| {"year": "1994", "tier": "T1", "problem_label": "1", "problem_type": null, "exam": "APMO", "problem": "Let $f: \\mathbb{R} \\rightarrow \\mathbb{R}$ be a function such that\n(i) For all $x, y \\in \\mathbb{R}$,\n\n$$\nf(x)+f(y)+1 \\geq f(x+y) \\geq f(x)+f(y)\n$$\n\n(ii) For all $x \\in[0,1), f(0) \\geq f(x)$,\n(iii) $-f(-1)=f(1)=1$.\n\nFind all such functions $f$.\nAnswer: $f(x)=\\lfloor x\\rfloor$, the largest integer that does not exceed $x$, is the only function.", "solution": "Plug $y \\rightarrow 1$ in (i):\n\n$$\nf(x)+f(1)+1 \\geq f(x+1) \\geq f(x)+f(1) \\Longleftrightarrow f(x)+1 \\leq f(x+1) \\leq f(x)+2\n$$\n\nNow plug $y \\rightarrow-1$ and $x \\rightarrow x+1$ in (i):\n\n$$\nf(x+1)+f(-1)+1 \\geq f(x) \\geq f(x+1)+f(-1) \\Longleftrightarrow f(x) \\leq f(x+1) \\leq f(x)+1\n$$\n\nHence $f(x+1)=f(x)+1$ and we only need to define $f(x)$ on $[0,1)$. Note that $f(1)=$ $f(0)+1 \\Longrightarrow f(0)=0$.\nCondition (ii) states that $f(x) \\leq 0$ in $[0,1)$.\nNow plug $y \\rightarrow 1-x$ in (i):\n\n$$\nf(x)+f(1-x)+1 \\leq f(x+(1-x)) \\leq f(x)+f(1-x) \\Longrightarrow f(x)+f(1-x) \\geq 0\n$$\n\nIf $x \\in(0,1)$ then $1-x \\in(0,1)$ as well, so $f(x) \\leq 0$ and $f(1-x) \\leq 0$, which implies $f(x)+f(1-x) \\leq 0$. Thus, $f(x)=f(1-x)=0$ for $x \\in(0,1)$. This combined with $f(0)=0$ and $f(x+1)=f(x)+1$ proves that $f(x)=\\lfloor x\\rfloor$, which satisfies the problem conditions, as since\n$x+y=\\lfloor x\\rfloor+\\lfloor y\\rfloor+\\{x\\}+\\{y\\}$ and $0 \\leq\\{x\\}+\\{y\\}<2 \\Longrightarrow\\lfloor x\\rfloor+\\lfloor y\\rfloor \\leq x+y<\\lfloor x\\rfloor+\\lfloor y\\rfloor+2$ implies\n\n$$\n\\lfloor x\\rfloor+\\lfloor y\\rfloor+1 \\geq\\lfloor x+y\\rfloor \\geq\\lfloor x\\rfloor+\\lfloor y\\rfloor .\n$$", "metadata": {"resource_path": "APMO/segmented/en-apmo1994_sol.jsonl", "problem_match": "# Problem 1", "solution_match": "# Solution\n\n"}} |
| {"year": "1994", "tier": "T1", "problem_label": "2", "problem_type": null, "exam": "APMO", "problem": "Given a nondegenerate triangle $A B C$, with circumcentre $O$, orthocentre $H$, and circumradius $R$, prove that $|O H|<3 R$.", "solution": "Embed $A B C$ in the complex plane, with $A, B$ and $C$ in the circle $|z|=R$, so $O$ is the origin. Represent each point by its lowercase letter. It is well known that $h=a+b+c$, so\n\n$$\nO H=|a+b+c| \\leq|a|+|b|+|c|=3 R .\n$$\n\nThe equality cannot occur because $a, b$, and $c$ are not collinear, so $O H<3 R$.", "metadata": {"resource_path": "APMO/segmented/en-apmo1994_sol.jsonl", "problem_match": "# Problem 2", "solution_match": "# Solution 1"}} |
