| {"year": "1993", "tier": "T1", "problem_label": "1", "problem_type": null, "exam": "APMO", "problem": "Let $A B C D$ be a quadrilateral such that all sides have equal length and angle $\\angle A B C$ is 60 degrees. Let $\\ell$ be a line passing through $D$ and not intersecting the quadrilateral (except at $D)$. Let $E$ and $F$ be the points of intersection of $\\ell$ with $A B$ and $B C$ respectively. Let $M$ be the point of intersection of $C E$ and $A F$.\nProve that $C A^{2}=C M \\times C E$.", "solution": "\n\nTriangles $A E D$ and $C D F$ are similar, because $A D \\| C F$ and $A E \\| C D$. Thus, since $A B C$ and $A C D$ are equilateral triangles,\n\n$$\n\\frac{A E}{C D}=\\frac{A D}{C F} \\Longleftrightarrow \\frac{A E}{A C}=\\frac{A C}{C F}\n$$\n\nThe last equality combined with\n\n$$\n\\angle E A C=180^{\\circ}-\\angle B A C=120^{\\circ}=\\angle A C F\n$$\n\nshows that triangles $E A C$ and $A C F$ are also similar. Therefore $\\angle C A M=\\angle C A F=\\angle A E C$, which implies that line $A C$ is tangent to the circumcircle of $A M E$. By the power of a point, $C A^{2}=C M \\cdot C E$, and we are done.", "metadata": {"resource_path": "APMO/segmented/en-apmo1993_sol.jsonl", "problem_match": "# Problem 1", "solution_match": "# Solution\n\n"}} |
| {"year": "1993", "tier": "T1", "problem_label": "2", "problem_type": null, "exam": "APMO", "problem": "Find the total number of different integer values the function\n\n$$\nf(x)=[x]+[2 x]+\\left[\\frac{5 x}{3}\\right]+[3 x]+[4 x]\n$$\n\ntakes for real numbers $x$ with $0 \\leq x \\leq 100$.\nNote: $[t]$ is the largest integer that does not exceed $t$.\nAnswer: 734.", "solution": "Note that, since $[x+n]=[x]+n$ for any integer $n$,\n\n$$\nf(x+3)=[x+3]+[2(x+3)]+\\left[\\frac{5(x+3)}{3}\\right]+[3(x+3)]+[4(x+3)]=f(x)+35,\n$$\n\none only needs to investigate the interval $[0,3)$.\nThe numbers in this interval at which at least one of the real numbers $x, 2 x, \\frac{5 x}{3}, 3 x, 4 x$ is an integer are\n\n- $0,1,2$ for $x$;\n- $\\frac{n}{2}, 0 \\leq n \\leq 5$ for $2 x$;\n- $\\frac{3 n}{5}, 0 \\leq n \\leq 4$ for $\\frac{5 x}{3}$;\n- $\\frac{n}{3}, 0 \\leq n \\leq 8$ for $3 x$;\n- $\\frac{n}{4}, 0 \\leq n \\leq 11$ for $4 x$.\n\nOf these numbers there are\n\n- 3 integers $(0,1,2)$;\n- 3 irreducible fractions with 2 as denominator (the numerators are $1,3,5$ );\n- 6 irreducible fractions with 3 as denominator (the numerators are $1,2,4,5,7,8$ );\n- 6 irreducible fractions with 4 as denominator (the numerators are $1,3,5,7,9,11,13,15$ );\n- 4 irreducible fractions with 5 as denominator (the numerators are 3,6,9,12).\n\nTherefore $f(x)$ increases 22 times per interval. Since $100=33 \\cdot 3+1$, there are $33 \\cdot 22$ changes of value in $[0,99)$. Finally, there are 8 more changes in [99,100]: 99, 100, $99 \\frac{1}{2}, 99 \\frac{1}{3}, 99 \\frac{2}{3}, 99 \\frac{1}{4}$, $99 \\frac{3}{4}, 99 \\frac{3}{5}$.