use the same construction involving a line separating the points into groups of $n$ points each, but give different proofs that this construction works. Although Solution 1 is very short, the Problem Selection Committee does not believe any of the solutions is easy to find and thus rates this as a problem of medium difficulty.", "solution": "When tracing a cyclic path through the $A_{i}$ in order, with straight line segments between consecutive points, let $\\theta_{i}$ be the exterior angle at $A_{i}$, with a sign convention that it is positive if the path turns left and negative if the path turns right. Then $\\sum_{i=1}^{2 n} \\theta_{i}=360 k^{\\circ}$ for some integer $k$. Let $\\phi_{i}=\\angle A_{i-1} A_{i} A_{i+1}($ indices $\\bmod 2 n)$, defined as in the problem; thus $\\phi_{i}=180^{\\circ}-\\left|\\theta_{i}\\right|$. Let $L$ be the set of $i$ for which the path turns left at $A_{i}$ and let $R$ be the set for which it turns right. Then $S=\\sum_{i \\in L} \\phi_{i}-\\sum_{i \\in R} \\phi_{i}=(180(|L|-|R|)-360 k)^{\\circ}$, which is a multiple of $360^{\\circ}$ since the number of points is even. We will show that the points can be labelled such that $S=0$, in which case $L$ and $R$ satisfy the required condition of the problem. Note that the value of $S$ is defined for a slightly larger class of configurations: it is OK for two points to coincide, as long as they are not consecutive, and OK for three points to be collinear, as long as $A_{i}, A_{i+1}$ and $A_{i+2}$ do not appear on a line in that order. In what follows it will be convenient, although not strictly necessary, to consider such configurations. Consider how $S$ changes if a single one of the $A_{i}$ is moved along some straight-line path (not passing through any $A_{j}$ and not lying on any line $A_{j} A_{k}$, but possibly crossing such lines). Because $S$ is a multiple of $360^{\\circ}$, and the angles change continuously, $S$ can only change when a point moves between $R$ and $L$. Furthermore, if $\\phi_{j}=0$ when $A_{j}$ moves between $R$ and $L, S$ is unchanged; it only changes if $\\phi_{j}=180^{\\circ}$ when $A_{j}$ moves between those sets. For any starting choice of points, we will now construct a new configuration, with labels such that $S=0$, that can be perturbed into the original one without any $\\phi_{i}$ passing through $180^{\\circ}$, so that $S=0$ for the original configuration with those labels as well. Take some line such that there are $n$ points on each side of that line. The new configuration has $n$ copies of a single point on each side of the line, and a path that alternates between sides of the line; all angles are 0 , so this configuration has $S=0$. Perturbing the points into their original positions, while keeping each point on its side of the line, no angle $\\phi_{i}$ can pass through $180^{\\circ}$, because no straight line can go from one side of the line to the other and back. So the perturbation process leaves $S=0$. Comment. More complicated variants of this solution are also possible; for example, a path defined using four quadrants of the plane rather than just two half-planes.", "metadata": {"resource_path": "IMO/segmented/en-IMO2019SL.jsonl"}} -{"year": "2019", "tier": "T0", "problem_label": "C6", "problem_type": "Combinatorics", "exam": "IMO", "problem": "Let $n>1$ be an integer. Suppose we are given $2 n$ points in a plane such that no three of them are collinear. The points are to be labelled $A_{1}, A_{2}, \\ldots, A_{2 n}$ in some order. We then consider the $2 n$ angles $\\angle A_{1} A_{2} A_{3}, \\angle A_{2} A_{3} A_{4}, \\ldots, \\angle A_{2 n-2} A_{2 n-1} A_{2 n}, \\angle A_{2 n-1} A_{2 n} A_{1}$, $\\angle A_{2 n} A_{1} A_{2}$. We measure each angle in the way that gives the smallest positive value (i.e. between $0^{\\circ}$ and $180^{\\circ}$ ). Prove that there exists an ordering of the given points such that the resulting $2 n$ angles can be separated into two groups with the sum of one group of angles equal to the sum of the other group. Comment. The first three solutions all use the same construction involving a line separating the points into groups of $n$ points each, but give different proofs that this construction works. Although Solution 1 is very short, the Problem Selection Committee does not believe any of the solutions is easy to find and thus rates this as a problem of medium difficulty.", "solution": "First, let $\\ell$ be a line in the plane such that there are $n$ points on one side and the other $n$ points on the other side. For convenience, assume $\\ell$ is horizontal (otherwise, we can rotate the plane). Then we can use the terms \"above\", \"below\", \"left\" and \"right\" in the usual way. We denote the $n$ points above the line in an arbitrary order as $P_{1}, P_{2}, \\ldots, P_{n}$, and the $n$ points below the line as $Q_{1}, Q_{2}, \\ldots, Q_{n}$. If we connect $P_{i}$ and $Q_{j}$ with a line segment, the line segment will intersect with the line $\\ell$. Denote the intersection as $I_{i j}$. If $P_{i}$ is connected to $Q_{j}$ and $Q_{k}$, where $jk$, then the sign of $\\angle Q_{j} P_{i} Q_{k}$ is taken to be the same as for $\\angle Q_{k} P_{i} Q_{j}$. Similarly, we can define the sign of $\\angle P_{j} Q_{i} P_{k}$ with $j1$ be an integer. Suppose we are given $2 n$ points in a plane such that no three of them are collinear. The points are to be labelled $A_{1}, A_{2}, \\ldots, A_{2 n}$ in some order. We then consider the $2 n$ angles $\\angle A_{1} A_{2} A_{3}, \\angle A_{2} A_{3} A_{4}, \\ldots, \\angle A_{2 n-2} A_{2 n-1} A_{2 n}, \\angle A_{2 n-1} A_{2 n} A_{1}$, $\\angle A_{2 n} A_{1} A_{2}$. We measure each angle in the way that gives the smallest positive value (i.e. between $0^{\\circ}$ and $180^{\\circ}$ ). Prove that there exists an ordering of the given points such that the resulting $2 n$ angles can be separated into two groups with the sum of one group of angles equal to the sum of the other group. Comment. The first three solutions all use the same construction involving a line separating the points into groups of $n$ points each, but give different proofs that this construction works. Although Solution 1 is very short, the Problem Selection Committee does not believe any of the solutions is easy to find and thus rates this as a problem of medium difficulty.", "solution": "We shall think instead of the problem as asking us to assign a weight $\\pm 1$ to each angle, such that the weighted sum of all the angles is zero. Given an ordering $A_{1}, \\ldots, A_{2 n}$ of the points, we shall assign weights according to the following recipe: walk in order from point to point, and assign the left turns +1 and the right turns -1 . This is the same weighting as in Solution 3, and as in that solution, the weighted sum is a multiple of $360^{\\circ}$. We now aim to show the following: Lemma. Transposing any two consecutive points in the ordering changes the weighted sum by $\\pm 360^{\\circ}$ or 0 . Knowing that, we can conclude quickly: if the ordering $A_{1}, \\ldots, A_{2 n}$ has weighted angle sum $360 k^{\\circ}$, then the ordering $A_{2 n}, \\ldots, A_{1}$ has weighted angle sum $-360 k^{\\circ}$ (since the angles are the same, but left turns and right turns are exchanged). We can reverse the ordering of $A_{1}$, $\\ldots, A_{2 n}$ by a sequence of transpositions of consecutive points, and in doing so the weighted angle sum must become zero somewhere along the way. We now prove that lemma: Proof. Transposing two points amounts to taking a section $A_{k} A_{k+1} A_{k+2} A_{k+3}$ as depicted, reversing the central line segment $A_{k+1} A_{k+2}$, and replacing its two neighbours with the dotted lines. ![](https://cdn.mathpix.com/cropped/2024_11_18_9ee8c2ac2c2f5002ff14g-048.jpg?height=424&width=1357&top_left_y=1156&top_left_x=354) Figure 1: Transposing two consecutive vertices: before (left) and afterwards (right) In each triangle, we alter the sum by $\\pm 180^{\\circ}$. Indeed, using (anticlockwise) directed angles modulo $360^{\\circ}$, we either add or subtract all three angles of each triangle. Hence both triangles together alter the sum by $\\pm 180 \\pm 180^{\\circ}$, which is $\\pm 360^{\\circ}$ or 0 .", "metadata": {"resource_path": "IMO/segmented/en-IMO2019SL.jsonl"}} -{"year": "2019", "tier": "T0", "problem_label": "C7", "problem_type": "Combinatorics", "exam": "IMO", "problem": "There are 60 empty boxes $B_{1}, \\ldots, B_{60}$ in a row on a table and an unlimited supply of pebbles. Given a positive integer $n$, Alice and Bob play the following game. In the first round, Alice takes $n$ pebbles and distributes them into the 60 boxes as she wishes. Each subsequent round consists of two steps: (a) Bob chooses an integer $k$ with $1 \\leqslant k \\leqslant 59$ and splits the boxes into the two groups $B_{1}, \\ldots, B_{k}$ and $B_{k+1}, \\ldots, B_{60}$. (b) Alice picks one of these two groups, adds one pebble to each box in that group, and removes one pebble from each box in the other group. Bob wins if, at the end of any round, some box contains no pebbles. Find the smallest $n$ such that Alice can prevent Bob from winning. (Czech Republic) Answer: $n=960$. In general, if there are $N>1$ boxes, the answer is $n=\\left\\lfloor\\frac{N}{2}+1\\right\\rfloor\\left\\lceil\\frac{N}{2}+1\\right\\rceil-1$. Common remarks. We present solutions for the general case of $N>1$ boxes, and write $M=\\left\\lfloor\\frac{N}{2}+1\\right\\rfloor\\left\\lceil\\frac{N}{2}+1\\right\\rceil-1$ for the claimed answer. For $1 \\leqslant kk$. We may then assume, without loss of generality, that Alice again picks the left group. Proof. Suppose Alice picks the right group in the second round. Then the combined effect of the two rounds is that each of the boxes $B_{k+1}, \\ldots, B_{\\ell}$ lost two pebbles (and the other boxes are unchanged). Hence this configuration is strictly dominated by that before the first round, and it suffices to consider only Alice's other response. Solution 1 (Alice). Alice initially distributes pebbles according to $V_{\\left\\lceil\\frac{N}{2}\\right\\rceil}$. Suppose the current configuration of pebbles dominates $V_{i}$. If Bob makes a $k$-move with $k \\geqslant i$ then Alice picks the left group, which results in a configuration that dominates $V_{i+1}$. Likewise, if Bob makes a $k$-move with $k4$, in the style of IMO 2014 Problem 6. (Thailand)", "solution": "We will show Alice needs to ask at most $4 n-7$ questions. Her strategy has the following phases. In what follows, $S$ is the set of towns that Alice, so far, does not know to have more than one outgoing road (so initially $|S|=n$ ). Phase 1. Alice chooses any two towns, say $A$ and $B$. Without loss of generality, suppose that the King of Hearts' answer is that the road goes from $A$ to $B$. At the end of this phase, Alice has asked 1 question. Phase 2. During this phase there is a single (variable) town $T$ that is known to have at least one incoming road but not yet known to have any outgoing roads. Initially, $T$ is $B$. Alice does the following $n-2$ times: she picks a town $X$ she has not asked about before, and asks the direction of the road between $T$ and $X$. If it is from $X$ to $T, T$ is unchanged; if it is from $T$ to $X, X$ becomes the new choice of town $T$, as the previous $T$ is now known to have an outgoing road. At the end of this phase, Alice has asked a total of $n-1$ questions. The final town $T$ is not yet known to have any outgoing roads, while every other town has exactly one outgoing road known. The undirected graph of roads whose directions are known is a tree. Phase 3. During this phase, Alice asks about the directions of all roads between $T$ and another town she has not previously asked about, stopping if she finds two outgoing roads from $T$. This phase involves at most $n-2$ questions. If she does not find two outgoing roads from $T$, she has answered her original question with at most $2 n-3 \\leqslant 4 n-7$ questions, so in what follows we suppose that she does find two outgoing roads, asking a total of $k$ questions in this phase, where $2 \\leqslant k \\leqslant n-2$ (and thus $n \\geqslant 4$ for what follows). For every question where the road goes towards $T$, the town at the other end is removed from $S$ (as it already had one outgoing road known), while the last question resulted in $T$ being removed from $S$. So at the end of this phase, $|S|=n-k+1$, while a total of $n+k-1$ questions have been asked. Furthermore, the undirected graph of roads within $S$ whose directions are known contains no cycles (as $T$ is no longer a member of $S$, all questions asked in this phase involved $T$ and the graph was a tree before this phase started). Every town in $S$ has exactly one outgoing road known (not necessarily to another town in $S$ ). Phase 4. During this phase, Alice repeatedly picks any pair of towns in $S$ for which she does not know the direction of the road between them. Because every town in $S$ has exactly one outgoing road known, this always results in the removal of one of those two towns from $S$. Because there are no cycles in the graph of roads of known direction within $S$, this can continue until there are at most 2 towns left in $S$. If it ends with $t$ towns left, $n-k+1-t$ questions were asked in this phase, so a total of $2 n-t$ questions have been asked. Phase 5. During this phase, Alice asks about all the roads from the remaining towns in $S$ that she has not previously asked about. She has definitely already asked about any road between those towns (if $t=2$ ). She must also have asked in one of the first two phases about at least one other road involving one of those towns (as those phases resulted in a tree with $n>2$ vertices). So she asks at most $t(n-t)-1$ questions in this phase. At the end of this phase, Alice knows whether any town has at most one outgoing road. If $t=1$, at most $3 n-3 \\leqslant 4 n-7$ questions were needed in total, while if $t=2$, at most $4 n-7$ questions were needed in total. Comment 1. The version of this problem originally submitted asked only for an upper bound of $5 n$, which is much simpler to prove. The Problem Selection Committee preferred a version with an asymptotically optimal constant. In the following comment, we will show that the constant is optimal. Comment 2. We will show that Alice cannot always find out by asking at most $4 n-3\\left(\\log _{2} n\\right)-$ 15 questions, if $n \\geqslant 8$. To show this, we suppose the King of Hearts is choosing the directions as he goes along, only picking the direction of a road when Alice asks about it for the first time. We provide a strategy for the King of Hearts that ensures that, after the given number of questions, the map is still consistent both with the existence of a town with at most one outgoing road, and with the nonexistence of such a town. His strategy has the following phases. When describing how the King of Hearts' answer to a question is determined below, we always assume he is being asked about a road for the first time (otherwise, he just repeats his previous answer for that road). This strategy is described throughout in graph-theoretic terms (vertices and edges rather than towns and roads). Phase 1. In this phase, we consider the undirected graph formed by edges whose directions are known. The phase terminates when there are exactly 8 connected components whose undirected graphs are trees. The following invariant is maintained: in a component with $k$ vertices whose undirected graph is a tree, every vertex has at most $\\left[\\log _{2} k\\right\\rfloor$ edges into it. - If the King of Hearts is asked about an edge between two vertices in the same component, or about an edge between two components at least one of which is not a tree, he chooses any direction for that edge arbitrarily. - If he is asked about an edge between a vertex in component $A$ that has $a$ vertices and is a tree and a vertex in component $B$ that has $b$ vertices and is a tree, suppose without loss of generality that $a \\geqslant b$. He then chooses the edge to go from $A$ to $B$. In this case, the new number of edges into any vertex is at most $\\max \\left\\{\\left\\lfloor\\log _{2} a\\right\\rfloor,\\left\\lfloor\\log _{2} b\\right\\rfloor+1\\right\\} \\leqslant\\left\\lfloor\\log _{2}(a+b)\\right\\rfloor$. In all cases, the invariant is preserved, and the number of tree components either remains unchanged or goes down by 1. Assuming Alice does not repeat questions, the process must eventually terminate with 8 tree components, and at least $n-8$ questions having been asked. Note that each tree component contains at least one vertex with no outgoing edges. Colour one such vertex in each tree component red. Phase 2. Let $V_{1}, V_{2}$ and $V_{3}$ be the three of the red vertices whose components are smallest (so their components together have at most $\\left\\lfloor\\frac{3}{8} n\\right\\rfloor$ vertices, with each component having at most $\\left\\lfloor\\frac{3}{8} n-2\\right\\rfloor$ vertices). Let sets $C_{1}, C_{2}, \\ldots$ be the connected components after removing the $V_{j}$. By construction, there are no edges with known direction between $C_{i}$ and $C_{j}$ for $i \\neq j$, and there are at least five such components. If at any point during this phase, the King of Hearts is asked about an edge within one of the $C_{i}$, he chooses an arbitrary direction. If he is asked about an edge between $C_{i}$ and $C_{j}$ for $i \\neq j$, he answers so that all edges go from $C_{i}$ to $C_{i+1}$ and $C_{i+2}$, with indices taken modulo the number of components, and chooses arbitrarily for other pairs. This ensures that all vertices other than the $V_{j}$ will have more than one outgoing edge. For edges involving one of the $V_{j}$ he answers as follows, so as to remain consistent for as long as possible with both possibilities for whether one of those vertices has at most one outgoing edge. Note that as they were red vertices, they have no outgoing edges at the start of this phase. For edges between two of the $V_{j}$, he answers that the edges go from $V_{1}$ to $V_{2}$, from $V_{2}$ to $V_{3}$ and from $V_{3}$ to $V_{1}$. For edges between $V_{j}$ and some other vertex, he always answers that the edge goes into $V_{j}$, except for the last such edge for which he is asked the question for any given $V_{j}$, for which he answers that the edge goes out of $V_{j}$. Thus, as long as at least one of the $V_{j}$ has not had the question answered for all the vertices that are not among the $V_{j}$, his answers are still compatible both with all vertices having more than one outgoing edge, and with that $V_{j}$ having only one outgoing edge. At the start of this phase, each of the $V_{j}$ has at most $\\left\\lfloor\\log _{2}\\left\\lfloor\\frac{3}{8} n-2\\right\\rfloor\\right\\rfloor<\\left(\\log _{2} n\\right)-1$ incoming edges. Thus, Alice cannot determine whether some vertex has only one outgoing edge within $3(n-$ $\\left.3-\\left(\\left(\\log _{2} n\\right)-1\\right)\\right)-1$ questions in this phase; that is, $4 n-3\\left(\\log _{2} n\\right)-15$ questions total. Comment 3. We can also improve the upper bound slightly, to $4 n-2\\left(\\log _{2} n\\right)+1$. (We do not know where the precise minimum number of questions lies between $4 n-3\\left(\\log _{2} n\\right)+O(1)$ and $4 n-2\\left(\\log _{2} n\\right)+$ $O(1)$.) Suppose $n \\geqslant 5$ (otherwise no questions are required at all). To do this, we replace Phases 1 and 2 of the given solution with a different strategy that also results in a spanning tree where one vertex $V$ is not known to have any outgoing edges, and all other vertices have exactly one outgoing edge known, but where there is more control over the numbers of incoming edges. In Phases 3 and 4 we then take more care about the order in which pairs of towns are chosen, to ensure that each of the remaining towns has already had a question asked about at least $\\log _{2} n+O(1)$ edges. Define trees $T_{m}$ with $2^{m}$ vertices, exactly one of which (the root) has no outgoing edges and the rest of which have exactly one outgoing edge, as follows: $T_{0}$ is a single vertex, while $T_{m}$ is constructed by joining the roots of two copies of $T_{m-1}$ with an edge in either direction. If $n=2^{m}$ we can readily ask $n-1$ questions, resulting in a tree $T_{m}$ for the edges with known direction: first ask about $2^{m-1}$ disjoint pairs of vertices, then about $2^{m-2}$ disjoint pairs of the roots of the resulting $T_{1}$ trees, and so on. For the general case, where $n$ is not a power of 2 , after $k$ stages of this process we have $\\left\\lfloor n / 2^{k}\\right\\rfloor$ trees, each of which is like $T_{k}$ but may have some extra vertices (but, however, a unique root). If there are an even number of trees, then ask about pairs of their roots. If there are an odd number (greater than 1) of trees, when a single $T_{k}$ is left over, ask about its root together with that of one of the $T_{k+1}$ trees. Say $m=\\left\\lfloor\\log _{2} n\\right\\rfloor$. The result of that process is a single $T_{m}$ tree, possibly with some extra vertices but still a unique root $V$. That root has at least $m$ incoming edges, and we may list vertices $V_{0}$, $\\ldots, V_{m-1}$ with edges to $V$, such that, for all $0 \\leqslant i1$ we have $D\\left(x_{i}, x_{j}\\right)>d$, since $$ \\left|x_{i}-x_{j}\\right|=\\left|x_{i+1}-x_{i}\\right|+\\left|x_{j}-x_{i+1}\\right| \\geqslant 2^{d}+2^{d}=2^{d+1} . $$ Now choose the minimal $i \\leqslant t$ and the maximal $j \\geqslant t+1$ such that $D\\left(x_{i}, x_{i+1}\\right)=$ $D\\left(x_{i+1}, x_{i+2}\\right)=\\cdots=D\\left(x_{j-1}, x_{j}\\right)=d$. Let $E$ be the set of all the $x_{s}$ with even indices $i \\leqslant s \\leqslant j, O$ be the set of those with odd indices $i \\leqslant s \\leqslant j$, and $R$ be the rest of the elements (so that $\\mathcal{S}$ is the disjoint union of $E, O$ and $R$ ). Set $\\mathcal{S}_{O}=R \\cup O$ and $\\mathcal{S}_{E}=R \\cup E$; we have $\\left|\\mathcal{S}_{O}\\right|<|\\mathcal{S}|$ and $\\left|\\mathcal{S}_{E}\\right|<|\\mathcal{S}|$, so $w\\left(\\mathcal{S}_{O}\\right), w\\left(\\mathcal{S}_{E}\\right) \\leqslant 1$ by the inductive hypothesis. Clearly, $r_{\\mathcal{S}_{O}}(x) \\leqslant r_{\\mathcal{S}}(x)$ and $r_{\\mathcal{S}_{E}}(x) \\leqslant r_{\\mathcal{S}}(x)$ for any $x \\in R$, and thus $$ \\begin{aligned} \\sum_{x \\in R} 2^{-r_{\\mathcal{S}}(x)} & =\\frac{1}{2} \\sum_{x \\in R}\\left(2^{-r_{\\mathcal{S}}(x)}+2^{-r_{\\mathcal{S}}(x)}\\right) \\\\ & \\leqslant \\frac{1}{2} \\sum_{x \\in R}\\left(2^{-r_{\\mathcal{S}_{O}}(x)}+2^{-r_{\\mathcal{S}_{E}}(x)}\\right) \\end{aligned} $$ On the other hand, for every $x \\in O$, there is no $y \\in \\mathcal{S}_{O}$ such that $D_{\\mathcal{S}_{O}}(x, y)=d$ (as all candidates from $\\mathcal{S}$ were in $E$ ). Hence, we have $r_{\\mathcal{S}_{O}}(x) \\leqslant r_{\\mathcal{S}}(x)-1$, and thus $$ \\sum_{x \\in O} 2^{-r \\mathcal{S}_{\\mathcal{S}}(x)} \\leqslant \\frac{1}{2} \\sum_{x \\in O} 2^{-r s_{O}(x)} $$ Similarly, for every $x \\in E$, we have $$ \\sum_{x \\in E} 2^{-r_{\\mathcal{S}}(x)} \\leqslant \\frac{1}{2} \\sum_{x \\in E} 2^{-r_{\\mathcal{S}_{E}}(x)} $$ We can then combine these to give $$ \\begin{aligned} w(S) & =\\sum_{x \\in R} 2^{-r_{\\mathcal{S}}(x)}+\\sum_{x \\in O} 2^{-r_{\\mathcal{S}}(x)}+\\sum_{x \\in E} 2^{-r_{\\mathcal{S}}(x)} \\\\ & \\leqslant \\frac{1}{2} \\sum_{x \\in R}\\left(2^{-r_{\\mathcal{S}_{O}}(x)}+2^{-r_{\\mathcal{S}_{E}}(x)}\\right)+\\frac{1}{2} \\sum_{x \\in O} 2^{-r_{\\mathcal{S}_{O}}(x)}+\\frac{1}{2} \\sum_{x \\in E} 2^{-r_{\\mathcal{S}_{E}}(x)} \\\\ & \\left.=\\frac{1}{2}\\left(\\sum_{x \\in \\mathcal{S}_{O}} 2^{-r_{S_{O}}(x)}+\\sum_{x \\in \\mathcal{S}_{E}} 2^{-r s_{S_{E}}(x)}\\right) \\quad \\text { (since } \\mathcal{S}_{O}=O \\cup R \\text { and } \\mathcal{S}_{E}=E \\cup R\\right) \\\\ & \\left.\\left.=\\frac{1}{2}\\left(w\\left(\\mathcal{S}_{O}\\right)+w\\left(\\mathcal{S}_{E}\\right)\\right)\\right) \\quad \\text { (by definition of } w(\\cdot)\\right) \\\\ & \\leqslant 1 \\quad \\text { (by the inductive hypothesis) } \\end{aligned} $$ which completes the induction. Comment 1. The sets $O$ and $E$ above are not the only ones we could have chosen. Indeed, we could instead have used the following definitions: Let $d$ be the maximal scale between two distinct elements of $\\mathcal{S}$; that is, $d=D\\left(x_{1}, x_{n}\\right)$. Let $O=\\left\\{x \\in \\mathcal{S}: D\\left(x, x_{n}\\right)=d\\right\\}$ (a 'left' part of the set) and let $E=\\left\\{x \\in \\mathcal{S}: D\\left(x_{1}, x\\right)=d\\right\\}$ (a 'right' part of the set). Note that these two sets are disjoint, and nonempty (since they contain $x_{1}$ and $x_{n}$ respectively). The rest of the proof is then the same as in Solution 1. Comment 2. Another possible set $\\mathcal{F}$ containing $2^{k}$ members could arise from considering a binary tree of height $k$, allocating a real number to each leaf, and trying to make the scale between the values of two leaves dependent only on the (graph) distance between them. The following construction makes this more precise. We build up sets $\\mathcal{F}_{k}$ recursively. Let $\\mathcal{F}_{0}=\\{0\\}$, and then let $\\mathcal{F}_{k+1}=\\mathcal{F}_{k} \\cup\\left\\{x+3 \\cdot 4^{k}: x \\in \\mathcal{F}_{k}\\right\\}$ (i.e. each half of $\\mathcal{F}_{k+1}$ is a copy of $F_{k}$ ). We have that $\\mathcal{F}_{k}$ is contained in the interval $\\left[0,4^{k+1}\\right.$ ), and so it follows by induction on $k$ that every member of $F_{k+1}$ has $k$ different scales in its own half of $F_{k+1}$ (by the inductive hypothesis), and only the single scale $2 k+1$ in the other half of $F_{k+1}$. Both of the constructions presented here have the property that every member of $\\mathcal{F}$ has exactly $k$ different scales in $\\mathcal{F}$. Indeed, it can be seen that this must hold (up to a slight perturbation) for any such maximal set. Suppose there were some element $x$ with only $k-1$ different scales in $\\mathcal{F}$ (and every other element had at most $k$ different scales). Then we take some positive real $\\epsilon$, and construct a new set $\\mathcal{F}^{\\prime}=\\{y: y \\in \\mathcal{F}, y \\leqslant x\\} \\cup\\{y+\\epsilon: y \\in \\mathcal{F}, y \\geqslant x\\}$. We have $\\left|\\mathcal{F}^{\\prime}\\right|=|\\mathcal{F}|+1$, and if $\\epsilon$ is sufficiently small then $\\mathcal{F}^{\\prime}$ will also satisfy the property that no member has more than $k$ different scales in $\\mathcal{F}^{\\prime}$. This observation might be used to motivate the idea of weighting members of an arbitrary set $\\mathcal{S}$ of reals according to how many different scales they have in $\\mathcal{S}$.", "metadata": {"resource_path": "IMO/segmented/en-IMO2019SL.jsonl"}} -{"year": "2019", "tier": "T0", "problem_label": "G1", "problem_type": "Geometry", "exam": "IMO", "problem": "Let $A B C$ be a triangle. Circle $\\Gamma$ passes through $A$, meets segments $A B$ and $A C$ again at points $D$ and $E$ respectively, and intersects segment $B C$ at $F$ and $G$ such that $F$ lies between $B$ and $G$. The tangent to circle $B D F$ at $F$ and the tangent to circle $C E G$ at $G$ meet at point $T$. Suppose that points $A$ and $T$ are distinct. Prove that line $A T$ is parallel to $B C$. (Nigeria)", "solution": "Notice that $\\angle T F B=\\angle F D A$ because $F T$ is tangent to circle $B D F$, and moreover $\\angle F D A=\\angle C G A$ because quadrilateral $A D F G$ is cyclic. Similarly, $\\angle T G B=\\angle G E C$ because $G T$ is tangent to circle $C E G$, and $\\angle G E C=\\angle C F A$. Hence, $$ \\angle T F B=\\angle C G A \\text { and } \\quad \\angle T G B=\\angle C F A $$ ![](https://cdn.mathpix.com/cropped/2024_11_18_9ee8c2ac2c2f5002ff14g-058.jpg?height=624&width=1326&top_left_y=930&top_left_x=365) Triangles $F G A$ and $G F T$ have a common side $F G$, and by (1) their angles at $F, G$ are the same. So, these triangles are congruent. So, their altitudes starting from $A$ and $T$, respectively, are equal and hence $A T$ is parallel to line $B F G C$. Comment. Alternatively, we can prove first that $T$ lies on $\\Gamma$. For example, this can be done by showing that $\\angle A F T=\\angle A G T$ using (1). Then the statement follows as $\\angle T A F=\\angle T G F=\\angle G F A$.", "metadata": {"resource_path": "IMO/segmented/en-IMO2019SL.jsonl"}} -{"year": "2019", "tier": "T0", "problem_label": "G2", "problem_type": "Geometry", "exam": "IMO", "problem": "Let $A B C$ be an acute-angled triangle and let $D, E$, and $F$ be the feet of altitudes from $A, B$, and $C$ to sides $B C, C A$, and $A B$, respectively. Denote by $\\omega_{B}$ and $\\omega_{C}$ the incircles of triangles $B D F$ and $C D E$, and let these circles be tangent to segments $D F$ and $D E$ at $M$ and $N$, respectively. Let line $M N$ meet circles $\\omega_{B}$ and $\\omega_{C}$ again at $P \\neq M$ and $Q \\neq N$, respectively. Prove that $M P=N Q$. (Vietnam)", "solution": "Denote the centres of $\\omega_{B}$ and $\\omega_{C}$ by $O_{B}$ and $O_{C}$, let their radii be $r_{B}$ and $r_{C}$, and let $B C$ be tangent to the two circles at $T$ and $U$, respectively. ![](https://cdn.mathpix.com/cropped/2024_11_18_9ee8c2ac2c2f5002ff14g-059.jpg?height=801&width=1014&top_left_y=659&top_left_x=521) From the cyclic quadrilaterals $A F D C$ and $A B D E$ we have $$ \\angle M D O_{B}=\\frac{1}{2} \\angle F D B=\\frac{1}{2} \\angle B A C=\\frac{1}{2} \\angle C D E=\\angle O_{C} D N, $$ so the right-angled triangles $D M O_{B}$ and $D N O_{C}$ are similar. The ratio of similarity between the two triangles is $$ \\frac{D N}{D M}=\\frac{O_{C} N}{O_{B} M}=\\frac{r_{C}}{r_{B}} . $$ Let $\\varphi=\\angle D M N$ and $\\psi=\\angle M N D$. The lines $F M$ and $E N$ are tangent to $\\omega_{B}$ and $\\omega_{C}$, respectively, so $$ \\angle M T P=\\angle F M P=\\angle D M N=\\varphi \\quad \\text { and } \\quad \\angle Q U N=\\angle Q N E=\\angle M N D=\\psi $$ (It is possible that $P$ or $Q$ coincides with $T$ or $U$, or lie inside triangles $D M T$ or $D U N$, respectively. To reduce case-sensitivity, we may use directed angles or simply ignore angles $M T P$ and $Q U N$.) In the circles $\\omega_{B}$ and $\\omega_{C}$ the lengths of chords $M P$ and $N Q$ are $$ M P=2 r_{B} \\cdot \\sin \\angle M T P=2 r_{B} \\cdot \\sin \\varphi \\quad \\text { and } \\quad N Q=2 r_{C} \\cdot \\sin \\angle Q U N=2 r_{C} \\cdot \\sin \\psi $$ By applying the sine rule to triangle $D N M$ we get $$ \\frac{D N}{D M}=\\frac{\\sin \\angle D M N}{\\sin \\angle M N D}=\\frac{\\sin \\varphi}{\\sin \\psi} $$ Finally, putting the above observations together, we get $$ \\frac{M P}{N Q}=\\frac{2 r_{B} \\sin \\varphi}{2 r_{C} \\sin \\psi}=\\frac{r_{B}}{r_{C}} \\cdot \\frac{\\sin \\varphi}{\\sin \\psi}=\\frac{D M}{D N} \\cdot \\frac{\\sin \\varphi}{\\sin \\psi}=\\frac{\\sin \\psi}{\\sin \\varphi} \\cdot \\frac{\\sin \\varphi}{\\sin \\psi}=1, $$ so $M P=N Q$ as required.", "metadata": {"resource_path": "IMO/segmented/en-IMO2019SL.jsonl"}} -{"year": "2019", "tier": "T0", "problem_label": "G4", "problem_type": "Geometry", "exam": "IMO", "problem": "Let $P$ be a point inside triangle $A B C$. Let $A P$ meet $B C$ at $A_{1}$, let $B P$ meet $C A$ at $B_{1}$, and let $C P$ meet $A B$ at $C_{1}$. Let $A_{2}$ be the point such that $A_{1}$ is the midpoint of $P A_{2}$, let $B_{2}$ be the point such that $B_{1}$ is the midpoint of $P B_{2}$, and let $C_{2}$ be the point such that $C_{1}$ is the midpoint of $P C_{2}$. Prove that points $A_{2}, B_{2}$, and $C_{2}$ cannot all lie strictly inside the circumcircle of triangle $A B C$. (Australia) ![](https://cdn.mathpix.com/cropped/2024_11_18_9ee8c2ac2c2f5002ff14g-062.jpg?height=763&width=715&top_left_y=578&top_left_x=662)", "solution": "Since $$ \\angle A P B+\\angle B P C+\\angle C P A=2 \\pi=(\\pi-\\angle A C B)+(\\pi-\\angle B A C)+(\\pi-\\angle C B A), $$ at least one of the following inequalities holds: $$ \\angle A P B \\geqslant \\pi-\\angle A C B, \\quad \\angle B P C \\geqslant \\pi-\\angle B A C, \\quad \\angle C P A \\geqslant \\pi-\\angle C B A . $$ Without loss of generality, we assume that $\\angle B P C \\geqslant \\pi-\\angle B A C$. We have $\\angle B P C>\\angle B A C$ because $P$ is inside $\\triangle A B C$. So $\\angle B P C \\geqslant \\max (\\angle B A C, \\pi-\\angle B A C)$ and hence $$ \\sin \\angle B P