| {"year": "1994", "tier": "T1", "problem_label": "2", "problem_type": null, "exam": "APMO", "problem": "Given a nondegenerate triangle $A B C$, with circumcentre $O$, orthocentre $H$, and circumradius $R$, prove that $|O H|<3 R$.", "solution": "Suppose with loss of generality that $\\angle A<90^{\\circ}$. Let $B D$ be an altitude. Then\n\n$$\nA H=\\frac{A D}{\\cos \\left(90^{\\circ}-C\\right)}=\\frac{A B \\cos A}{\\sin C}=2 R \\cos A\n$$\n\nBy the triangle inequality,\n\n$$\nO H<A O+A H=R+2 R \\cos A<3 R\n$$\n\nComment: With a bit more work, if $a, b, c$ are the sidelengths of $A B C$, one can show that\n\n$$\nO H^{2}=9 R^{2}-a^{2}-b^{2}-c^{2} .\n$$\n\nIn fact, using vectors in a coordinate system with $O$ as origin, by the Euler line\n\n$$\n\\overrightarrow{O H}=3 \\overrightarrow{O G}=3 \\cdot \\frac{\\overrightarrow{O A}+\\overrightarrow{O B}+\\overrightarrow{O C}}{3}=\\overrightarrow{O A}+\\overrightarrow{O B}+\\overrightarrow{O C}\n$$\n\nso\n\n$$\nO H^{2}=\\overrightarrow{O H} \\cdot \\overrightarrow{O H}=(\\overrightarrow{O A}+\\overrightarrow{O B}+\\overrightarrow{O C}) \\cdot(\\overrightarrow{O A}+\\overrightarrow{O B}+\\overrightarrow{O C})\n$$\n\nExpanding and using the fact that $\\overrightarrow{O X} \\cdot \\overrightarrow{O X}=O X^{2}=R^{2}$ for $X \\in\\{A, B, C\\}$, as well as\n$\\overrightarrow{O A} \\cdot \\overrightarrow{O B}=O A \\cdot O B \\cdot \\cos \\angle A O B=R^{2} \\cos 2 C=R^{2}\\left(1-2 \\sin ^{2} C\\right)=R^{2}\\left(1-2\\left(\\frac{c}{2 R}\\right)^{2}\\right)=R^{2}-\\frac{c^{2}}{2}$, we find that\n\n$$\n\\begin{aligned}\nO H^{2} & =\\overrightarrow{O A} \\cdot \\overrightarrow{O A}+\\overrightarrow{O B} \\cdot \\overrightarrow{O B}+\\overrightarrow{O C} \\cdot \\overrightarrow{O C}+2 \\overrightarrow{O A} \\cdot \\overrightarrow{O B}+2 \\overrightarrow{O A} \\cdot \\overrightarrow{O C}+2 \\overrightarrow{O B} \\cdot \\overrightarrow{O C} \\\\\n& =3 R^{2}+\\left(2 R^{2}-c^{2}\\right)+\\left(2 R^{2}-b^{2}\\right)+\\left(2 R^{2}-a^{2}\\right) \\\\\n& =9 R^{2}-a^{2}-b^{2}-c^{2}\n\\end{aligned}\n$$\n\nas required.\nThis proves that $O H^{2}<9 R^{2} \\Longrightarrow O H<3 R$, and since $a, b, c$ can be arbitrarily small (fix the circumcircle and choose $A, B, C$ arbitrarily close in this circle), the bound is sharp.", "metadata": {"resource_path": "APMO/segmented/en-apmo1994_sol.jsonl", "problem_match": "# Problem 2", "solution_match": "# Solution 2"}} |
| {"year": "1994", "tier": "T1", "problem_label": "3", "problem_type": null, "exam": "APMO", "problem": "Let $n$ be an integer of the form $a^{2}+b^{2}$, where $a$ and $b$ are relatively prime integers and such that if $p$ is a prime, $p \\leq \\sqrt{n}$, then $p$ divides $a b$. Determine all such $n$.\n\nAnswer: $n=2,5,13$.", "solution": "A prime $p$ divides $a b$ if and only if divides either $a$ or $b$. If $n=a^{2}+b^{2}$ is a composite then it has a prime divisor $p \\leq \\sqrt{n}$, and if $p$ divides $a$ it divides $b$ and vice-versa, which is not possible because $a$ and $b$ are coprime. Therefore $n$ is a prime.\nSuppose without loss of generality that $a \\geq b$ and consider $a-b$. Note that $a^{2}+b^{2}=(a-b)^{2}+2 a b$.\n\n- If $a=b$ then $a=b=1$ because $a$ and $b$ are coprime. $n=2$ is a solution.