\nThe total is then $33 \\cdot 22+8=734$.\nComment: A more careful inspection shows that the range of $f$ are the numbers congruent modulo 35 to one of\n\n$$\n0,1,2,4,5,6,7,11,12,13,14,16,17,18,19,23,24,25,26,28,29,30\n$$\n\nin the interval $[0, f(100)]=[0,1166]$. Since $1166 \\equiv 11(\\bmod 35)$, this comprises 33 cycles plus the 8 numbers in the previous list.", "metadata": {"resource_path": "APMO/segmented/en-apmo1993_sol.jsonl", "problem_match": "# Problem 2", "solution_match": "# Solution\n\n"}} |
| {"year": "1993", "tier": "T1", "problem_label": "3", "problem_type": null, "exam": "APMO", "problem": "Let\n\n$$\nf(x)=a_{n} x^{n}+a_{n-1} x^{n-1}+\\cdots+a_{0} \\quad \\text { and } \\quad g(x)=c_{n+1} x^{n+1}+c_{n} x^{n}+\\cdots+c_{0}\n$$\n\nbe non-zero polynomials with real coefficients such that $g(x)=(x+r) f(x)$ for some real number $r$. If $a=\\max \\left(\\left|a_{n}\\right|, \\ldots,\\left|a_{0}\\right|\\right)$ and $c=\\max \\left(\\left|c_{n+1}\\right|, \\ldots,\\left|c_{0}\\right|\\right)$, prove that $\\frac{a}{c} \\leq n+1$.", "solution": "Expanding $(x+r) f(x)$, we find that $c_{n+1}=a_{n}, c_{k}=a_{k-1}+r a_{k}$ for $k=1,2, \\ldots, n$, and $c_{0}=r a_{0}$. Consider three cases:\n\n- $r=0$. Then $c_{0}=0$ and $c_{k}=a_{k-1}$ for $k=1,2, \\ldots, n$, and $a=c \\Longrightarrow \\frac{a}{c}=1 \\leq n+1$.\n- $|r| \\geq 1$. Then\n\n$$\n\\begin{gathered}\n\\left|a_{0}\\right|=\\left|\\frac{c_{0}}{r}\\right| \\leq c \\\\\n\\left|a_{1}\\right|=\\left|\\frac{c_{1}-a_{0}}{r}\\right| \\leq\\left|c_{1}\\right|+\\left|a_{0}\\right| \\leq 2 c\n\\end{gathered}\n$$\n\nand inductively if $\\left|a_{k}\\right| \\leq(k+1) c$\n\n$$\n\\left|a_{k+1}\\right|=\\left|\\frac{c_{k+1}-a_{k}}{r}\\right| \\leq\\left|c_{k+1}\\right|+\\left|a_{k}\\right| \\leq c+(k+1) c=(k+2) c\n$$\n\nTherefore, $\\left|a_{k}\\right| \\leq(k+1) c \\leq(n+1) c$ for all $k$, and $a \\leq(n+1) c \\Longleftrightarrow \\frac{a}{c} \\leq n+1$.\n\n- $0<|r|<1$. Now work backwards: $\\left|a_{n}\\right|=\\left|c_{n+1}\\right| \\leq c$,\n\n$$\n\\left|a_{n-1}\\right|=\\left|c_{n}-r a_{n}\\right| \\leq\\left|c_{n}\\right|+\\left|r a_{n}\\right|<c+c=2 c,\n$$\n\nand inductively if $\\left|a_{n-k}\\right| \\leq(k+1) c$\n\n$$\n\\left|a_{n-k-1}\\right|=\\left|c_{n-k}-r a_{n-k}\\right| \\leq\\left|c_{n-k}\\right|+\\left|r a_{n-k}\\right|<c+(k+1) c=(k+2) c .\n$$\n\nTherefore, $\\left|a_{n-k}\\right| \\leq(k+1) c \\leq(n+1) c$ for all $k$, and $a \\leq(n+1) c$ again.", "metadata": {"resource_path": "APMO/segmented/en-apmo1993_sol.jsonl", "problem_match": "# Problem 3", "solution_match": "# Solution\n\n"}} |