C \\leqslant \\sin \\angle B A C . $$ Let the rays $A P, B P$, and $C P$ cross the circumcircle $\\Omega$ again at $A_{3}, B_{3}$, and $C_{3}$, respectively. We will prove that at least one of the ratios $\\frac{P B_{1}}{B_{1} B_{3}}$ and $\\frac{P C_{1}}{C_{1} C_{3}}$ is at least 1 , which yields that one of the points $B_{2}$ and $C_{2}$ does not lie strictly inside $\\Omega$. Because $A, B, C, B_{3}$ lie on a circle, the triangles $C B_{1} B_{3}$ and $B B_{1} A$ are similar, so $$ \\frac{C B_{1}}{B_{1} B_{3}}=\\frac{B B_{1}}{B_{1} A} $$ Applying the sine rule we obtain $$ \\frac{P B_{1}}{B_{1} B_{3}}=\\frac{P B_{1}}{C B_{1}} \\cdot \\frac{C B_{1}}{B_{1} B_{3}}=\\frac{P B_{1}}{C B_{1}} \\cdot \\frac{B B_{1}}{B_{1} A}=\\frac{\\sin \\angle A C P}{\\sin \\angle B P C} \\cdot \\frac{\\sin \\angle B A C}{\\sin \\angle P B A} . $$ Similarly, $$ \\frac{P C_{1}}{C_{1} C_{3}}=\\frac{\\sin \\angle P B A}{\\sin \\angle B P C} \\cdot \\frac{\\sin \\angle B A C}{\\sin \\angle A C P} $$ Multiplying these two equations we get $$ \\frac{P B_{1}}{B_{1} B_{3}} \\cdot \\frac{P C_{1}}{C_{1} C_{3}}=\\frac{\\sin ^{2} \\angle B A C}{\\sin ^{2} \\angle B P C} \\geqslant 1 $$ using (*), which yields the desired conclusion. Comment. It also cannot happen that all three points $A_{2}, B_{2}$, and $C_{2}$ lie strictly outside $\\Omega$. The same proof works almost literally, starting by assuming without loss of generality that $\\angle B P C \\leqslant \\pi-\\angle B A C$ and using $\\angle B P C>\\angle B A C$ to deduce that $\\sin \\angle B P C \\geqslant \\sin \\angle B A C$. It is possible for $A_{2}, B_{2}$, and $C_{2}$ all to lie on the circumcircle; from the above solution we may derive that this happens if and only if $P$ is the orthocentre of the triangle $A B C$, (which lies strictly inside $A B C$ if and only if $A B C$ is acute).", "metadata": {"resource_path": "IMO/segmented/en-IMO2019SL.jsonl"}} -{"year": "2019", "tier": "T0", "problem_label": "G4", "problem_type": "Geometry", "exam": "IMO", "problem": "Let $P$ be a point inside triangle $A B C$. Let $A P$ meet $B C$ at $A_{1}$, let $B P$ meet $C A$ at $B_{1}$, and let $C P$ meet $A B$ at $C_{1}$. Let $A_{2}$ be the point such that $A_{1}$ is the midpoint of $P A_{2}$, let $B_{2}$ be the point such that $B_{1}$ is the midpoint of $P B_{2}$, and let $C_{2}$ be the point such that $C_{1}$ is the midpoint of $P C_{2}$. Prove that points $A_{2}, B_{2}$, and $C_{2}$ cannot all lie strictly inside the circumcircle of triangle $A B C$. (Australia) ![](https://cdn.mathpix.com/cropped/2024_11_18_9ee8c2ac2c2f5002ff14g-062.jpg?height=763&width=715&top_left_y=578&top_left_x=662)", "solution": "Define points $A_{3}, B_{3}$, and $C_{3}$ as in Observe that $\\triangle P B C_{3} \\sim \\triangle P C B_{3}$. Let $X$ be the point on side $P B_{3}$ that corresponds to point $C_{1}$ on side $P C_{3}$ under this similarity. In other words, $X$ lies on segment $P B_{3}$ and satisfies $P X: X B_{3}=P C_{1}: C_{1} C_{3}$. It follows that $$ \\angle X C P=\\angle P B C_{1}=\\angle B_{3} B A=\\angle B_{3} C B_{1} . $$ Hence lines $C X$ and $C B_{1}$ are isogonal conjugates in $\\triangle P C B_{3}$. ![](https://cdn.mathpix.com/cropped/2024_11_18_9ee8c2ac2c2f5002ff14g-063.jpg?height=812&width=807&top_left_y=1207&top_left_x=533) Let $Y$ be the foot of the bisector of $\\angle B_{3} C P$ in $\\triangle P C B_{3}$. Since $P C_{1}0$ and $\\alpha+\\beta+\\gamma=1$. Then $$ A_{1}=\\frac{\\beta B+\\gamma C}{\\beta+\\gamma}=\\frac{1}{1-\\alpha} P-\\frac{\\alpha}{1-\\alpha} A, $$ so $$ A_{2}=2 A_{1}-P=\\frac{1+\\alpha}{1-\\alpha} P-\\frac{2 \\alpha}{1-\\alpha} A $$ Hence $$ \\left|A_{2}\\right|^{2}=\\left(\\frac{1+\\alpha}{1-\\alpha}\\right)^{2}|P|^{2}+\\left(\\frac{2 \\alpha}{1-\\alpha}\\right)^{2}|A|^{2}-\\frac{4 \\alpha(1+\\alpha)}{(1-\\alpha)^{2}} A \\cdot P $$ Using $|A|^{2}=1$ we obtain $$ \\frac{(1-\\alpha)^{2}}{2(1+\\alpha)}\\left|A_{2}\\right|^{2}=\\frac{1+\\alpha}{2}|P|^{2}+\\frac{2 \\alpha^{2}}{1+\\alpha}-2 \\alpha A \\cdot P $$ Likewise $$ \\frac{(1-\\beta)^{2}}{2(1+\\beta)}\\left|B_{2}\\right|^{2}=\\frac{1+\\beta}{2}|P|^{2}+\\frac{2 \\beta^{2}}{1+\\beta}-2 \\beta B \\cdot P $$ and $$ \\frac{(1-\\gamma)^{2}}{2(1+\\gamma)}\\left|C_{2}\\right|^{2}=\\frac{1+\\gamma}{2}|P|^{2}+\\frac{2 \\gamma^{2}}{1+\\gamma}-2 \\gamma C \\cdot P $$ Summing (1), (2) and (3) we obtain on the LHS the positive linear combination $$ \\text { LHS }=\\frac{(1-\\alpha)^{2}}{2(1+\\alpha)}\\left|A_{2}\\right|^{2}+\\frac{(1-\\beta)^{2}}{2(1+\\beta)}\\left|B_{2}\\right|^{2}+\\frac{(1-\\gamma)^{2}}{2(1+\\gamma)}\\left|C_{2}\\right|^{2} $$ and on the RHS the quantity $$ \\left(\\frac{1+\\alpha}{2}+\\frac{1+\\beta}{2}+\\frac{1+\\gamma}{2}\\right)|P|^{2}+\\left(\\frac{2 \\alpha^{2}}{1+\\alpha}+\\frac{2 \\beta^{2}}{1+\\beta}+\\frac{2 \\gamma^{2}}{1+\\gamma}\\right)-2(\\alpha A \\cdot P+\\beta B \\cdot P+\\gamma C \\cdot P) . $$ The first term is $2|P|^{2}$ and the last term is $-2 P \\cdot P$, so $$ \\begin{aligned} \\text { RHS } & =\\left(\\frac{2 \\alpha^{2}}{1+\\alpha}+\\frac{2 \\beta^{2}}{1+\\beta}+\\frac{2 \\gamma^{2}}{1+\\gamma}\\right) \\\\ & =\\frac{3 \\alpha-1}{2}+\\frac{(1-\\alpha)^{2}}{2(1+\\alpha)}+\\frac{3 \\beta-1}{2}+\\frac{(1-\\beta)^{2}}{2(1+\\beta)}+\\frac{3 \\gamma-1}{2}+\\frac{(1-\\gamma)^{2}}{2(1+\\gamma)} \\\\ & =\\frac{(1-\\alpha)^{2}}{2(1+\\alpha)}+\\frac{(1-\\beta)^{2}}{2(1+\\beta)}+\\frac{(1-\\gamma)^{2}}{2(1+\\gamma)} \\end{aligned} $$ Here we used the fact that $$ \\frac{3 \\alpha-1}{2}+\\frac{3 \\beta-1}{2}+\\frac{3 \\gamma-1}{2}=0 . $$ We have shown that a linear combination of $\\left|A_{1}\\right|^{2},\\left|B_{1}\\right|^{2}$, and $\\left|C_{1}\\right|^{2}$ with positive coefficients is equal to the sum of the coefficients. Therefore at least one of $\\left|A_{1}\\right|^{2},\\left|B_{1}\\right|^{2}$, and $\\left|C_{1}\\right|^{2}$ must be at least 1 , as required. Comment. This proof also works when $P$ is any point for which $\\alpha, \\beta, \\gamma>-1, \\alpha+\\beta+\\gamma=1$, and $\\alpha, \\beta, \\gamma \\neq 1$. (In any cases where $\\alpha=1$ or $\\beta=1$ or $\\gamma=1$, some points in the construction are not defined.) This page is intentionally left blank", "metadata": {"resource_path": "IMO/segmented/en-IMO2019SL.jsonl"}} -{"year": "2019", "tier": "T0", "problem_label": "G5", "problem_type": "Geometry", "exam": "IMO", "problem": "Let $A B C D E$ be a convex pentagon with $C D=D E$ and $\\angle E D C \\neq 2 \\cdot \\angle A D B$. Suppose that a point $P$ is located in the interior of the pentagon such that $A P=A E$ and $B P=B C$. Prove that $P$ lies on the diagonal $C E$ if and only if area $(B C D)+\\operatorname{area}(A D E)=$ $\\operatorname{area}(A B D)+\\operatorname{area}(A B P)$. (Hungary)", "solution": "Let $P^{\\prime}$ be the reflection of $P$ across line $A B$, and let $M$ and $N$ be the midpoints of $P^{\\prime} E$ and $P^{\\prime} C$ respectively. Convexity ensures that $P^{\\prime}$ is distinct from both $E$ and $C$, and hence from both $M$ and $N$. We claim that both the area condition and the collinearity condition in the problem are equivalent to the condition that the (possibly degenerate) right-angled triangles $A P^{\\prime} M$ and $B P^{\\prime} N$ are directly similar (equivalently, $A P^{\\prime} E$ and $B P^{\\prime} C$ are directly similar). ![](https://cdn.mathpix.com/cropped/2024_11_18_9ee8c2ac2c2f5002ff14g-066.jpg?height=535&width=549&top_left_y=806&top_left_x=756) For the equivalence with the collinearity condition, let $F$ denote the foot of the perpendicular from $P^{\\prime}$ to $A B$, so that $F$ is the midpoint of $P P^{\\prime}$. We have that $P$ lies on $C E$ if and only if $F$ lies on $M N$, which occurs if and only if we have the equality $\\angle A F M=\\angle B F N$ of signed angles modulo $\\pi$. By concyclicity of $A P^{\\prime} F M$ and $B F P^{\\prime} N$, this is equivalent to $\\angle A P^{\\prime} M=\\angle B P^{\\prime} N$, which occurs if and only if $A P^{\\prime} M$ and $B P^{\\prime} N$ are directly similar. ![](https://cdn.mathpix.com/cropped/2024_11_18_9ee8c2ac2c2f5002ff14g-066.jpg?height=306&width=441&top_left_y=1663&top_left_x=813) For the other equivalence with the area condition, we have the equality of signed areas $\\operatorname{area}(A B D)+\\operatorname{area}(A B P)=\\operatorname{area}\\left(A P^{\\prime} B D\\right)=\\operatorname{area}\\left(A P^{\\prime} D\\right)+\\operatorname{area}\\left(B D P^{\\prime}\\right)$. Using the identity $\\operatorname{area}(A D E)-\\operatorname{area}\\left(A P^{\\prime} D\\right)=\\operatorname{area}(A D E)+\\operatorname{area}\\left(A D P^{\\prime}\\right)=2$ area $(A D M)$, and similarly for $B$, we find that the area condition is equivalent to the equality $$ \\operatorname{area}(D A M)=\\operatorname{area}(D B N) . $$ Now note that $A$ and $B$ lie on the perpendicular bisectors of $P^{\\prime} E$ and $P^{\\prime} C$, respectively. If we write $G$ and $H$ for the feet of the perpendiculars from $D$ to these perpendicular bisectors respectively, then this area condition can be rewritten as $$ M A \\cdot G D=N B \\cdot H D $$ (In this condition, we interpret all lengths as signed lengths according to suitable conventions: for instance, we orient $P^{\\prime} E$ from $P^{\\prime}$ to $E$, orient the parallel line $D H$ in the same direction, and orient the perpendicular bisector of $P^{\\prime} E$ at an angle $\\pi / 2$ clockwise from the oriented segment $P^{\\prime} E$ - we adopt the analogous conventions at $B$.) ![](https://cdn.mathpix.com/cropped/2024_11_18_9ee8c2ac2c2f5002ff14g-067.jpg?height=529&width=575&top_left_y=204&top_left_x=758) To relate the signed lengths $G D$ and $H D$ to the triangles $A P^{\\prime} M$ and $B P^{\\prime} N$, we use the following calculation. Claim. Let $\\Gamma$ denote the circle centred on $D$ with both $E$ and $C$ on the circumference, and $h$ the power of $P^{\\prime}$ with respect to $\\Gamma$. Then we have the equality $$ G D \\cdot P^{\\prime} M=H D \\cdot P^{\\prime} N=\\frac{1}{4} h \\neq 0 . $$ Proof. Firstly, we have $h \\neq 0$, since otherwise $P^{\\prime}$ would lie on $\\Gamma$, and hence the internal angle bisectors of $\\angle E D P^{\\prime}$ and $\\angle P^{\\prime} D C$ would pass through $A$ and $B$ respectively. This would violate the angle inequality $\\angle E D C \\neq 2 \\cdot \\angle A D B$ given in the question. Next, let $E^{\\prime}$ denote the second point of intersection of $P^{\\prime} E$ with $\\Gamma$, and let $E^{\\prime \\prime}$ denote the point on $\\Gamma$ diametrically opposite $E^{\\prime}$, so that $E^{\\prime \\prime} E$ is perpendicular to $P^{\\prime} E$. The point $G$ lies on the perpendicular bisectors of the sides $P^{\\prime} E$ and $E E^{\\prime \\prime}$ of the right-angled triangle $P^{\\prime} E E^{\\prime \\prime}$; it follows that $G$ is the midpoint of $P^{\\prime} E^{\\prime \\prime}$. Since $D$ is the midpoint of $E^{\\prime} E^{\\prime \\prime}$, we have that $G D=\\frac{1}{2} P^{\\prime} E^{\\prime}$. Since $P^{\\prime} M=\\frac{1}{2} P^{\\prime} E$, we have $G D \\cdot P^{\\prime} M=\\frac{1}{4} P^{\\prime} E^{\\prime} \\cdot P^{\\prime} E=\\frac{1}{4} h$. The other equality $H D \\cdot P^{\\prime} N$ follows by exactly the same argument. ![](https://cdn.mathpix.com/cropped/2024_11_18_9ee8c2ac2c2f5002ff14g-067.jpg?height=727&width=672&top_left_y=1664&top_left_x=726) From this claim, we see that the area condition is equivalent to the equality $$ \\left(M A: P^{\\prime} M\\right)=\\left(N B: P^{\\prime} N\\right) $$ of ratios of signed lengths, which is equivalent to direct similarity of $A P^{\\prime} M$ and $B P^{\\prime} N$, as desired.", "metadata": {"resource_path": "IMO/segmented/en-IMO2019SL.jsonl"}} -{"year": "2019", "tier": "T0", "problem_label": "G5", "problem_type": "Geometry", "exam": "IMO", "problem": "Let $A B C D E$ be a convex pentagon with $C D=D E$ and $\\angle E D C \\neq 2 \\cdot \\angle A D B$. Suppose that a point $P$ is located in the interior of the pentagon such that $A P=A E$ and $B P=B C$. Prove that $P$ lies on the diagonal $C E$ if and only if area $(B C D)+\\operatorname{area}(A D E)=$ $\\operatorname{area}(A B D)+\\operatorname{area}(A B P)$. (Hungary)", "solution": "Along the perpendicular bisector of $C E$, define the linear function $$ f(X)=\\operatorname{area}(B C X)+\\operatorname{area}(A X E)-\\operatorname{area}(A B X)-\\operatorname{area}(A B P), $$ where, from now on, we always use signed areas. Thus, we want to show that $C, P, E$ are collinear if and only if $f(D)=0$. ![](https://cdn.mathpix.com/cropped/2024_11_18_9ee8c2ac2c2f5002ff14g-068.jpg?height=433&width=512&top_left_y=503&top_left_x=772) Let $P^{\\prime}$ be the reflection of $P$ across line $A B$. The point $P^{\\prime}$ does not lie on the line $C E$. To see this, we let $A^{\\prime \\prime}$ and $B^{\\prime \\prime}$ be the points obtained from $A$ and $B$ by dilating with scale factor 2 about $P^{\\prime}$, so that $P$ is the orthogonal projection of $P^{\\prime}$ onto $A^{\\prime \\prime} B^{\\prime \\prime}$. Since $A$ lies on the perpendicular bisector of $P^{\\prime} E$, the triangle $A^{\\prime \\prime} E P^{\\prime}$ is right-angled at $E$ (and $B^{\\prime \\prime} C P^{\\prime}$ similarly). If $P^{\\prime}$ were to lie on $C E$, then the lines $A^{\\prime \\prime} E$ and $B^{\\prime \\prime} C$ would be perpendicular to $C E$ and $A^{\\prime \\prime}$ and $B^{\\prime \\prime}$ would lie on the opposite side of $C E$ to $D$. It follows that the line $A^{\\prime \\prime} B^{\\prime \\prime}$ does not meet triangle $C D E$, and hence point $P$ does not lie inside $C D E$. But then $P$ must lie inside $A B C E$, and it is clear that such a point cannot reflect to a point $P^{\\prime}$ on $C E$. We thus let $O$ be the centre of the circle $C E P^{\\prime}$. The lines $A O$ and $B O$ are the perpendicular bisectors of $E P^{\\prime}$ and $C P^{\\prime}$, respectively, so $$ \\begin{aligned} \\operatorname{area}(B C O)+\\operatorname{area}(A O E) & =\\operatorname{area}\\left(O P^{\\prime} B\\right)+\\operatorname{area}\\left(P^{\\prime} O A\\right)=\\operatorname{area}\\left(P^{\\prime} B O A\\right) \\\\ & =\\operatorname{area}(A B O)+\\operatorname{area}\\left(B A P^{\\prime}\\right)=\\operatorname{area}(A B O)+\\operatorname{area}(A B P), \\end{aligned} $$ and hence $f(O)=0$. Notice that if point $O$ coincides with $D$ then points $A, B$ lie in angle domain $C D E$ and $\\angle E O C=2 \\cdot \\angle A O B$, which is not allowed. So, $O$ and $D$ must be distinct. Since $f$ is linear and vanishes at $O$, it follows that $f(D)=0$ if and only if $f$ is constant zero - we want to show this occurs if and only if $C, P, E$ are collinear. ![](https://cdn.mathpix.com/cropped/2024_11_18_9ee8c2ac2c2f5002ff14g-068.jpg?height=556&width=1254&top_left_y=1906&top_left_x=408) In the one direction, suppose firstly that $C, P, E$ are not collinear, and let $T$ be the centre of the circle $C E P$. The same calculation as above provides $$ \\operatorname{area}(B C T)+\\operatorname{area}(A T E)=\\operatorname{area}(P B T A)=\\operatorname{area}(A B T)-\\operatorname{area}(A B P) $$ $$ f(T)=-2 \\operatorname{area}(A B P) \\neq 0 $$ Hence, the linear function $f$ is nonconstant with its zero is at $O$, so that $f(D) \\neq 0$. In the other direction, suppose that the points $C, P, E$ are collinear. We will show that $f$ is constant zero by finding a second point (other than $O$ ) at which it vanishes. ![](https://cdn.mathpix.com/cropped/2024_11_18_9ee8c2ac2c2f5002ff14g-069.jpg?height=410&width=1098&top_left_y=366&top_left_x=479) Let $Q$ be the reflection of $P$ across the midpoint of $A B$, so $P A Q B$ is a parallelogram. It is easy to see that $Q$ is on the perpendicular bisector of $C E$; for instance if $A^{\\prime}$ and $B^{\\prime}$ are the points produced from $A$ and $B$ by dilating about $P$ with scale factor 2, then the projection of $Q$ to $C E$ is the midpoint of the projections of $A^{\\prime}$ and $B^{\\prime}$, which are $E$ and $C$ respectively. The triangles $B C Q$ and $A Q E$ are indirectly congruent, so $$ f(Q)=(\\operatorname{area}(B C Q)+\\operatorname{area}(A Q E))-(\\operatorname{area}(A B Q)-\\operatorname{area}(B A P))=0-0=0 $$ The points $O$ and $Q$ are distinct. To see this, consider the circle $\\omega$ centred on $Q$ with $P^{\\prime}$ on the circumference; since triangle $P P^{\\prime} Q$ is right-angled at $P^{\\prime}$, it follows that $P$ lies outside $\\omega$. On the other hand, $P$ lies between $C$ and $E$ on the line $C P E$. It follows that $C$ and $E$ cannot both lie on $\\omega$, so that $\\omega$ is not the circle $C E P^{\\prime}$ and $Q \\neq O$. Since $O$ and $Q$ are distinct zeroes of the linear function $f$, we have $f(D)=0$ as desired. Comment 1. The condition $\\angle E D C \\neq 2 \\cdot \\angle A D B$ cannot be omitted. If $D$ is the centre of circle $C E P^{\\prime}$, then the condition on triangle areas is satisfied automatically, without having $P$ on line $C E$. Comment 2. The \"only if\" part of this problem is easier than the \"if\" part. For example, in the second part of Solution 2, the triangles $E A Q$ and $Q B C$ are indirectly congruent, so the sum of their areas is 0 , and $D C Q E$ is a kite. Now one can easily see that $\\angle(A Q, D E)=\\angle(C D, C B)$ and $\\angle(B Q, D C)=\\angle(E D, E A)$, whence area $(B C D)=\\operatorname{area}(A Q D)+\\operatorname{area}(E Q A)$ and $\\operatorname{area}(A D E)=$ $\\operatorname{area}(B D Q)+\\operatorname{area}(B Q C)$, which yields the result. Comment 3. The origin of the problem is the following observation. Let $A B D H$ be a tetrahedron and consider the sphere $\\mathcal{S}$ that is tangent to the four face planes, internally to planes $A D H$ and $B D H$ and externally to $A B D$ and $A B H$ (or vice versa). It is known that the sphere $\\mathcal{S}$ exists if and only if area $(A D H)+\\operatorname{area}(B D H) \\neq \\operatorname{area}(A B H)+\\operatorname{area}(A B D)$; this relation comes from the usual formula for the volume of the tetrahedron. Let $T, T_{a}, T_{b}, T_{d}$ be the points of tangency between the sphere and the four planes, as shown in the picture. Rotate the triangle $A B H$ inward, the triangles $B D H$ and $A D H$ outward, into the triangles $A B P, B D C$ and $A D E$, respectively, in the plane $A B D$. Notice that the points $T_{d}, T_{a}, T_{b}$ are rotated to $T$, so we have $H T_{a}=H T_{b}=H T_{d}=P T=C T=E T$. Therefore, the point $T$ is the centre of the circle $C E P$. Hence, if the sphere exists then $C, E, P$ cannot be collinear. If the condition $\\angle E D C \\neq 2 \\cdot \\angle A D B$ is replaced by the constraint that the angles $\\angle E D A, \\angle A D B$ and $\\angle B D C$ satisfy the triangle inequality, it enables reconstructing the argument with the tetrahedron and the tangent sphere. ![](https://cdn.mathpix.com/cropped/2024_11_18_9ee8c2ac2c2f5002ff14g-070.jpg?height=1069&width=1306&top_left_y=825&top_left_x=381)", "metadata": {"resource_path": "IMO/segmented/en-IMO2019SL.jsonl"}} -{"year": "2019", "tier": "T0", "problem_label": "G6", "problem_type": "Geometry", "exam": "IMO", "problem": "Let $I$ be the incentre of acute-angled triangle $A B C$. Let the incircle meet $B C, C A$, and $A B$ at $D, E$, and $F$, respectively. Let line $E F$ intersect the circumcircle of the triangle at $P$ and $Q$, such that $F$ lies between $E$ and $P$. Prove that $\\angle D P A+\\angle A Q D=\\angle Q I P$. (Slovakia)", "solution": "Let $N$ and $M$ be the midpoints of the arcs $\\widehat{B C}$ of the circumcircle, containing and opposite vertex $A$, respectively. By $\\angle F A E=\\angle B A C=\\angle B N C$, the right-angled kites $A F I E$ and $N B M C$ are similar. Consider the spiral similarity $\\varphi$ (dilation in case of $A B=A C$ ) that moves $A F I E$ to $N B M C$. The directed angle in which $\\varphi$ changes directions is $\\angle(A F, N B)$, same as $\\angle(A P, N P)$ and $\\angle(A Q, N Q)$; so lines $A P$ and $A Q$ are mapped to lines $N P$ and $N Q$, respectively. Line $E F$ is mapped to $B C$; we can see that the intersection points $P=E F \\cap A P$ and $Q=E F \\cap A Q$ are mapped to points $B C \\cap N P$ and $B C \\cap N Q$, respectively. Denote these points by $P^{\\prime}$ and $Q^{\\prime}$, respectively. ![](https://cdn.mathpix.com/cropped/2024_11_18_9ee8c2ac2c2f5002ff14g-071.jpg?height=689&width=1512&top_left_y=883&top_left_x=272) Let $L$ be the midpoint of $B C$. We claim that points $P, Q, D, L$ are concyclic (if $D=L$ then line $B C$ is tangent to circle $P Q D$ ). Let $P Q$ and $B C$ meet at $Z$. By applying Menelaus' theorem to triangle $A B C$ and line $E F Z$, we have $$ \\frac{B D}{D C}=\\frac{B F}{F A} \\cdot \\frac{A E}{E C}=-\\frac{B Z}{Z C}, $$ so the pairs $B, C$ and $D, Z$ are harmonic. It is well-known that this implies $Z B \\cdot Z C=Z D \\cdot Z L$. (The inversion with pole $Z$ that swaps $B$ and $C$ sends $Z$ to infinity and $D$ to the midpoint of $B C$, because the cross-ratio is preserved.) Hence, $Z D \\cdot Z L=Z B \\cdot Z C=Z P \\cdot Z Q$ by the power of $Z$ with respect to the circumcircle; this proves our claim. By $\\angle M P P^{\\prime}=\\angle M Q Q^{\\prime}=\\angle M L P^{\\prime}=\\angle M L Q^{\\prime}=90^{\\circ}$, the quadrilaterals $M L P P^{\\prime}$ and $M L Q Q^{\\prime}$ are cyclic. Then the problem statement follows by $$ \\begin{aligned} \\angle D P A+\\angle A Q D & =360^{\\circ}-\\angle P A Q-\\angle Q D P=360^{\\circ}-\\angle P N Q-\\angle Q L P \\\\ & =\\angle L P N+\\angle N Q L=\\angle P^{\\prime} M L+\\angle L M Q^{\\prime}=\\angle P^{\\prime} M Q^{\\prime}=\\angle P I Q . \\end{aligned} $$", "metadata": {"resource_path": "IMO/segmented/en-IMO2019SL.jsonl"}} -{"year": "2019", "tier": "T0", "problem_label": "G6", "problem_type": "Geometry", "exam": "IMO", "problem": "Let $I$ be the incentre of acute-angled triangle $A B C$. Let the incircle meet $B C, C A$, and $A B$ at $D, E$, and $F$, respectively. Let line $E F$ intersect the circumcircle of the triangle at $P$ and $Q$, such that $F$ lies between $E$ and $P$. Prove that $\\angle D P A+\\angle A Q D=\\angle Q I P$. (Slovakia)", "solution": "Define the point $M$ and the same spiral similarity $\\varphi$ as in the previous solution. (The point $N$ is not necessary.) It is well-known that the centre of the spiral similarity that maps $F, E$ to $B, C$ is the Miquel point of the lines $F E, B C, B F$ and $C E$; that is, the second intersection of circles $A B C$ and $A E F$. Denote that point by $S$. By $\\varphi(F)=B$ and $\\varphi(E)=C$ the triangles $S B F$ and $S C E$ are similar, so we have $$ \\frac{S B}{S C}=\\frac{B F}{C E}=\\frac{B D}{C D} $$ By the converse of the angle bisector theorem, that indicates that line $S D$ bisects $\\angle B S C$ and hence passes through $M$. Let $K$ be the intersection point of lines $E F$ and $S I$. Notice that $\\varphi$ sends points $S, F, E, I$ to $S, B, C, M$, so $\\varphi(K)=\\varphi(F E \\cap S I)=B C \\cap S M=D$. By $\\varphi(I)=M$, we have $K D \\| I M$. ![](https://cdn.mathpix.com/cropped/2024_11_18_9ee8c2ac2c2f5002ff14g-072.jpg?height=669&width=706&top_left_y=839&top_left_x=675) We claim that triangles $S P I$ and $S D Q$ are similar, and so are triangles $S P D$ and $S I Q$. Let ray $S I$ meet the circumcircle again at $L$. Note that the segment $E F$ is perpendicular to the angle bisector $A M$. Then by $\\angle A M L=\\angle A S L=\\angle A S I=90^{\\circ}$, we have $M L \\| P Q$. Hence, $\\widetilde{P L}=\\widetilde{M Q}$ and therefore $\\angle P S L=\\angle M S Q=\\angle D S Q$. By $\\angle Q P S=\\angle Q M S$, the triangles $S P K$ and $S M Q$ are similar. Finally, $$ \\frac{S P}{S I}=\\frac{S P}{S K} \\cdot \\frac{S K}{S I}=\\frac{S M}{S Q} \\cdot \\frac{S D}{S M}=\\frac{S D}{S Q} $$ shows that triangles $S P I$ and $S D Q$ are similar. The second part of the claim can be proved analogously. Now the problem statement can be proved by $$ \\angle D P A+\\angle A Q D=\\angle D P S+\\angle S Q D=\\angle Q I S+\\angle S I P=\\angle Q I P . $$", "metadata": {"resource_path": "IMO/segmented/en-IMO2019SL.jsonl"}} -{"year": "2019", "tier": "T0", "problem_label": "G6", "problem_type": "Geometry", "exam": "IMO", "problem": "Let $I$ be the incentre of acute-angled triangle $A B C$. Let the incircle meet $B C, C A$, and $A B$ at $D, E$, and $F$, respectively. Let line $E F$ intersect the circumcircle of the triangle at $P$ and $Q$, such that $F$ lies between $E$ and $P$. Prove that $\\angle D P A+\\angle A Q D=\\angle Q I P$. (Slovakia)", "solution": "Denote the circumcircle of triangle $A B C$ by $\\Gamma$, and let rays $P D$ and $Q D$ meet $\\Gamma$ again at $V$ and $U$, respectively. We will show that $A U \\perp I P$ and $A V \\perp I Q$. Then the problem statement will follow as $$ \\angle D P A+\\angle A Q D=\\angle V U A+\\angle A V U=180^{\\circ}-\\angle U A V=\\angle Q I P . $$ Let $M$ be the midpoint of arc $\\widehat{B U V C}$ and let $N$ be the midpoint of arc $\\widehat{C A B}$; the lines $A I M$ and $A N$ being the internal and external bisectors of angle $B A C$, respectively, are perpendicular. Let the tangents drawn to $\\Gamma$ at $B$ and $C$ meet at $R$; let line $P Q$ meet $A U, A I, A V$ and $B C$ at $X, T, Y$ and $Z$, respectively. As in Solution 1, we observe that the pairs $B, C$ and $D, Z$ are harmonic. Projecting these points from $Q$ onto the circumcircle, we can see that $B, C$ and $U, P$ are also harmonic. Analogously, the pair $V, Q$ is harmonic with $B, C$. Consider the inversion about the circle with centre $R$, passing through $B$ and $C$. Points $B$ and $C$ are fixed points, so this inversion exchanges every point of $\\Gamma$ by its harmonic pair with respect to $B, C$. In particular, the inversion maps points $B, C, N, U, V$ to points $B, C, M, P, Q$, respectively. Combine the inversion with projecting $\\Gamma$ from $A$ to line $P Q$; the points $B, C, M, P, Q$ are projected to $F, E, T, P, Q$, respectively. ![](https://cdn.mathpix.com/cropped/2024_11_18_9ee8c2ac2c2f5002ff14g-073.jpg?height=1006&width=1384&top_left_y=1082&top_left_x=339) The combination of these two transformations is projective map from the lines $A B, A C$, $A N, A U, A V$ to $I F, I E, I T, I P, I Q$, respectively. On the other hand, we have $A B \\perp I F$, $A C \\perp I E$ and $A N \\perp A T$, so the corresponding lines in these two pencils are perpendicular. This proves $A U \\perp I P$ and $A V \\perp I Q$, and hence completes the solution.", "metadata": {"resource_path": "IMO/segmented/en-IMO2019SL.jsonl"}} -{"year": "2019", "tier": "T0", "problem_label": "G7", "problem_type": "Geometry", "exam": "IMO", "problem": "The incircle $\\omega$ of acute-angled scalene triangle $A B C$ has centre $I$ and meets sides $B C$, $C A$, and $A B$ at $D, E$, and $F$, respectively. The line through $D$ perpendicular to $E F$ meets $\\omega$ again at $R$. Line $A R$ meets $\\omega$ again at $P$. The circumcircles of triangles $P C E$ and $P B F$ meet again at $Q \\neq P$. Prove that lines $D I$ and $P Q$ meet on the external bisector of angle $B A C$. (India) Common remarks. Throughout the solution, $\\angle(a, b)$ denotes the directed angle between lines $a$ and $b$, measured modulo $\\pi$.", "solution": "Step 1. The external bisector of $\\angle B A C$ is the line through $A$ perpendicular to $I A$. Let $D I$ meet this line at $L$ and let $D I$ meet $\\omega$ at $K$. Let $N$ be the midpoint of $E F$, which lies on $I A$ and is the pole of line $A L$ with respect to $\\omega$. Since $A N \\cdot A I=A E^{2}=A R \\cdot A P$, the points $R$, $N, I$, and $P$ are concyclic. As $I R=I P$, the line $N I$ is the external bisector of $\\angle P N R$, so $P N$ meets $\\omega$ again at the point symmetric to $R$ with respect to $A N-i . e$ at $K$. Let $D N$ cross $\\omega$ again at $S$. Opposite sides of any quadrilateral inscribed in the circle $\\omega$ meet on the polar line of the intersection of the diagonals with respect to $\\omega$. Since $L$ lies on the polar line $A L$ of $N$ with respect to $\\omega$, the line $P S$ must pass through $L$. Thus it suffices to prove that the points $S, Q$, and $P$ are collinear. ![](https://cdn.mathpix.com/cropped/2024_11_18_9ee8c2ac2c2f5002ff14g-074.jpg?height=875&width=1350&top_left_y=1176&top_left_x=361) Step 2. Let $\\Gamma$ be the circumcircle of $\\triangle B I C$. Notice that $$ \\begin{aligned} & \\angle(B Q, Q C)=\\angle(B Q, Q P)+\\angle(P Q, Q C)=\\angle(B F, F P)+\\angle(P E, E C) \\\\ &=\\angle(E F, E P)+\\angle(F P, F E)=\\angle(F P, E P)=\\angle(D F, D E)=\\angle(B I, I C) \\end{aligned} $$ so $Q$ lies on $\\Gamma$. Let $Q P$ meet $\\Gamma$ again at $T$. It will now suffice to prove that $S, P$, and $T$ are collinear. Notice that $\\angle(B I, I T)=\\angle(B Q, Q T)=\\angle(B F, F P)=\\angle(F K, K P)$. Note $F D \\perp F K$ and $F D \\perp B I$ so $F K \\| B I$ and hence $I T$ is parallel to the line $K N P$. Since $D I=I K$, the line $I T$ crosses $D N$ at its midpoint $M$. Step 3. Let $F^{\\prime}$ and $E^{\\prime}$ be the midpoints of $D E$ and $D F$, respectively. Since $D E^{\\prime} \\cdot E^{\\prime} F=D E^{\\prime 2}=$ $B E^{\\prime} \\cdot E^{\\prime} I$, the point $E^{\\prime}$ lies on the radical axis of $\\omega$ and $\\Gamma$; the same holds for $F^{\\prime}$. Therefore, this radical axis is $E^{\\prime} F^{\\prime}$, and it passes through $M$. Thus $I M \\cdot M T=D M \\cdot M S$, so $S, I, D$, and $T$ are concyclic. This shows $\\angle(D S, S T)=\\angle(D I, I T)=\\angle(D K, K P)=\\angle(D S, S P)$, whence the points $S, P$, and $T$ are collinear, as desired. ![](https://cdn.mathpix.com/cropped/2024_11_18_9ee8c2ac2c2f5002ff14g-075.jpg?height=1089&width=992&top_left_y=198&top_left_x=532) Comment. Here is a longer alternative proof in step 1 that $P, S$, and $L$ are collinear, using a circular inversion instead of the fact that opposite sides of a quadrilateral inscribed in a circle $\\omega$ meet on the polar line with respect to $\\omega$ of the intersection of the diagonals. Let $G$ be the foot of the altitude from $N$ to the line $D I K L$. Observe that $N, G, K, S$ are concyclic (opposite right angles) so $$ \\angle D I P=2 \\angle D K P=\\angle G K N+\\angle D S P=\\angle G S N+\\angle N S P=\\angle G S P, $$ hence $I, G, S, P$ are concyclic. We have $I G \\cdot I L=I N \\cdot I A=r^{2}$ since $\\triangle I G N \\sim \\triangle I A L$. Inverting the circle $I G S P$ in circle $\\omega$, points $P$ and $S$ are fixed and $G$ is taken to $L$ so we find that $P, S$, and $L$ are collinear.", "metadata": {"resource_path": "IMO/segmented/en-IMO2019SL.jsonl"}} -{"year": "2019", "tier": "T0", "problem_label": "G7", "problem_type": "Geometry", "exam": "IMO", "problem": "The incircle $\\omega$ of acute-angled scalene triangle $A B C$ has centre $I$ and meets sides $B C$, $C A$, and $A B$ at $D, E$, and $F$, respectively. The line through $D$ perpendicular to $E F$ meets $\\omega$ again at $R$. Line $A R$ meets $\\omega$ again at $P$. The circumcircles of triangles $P C E$ and $P B F$ meet again at $Q \\neq P$. Prove that lines $D I$ and $P Q$ meet on the external bisector of angle $B A C$. (India) Common remarks. Throughout the solution, $\\angle(a, b)$ denotes the directed angle between lines $a$ and $b$, measured modulo $\\pi$.", "solution": "We start as in Step 1. Let $A R$ meet the circumcircle $\\Omega$ of $A B C$ again at $X$. The lines $A R$ and $A K$ are isogonal in the angle $B A C$; it is well known that in this case $X$ is the tangency point of $\\Omega$ with the $A$-mixtilinear circle. It is also well known that for this point $X$, the line $X I$ crosses $\\Omega$ again at the midpoint $M^{\\prime}$ of arc $B A C$. Step 2. Denote the circles $B F P$ and $C E P$ by $\\Omega_{B}$ and $\\Omega_{C}$, respectively. Let $\\Omega_{B}$ cross $A R$ and $E F$ again at $U$ and $Y$, respectively. We have $$ \\angle(U B, B F)=\\angle(U P, P F)=\\angle(R P, P F)=\\angle(R F, F A) $$ so $U B \\| R F$. ![](https://cdn.mathpix.com/cropped/2024_11_18_9ee8c2ac2c2f5002ff14g-076.jpg?height=1009&width=1474&top_left_y=883&top_left_x=248) Next, we show that the points $B, I, U$, and $X$ are concyclic. Since $$ \\angle(U B, U X)=\\angle(R F, R X)=\\angle(A F, A R)+\\angle(F R, F A)=\\angle\\left(M^{\\prime} B, M^{\\prime} X\\right)+\\angle(D R, D F) $$ it suffices to prove $\\angle(I B, I X)=\\angle\\left(M^{\\prime} B, M^{\\prime} X\\right)+\\angle(D R, D F)$, or $\\angle\\left(I B, M^{\\prime} B\\right)=\\angle(D R, D F)$. But both angles equal $\\angle(C I, C B)$, as desired. (This is where we used the fact that $M^{\\prime}$ is the midpoint of $\\operatorname{arc} B A C$ of $\\Omega$.) It follows now from circles $B U I X$ and $B P U F Y$ that $$ \\begin{aligned} \\angle(I U, U B)=\\angle(I X, B X)=\\angle\\left(M^{\\prime} X, B X\\right)= & \\frac{\\pi-\\angle A}{2} \\\\ & =\\angle(E F, A F)=\\angle(Y F, B F)=\\angle(Y U, B U) \\end{aligned} $$ so the points $Y, U$, and $I$ are collinear. Let $E F$ meet $B C$ at $W$. We have $$ \\angle(I Y, Y W)=\\angle(U Y, F Y)=\\angle(U B, F B)=\\angle(R F, A F)=\\angle(C I, C W) $$ so the points $W, Y, I$, and $C$ are concyclic. Similarly, if $V$ and $Z$ are the second meeting points of $\\Omega_{C}$ with $A R$ and $E F$, we get that the 4-tuples $(C, V, I, X)$ and $(B, I, Z, W)$ are both concyclic. Step 3. Let $Q^{\\prime}=C Y \\cap B Z$. We will show that $Q^{\\prime}=Q$. First of all, we have $$ \\begin{aligned} & \\angle\\left(Q^{\\prime} Y, Q^{\\prime} B\\right)=\\angle(C Y, Z B)=\\angle(C Y, Z Y)+\\angle(Z Y, B Z) \\\\ & =\\angle(C I, I W)+\\angle(I W, I B)=\\angle(C I, I B)=\\frac{\\pi-\\angle A}{2}=\\angle(F Y, F B), \\end{aligned} $$ so $Q^{\\prime} \\in \\Omega_{B}$. Similarly, $Q^{\\prime} \\in \\Omega_{C}$. Thus $Q^{\\prime} \\in \\Omega_{B} \\cap \\Omega_{C}=\\{P, Q\\}$ and it remains to prove that $Q^{\\prime} \\neq P$. If we had $Q^{\\prime}=P$, we would have $\\angle(P Y, P Z)=\\angle\\left(Q^{\\prime} Y, Q^{\\prime} Z\\right)=\\angle(I C, I B)$. This would imply $$ \\angle(P Y, Y F)+\\angle(E Z, Z P)=\\angle(P Y, P Z)=\\angle(I C, I B)=\\angle(P E, P F), $$ so circles $\\Omega_{B}$ and $\\Omega_{C}$ would be tangent at $P$. That is excluded in the problem conditions, so $Q^{\\prime}=Q$. ![](https://cdn.mathpix.com/cropped/2024_11_18_9ee8c2ac2c2f5002ff14g-077.jpg?height=966&width=1468&top_left_y=1070&top_left_x=251) Step 4. Now we are ready to show that $P, Q$, and $S$ are collinear. Notice that $A$ and $D$ are the poles of $E W$ and $D W$ with respect to $\\omega$, so $W$ is the pole of $A D$. Hence, $W I \\perp A D$. Since $C I \\perp D E$, this yields $\\angle(I C, W I)=\\angle(D E, D A)$. On the other hand, $D A$ is a symmedian in $\\triangle D E F$, so $\\angle(D E, D A)=\\angle(D N, D F)=\\angle(D S, D F)$. Therefore, $$ \\begin{aligned} \\angle(P S, P F)=\\angle(D S, D F)=\\angle(D E, D A)= & \\angle(I C, I W) \\\\ & =\\angle(Y C, Y W)=\\angle(Y Q, Y F)=\\angle(P Q, P F), \\end{aligned} $$ which yields the desired collinearity.", "metadata": {"resource_path": "IMO/segmented/en-IMO2019SL.jsonl"}} -{"year": "2019", "tier": "T0", "problem_label": "G8", "problem_type": "Geometry", "exam": "IMO", "problem": "Let $\\mathcal{L}$ be the set of all lines in the plane and let $f$ be a function that assigns to each line $\\ell \\in \\mathcal{L}$ a point $f(\\ell)$ on $\\ell$. Suppose that for any point $X$, and for any three lines $\\ell_{1}, \\ell_{2}, \\ell_{3}$ passing through $X$, the points $f\\left(\\ell_{1}\\right), f\\left(\\ell_{2}\\right), f\\left(\\ell_{3}\\right)$ and $X$ lie on a circle. Prove that there is a unique point $P$ such that $f(\\ell)=P$ for any line $\\ell$ passing through $P$. (Australia) Common remarks. The condition on $f$ is equivalent to the following: There is some function $g$ that assigns to each point $X$ a circle $g(X)$ passing through $X$ such that for any line $\\ell$ passing through $X$, the point $f(\\ell)$ lies on $g(X)$. (The function $g$ may not be uniquely defined for all points, if some points $X$ have at most one value of $f(\\ell)$ other than $X$; for such points, an arbitrary choice is made.) If there were two points $P$ and $Q$ with the given property, $f(P Q)$ would have to be both $P$ and $Q$, so there is at most one such point, and it will suffice to show that such a point exists.", "solution": "We provide a complete characterisation of the functions satisfying the given condition. Write $\\angle\\left(\\ell_{1}, \\ell_{2}\\right)$ for the directed angle modulo $180^{\\circ}$ between the lines $\\ell_{1}$ and $\\ell_{2}$. Given a point $P$ and an angle $\\alpha \\in\\left(0,180^{\\circ}\\right)$, for each line $\\ell$, let $\\ell^{\\prime}$ be the line through $P$ satisfying $\\angle\\left(\\ell^{\\prime}, \\ell\\right)=\\alpha$, and let $h_{P, \\alpha}(\\ell)$ be the intersection point of $\\ell$ and $\\ell^{\\prime}$. We will prove that there is some pair $(P, \\alpha)$ such that $f$ and $h_{P, \\alpha}$ are the same function. Then $P$ is the unique point in the problem statement. Given an angle $\\alpha$ and a point $P$, let a line $\\ell$ be called $(P, \\alpha)$-good if $f(\\ell)=h_{P, \\alpha}(\\ell)$. Let a point $X \\neq P$ be called $(P, \\alpha)$-good if the circle $g(X)$ passes through $P$ and some point $Y \\neq P, X$ on $g(X)$ satisfies $\\angle(P Y, Y X)=\\alpha$. It follows from this definition that if $X$ is $(P, \\alpha)$ good then every point $Y \\neq P, X$ of $g(X)$ satisfies this angle condition, so $h_{P, \\alpha}(X Y)=Y$ for every $Y \\in g(X)$. Equivalently, $f(\\ell) \\in\\left\\{X, h_{P, \\alpha}(\\ell)\\right\\}$ for each line $\\ell$ passing through $X$. This shows the following lemma. Lemma 1. If $X$ is $(P, \\alpha)$-good and $\\ell$ is a line passing through $X$ then either $f(\\ell)=X$ or $\\ell$ is $(P, \\alpha)$-good. Lemma 2. If $X$ and $Y$ are different $(P, \\alpha)$-good points, then line $X Y$ is $(P, \\alpha)$-good. Proof. If $X Y$ is not $(P, \\alpha)$-good then by the previous Lemma, $f(X Y)=X$ and similarly $f(X Y)=Y$, but clearly this is impossible as $X \\neq Y$. Lemma 3. If $\\ell_{1}$ and $\\ell_{2}$ are different $(P, \\alpha)$-good lines which intersect at $X \\neq P$, then either $f\\left(\\ell_{1}\\right)=X$ or $f\\left(\\ell_{2}\\right)=X$ or $X$ is $(P, \\alpha)$-good. Proof. If $f\\left(\\ell_{1}\\right), f\\left(\\ell_{2}\\right) \\neq X$, then $g(X)$ is the circumcircle of $X, f\\left(\\ell_{1}\\right)$ and $f\\left(\\ell_{2}\\right)$. Since $\\ell_{1}$ and $\\ell_{2}$ are $(P, \\alpha)$-good lines, the angles $$ \\angle\\left(P f\\left(\\ell_{1}\\right), f\\left(\\ell_{1}\\right) X\\right)=\\angle\\left(P f\\left(\\ell_{2}\\right), f\\left(\\ell_{2}\\right) X\\right)=\\alpha $$ so $P$ lies on $g(X)$. Hence, $X$ is $(P, \\alpha)$-good. Lemma 4. If $\\ell_{1}, \\ell_{2}$ and $\\ell_{3}$ are different $(P, \\alpha)$ good lines which intersect at $X \\neq P$, then $X$ is $(P, \\alpha)$-good. Proof. This follows from the previous Lemma since at most one of the three lines $\\ell_{i}$ can satisfy $f\\left(\\ell_{i}\\right)=X$ as the three lines are all $(P, \\alpha)$-good. Lemma 5. If $A B C$ is a triangle such that $A, B, C, f(A B), f(A C)$ and $f(B C)$ are all different points, then there is some point $P$ and some angle $\\alpha$ such that $A, B$ and $C$ are $(P, \\alpha)$-good points and $A B, B C$ and $C A$ are $(P, \\alpha)-\\operatorname{good}$ lines. ![](https://cdn.mathpix.com/cropped/2024_11_18_9ee8c2ac2c2f5002ff14g-079.jpg?height=415&width=612&top_left_y=207&top_left_x=728) Proof. Let $D, E, F$ denote the points $f(B C), f(A C), f(A B)$, respectively. Then $g(A)$, $g(B)$ and $g(C)$ are the circumcircles of $A E F, B D F$ and $C D E$, respectively. Let $P \\neq F$ be the second intersection of circles $g(A)$ and $g(B)$ (or, if these circles are tangent at $F$, then $P=F$ ). By Miquel's theorem (or an easy angle chase), $g(C)$ also passes through $P$. Then by the cyclic quadrilaterals, the directed angles $$ \\angle(P D, D C)=\\angle(P F, F B)=\\angle(P E, E A)=\\alpha $$ for some angle $\\alpha$. Hence, lines $A B, B C$ and $C A$ are all $(P, \\alpha)$-good, so by Lemma $3, A, B$ and $C$ are $(P, \\alpha)$-good. (In the case where $P=D$, the line $P D$ in the equation above denotes the line which is tangent to $g(B)$ at $P=D$. Similar definitions are used for $P E$ and $P F$ in the cases where $P=E$ or $P=F$.) Consider the set $\\Omega$ of all points $(x, y)$ with integer coordinates $1 \\leqslant x, y \\leqslant 1000$, and consider the set $L_{\\Omega}$ of all horizontal, vertical and diagonal lines passing through at least one point in $\\Omega$. A simple counting argument shows that there are 5998 lines in $L_{\\Omega}$. For each line $\\ell$ in $L_{\\Omega}$ we colour the point $f(\\ell)$ red. Then there are at most 5998 red points. Now we partition the points in $\\Omega$ into 10000 ten by ten squares. Since there are at most 5998 red points, at least one of these squares $\\Omega_{10}$ contains no red points. Let $(m, n)$ be the bottom left point in $\\Omega_{10}$. Then the triangle with vertices $(m, n),(m+1, n)$ and $(m, n+1)$ satisfies the condition of Lemma 5 , so these three vertices are all $(P, \\alpha)$-good for some point $P$ and angle $\\alpha$, as are the lines joining them. From this point on, we will simply call a point or line good if it is $(P, \\alpha)$-good for this particular pair $(P, \\alpha)$. Now by Lemma 1, the line $x=m+1$ is good, as is the line $y=n+1$. Then Lemma 3 implies that $(m+1, n+1)$ is good. By applying these two lemmas repeatedly, we can prove that the line $x+y=m+n+2$ is good, then the points $(m, n+2)$ and $(m+2, n)$ then the lines $x=m+2$ and $y=n+2$, then the points $(m+2, n+1),(m+1, n+2)$ and $(m+2, n+2)$ and so on until we have prove that all points in $\\Omega_{10}$ are good. Now we will use this to prove that every point $S \\neq P$ is good. Since $g(S)$ is a circle, it passes through at most two points of $\\Omega_{10}$ on any vertical line, so at most 20 points in total. Moreover, any line $\\ell$ through $S$ intersects at most 10 points in $\\Omega_{10}$. Hence, there are at least eight lines $\\ell$ through $S$ which contain a point $Q$ in $\\Omega_{10}$ which is not on $g(S)$. Since $Q$ is not on $g(S)$, the point $f(\\ell) \\neq Q$. Hence, by Lemma 1 , the line $\\ell$ is good. Hence, at least eight good lines pass through $S$, so by Lemma 4, the point $S$ is good. Hence, every point $S \\neq P$ is good, so by Lemma 2 , every line is good. In particular, every line $\\ell$ passing through $P$ is good, and therefore satisfies $f(\\ell)=P$, as required.", "metadata": {"resource_path": "IMO/segmented/en-IMO2019SL.jsonl"}} -{"year": "2019", "tier": "T0", "problem_label": "G8", "problem_type": "Geometry", "exam": "IMO", "problem": "Let $\\mathcal{L}$ be the set of all lines in the plane and let $f$ be a function that assigns to each line $\\ell \\in \\mathcal{L}$ a point $f(\\ell)$ on $\\ell$. Suppose that for any point $X$, and for any three lines $\\ell_{1}, \\ell_{2}, \\ell_{3}$ passing through $X$, the points $f\\left(\\ell_{1}\\right), f\\left(\\ell_{2}\\right), f\\left(\\ell_{3}\\right)$ and $X$ lie on a circle. Prove that there is a unique point $P$ such that $f(\\ell)=P$ for any line $\\ell$ passing through $P$. (Australia) Common remarks. The condition on $f$ is equivalent to the following: There is some function $g$ that assigns to each point $X$ a circle $g(X)$ passing through $X$ such that for any line $\\ell$ passing through $X$, the point $f(\\ell)$ lies on $g(X)$. (The function $g$ may not be uniquely defined for all points, if some points $X$ have at most one value of $f(\\ell)$ other than $X$; for such points, an arbitrary choice is made.) If there were two points $P$ and $Q$ with the given property, $f(P Q)$ would have to be both $P$ and $Q$, so there is at most one such point, and it will suffice to show that such a point exists.", "solution": "Note that for any distinct points $X, Y$, the circles $g(X)$ and $g(Y)$ meet on $X Y$ at the point $f(X Y) \\in g(X) \\cap g(Y) \\cap(X Y)$. We write $s(X, Y)$ for the second intersection point of circles $g(X)$ and $g(Y)$. Lemma 1. Suppose that $X, Y$ and $Z$ are not collinear, and that $f(X Y) \\notin\\{X, Y\\}$ and similarly for $Y Z$ and $Z X$. Then $s(X, Y)=s(Y, Z)=s(Z, X)$. Proof. The circles $g(X), g(Y)$ and $g(Z)$ through the vertices of triangle $X Y Z$ meet pairwise on the corresponding edges (produced). By Miquel's theorem, the second points of intersection of any two of the circles coincide. (See the diagram for Lemma 5 of Solution 1.) Now pick any line $\\ell$ and any six different points $Y_{1}, \\ldots, Y_{6}$ on $\\ell \\backslash\\{f(\\ell)\\}$. Pick a point $X$ not on $\\ell$ or any of the circles $g\\left(Y_{i}\\right)$. Reordering the indices if necessary, we may suppose that $Y_{1}, \\ldots, Y_{4}$ do not lie on $g(X)$, so that $f\\left(X Y_{i}\\right) \\notin\\left\\{X, Y_{i}\\right\\}$ for $1 \\leqslant i \\leqslant 4$. By applying the above lemma to triangles $X Y_{i} Y_{j}$ for $1 \\leqslant i0$, there is a point $P_{\\epsilon}$ with $g\\left(P_{\\epsilon}\\right)$ of radius at most $\\epsilon$. Then there is a point $P$ with the given property. Proof. Consider a sequence $\\epsilon_{i}=2^{-i}$ and corresponding points $P_{\\epsilon_{i}}$. Because the two circles $g\\left(P_{\\epsilon_{i}}\\right)$ and $g\\left(P_{\\epsilon_{j}}\\right)$ meet, the distance between $P_{\\epsilon_{i}}$ and $P_{\\epsilon_{j}}$ is at most $2^{1-i}+2^{1-j}$. As $\\sum_{i} \\epsilon_{i}$ converges, these points converge to some point $P$. For all $\\epsilon>0$, the point $P$ has distance at most $2 \\epsilon$ from $P_{\\epsilon}$, and all circles $g(X)$ pass through a point with distance at most $2 \\epsilon$ from $P_{\\epsilon}$, so distance at most $4 \\epsilon$ from $P$. A circle that passes distance at most $4 \\epsilon$ from $P$ for all $\\epsilon>0$ must pass through $P$, so by Lemma 1 the point $P$ has the given property. Lemma 3. Suppose no two of the circles $g(X)$ cut. Then there is a point $P$ with the given property. Proof. Consider a circle $g(X)$ with centre $Y$. The circle $g(Y)$ must meet $g(X)$ without cutting it, so has half the radius of $g(X)$. Repeating this argument, there are circles with arbitrarily small radius and the result follows by Lemma 2. Lemma 4. Suppose there are six different points $A, B_{1}, B_{2}, B_{3}, B_{4}, B_{5}$ such that no three are collinear, no four are concyclic, and all the circles $g\\left(B_{i}\\right)$ cut pairwise at $A$. Then there is a point $P$ with the given property. Proof. Consider some line $\\ell$ through $A$ that does not pass through any of the $B_{i}$ and is not tangent to any of the $g\\left(B_{i}\\right)$. Fix some direction along that line, and let $X_{\\epsilon}$ be the point on $\\ell$ that has distance $\\epsilon$ from $A$ in that direction. In what follows we consider only those $\\epsilon$ for which $X_{\\epsilon}$ does not lie on any $g\\left(B_{i}\\right)$ (this restriction excludes only finitely many possible values of $\\epsilon$ ). Consider the circle $g\\left(X_{\\epsilon}\\right)$. Because no four of the $B_{i}$ are concyclic, at most three of them lie on this circle, so at least two of them do not. There must be some sequence of $\\epsilon \\rightarrow 0$ such that it is the same two of the $B_{i}$ for all $\\epsilon$ in that sequence, so now restrict attention to that sequence, and suppose without loss of generality that $B_{1}$ and $B_{2}$ do not lie on $g\\left(X_{\\epsilon}\\right)$ for any $\\epsilon$ in that sequence. Then $f\\left(X_{\\epsilon} B_{1}\\right)$ is not $B_{1}$, so must be the other point of intersection of $X_{\\epsilon} B_{1}$ with $g\\left(B_{1}\\right)$, and the same applies with $B_{2}$. Now consider the three points $X_{\\epsilon}, f\\left(X_{\\epsilon} B_{1}\\right)$ and $f\\left(X_{\\epsilon} B_{2}\\right)$. As $\\epsilon \\rightarrow 0$, the angle at $X_{\\epsilon}$ tends to $\\angle B_{1} A B_{2}$ or $180^{\\circ}-\\angle B_{1} A B_{2}$, which is not 0 or $180^{\\circ}$ because no three of the points were collinear. All three distances between those points are bounded above by constant multiples of $\\epsilon$ (in fact, if the triangle is scaled by a factor of $1 / \\epsilon$, it tends to a fixed triangle). Thus the circumradius of those three points, which is the radius of $g\\left(X_{\\epsilon}\\right)$, is also bounded above by a constant multiple of $\\epsilon$, and so the result follows by Lemma 2 . Lemma 5. Suppose there are two points $A$ and $B$ such that $g(A)$ and $g(B)$ cut. Then there is a point $P$ with the given property. Proof. Suppose