\n- If $a-b=1$ then $a$ and $b$ are coprime and $a^{2}+b^{2}=(a-b)^{2}+2 a b=2 a b+1=2 b(b+1)+1=$ $2 b^{2}+2 b+1$. So any prime factor of any number smaller than $\\sqrt{2 b^{2}+2 b+1}$ is a divisor of $a b=b(b+1)$.\nOne can check that $b=1$ and $b=2$ yields the solutions $n=1^{2}+2^{2}=5$ (the only prime $p$ is 2 ) and $n=2^{2}+3^{2}=13$ (the only primes $p$ are 2 and 3 ). Suppose that $b>2$.\nConsider, for instance, the prime factors of $b-1 \\leq \\sqrt{2 b^{2}+2 b+1}$, which is coprime with $b$. Any prime must then divide $a=b+1$. Then it divides $(b+1)-(b-1)=2$, that is, $b-1$ can only have 2 as a prime factor, that is, $b-1$ is a power of 2 , and since $b-1 \\geq 2$, $b$ is odd.\nSince $2 b^{2}+2 b+1-(b+2)^{2}=b^{2}-2 b-3=(b-3)(b+1) \\geq 0$, we can also consider any prime divisor of $b+2$. Since $b$ is odd, $b$ and $b+2$ are also coprime, so any prime divisor of $b+2$ must divide $a=b+1$. But $b+1$ and $b+2$ are also coprime, so there can be no such primes. This is a contradiction, and $b \\geq 3$ does not yield any solutions.\n- If $a-b>1$, consider a prime divisor $p$ of $a-b=\\sqrt{a^{2}-2 a b+b^{2}}<\\sqrt{a^{2}+b^{2}}$. Since $p$ divides one of $a$ and $b, p$ divides both numbers (just add or subtract $a-b$ accordingly.) This is a contradiction.\n\nHence the only solutions are $n=2,5,13$.", "metadata": {"resource_path": "APMO/segmented/en-apmo1994_sol.jsonl", "problem_match": "# Problem 3", "solution_match": "# Solution\n\n"}} |
| {"year": "1994", "tier": "T1", "problem_label": "4", "problem_type": null, "exam": "APMO", "problem": "Is there an infinite set of points in the plane such that no three points are collinear, and the distance between any two points is rational?\nAnswer: Yes.", "solution": "The answer is yes and we present the following construction: the idea is considering points in the unit circle of the form $P_{n}=(\\cos (2 n \\theta), \\sin (2 n \\theta))$ for an appropriate $\\theta$. Then the distance $P_{m} P_{n}$ is the length of the chord with central angle $(2 m-2 n) \\theta \\bmod \\pi$, that is, $2|\\sin ((m-n) \\theta)|$. Our task is then finding $\\theta$ such that (i) $\\sin (k \\theta)$ is rational for all $k \\in \\mathbb{Z}$; (ii) points $P_{n}$ are all distinct. We claim that $\\theta \\in(0, \\pi / 2)$ such that $\\cos \\theta=\\frac{3}{5}$ and therefore $\\sin \\theta=\\frac{4}{5}$ does the job. Proof of (i): We know that $\\sin ((n+1) \\theta)+\\sin ((n-1) \\theta)=2 \\sin (n \\theta) \\cos \\theta$, so if $\\sin ((n-1) \\theta$ and $\\sin (n \\theta)$ are both rational then $\\sin ((n+1) \\theta)$ also is. Since $\\sin (0 \\theta)=0$ and $\\sin \\theta$ are rational, an induction shows that $\\sin (n \\theta)$ is rational for $n \\in \\mathbb{Z}_{>0}$; the result is also true if $n$ is negative because $\\sin$ is an odd function.\nProof of (ii): $P_{m}=P_{n} \\Longleftrightarrow 2 n \\theta=2 m \\theta+2 k \\pi$ for some $k \\in \\mathbb{Z}$, which implies $\\sin ((n-m) \\theta)=$ $\\sin (k \\pi)=0$. We show that $\\sin (k \\theta) \\neq 0$ for all $k \\neq 0$.