| {"year": "1993", "tier": "T1", "problem_label": "4", "problem_type": null, "exam": "APMO", "problem": "Determine all positive integers $n$ for which the equation\n\n$$\nx^{n}+(2+x)^{n}+(2-x)^{n}=0\n$$\n\nhas an integer as a solution.\n\n## Answer: $n=1$.", "solution": "If $n$ is even, $x^{n}+(2+x)^{n}+(2-x)^{n}>0$, so $n$ is odd.\nFor $n=1$, the equation reduces to $x+(2+x)+(2-x)=0$, which has the unique solution $x=-4$.\nFor $n>1$, notice that $x$ is even, because $x, 2-x$, and $2+x$ have all the same parity. Let $x=2 y$, so the equation reduces to\n\n$$\ny^{n}+(1+y)^{n}+(1-y)^{n}=0 .\n$$\n\nLooking at this equation modulo 2 yields that $y+(1+y)+(1-y)=y+2$ is even, so $y$ is even. Using the factorization\n\n$$\na^{n}+b^{n}=(a+b)\\left(a^{n-1}-a^{n-2} b+\\cdots+b^{n-1}\\right) \\quad \\text { for } n \\text { odd, }\n$$\n\nwhich has a sum of $n$ terms as the second factor, the equation is now equivalent to\n\n$$\ny^{n}+(1+y+1-y)\\left((1+y)^{n-1}-(1+y)^{n-2}(1-y)+\\cdots+(1-y)^{n-1}\\right)=0\n$$\n\nor\n\n$$\ny^{n}=-2\\left((1+y)^{n-1}-(1+y)^{n-2}(1-y)+\\cdots+(1-y)^{n-1}\\right) .\n$$\n\nEach of the $n$ terms in the second factor is odd, and $n$ is odd, so the second factor is odd. Therefore, $y^{n}$ has only one factor 2 , which is a contradiction to the fact that, $y$ being even, $y^{n}$ has at least $n>1$ factors 2 . Hence there are no solutions if $n>1$.", "metadata": {"resource_path": "APMO/segmented/en-apmo1993_sol.jsonl", "problem_match": "# Problem 4", "solution_match": "# Solution\n\n"}} |
| {"year": "1993", "tier": "T1", "problem_label": "5", "problem_type": null, "exam": "APMO", "problem": "Let $P_{1}, P_{2}, \\ldots, P_{1993}=P_{0}$ be distinct points in the $x y$-plane with the following properties:\n(i) both coordinates of $P_{i}$ are integers, for $i=1,2, \\ldots, 1993$;\n(ii) there is no point other than $P_{i}$ and $P_{i+1}$ on the line segment joining $P_{i}$ with $P_{i+1}$ whose coordinates are both integers, for $i=0,1, \\ldots, 1992$.\n\nProve that for some $i, 0 \\leq i \\leq 1992$, there exists a point $Q$ with coordinates $\\left(q_{x}, q_{y}\\right)$ on the line segment joining $P_{i}$ with $P_{i+1}$ such that both $2 q_{x}$ and $2 q_{y}$ are odd integers.", "solution": "Call a point $(x, y) \\in \\mathbb{Z}^{2}$ even or odd according to the parity of $x+y$. Since there are an odd number of points, there are two points $P_{i}=(a, b)$ and $P_{i+1}=(c, d), 0 \\leq i \\leq 1992$ with the same parity. This implies that $a+b+c+d$ is even. We claim that the midpoint of $P_{i} P_{i+1}$ is the desired point $Q$.\nIn fact, since $a+b+c+d=(a+c)+(b+d)$ is even, $a$ and $c$ have the same parity if and only if $b$ and $d$ also have the same parity. If both happen then the midpoint of $P_{i} P_{i+1}, Q=\\left(\\frac{a+c}{2}, \\frac{b+d}{2}\\right)$, has integer coordinates, which violates condition (ii). Then $a$ and $c$, as well as $b$ and $d$, have different parities, and $2 q_{x}=a+c$ and $2 q_{y}=b+d$ are both odd integers.", "metadata": {"resource_path": "APMO/segmented/en-apmo1993_sol.jsonl", "problem_match": "# Problem 5", "solution_match": "# Solution\n\n"}} |
|
|