that $g(A)$ and $g(B)$ cut at $C$ and $D$. One of those points, without loss of generality $C$, must be $f(A B)$, and so lie on the line $A B$. We now consider two cases, according to whether $D$ also lies on that line.", "metadata": {"resource_path": "IMO/segmented/en-IMO2019SL.jsonl"}} -{"year": "2019", "tier": "T0", "problem_label": "G8", "problem_type": "Geometry", "exam": "IMO", "problem": "Let $\\mathcal{L}$ be the set of all lines in the plane and let $f$ be a function that assigns to each line $\\ell \\in \\mathcal{L}$ a point $f(\\ell)$ on $\\ell$. Suppose that for any point $X$, and for any three lines $\\ell_{1}, \\ell_{2}, \\ell_{3}$ passing through $X$, the points $f\\left(\\ell_{1}\\right), f\\left(\\ell_{2}\\right), f\\left(\\ell_{3}\\right)$ and $X$ lie on a circle. Prove that there is a unique point $P$ such that $f(\\ell)=P$ for any line $\\ell$ passing through $P$. (Australia) Common remarks. The condition on $f$ is equivalent to the following: There is some function $g$ that assigns to each point $X$ a circle $g(X)$ passing through $X$ such that for any line $\\ell$ passing through $X$, the point $f(\\ell)$ lies on $g(X)$. (The function $g$ may not be uniquely defined for all points, if some points $X$ have at most one value of $f(\\ell)$ other than $X$; for such points, an arbitrary choice is made.) If there were two points $P$ and $Q$ with the given property, $f(P Q)$ would have to be both $P$ and $Q$, so there is at most one such point, and it will suffice to show that such a point exists.", "solution": "For any point $X$, denote by $t(X)$ the line tangent to $g(X)$ at $X$; notice that $f(t(X))=X$, so $f$ is surjective. Step 1: We find a point $P$ for which there are at least two different lines $p_{1}$ and $p_{2}$ such that $f\\left(p_{i}\\right)=P$. Choose any point $X$. If $X$ does not have this property, take any $Y \\in g(X) \\backslash\\{X\\}$; then $f(X Y)=Y$. If $Y$ does not have the property, $t(Y)=X Y$, and the circles $g(X)$ and $g(Y)$ meet again at some point $Z$. Then $f(X Z)=Z=f(Y Z)$, so $Z$ has the required property. We will show that $P$ is the desired point. From now on, we fix two different lines $p_{1}$ and $p_{2}$ with $f\\left(p_{1}\\right)=f\\left(p_{2}\\right)=P$. Assume for contradiction that $f(\\ell)=Q \\neq P$ for some line $\\ell$ through $P$. We fix $\\ell$, and note that $Q \\in g(P)$. Step 2: We prove that $P \\in g(Q)$. Take an arbitrary point $X \\in \\ell \\backslash\\{P, Q\\}$. Two cases are possible for the position of $t(X)$ in relation to the $p_{i}$; we will show that each case (and subcase) occurs for only finitely many positions of $X$, yielding a contradiction. Case 2.1: $t(X)$ is parallel to one of the $p_{i}$; say, to $p_{1}$. Let $t(X) \\operatorname{cross} p_{2}$ at $R$. Then $g(R)$ is the circle $(P R X)$, as $f(R P)=P$ and $f(R X)=X$. Let $R Q$ cross $g(R)$ again at $S$. Then $f(R Q) \\in\\{R, S\\} \\cap g(Q)$, so $g(Q)$ contains one of the points $R$ and $S$. If $R \\in g(Q)$, then $R$ is one of finitely many points in the intersection $g(Q) \\cap p_{2}$, and each of them corresponds to a unique position of $X$, since $R X$ is parallel to $p_{1}$. If $S \\in g(Q)$, then $\\angle(Q S, S P)=\\angle(R S, S P)=\\angle(R X, X P)=\\angle\\left(p_{1}, \\ell\\right)$, so $\\angle(Q S, S P)$ is constant for all such points $X$, and all points $S$ obtained in such a way lie on one circle $\\gamma$ passing through $P$ and $Q$. Since $g(Q)$ does not contain $P$, it is different from $\\gamma$, so there are only finitely many points $S$. Each of them uniquely determines $R$ and thus $X$. ![](https://cdn.mathpix.com/cropped/2024_11_18_9ee8c2ac2c2f5002ff14g-083.jpg?height=843&width=909&top_left_y=207&top_left_x=568) So, Case 2.1 can occur for only finitely many points $X$. Case 2.2: $t(X)$ crosses $p_{1}$ and $p_{2}$ at $R_{1}$ and $R_{2}$, respectively. Clearly, $R_{1} \\neq R_{2}$, as $t(X)$ is the tangent to $g(X)$ at $X$, and $g(X)$ meets $\\ell$ only at $X$ and $Q$. Notice that $g\\left(R_{i}\\right)$ is the circle $\\left(P X R_{i}\\right)$. Let $R_{i} Q$ meet $g\\left(R_{i}\\right)$ again at $S_{i}$; then $S_{i} \\neq Q$, as $g\\left(R_{i}\\right)$ meets $\\ell$ only at $P$ and $X$. Then $f\\left(R_{i} Q\\right) \\in\\left\\{R_{i}, S_{i}\\right\\}$, and we distinguish several subcases. ![](https://cdn.mathpix.com/cropped/2024_11_18_9ee8c2ac2c2f5002ff14g-083.jpg?height=824&width=866&top_left_y=1501&top_left_x=595) Subcase 2.2.1: $f\\left(R_{1} Q\\right)=S_{1}, f\\left(R_{2} Q\\right)=S_{2}$; so $S_{1}, S_{2} \\in g(Q)$. In this case we have $0=\\angle\\left(R_{1} X, X P\\right)+\\angle\\left(X P, R_{2} X\\right)=\\angle\\left(R_{1} S_{1}, S_{1} P\\right)+\\angle\\left(S_{2} P, S_{2} R_{2}\\right)=$ $\\angle\\left(Q S_{1}, S_{1} P\\right)+\\angle\\left(S_{2} P, S_{2} Q\\right)$, which shows $P \\in g(Q)$. Subcase 2.2.2: $f\\left(R_{1} Q\\right)=R_{1}, f\\left(R_{2} Q\\right)=R_{2}$; so $R_{1}, R_{2} \\in g(Q)$. This can happen for at most four positions of $X$ - namely, at the intersections of $\\ell$ with a line of the form $K_{1} K_{2}$, where $K_{i} \\in g(Q) \\cap p_{i}$. Subcase 2.2.3: $f\\left(R_{1} Q\\right)=S_{1}, f\\left(R_{2} Q\\right)=R_{2}$ (the case $f\\left(R_{1} Q\\right)=R_{1}, f\\left(R_{2} Q\\right)=S_{2}$ is similar). In this case, there are at most two possible positions for $R_{2}$ - namely, the meeting points of $g(Q)$ with $p_{2}$. Consider one of them. Let $X$ vary on $\\ell$. Then $R_{1}$ is the projection of $X$ to $p_{1}$ via $R_{2}, S_{1}$ is the projection of $R_{1}$ to $g(Q)$ via $Q$. Finally, $\\angle\\left(Q S_{1}, S_{1} X\\right)=\\angle\\left(R_{1} S_{1}, S_{1} X\\right)=$ $\\angle\\left(R_{1} P, P X\\right)=\\angle\\left(p_{1}, \\ell\\right) \\neq 0$, so $X$ is obtained by a fixed projective transform $g(Q) \\rightarrow \\ell$ from $S_{1}$. So, if there were three points $X$ satisfying the conditions of this subcase, the composition of the three projective transforms would be the identity. But, if we apply it to $X=Q$, we successively get some point $R_{1}^{\\prime}$, then $R_{2}$, and then some point different from $Q$, a contradiction. Thus Case 2.2 also occurs for only finitely many points $X$, as desired. Step 3: We show that $f(P Q)=P$, as desired. The argument is similar to that in Step 2, with the roles of $Q$ and $X$ swapped. Again, we show that there are only finitely many possible positions for a point $X \\in \\ell \\backslash\\{P, Q\\}$, which is absurd. Case 3.1: $t(Q)$ is parallel to one of the $p_{i}$; say, to $p_{1}$. Let $t(Q)$ cross $p_{2}$ at $R$; then $g(R)$ is the circle $(P R Q)$. Let $R X \\operatorname{cross} g(R)$ again at $S$. Then $f(R X) \\in\\{R, S\\} \\cap g(X)$, so $g(X)$ contains one of the points $R$ and $S$. ![](https://cdn.mathpix.com/cropped/2024_11_18_9ee8c2ac2c2f5002ff14g-084.jpg?height=729&width=909&top_left_y=1126&top_left_x=568) Subcase 3.1.1: $S=f(R X) \\in g(X)$. We have $\\angle(t(X), Q X)=\\angle(S X, S Q)=\\angle(S R, S Q)=\\angle(P R, P Q)=\\angle\\left(p_{2}, \\ell\\right)$. Hence $t(X) \\| p_{2}$. Now we recall Case 2.1: we let $t(X)$ cross $p_{1}$ at $R^{\\prime}$, so $g\\left(R^{\\prime}\\right)=\\left(P R^{\\prime} X\\right)$, and let $R^{\\prime} Q$ meet $g\\left(R^{\\prime}\\right)$ again at $S^{\\prime}$; notice that $S^{\\prime} \\neq Q$. Excluding one position of $X$, we may assume that $R^{\\prime} \\notin g(Q)$, so $R^{\\prime} \\neq f\\left(R^{\\prime} Q\\right)$. Therefore, $S^{\\prime}=f\\left(R^{\\prime} Q\\right) \\in g(Q)$. But then, as in Case 2.1, we get $\\angle(t(Q), P Q)=\\angle\\left(Q S^{\\prime}, S^{\\prime} P\\right)=\\angle\\left(R^{\\prime} X, X P\\right)=\\angle\\left(p_{2}, \\ell\\right)$. This means that $t(Q)$ is parallel to $p_{2}$, which is impossible. Subcase 3.1.2: $R=f(R X) \\in g(X)$. In this case, we have $\\angle(t(X), \\ell)=\\angle(R X, R Q)=\\angle\\left(R X, p_{1}\\right)$. Again, let $R^{\\prime}=t(X) \\cap p_{1}$; this point exists for all but at most one position of $X$. Then $g\\left(R^{\\prime}\\right)=\\left(R^{\\prime} X P\\right)$; let $R^{\\prime} Q$ meet $g\\left(R^{\\prime}\\right)$ again at $S^{\\prime}$. Due to $\\angle\\left(R^{\\prime} X, X R\\right)=\\angle(Q X, Q R)=\\angle\\left(\\ell, p_{1}\\right), R^{\\prime}$ determines $X$ in at most two ways, so for all but finitely many positions of $X$ we have $R^{\\prime} \\notin g(Q)$. Therefore, for those positions we have $S^{\\prime}=f\\left(R^{\\prime} Q\\right) \\in g(Q)$. But then $\\angle\\left(R X, p_{1}\\right)=\\angle\\left(R^{\\prime} X, X P\\right)=\\angle\\left(R^{\\prime} S^{\\prime}, S^{\\prime} P\\right)=$ $\\angle\\left(Q S^{\\prime}, S^{\\prime} P\\right)=\\angle(t(Q), Q P)$ is fixed, so this case can hold only for one specific position of $X$ as well. Thus, in Case 3.1, there are only finitely many possible positions of $X$, yielding a contradiction. Case 3.2: $t(Q)$ crosses $p_{1}$ and $p_{2}$ at $R_{1}$ and $R_{2}$, respectively. By Step $2, R_{1} \\neq R_{2}$. Notice that $g\\left(R_{i}\\right)$ is the circle $\\left(P Q R_{i}\\right)$. Let $R_{i} X$ meet $g\\left(R_{i}\\right)$ at $S_{i}$; then $S_{i} \\neq X$. Then $f\\left(R_{i} X\\right) \\in\\left\\{R_{i}, S_{i}\\right\\}$, and we distinguish several subcases. ![](https://cdn.mathpix.com/cropped/2024_11_18_9ee8c2ac2c2f5002ff14g-085.jpg?height=854&width=866&top_left_y=424&top_left_x=589) Subcase 3.2.1: $f\\left(R_{1} X\\right)=S_{1}$ and $f\\left(R_{2} X\\right)=S_{2}$, so $S_{1}, S_{2} \\in g(X)$. As in Subcase 2.2.1, we have $0=\\angle\\left(R_{1} Q, Q P\\right)+\\angle\\left(Q P, R_{2} Q\\right)=\\angle\\left(X S_{1}, S_{1} P\\right)+\\angle\\left(S_{2} P, S_{2} X\\right)$, which shows $P \\in g(X)$. But $X, Q \\in g(X)$ as well, so $g(X)$ meets $\\ell$ at three distinct points, which is absurd. Subcase 3.2.2: $f\\left(R_{1} X\\right)=R_{1}, f\\left(R_{2} X\\right)=R_{2}$, so $R_{1}, R_{2} \\in g(X)$. Now three distinct collinear points $R_{1}, R_{2}$, and $Q$ belong to $g(X)$, which is impossible. Subcase 3.2.3: $f\\left(R_{1} X\\right)=S_{1}, f\\left(R_{2} X\\right)=R_{2}$ (the case $f\\left(R_{1} X\\right)=R_{1}, f\\left(R_{2} X\\right)=S_{2}$ is similar). We have $\\angle\\left(X R_{2}, R_{2} Q\\right)=\\angle\\left(X S_{1}, S_{1} Q\\right)=\\angle\\left(R_{1} S_{1}, S_{1} Q\\right)=\\angle\\left(R_{1} P, P Q\\right)=\\angle\\left(p_{1}, \\ell\\right)$, so this case can occur for a unique position of $X$. Thus, in Case 3.2 , there is only a unique position of $X$, again yielding the required contradiction.", "metadata": {"resource_path": "IMO/segmented/en-IMO2019SL.jsonl"}} -{"year": "2019", "tier": "T0", "problem_label": "G8", "problem_type": "Geometry", "exam": "IMO", "problem": "Let $\\mathcal{L}$ be the set of all lines in the plane and let $f$ be a function that assigns to each line $\\ell \\in \\mathcal{L}$ a point $f(\\ell)$ on $\\ell$. Suppose that for any point $X$, and for any three lines $\\ell_{1}, \\ell_{2}, \\ell_{3}$ passing through $X$, the points $f\\left(\\ell_{1}\\right), f\\left(\\ell_{2}\\right), f\\left(\\ell_{3}\\right)$ and $X$ lie on a circle. Prove that there is a unique point $P$ such that $f(\\ell)=P$ for any line $\\ell$ passing through $P$. (Australia) Common remarks. The condition on $f$ is equivalent to the following: There is some function $g$ that assigns to each point $X$ a circle $g(X)$ passing through $X$ such that for any line $\\ell$ passing through $X$, the point $f(\\ell)$ lies on $g(X)$. (The function $g$ may not be uniquely defined for all points, if some points $X$ have at most one value of $f(\\ell)$ other than $X$; for such points, an arbitrary choice is made.) If there were two points $P$ and $Q$ with the given property, $f(P Q)$ would have to be both $P$ and $Q$, so there is at most one such point, and it will suffice to show that such a point exists.", "solution": "We will get an upper bound on $n$ from the speed at which $v_{2}\\left(L_{n}\\right)$ grows. From $$ L_{n}=\\left(2^{n}-1\\right)\\left(2^{n}-2\\right) \\cdots\\left(2^{n}-2^{n-1}\\right)=2^{1+2+\\cdots+(n-1)}\\left(2^{n}-1\\right)\\left(2^{n-1}-1\\right) \\cdots\\left(2^{1}-1\\right) $$ we read $$ v_{2}\\left(L_{n}\\right)=1+2+\\cdots+(n-1)=\\frac{n(n-1)}{2} . $$ On the other hand, $v_{2}(m!)$ is expressed by the Legendre formula as $$ v_{2}(m!)=\\sum_{i=1}^{\\infty}\\left\\lfloor\\frac{m}{2^{i}}\\right\\rfloor . $$ As usual, by omitting the floor functions, $$ v_{2}(m!)<\\sum_{i=1}^{\\infty} \\frac{m}{2^{i}}=m $$ Thus, $L_{n}=m$ ! implies the inequality $$ \\frac{n(n-1)}{2}1.3 \\cdot 10^{12}$. For $n \\geqslant 7$ we prove (3) by the following inequalities: $$ \\begin{aligned} \\left(\\frac{n(n-1)}{2}\\right)! & =15!\\cdot 16 \\cdot 17 \\cdots \\frac{n(n-1)}{2}>2^{36} \\cdot 16^{\\frac{n(n-1)}{2}-15} \\\\ & =2^{2 n(n-1)-24}=2^{n^{2}} \\cdot 2^{n(n-2)-24}>2^{n^{2}} . \\end{aligned} $$ Putting together (2) and (3), for $n \\geqslant 6$ we get a contradiction, since $$ L_{n}<2^{n^{2}}<\\left(\\frac{n(n-1)}{2}\\right)!a^{3} . $$ So $3 a^{3} \\geqslant(a b c)^{2}>a^{3}$ and hence $3 a \\geqslant b^{2} c^{2}>a$. Now $b^{3}+c^{3}=a^{2}\\left(b^{2} c^{2}-a\\right) \\geqslant a^{2}$, and so $$ 18 b^{3} \\geqslant 9\\left(b^{3}+c^{3}\\right) \\geqslant 9 a^{2} \\geqslant b^{4} c^{4} \\geqslant b^{3} c^{5} $$ so $18 \\geqslant c^{5}$ which yields $c=1$. Now, note that we must have $a>b$, as otherwise we would have $2 b^{3}+1=b^{4}$ which has no positive integer solutions. So $$ a^{3}-b^{3} \\geqslant(b+1)^{3}-b^{3}>1 $$ and $$ 2 a^{3}>1+a^{3}+b^{3}>a^{3} $$ which implies $2 a^{3}>a^{2} b^{2}>a^{3}$ and so $2 a>b^{2}>a$. Therefore $$ 4\\left(1+b^{3}\\right)=4 a^{2}\\left(b^{2}-a\\right) \\geqslant 4 a^{2}>b^{4} $$ so $4>b^{3}(b-4)$; that is, $b \\leqslant 4$. Now, for each possible value of $b$ with $2 \\leqslant b \\leqslant 4$ we obtain a cubic equation for $a$ with constant coefficients. These are as follows: $$ \\begin{array}{rlrl} b=2: & & & a^{3}-4 a^{2}+9=0 \\\\ b=3: & & a^{3}-9 a^{2}+28=0 \\\\ b=4: & & a^{3}-16 a^{2}+65=0 . \\end{array} $$ The only case with an integer solution for $a$ with $b \\leqslant a$ is $b=2$, leading to $(a, b, c)=(3,2,1)$. Comment 1.1. Instead of writing down each cubic equation explicitly, we could have just observed that $a^{2} \\mid b^{3}+1$, and for each choice of $b$ checked each square factor of $b^{3}+1$ for $a^{2}$. We could also have observed that, with $c=1$, the relation $18 b^{3} \\geqslant b^{4} c^{4}$ becomes $b \\leqslant 18$, and we can simply check all possibilities for $b$ (instead of working to prove that $b \\leqslant 4$ ). This check becomes easier after using the factorisation $b^{3}+1=(b+1)\\left(b^{2}-b+1\\right)$ and observing that no prime besides 3 can divide both of the factors. Comment 1.2. Another approach to finish the problem after establishing that $c \\leqslant 1$ is to set $k=b^{2} c^{2}-a$, which is clearly an integer and must be positive as it is equal to $\\left(b^{3}+c^{3}\\right) / a^{2}$. Then we divide into cases based on whether $k=1$ or $k \\geqslant 2$; in the first case, we have $b^{3}+1=a^{2}=\\left(b^{2}-1\\right)^{2}$ whose only positive root is $b=2$, and in the second case we have $b^{2} \\leqslant 3 a$, and so $$ b^{4} \\leqslant(3 a)^{2} \\leqslant \\frac{9}{2}\\left(k a^{2}\\right)=\\frac{9}{2}\\left(b^{3}+1\\right), $$ which implies that $b \\leqslant 4$.", "metadata": {"resource_path": "IMO/segmented/en-IMO2019SL.jsonl"}} -{"year": "2019", "tier": "T0", "problem_label": "N2", "problem_type": "Number Theory", "exam": "IMO", "problem": "Find all triples $(a, b, c)$ of positive integers such that $a^{3}+b^{3}+c^{3}=(a b c)^{2}$. (Nigeria) Answer: The solutions are $(1,2,3)$ and its permutations. Common remarks. Note that the equation is symmetric. In all solutions, we will assume without loss of generality that $a \\geqslant b \\geqslant c$, and prove that the only solution is $(a, b, c)=(3,2,1)$. The first two solutions all start by proving that $c=1$.", "solution": "Again, we will start by proving that $c=1$. Suppose otherwise that $c \\geqslant 2$. We have $a^{3}+b^{3}+c^{3} \\leqslant 3 a^{3}$, so $b^{2} c^{2} \\leqslant 3 a$. Since $c \\geqslant 2$, this tells us that $b \\leqslant \\sqrt{3 a / 4}$. As the right-hand side of the original