\nWe prove a stronger result: let $\\sin (k \\theta)=\\frac{a_{k}}{5^{k}}$. Then\n\n$$\n\\begin{aligned}\n\\sin ((k+1) \\theta)+\\sin ((k-1) \\theta)=2 \\sin (k \\theta) \\cos \\theta & \\Longleftrightarrow \\frac{a_{k+1}}{5^{k+1}}+\\frac{a_{k-1}}{5^{k-1}}=2 \\cdot \\frac{a_{k}}{5^{k}} \\cdot \\frac{3}{5} \\\\\n& \\Longleftrightarrow a_{k+1}=6 a_{k}-25 a_{k-1}\n\\end{aligned}\n$$\n\nSince $a_{0}=0$ and $a_{1}=4, a_{k}$ is an integer for $k \\geq 0$, and $a_{k+1} \\equiv a_{k}(\\bmod 5)$ for $k \\geq 1$ (note that $a_{-1}=-\\frac{4}{25}$ is not an integer!). Thus $a_{k} \\equiv 4(\\bmod 5)$ for all $k \\geq 1$, and $\\sin (k \\theta)=\\frac{a_{k}}{5^{k}}$ is an irreducible fraction with $5^{k}$ as denominator and $a_{k} \\equiv 4(\\bmod 5)$. This proves (ii) and we are done.", "metadata": {"resource_path": "APMO/segmented/en-apmo1994_sol.jsonl", "problem_match": "# Problem 4", "solution_match": "# Solution 1"}} |
| {"year": "1994", "tier": "T1", "problem_label": "4", "problem_type": null, "exam": "APMO", "problem": "Is there an infinite set of points in the plane such that no three points are collinear, and the distance between any two points is rational?\nAnswer: Yes.", "solution": "We present a different construction. Consider the (collinear) points\n\n$$\nP_{k}=\\left(1, \\frac{x_{k}}{y_{k}}\\right),\n$$\n\nsuch that the distance $O P_{k}$ from the origin $O$,\n\n$$\nO P_{k}=\\frac{\\sqrt{x_{k}^{2}+y_{k}^{2}}}{y_{k}}\n$$\n\nis rational, and $x_{k}$ and $y_{k}$ are integers. Clearly, $P_{i} P_{j}=\\left|\\frac{x_{i}}{y_{i}}-\\frac{x_{j}}{y_{j}}\\right|$ is rational.\nPerform an inversion with center $O$ and unit radius. It maps the line $x=1$, which contains all points $P_{k}$, to a circle (minus the origin). Let $Q_{k}$ be the image of $P_{k}$ under this inversion. Then\n\n$$\nQ_{i} Q_{j}=\\frac{1^{2} P_{i} P_{j}}{O P_{i} \\cdot O P_{j}}\n$$\n\nis rational and we are done if we choose $x_{k}$ and $y_{k}$ accordingly. But this is not hard, as we can choose the legs of a Pythagorean triple, say\n\n$$\nx_{k}=k^{2}-1, \\quad y_{k}=2 k\n$$\n\nThis implies $O P_{k}=\\frac{k^{2}+1}{2 k}$, and then\n\n$$\nQ_{i} Q_{j}=\\frac{\\left|\\frac{i^{2}-1}{i}-\\frac{j^{2}-1}{j}\\right|}{\\frac{i^{2}+1}{2 i} \\cdot \\frac{j^{2}+1}{2 j}}=\\frac{|4(i-j)(i j+1)|}{\\left(i^{2}+1\\right)\\left(j^{2}+1\\right)}\n$$", "metadata": {"resource_path": "APMO/segmented/en-apmo1994_sol.jsonl", "problem_match": "# Problem 4", "solution_match": "# Solution 2"}} |
| {"year": "1994", "tier": "T1", "problem_label": "5", "problem_type": null, "exam": "APMO", "problem": "You are given three lists $A, B$, and $C$. List $A$ contains the numbers of the form $10^{k}$ in base 10, with $k$ any integer greater than or equal to 1 . Lists $B$ and $C$ contain the same numbers translated into base 2 and 5 respectively:\n\n| $A$ | $B$ | $C$ |\n| :--- | :--- | :--- |\n| 10 | 1010 | 20 |\n| 100 | 1100100 | 400 |\n| 1000 | 1111101000 | 13000 |\n| $\\vdots$ | $\\vdots$ | $\\vdots$ |\n\nProve that for every integer $n>1$, there is exactly one number in exactly one of the lists $B$ or $C$ that has exactly $n$ digits.", "solution": "Let $b_{k}$ and $c_{k}$ be the number of digits in the $k$ th term in lists $B$ and $C$, respectively. Then\n\n$$\n2^{b_{k}-1} \\leq 10^{k}<2^{b_{k}} \\Longleftrightarrow \\log _{2} 10^{k}<b_{k} \\leq \\log _{2} 10^{k}+1 \\Longleftrightarrow b_{k}=\\left\\lfloor k \\cdot \\log _{2} 10\\right\\rfloor+1\n$$\n\nand, similarly\n\n$$\nc_{k}=\\left\\lfloor k \\cdot \\log _{5} 10\\right\\rfloor+1\n$$\n\nBeatty's theorem states that if $\\alpha$ and $\\beta$ are irrational positive numbers such that\n\n$$\n\\frac{1}{\\alpha}+\\frac{1}{\\beta}=1\n$$\n\nthen the sequences $\\lfloor k \\alpha\\rfloor$ and $\\lfloor k \\beta\\rfloor, k=1,2, \\ldots$, partition the positive integers.\nThen, since\n\n$$\n\\frac{1}{\\log _{2} 10}+\\frac{1}{\\log _{5} 10}=\\log _{10} 2+\\log _{10} 5=\\log _{10}(2 \\cdot 5)=1\n$$\n\nthe sequences $b_{k}-1$ and $c_{k}-1$ partition the positive integers, and therefore each integer greater than 1 appears in $b_{k}$ or $c_{k}$ exactly once. We are done.\nComment: For the sake of completeness, a proof of Beatty's theorem follows.\nLet $x_{n}=\\alpha n$ and $y_{n}=\\beta n, n \\geq 1$ integer. Note that, since $\\alpha m=\\beta n$ implies that $\\frac{\\alpha}{\\beta}$ is rational but\n\n$$\n\\frac{\\alpha}{\\beta}=\\alpha \\cdot \\frac{1}{\\beta}=\\alpha\\left(1-\\frac{1}{\\alpha}\\right)=\\alpha-1\n$$\n\nis irrational, the sequences have no common terms, and all terms in both sequences are irrational.\nThe theorem is equivalent to proving that exactly one term of either $x_{n}$ of $y_{n}$ lies in the interval $(N, N+1)$ for each $N$ positive integer. For that purpose we count the number of terms of the union of the two sequences in the interval $(0, N)$ : since $n \\alpha<N \\Longleftrightarrow n<\\frac{N}{\\alpha}$, there are $\\left\\lfloor\\frac{N}{\\alpha}\\right\\rfloor$ terms of $x_{n}$ in the interval and, similarly, $\\left\\lfloor\\frac{N}{\\beta}\\right\\rfloor$ terms of $y_{n}$ in the same interval. Since the sequences are disjoint, the total of numbers is\n\n$$\nT(N)=\\left\\lfloor\\frac{N}{\\alpha}\\right\\rfloor+\\left\\lfloor\\frac{N}{\\beta}\\right\\rfloor\n$$\n\nHowever, $x-1<\\lfloor x\\rfloor<x$ for nonintegers $x$, so\n\n$$\n\\begin{aligned}\n\\frac{N}{\\alpha}-1+\\frac{N}{\\beta}-1<T(N)<\\frac{N}{\\alpha}+\\frac{N}{\\beta} & \\Longleftrightarrow N\\left(\\frac{1}{\\alpha}+\\frac{1}{\\beta}\\right)-2<T(N)<N\\left(\\frac{1}{\\alpha}+\\frac{1}{\\beta}\\right) \\\\\n& \\Longleftrightarrow N-2<T(N)<N,\n\\end{aligned}\n$$\n\nthat is, $T(N)=N-1$.\nTherefore the number of terms in $(N, N+1)$ is $T(N+1)-T(N)=N-(N-1)=1$, and the result follows.", "metadata": {"resource_path": "APMO/segmented/en-apmo1994_sol.jsonl", "problem_match": "# Problem 5", "solution_match": "# Solution\n\n"}} |
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