equation is a multiple of $a^{2}$, we have $a^{2} \\leqslant 2 b^{3} \\leqslant 2(3 a / 4)^{3 / 2}$. In other words, $a \\leqslant \\frac{27}{16}<2$, which contradicts the assertion that $a \\geqslant c \\geqslant 2$. So there are no solutions in this case, and so we must have $c=1$. Now, the original equation becomes $a^{3}+b^{3}+1=a^{2} b^{2}$. Observe that $a \\geqslant 2$, since otherwise $a=b=1$ as $a \\geqslant b$. The right-hand side is a multiple of $a^{2}$, so the left-hand side must be as well. Thus, $b^{3}+1 \\geqslant$ $a^{2}$. Since $a \\geqslant b$, we also have $$ b^{2}=a+\\frac{b^{3}+1}{a^{2}} \\leqslant 2 a+\\frac{1}{a^{2}} $$ and so $b^{2} \\leqslant 2 a$ since $b^{2}$ is an integer. Thus $(2 a)^{3 / 2}+1 \\geqslant b^{3}+1 \\geqslant a^{2}$, from which we deduce $a \\leqslant 8$. Now, for each possible value of $a$ with $2 \\leqslant a \\leqslant 8$ we obtain a cubic equation for $b$ with constant coefficients. These are as follows: $$ \\begin{array}{ll} a=2: & \\\\ b^{3}-4 b^{2}+9=0 \\\\ a=3: & \\\\ b^{3}-9 b^{2}+28=0 \\\\ a=4: & \\\\ a=5: 16 b^{2}+65=0 \\\\ a=6: & \\\\ b^{3}-25 b^{2}+126=0 \\\\ a=7: & \\\\ b^{3}-36 b^{2}+217=0 \\\\ a=8: & \\\\ b^{3}-49 b^{2}+344=0 \\\\ b^{3}-64 b^{2}+513=0 . \\end{array} $$ The only case with an integer solution for $b$ with $a \\geqslant b$ is $a=3$, leading to $(a, b, c)=(3,2,1)$. Comment 2.1. As in Solution 1, instead of writing down each cubic equation explicitly, we could have just observed that $b^{2} \\mid a^{3}+1$, and for each choice of $a$ checked each square factor of $a^{3}+1$ for $b^{2}$. Comment 2.2. This solution does not require initially proving that $c=1$, in which case the bound would become $a \\leqslant 108$. The resulting cases could, in principle, be checked by a particularly industrious student.", "metadata": {"resource_path": "IMO/segmented/en-IMO2019SL.jsonl"}} -{"year": "2019", "tier": "T0", "problem_label": "N2", "problem_type": "Number Theory", "exam": "IMO", "problem": "Find all triples $(a, b, c)$ of positive integers such that $a^{3}+b^{3}+c^{3}=(a b c)^{2}$. (Nigeria) Answer: The solutions are $(1,2,3)$ and its permutations. Common remarks. Note that the equation is symmetric. In all solutions, we will assume without loss of generality that $a \\geqslant b \\geqslant c$, and prove that the only solution is $(a, b, c)=(3,2,1)$. The first two solutions all start by proving that $c=1$.", "solution": "Set $k=\\left(b^{3}+c^{3}\\right) / a^{2} \\leqslant 2 a$, and rewrite the original equation as $a+k=(b c)^{2}$. Since $b^{3}$ and $c^{3}$ are positive integers, we have $(b c)^{3} \\geqslant b^{3}+c^{3}-1=k a^{2}-1$, so $$ a+k \\geqslant\\left(k a^{2}-1\\right)^{2 / 3} $$ As in Comment 1.2, $k$ is a positive integer; for each value of $k \\geqslant 1$, this gives us a polynomial inequality satisfied by $a$ : $$ k^{2} a^{4}-a^{3}-5 k a^{2}-3 k^{2} a-\\left(k^{3}-1\\right) \\leqslant 0 . $$ We now prove that $a \\leqslant 3$. Indeed, $$ 0 \\geqslant \\frac{k^{2} a^{4}-a^{3}-5 k a^{2}-3 k^{2} a-\\left(k^{3}-1\\right)}{k^{2}} \\geqslant a^{4}-a^{3}-5 a^{2}-3 a-k \\geqslant a^{4}-a^{3}-5 a^{2}-5 a $$ which fails when $a \\geqslant 4$. This leaves ten triples with $3 \\geqslant a \\geqslant b \\geqslant c \\geqslant 1$, which may be checked manually to give $(a, b, c)=(3,2,1)$.", "metadata": {"resource_path": "IMO/segmented/en-IMO2019SL.jsonl"}} -{"year": "2019", "tier": "T0", "problem_label": "N2", "problem_type": "Number Theory", "exam": "IMO", "problem": "Find all triples $(a, b, c)$ of positive integers such that $a^{3}+b^{3}+c^{3}=(a b c)^{2}$. (Nigeria) Answer: The solutions are $(1,2,3)$ and its permutations. Common remarks. Note that the equation is symmetric. In all solutions, we will assume without loss of generality that $a \\geqslant b \\geqslant c$, and prove that the only solution is $(a, b, c)=(3,2,1)$. The first two solutions all start by proving that $c=1$.", "solution": "Again, observe that $b^{3}+c^{3}=a^{2}\\left(b^{2} c^{2}-a\\right)$, so $b \\leqslant a \\leqslant b^{2} c^{2}-1$. We consider the function $f(x)=x^{2}\\left(b^{2} c^{2}-x\\right)$. It can be seen that that on the interval $\\left[0, b^{2} c^{2}-1\\right]$ the function $f$ is increasing if $x<\\frac{2}{3} b^{2} c^{2}$ and decreasing if $x>\\frac{2}{3} b^{2} c^{2}$. Consequently, it must be the case that $$ b^{3}+c^{3}=f(a) \\geqslant \\min \\left(f(b), f\\left(b^{2} c^{2}-1\\right)\\right) $$ First, suppose that $b^{3}+c^{3} \\geqslant f\\left(b^{2} c^{2}-1\\right)$. This may be written $b^{3}+c^{3} \\geqslant\\left(b^{2} c^{2}-1\\right)^{2}$, and so $$ 2 b^{3} \\geqslant b^{3}+c^{3} \\geqslant\\left(b^{2} c^{2}-1\\right)^{2}>b^{4} c^{4}-2 b^{2} c^{2} \\geqslant b^{4} c^{4}-2 b^{3} c^{4} $$ Thus, $(b-2) c^{4}<2$, and the only solutions to this inequality have $(b, c)=(2,2)$ or $b \\leqslant 3$ and $c=1$. It is easy to verify that the only case giving a solution for $a \\geqslant b$ is $(a, b, c)=(3,2,1)$. Otherwise, suppose that $b^{3}+c^{3}=f(a) \\geqslant f(b)$. Then, we have $$ 2 b^{3} \\geqslant b^{3}+c^{3}=a^{2}\\left(b^{2} c^{2}-a\\right) \\geqslant b^{2}\\left(b^{2} c^{2}-b\\right) . $$ Consequently $b c^{2} \\leqslant 3$, with strict inequality in the case that $b \\neq c$. Hence $c=1$ and $b \\leqslant 2$. Both of these cases have been considered already, so we are done. Comment 4.1. Instead of considering which of $f(b)$ and $f\\left(b^{2} c^{2}-1\\right)$ is less than $f(a)$, we may also proceed by explicitly dividing into cases based on whether $a \\geqslant \\frac{2}{3} b^{2} c^{2}$ or $a<\\frac{2}{3} b^{2} c^{2}$. The first case may now be dealt with as follows. We have $b^{3} c^{3}+1 \\geqslant b^{3}+c^{3}$ as $b^{3}$ and $c^{3}$ are positive integers, so we have $$ b^{3} c^{3}+1 \\geqslant b^{3}+c^{3} \\geqslant a^{2} \\geqslant \\frac{4}{9} b^{4} c^{4} $$ This implies $b c \\leqslant 2$, and hence $c=1$ and $b \\leqslant 2$.", "metadata": {"resource_path": "IMO/segmented/en-IMO2019SL.jsonl"}} -{"year": "2019", "tier": "T0", "problem_label": "N3", "problem_type": "Number Theory", "exam": "IMO", "problem": "We say that a set $S$ of integers is rootiful if, for any positive integer $n$ and any $a_{0}, a_{1}, \\ldots, a_{n} \\in S$, all integer roots of the polynomial $a_{0}+a_{1} x+\\cdots+a_{n} x^{n}$ are also in $S$. Find all rootiful sets of integers that contain all numbers of the form $2^{a}-2^{b}$ for positive integers $a$ and $b$. (Czech Republic) Answer: The set $\\mathbb{Z}$ of all integers is the only such rootiful set.", "solution": "The set $\\mathbb{Z}$ of all integers is clearly rootiful. We shall prove that any rootiful set $S$ containing all the numbers of the form $2^{a}-2^{b}$ for $a, b \\in \\mathbb{Z}_{>0}$ must be all of $\\mathbb{Z}$. First, note that $0=2^{1}-2^{1} \\in S$ and $2=2^{2}-2^{1} \\in S$. Now, $-1 \\in S$, since it is a root of $2 x+2$, and $1 \\in S$, since it is a root of $2 x^{2}-x-1$. Also, if $n \\in S$ then $-n$ is a root of $x+n$, so it suffices to prove that all positive integers must be in $S$. Now, we claim that any positive integer $n$ has a multiple in $S$. Indeed, suppose that $n=2^{\\alpha} \\cdot t$ for $\\alpha \\in \\mathbb{Z}_{\\geqslant 0}$ and $t$ odd. Then $t \\mid 2^{\\phi(t)}-1$, so $n \\mid 2^{\\alpha+\\phi(t)+1}-2^{\\alpha+1}$. Moreover, $2^{\\alpha+\\phi(t)+1}-2^{\\alpha+1} \\in S$, and so $S$ contains a multiple of every positive integer $n$. We will now prove by induction that all positive integers are in $S$. Suppose that $0,1, \\ldots, n-$ $1 \\in S$; furthermore, let $N$ be a multiple of $n$ in $S$. Consider the base- $n$ expansion of $N$, say $N=a_{k} n^{k}+a_{k-1} n^{k-1}+\\cdots+a_{1} n+a_{0}$. Since $0 \\leqslant a_{i}0}$. We show that, in fact, every integer $k$ with $|k|>2$ can be expressed as a root of a polynomial whose coefficients are of the form $2^{a}-2^{b}$. Observe that it suffices to consider the case where $k$ is positive, as if $k$ is a root of $a_{n} x^{n}+\\cdots+a_{1} x+a_{0}=0$, then $-k$ is a root of $(-1)^{n} a_{n} x^{n}+\\cdots-$ $a_{1} x+a_{0}=0$. Note that $$ \\left(2^{a_{n}}-2^{b_{n}}\\right) k^{n}+\\cdots+\\left(2^{a_{0}}-2^{b_{0}}\\right)=0 $$ is equivalent to $$ 2^{a_{n}} k^{n}+\\cdots+2^{a_{0}}=2^{b_{n}} k^{n}+\\cdots+2^{b_{0}} $$ Hence our aim is to show that two numbers of the form $2^{a_{n}} k^{n}+\\cdots+2^{a_{0}}$ are equal, for a fixed value of $n$. We consider such polynomials where every term $2^{a_{i}} k^{i}$ is at most $2 k^{n}$; in other words, where $2 \\leqslant 2^{a_{i}} \\leqslant 2 k^{n-i}$, or, equivalently, $1 \\leqslant a_{i} \\leqslant 1+(n-i) \\log _{2} k$. Therefore, there must be $1+\\left\\lfloor(n-i) \\log _{2} k\\right\\rfloor$ possible choices for $a_{i}$ satisfying these constraints. The number of possible polynomials is then $$ \\prod_{i=0}^{n}\\left(1+\\left\\lfloor(n-i) \\log _{2} k\\right\\rfloor\\right) \\geqslant \\prod_{i=0}^{n-1}(n-i) \\log _{2} k=n!\\left(\\log _{2} k\\right)^{n} $$ where the inequality holds as $1+\\lfloor x\\rfloor \\geqslant x$. As there are $(n+1)$ such terms in the polynomial, each at most $2 k^{n}$, such a polynomial must have value at most $2 k^{n}(n+1)$. However, for large $n$, we have $n!\\left(\\log _{2} k\\right)^{n}>2 k^{n}(n+1)$. Therefore there are more polynomials than possible values, so some two must be equal, as required.", "metadata": {"resource_path": "IMO/segmented/en-IMO2019SL.jsonl"}} -{"year": "2019", "tier": "T0", "problem_label": "N4", "problem_type": "Number Theory", "exam": "IMO", "problem": "Let $\\mathbb{Z}_{>0}$ be the set of positive integers. A positive integer constant $C$ is given. Find all functions $f: \\mathbb{Z}_{>0} \\rightarrow \\mathbb{Z}_{>0}$ such that, for all positive integers $a$ and $b$ satisfying $a+b>C$, $$ a+f(b) \\mid a^{2}+b f(a) $$ (Croatia) Answer: The functions satisfying (*) are exactly the functions $f(a)=k a$ for some constant $k \\in \\mathbb{Z}_{>0}$ (irrespective of the value of $C$ ). Common remarks. It is easy to verify that the functions $f(a)=k a$ satisfy (*). Thus, in the proofs below, we will only focus on the converse implication: that condition (*) implies that $f=k a$. A common minor part of these solutions is the derivation of some relatively easy bounds on the function $f$. An upper bound is easily obtained by setting $a=1$ in (*), giving the inequality $$ f(b) \\leqslant b \\cdot f(1) $$ for all sufficiently large $b$. The corresponding lower bound is only marginally more difficult to obtain: substituting $b=1$ in the original equation shows that $$ a+f(1) \\mid\\left(a^{2}+f(a)\\right)-(a-f(1)) \\cdot(a+f(1))=f(1)^{2}+f(a) $$ for all sufficiently large $a$. It follows from this that one has the lower bound $$ f(a) \\geqslant a+f(1) \\cdot(1-f(1)) $$ again for all sufficiently large $a$. Each of the following proofs makes use of at least one of these bounds.", "solution": "First, we show that $b \\mid f(b)^{2}$ for all $b$. To do this, we choose a large positive integer $n$ so that $n b-f(b) \\geqslant C$. Setting $a=n b-f(b)$ in (*) then shows that $$ n b \\mid(n b-f(b))^{2}+b f(n b-f(b)) $$ so that $b \\mid f(b)^{2}$ as claimed. Now in particular we have that $p \\mid f(p)$ for every prime $p$. If we write $f(p)=k(p) \\cdot p$, then the bound $f(p) \\leqslant f(1) \\cdot p$ (valid for $p$ sufficiently large) shows that some value $k$ of $k(p)$ must be attained for infinitely many $p$. We will show that $f(a)=k a$ for all positive integers $a$. To do this, we substitute $b=p$ in (*), where $p$ is any sufficiently large prime for which $k(p)=k$, obtaining $$ a+k p \\mid\\left(a^{2}+p f(a)\\right)-a(a+k p)=p f(a)-p k a . $$ For suitably large $p$ we have $\\operatorname{gcd}(a+k p, p)=1$, and hence we have $$ a+k p \\mid f(a)-k a $$ But the only way this can hold for arbitrarily large $p$ is if $f(a)-k a=0$. This concludes the proof. Comment. There are other ways to obtain the divisibility $p \\mid f(p)$ for primes $p$, which is all that is needed in this proof. For instance, if $f(p)$ were not divisible by $p$ then the arithmetic progression $p^{2}+b f(p)$ would attain prime values for infinitely many $b$ by Dirichlet's Theorem: hence, for these pairs p , b, we would have $p+f(b)=p^{2}+b f(p)$. Substituting $a \\mapsto b$ and $b \\mapsto p$ in $(*)$ then shows that $\\left(f(p)^{2}-p^{2}\\right)(p-1)$ is divisible by $b+f(p)$ and hence vanishes, which is impossible since $p \\nmid f(p)$ by assumption.", "metadata": {"resource_path": "IMO/segmented/en-IMO2019SL.jsonl"}} -{"year": "2019", "tier": "T0", "problem_label": "N4", "problem_type": "Number Theory", "exam": "IMO", "problem": "Let $\\mathbb{Z}_{>0}$ be the set of positive integers. A positive integer constant $C$ is given. Find all functions $f: \\mathbb{Z}_{>0} \\rightarrow \\mathbb{Z}_{>0}$ such that, for all positive integers $a$ and $b$ satisfying $a+b>C$, $$ a+f(b) \\mid a^{2}+b f(a) $$ (Croatia) Answer: The functions satisfying (*) are exactly the functions $f(a)=k a$ for some constant $k \\in \\mathbb{Z}_{>0}$ (irrespective of the value of $C$ ). Common remarks. It is easy to verify that the functions $f(a)=k a$ satisfy (*). Thus, in the proofs below, we will only focus on the converse implication: that condition (*) implies that $f=k a$. A common minor part of these solutions is the derivation of some relatively easy bounds on the function $f$. An upper bound is easily obtained by setting $a=1$ in (*), giving the inequality $$ f(b) \\leqslant b \\cdot f(1) $$ for all sufficiently large $b$. The corresponding lower bound is only marginally more difficult to obtain: substituting $b=1$ in the original equation shows that $$ a+f(1) \\mid\\left(a^{2}+f(a)\\right)-(a-f(1)) \\cdot(a+f(1))=f(1)^{2}+f(a) $$ for all sufficiently large $a$. It follows from this that one has the lower bound $$ f(a) \\geqslant a+f(1) \\cdot(1-f(1)) $$ again for all sufficiently large $a$. Each of the following proofs makes use of at least one of these bounds.", "solution": "First, we substitute $b=1$ in (*) and rearrange to find that $$ \\frac{f(a)+f(1)^{2}}{a+f(1)}=f(1)-a+\\frac{a^{2}+f(a)}{a+f(1)} $$ is a positive integer for sufficiently large $a$. Since $f(a) \\leqslant a f(1)$, for all sufficiently large $a$, it follows that $\\frac{f(a)+f(1)^{2}}{a+f(1)} \\leqslant f(1)$ also and hence there is a positive integer $k$ such that $\\frac{f(a)+f(1)^{2}}{a+f(1)}=k$ for infinitely many values of $a$. In other words, $$ f(a)=k a+f(1) \\cdot(k-f(1)) $$ for infinitely many $a$. Fixing an arbitrary choice of $a$ in (*), we have that $$ \\frac{a^{2}+b f(a)}{a+k b+f(1) \\cdot(k-f(1))} $$ is an integer for infinitely many $b$ (the same $b$ as above, maybe with finitely many exceptions). On the other hand, for $b$ taken sufficiently large, this quantity becomes arbitrarily close to $\\frac{f(a)}{k}$; this is only possible if $\\frac{f(a)}{k}$ is an integer and $$ \\frac{a^{2}+b f(a)}{a+k b+f(1) \\cdot(k-f(1))}=\\frac{f(a)}{k} $$ for infinitely many $b$. This rearranges to $$ \\frac{f(a)}{k} \\cdot(a+f(1) \\cdot(k-f(1)))=a^{2} $$ Hence $a^{2}$ is divisible by $a+f(1) \\cdot(k-f(1))$, and hence so is $f(1)^{2}(k-f(1))^{2}$. The only way this can occur for all $a$ is if $k=f(1)$, in which case $(* *)$ provides that $f(a)=k a$ for all $a$, as desired.", "metadata": {"resource_path": "IMO/segmented/en-IMO2019SL.jsonl"}} -{"year": "2019", "tier": "T0", "problem_label": "N4", "problem_type": "Number Theory", "exam": "IMO", "problem": "Let $\\mathbb{Z}_{>0}$ be the set of positive integers. A positive integer constant $C$ is given. Find all functions $f: \\mathbb{Z}_{>0} \\rightarrow \\mathbb{Z}_{>0}$ such that, for all positive integers $a$ and $b$ satisfying $a+b>C$, $$ a+f(b) \\mid a^{2}+b f(a) $$ (Croatia) Answer: The functions satisfying (*) are exactly the functions $f(a)=k a$ for some constant $k \\in \\mathbb{Z}_{>0}$ (irrespective of the value of $C$ ). Common remarks. It is easy to verify that the functions $f(a)=k a$ satisfy (*). Thus, in the proofs below, we will only focus on the converse implication: that condition (*) implies that $f=k a$. A common minor part of these solutions is the derivation of some relatively easy bounds on the function $f$. An upper bound is easily obtained by setting $a=1$ in (*), giving the inequality $$ f(b) \\leqslant b \\cdot f(1) $$ for all sufficiently large $b$. The corresponding lower bound is only marginally more difficult to obtain: substituting $b=1$ in the original equation shows that $$ a+f(1) \\mid\\left(a^{2}+f(a)\\right)-(a-f(1)) \\cdot(a+f(1))=f(1)^{2}+f(a) $$ for all sufficiently large $a$. It follows from this that one has the lower bound $$ f(a) \\geqslant a+f(1) \\cdot(1-f(1)) $$ again for all sufficiently large $a$. Each of the following proofs makes use of at least one of these bounds.", "solution": "Fix any two distinct positive integers $a$ and $b$. From (*) it follows that the two integers $$ \\left(a^{2}+c f(a)\\right) \\cdot(b+f(c)) \\text { and }\\left(b^{2}+c f(b)\\right) \\cdot(a+f(c)) $$ are both multiples of $(a+f(c)) \\cdot(b+f(c))$ for all sufficiently large $c$. Taking an appropriate linear combination to eliminate the $c f(c)$ term, we find after expanding out that the integer $$ \\left[a^{2} f(b)-b^{2} f(a)\\right] \\cdot f(c)+[(b-a) f(a) f(b)] \\cdot c+[a b(a f(b)-b f(a))] $$ is also a multiple of $(a+f(c)) \\cdot(b+f(c))$. But as $c$ varies, $(\\dagger)$ is bounded above by a positive multiple of $c$ while $(a+f(c)) \\cdot(b+f(c))$ is bounded below by a positive multiple of $c^{2}$. The only way that such a divisibility can hold is if in fact $$ \\left[a^{2} f(b)-b^{2} f(a)\\right] \\cdot f(c)+[(b-a) f(a) f(b)] \\cdot c+[a b(a f(b)-b f(a))]=0 $$ for sufficiently large $c$. Since the coefficient of $c$ in this linear relation is nonzero, it follows that there are constants $k, \\ell$ such that $f(c)=k c+\\ell$ for all sufficiently large $c$; the constants $k$ and $\\ell$ are necessarily integers. The value of $\\ell$ satisfies $$ \\left[a^{2} f(b)-b^{2} f(a)\\right] \\cdot \\ell+[a b(a f(b)-b f(a))]=0 $$ and hence $b \\mid \\ell a^{2} f(b)$ for all $a$ and $b$. Taking $b$ sufficiently large so that $f(b)=k b+\\ell$, we thus have that $b \\mid \\ell^{2} a^{2}$ for all sufficiently large $b$; this implies that $\\ell=0$. From ( $\\dagger \\dagger \\dagger$ ) it then follows that $\\frac{f(a)}{a}=\\frac{f(b)}{b}$ for all $a \\neq b$, so that there is a constant $k$ such that $f(a)=k a$ for all $a$ ( $k$ is equal to the constant defined earlier).", "metadata": {"resource_path": "IMO/segmented/en-IMO2019SL.jsonl"}} -{"year": "2019", "tier": "T0", "problem_label": "N4", "problem_type": "Number Theory", "exam": "IMO", "problem": "Let $\\mathbb{Z}_{>0}$ be the set of positive integers. A positive integer constant $C$ is given. Find all functions $f: \\mathbb{Z}_{>0} \\rightarrow \\mathbb{Z}_{>0}$ such that, for all positive integers $a$ and $b$ satisfying $a+b>C$, $$ a+f(b) \\mid a^{2}+b f(a) $$ (Croatia) Answer: The functions satisfying (*) are exactly the functions $f(a)=k a$ for some constant $k \\in \\mathbb{Z}_{>0}$ (irrespective of the value of $C$ ). Common remarks. It is easy to verify that the functions $f(a)=k a$ satisfy (*). Thus, in the proofs below, we will only focus on the converse implication: that condition (*) implies that $f=k a$. A common minor part of these solutions is the derivation of some relatively easy bounds on the function $f$. An upper bound is easily obtained by setting $a=1$ in (*), giving the inequality $$ f(b) \\leqslant b \\cdot f(1) $$ for all sufficiently large $b$. The corresponding lower bound is only marginally more difficult to obtain: substituting $b=1$ in the original equation shows that $$ a+f(1) \\mid\\left(a^{2}+f(a)\\right)-(a-f(1)) \\cdot(a+f(1))=f(1)^{2}+f(a) $$ for all sufficiently large $a$. It follows from this that one has the lower bound $$ f(a) \\geqslant a+f(1) \\cdot(1-f(1)) $$ again for all sufficiently large $a$. Each of the following proofs makes use of at least one of these bounds.", "solution": "Let $\\Gamma$ denote the set of all points $(a, f(a))$, so that $\\Gamma$ is an infinite subset of the upper-right quadrant of the plane. For a point $A=(a, f(a))$ in $\\Gamma$, we define a point $A^{\\prime}=\\left(-f(a),-f(a)^{2} / a\\right)$ in the lower-left quadrant of the plane, and let $\\Gamma^{\\prime}$ denote the set of all such points $A^{\\prime}$. ![](https://cdn.mathpix.com/cropped/2024_11_18_9ee8c2ac2c2f5002ff14g-095.jpg?height=552&width=652&top_left_y=449&top_left_x=702) Claim. For any point $A \\in \\Gamma$, the set $\\Gamma$ is contained in finitely many lines through the point $A^{\\prime}$. Proof. Let $A=(a, f(a))$. The functional equation (with $a$ and $b$ interchanged) can be rewritten as $b+f(a) \\mid a f(b)-b f(a)$, so that all but finitely many points in $\\Gamma$ are contained in one of the lines with equation $$ a y-f(a) x=m(x+f(a)) $$ for $m$ an integer. Geometrically, these are the lines through $A^{\\prime}=\\left(-f(a),-f(a)^{2} / a\\right)$ with gradient $\\frac{f(a)+m}{a}$. Since $\\Gamma$ is contained, with finitely many exceptions, in the region $0 \\leqslant y \\leqslant$ $f(1) \\cdot x$ and the point $A^{\\prime}$ lies strictly in the lower-left quadrant of the plane, there are only finitely many values of $m$ for which this line meets $\\Gamma$. This concludes the proof of the claim. Now consider any distinct points $A, B \\in \\Gamma$. It is clear that $A^{\\prime}$ and $B^{\\prime}$ are distinct. A line through $A^{\\prime}$ and a line through $B^{\\prime}$ only meet in more than one point if these two lines are equal to the line $A^{\\prime} B^{\\prime}$. It then follows from the above claim that the line $A^{\\prime} B^{\\prime}$ must contain all but finitely many points of $\\Gamma$. If $C$ is another point of $\\Gamma$, then the line $A^{\\prime} C^{\\prime}$ also passes through all but finitely many points of $\\Gamma$, which is only possible if $A^{\\prime} C^{\\prime}=A^{\\prime} B^{\\prime}$. We have thus seen that there is a line $\\ell$ passing through all points of $\\Gamma^{\\prime}$ and through all but finitely many points of $\\Gamma$. We claim that this line passes through the origin $O$ and passes through every point of $\\Gamma$. To see this, note that by construction $A, O, A^{\\prime}$ are collinear for every point $A \\in \\Gamma$. Since $\\ell=A A^{\\prime}$ for all but finitely many points $A \\in \\Gamma$, it thus follows that $O \\in \\ell$. Thus any $A \\in \\Gamma$ lies on the line $\\ell=A^{\\prime} O$. Since $\\Gamma$ is contained in a line through $O$, it follows that there is a real constant $k$ (the gradient of $\\ell$ ) such that $f(a)=k a$ for all $a$. The number $k$ is, of course, a positive integer. Comment. Without the $a+b>C$ condition, this problem is approachable by much more naive methods. For instance, using the given divisibility for $a, b \\in\\{1,2,3\\}$ one can prove by a somewhat tedious case-check that $f(2)=2 f(1)$ and $f(3)=3 f(1)$; this then forms the basis of an induction establishing that $f(n)=n f(1)$ for all $n$.", "metadata": {"resource_path": "IMO/segmented/en-IMO2019SL.jsonl"}} -{"year": "2019", "tier": "T0", "problem_label": "N6", "problem_type": "Number Theory", "exam": "IMO", "problem": "Let $H=\\left\\{\\lfloor i \\sqrt{2}\\rfloor: i \\in \\mathbb{Z}_{>0}\\right\\}=\\{1,2,4,5,7, \\ldots\\}$, and let $n$ be a positive integer. Prove that there exists a constant $C$ such that, if $A \\subset\\{1,2, \\ldots, n\\}$ satisfies $|A| \\geqslant C \\sqrt{n}$, then there exist $a, b \\in A$ such that $a-b \\in H$. (Brazil) Common remarks. In all solutions, we will assume that $A$ is a set such that $\\{a-b: a, b \\in A\\}$ is disjoint from $H$, and prove that $|A|1-\\frac{1}{\\sqrt{2}} . $$ To see why, observe that $n \\in H$ if and only if $00}$. In other words, $01$ since $a_{i}-a_{1} \\notin H$. Furthermore, we must have $\\left\\{a_{i} / \\sqrt{2}\\right\\}<\\left\\{a_{j} / \\sqrt{2}\\right\\}$ whenever $i1 / \\sqrt{2}>$ $1-1 / \\sqrt{2}$, contradicting (1). Now, we have a sequence $0=a_{1}\\frac{1}{2 d \\sqrt{2}} $$ To see why this is the case, let $h=\\lfloor d / \\sqrt{2}\\rfloor$, so $\\{d / \\sqrt{2}\\}=d / \\sqrt{2}-h$. Then $$ \\left\\{\\frac{d}{\\sqrt{2}}\\right\\}\\left(\\frac{d}{\\sqrt{2}}+h\\right)=\\frac{d^{2}-2 h^{2}}{2} \\geqslant \\frac{1}{2} $$ since the numerator is a positive integer. Because $d / \\sqrt{2}+h<2 d / \\sqrt{2}$, inequality (2) follows. Let $d_{i}=a_{i+1}-a_{i}$, for $1 \\leqslant i\\sum_{i}\\left\\{\\frac{d_{i}}{\\sqrt{2}}\\right\\}>\\frac{1}{2 \\sqrt{2}} \\sum_{i} \\frac{1}{d_{i}} \\geqslant \\frac{(k-1)^{2}}{2 \\sqrt{2}} \\frac{1}{\\sum_{i} d_{i}}>\\frac{(k-1)^{2}}{2 \\sqrt{2}} \\cdot \\frac{1}{n} . $$ Here, the first inequality holds because $\\left\\{a_{k} / \\sqrt{2}\\right\\}<1-1 / \\sqrt{2}$, the second follows from (2), the third follows from an easy application of the AM-HM inequality (or Cauchy-Schwarz), and the fourth follows from the fact that $\\sum_{i} d_{i}=a_{k}k-1 $$ which provides the required bound on $k$.", "metadata": {"resource_path": "IMO/segmented/en-IMO2019SL.jsonl"}} -{"year": "2019", "tier": "T0", "problem_label": "N6", "problem_type": "Number Theory", "exam": "IMO", "problem": "Let $H=\\left\\{\\lfloor i \\sqrt{2}\\rfloor: i \\in \\mathbb{Z}_{>0}\\right\\}=\\{1,2,4,5,7, \\ldots\\}$, and let $n$ be a positive integer. Prove that there exists a constant $C$ such that, if $A \\subset\\{1,2, \\ldots, n\\}$ satisfies $|A| \\geqslant C \\sqrt{n}$, then there exist $a, b \\in A$ such that $a-b \\in H$. (Brazil) Common remarks. In all solutions, we will assume that $A$ is a set such that $\\{a-b: a, b \\in A\\}$ is disjoint from $H$, and prove that $|A|0}\\right\\}$ is the complementary Beatty sequence to $H$ (in other words, $H$ and $J$ are disjoint with $H \\cup J=\\mathbb{Z}_{>0}$ ). Write $A=\\left\\{a_{1}0}$. For any $j>i$, we have $a_{j}-a_{i}=\\left\\lfloor\\alpha b_{j}\\right\\rfloor-\\left\\lfloor\\alpha b_{i}\\right\\rfloor$. Because $a_{j}-a_{i} \\in J$, we also have $a_{j}-a_{i}=\\lfloor\\alpha t\\rfloor$ for some positive integer $t$. Thus, $\\lfloor\\alpha t\\rfloor=\\left\\lfloor\\alpha b_{j}\\right\\rfloor-\\left\\lfloor\\alpha b_{i}\\right\\rfloor$. The right hand side must equal either $\\left\\lfloor\\alpha\\left(b_{j}-b_{i}\\right)\\right\\rfloor$ or $\\left\\lfloor\\alpha\\left(b_{j}-b_{i}\\right)\\right\\rfloor-1$, the latter of which is not a member of $J$ as $\\alpha>2$. Therefore, $t=b_{j}-b_{i}$ and so we have $\\left\\lfloor\\alpha b_{j}\\right\\rfloor-\\left\\lfloor\\alpha b_{i}\\right\\rfloor=\\left\\lfloor\\alpha\\left(b_{j}-b_{i}\\right)\\right\\rfloor$. For $1 \\leqslant i1 /(2 d \\sqrt{2})$ for positive integers $d)$ proves that $1>\\left((k-1)^{2} /(2 \\sqrt{2})\\right)(\\alpha / n)$, which again rearranges to give $$ \\sqrt{2 \\sqrt{2}-2} \\cdot \\sqrt{n}>k-1 $$ Comment. The use of Beatty sequences in Solution 2 is essentially a way to bypass (1). Both Solutions 1 and 2 use the fact that $\\sqrt{2}<2$; the statement in the question would still be true if $\\sqrt{2}$ did not have this property (for instance, if it were replaced with $\\alpha$ ), but any argument along the lines of Solutions 1 or 2 would be more complicated.", "metadata": {"resource_path": "IMO/segmented/en-IMO2019SL.jsonl"}} -{"year": "2019", "tier": "T0", "problem_label": "N6", "problem_type": "Number Theory", "exam": "IMO", "problem": "Let $H=\\left\\{\\lfloor i \\sqrt{2}\\rfloor: i \\in \\mathbb{Z}_{>0}\\right\\}=\\{1,2,4,5,7, \\ldots\\}$, and let $n$ be a positive integer. Prove that there exists a constant $C$ such that, if $A \\subset\\{1,2, \\ldots, n\\}$ satisfies $|A| \\geqslant C \\sqrt{n}$, then there exist $a, b \\in A$ such that $a-b \\in H$. (Brazil) Common remarks. In all solutions, we will assume that $A$ is a set such that $\\{a-b: a, b \\in A\\}$ is disjoint from $H$, and prove that $|A|0} \\backslash H$, so all differences between elements of $A$ are in $J$. We start by making the following observation. Suppose we have a set $B \\subseteq\\{1,2, \\ldots, n\\}$ such that all of the differences between elements of $B$ are in $H$. Then $|A| \\cdot|B| \\leqslant 2 n$. To see why, observe that any two sums of the form $a+b$ with $a \\in A, b \\in B$ are different; otherwise, we would have $a_{1}+b_{1}=a_{2}+b_{2}$, and so $\\left|a_{1}-a_{2}\\right|=\\left|b_{2}-b_{1}\\right|$. However, then the left hand side is in $J$ whereas the right hand side is in $H$. Thus, $\\{a+b: a \\in A, b \\in B\\}$ is a set of size $|A| \\cdot|B|$ all of whose elements are no greater than $2 n$, yielding the claimed inequality. With this in mind, it suffices to construct a set $B$, all of whose differences are in $H$ and whose size is at least $C^{\\prime} \\sqrt{n}$ for some constant $C^{\\prime}>0$. To do so, we will use well-known facts about the negative Pell equation $X^{2}-2 Y^{2}=-1$; in particular, that there are infinitely many solutions and the values of $X$ are given by the recurrence $X_{1}=1, X_{2}=7$ and $X_{m}=6 X_{m-1}-X_{m-2}$. Therefore, we may choose $X$ to be a solution with $\\sqrt{n} / 6\\left(\\frac{X}{\\sqrt{2}}-Y\\right) \\geqslant \\frac{-3}{\\sqrt{2 n}}, $$ from which it follows that $\\{X / \\sqrt{2}\\}>1-(3 / \\sqrt{2 n})$. Combined with (1), this shows that all differences between elements of $B$ are in $H$. Comment. Some of the ideas behind Solution 3 may be used to prove that the constant $C=\\sqrt{2 \\sqrt{2}-2}$ (from Solutions 1 and 2) is optimal, in the sense that there are arbitrarily large values of $n$ and sets $A_{n} \\subseteq\\{1,2, \\ldots, n\\}$ of size roughly $C \\sqrt{n}$, all of whose differences are contained in $J$. To see why, choose $X$ to come from a sufficiently large solution to the Pell equation $X^{2}-2 Y^{2}=1$, so $\\{X / \\sqrt{2}\\} \\approx 1 /(2 X \\sqrt{2})$. In particular, $\\{X\\},\\{2 X\\}, \\ldots,\\{[2 X \\sqrt{2}(1-1 / \\sqrt{2})\\rfloor X\\}$ are all less than $1-1 / \\sqrt{2}$. Thus, by (1) any positive integer of the form $i X$ for $1 \\leqslant i \\leqslant\\lfloor 2 X \\sqrt{2}(1-1 / \\sqrt{2})\\rfloor$ lies in $J$. Set $n \\approx 2 X^{2} \\sqrt{2}(1-1 / \\sqrt{2})$. We now have a set $A=\\{i X: i \\leqslant\\lfloor 2 X \\sqrt{2}(1-1 / \\sqrt{2})\\rfloor\\}$ containing roughly $2 X \\sqrt{2}(1-1 / \\sqrt{2})$ elements less than or equal to $n$ such that all of the differences lie in $J$, and we can see that $|A| \\approx C \\sqrt{n}$ with $C=\\sqrt{2 \\sqrt{2}-2}$.", "metadata": {"resource_path": "IMO/segmented/en-IMO2019SL.jsonl"}} -{"year": "2019", "tier": "T0", "problem_label": "N6", "problem_type": "Number Theory", "exam": "IMO", "problem": "Let $H=\\left\\{\\lfloor i \\sqrt{2}\\rfloor: i \\in \\mathbb{Z}_{>0}\\right\\}=\\{1,2,4,5,7, \\ldots\\}$, and let $n$ be a positive integer. Prove that there exists a constant $C$ such that, if $A \\subset\\{1,2, \\ldots, n\\}$ satisfies $|A| \\geqslant C \\sqrt{n}$, then there exist $a, b \\in A$ such that $a-b \\in H$. (Brazil) Common remarks. In all solutions, we will assume that $A$ is a set such that $\\{a-b: a, b \\in A\\}$ is disjoint from $H$, and prove that $|A|\\sqrt{Y}$ elements; if that subsequence is decreasing, we may reflect (using (4) or (5)) to ensure that it is increasing. Call the subsequence $\\left\\{y_{1} \\sqrt{2}\\right\\},\\left\\{y_{2} \\sqrt{2}\\right\\}, \\ldots,\\left\\{y_{t} \\sqrt{2}\\right\\}$ for $t>\\sqrt{Y}$. Now, set $B=\\left\\{\\left\\lfloor y_{i} \\sqrt{2}\\right\\rfloor: 1 \\leqslant i \\leqslant t\\right\\}$. We have $\\left\\lfloor y_{j} \\sqrt{2}\\right\\rfloor-\\left\\lfloor y_{i} \\sqrt{2}\\right\\rfloor=\\left\\lfloor\\left(y_{j}-y_{i}\\right) \\sqrt{2}\\right\\rfloor$ for $i\\sqrt{Y}>\\sqrt{n} / \\sqrt{3 \\sqrt{2}}$, this is the required set. Comment. Any solution to this problem will need to use the fact that $\\sqrt{2}$ cannot be approximated well by rationals, either directly or implicitly (for example, by using facts about solutions to Pelllike equations). If $\\sqrt{2}$ were replaced by a value of $\\theta$ with very good rational approximations (from below), then an argument along the lines of Solution 3 would give long arithmetic progressions in $\\{\\lfloor i \\theta\\rfloor: 0 \\leqslant i\\left\\lceil\\frac{c^{2}}{b}\\right\\rceil-1=c k+k^{2}-1 \\geqslant c k $$ and $$ \\frac{(c-k)(c+k)}{b}<\\frac{c^{2}}{b} \\leqslant\\left\\lceil\\frac{c^{2}}{b}\\right\\rceil=(c+k) k $$ it can be seen that $c>b k>c-k$, so $$ c=k b+r \\quad \\text { with some } 00$ and $0a$, so $$ \\begin{aligned} & c^{2}-1