{"year": "2011", "tier": "T1", "problem_label": "1", "problem_type": null, "exam": "RMM", "problem": "Prove that there exist two functions\n\n$$\nf, g: \\mathbb{R} \\rightarrow \\mathbb{R}\n$$\n\nsuch that $f \\circ g$ is strictly decreasing, while $g \\circ f$ is strictly increasing.\n(Poland) Andrzej KomisArsKi \\& Marcin Kuczma\n\n#", "solution": "Let\n\n$$\n\\begin{aligned}\n& \\cdot A=\\bigcup_{k \\in \\mathbb{Z}}\\left(\\left[-2^{2 k+1},-2^{2 k}\\right) \\bigcup\\left(2^{2 k}, 2^{2 k+1}\\right]\\right) \\\\\n& \\cdot B=\\bigcup_{k \\in \\mathbb{Z}}\\left(\\left[-2^{2 k},-2^{2 k-1}\\right) \\bigcup\\left(2^{2 k-1}, 2^{2 k}\\right]\\right)\n\\end{aligned}\n$$\n\nThus $A=2 B, B=2 A, A=-A, B=-B, A \\cap B=\\varnothing$, and finally $A \\cup B \\cup\\{0\\}=\\mathbb{R}$. Let us take\n\n$$\nf(x)=\\left\\{\\begin{array}{lll}\nx & \\text { for } & x \\in A \\\\\n-x & \\text { for } & x \\in B \\\\\n0 & \\text { for } & x=0\n\\end{array}\\right.\n$$\n\nTake $g(x)=2 f(x)$. Thus $f(g(x))=f(2 f(x))=-2 x$ and $g(f(x))=2 f(f(x))=2 x$.", "metadata": {"resource_path": "RMM/segmented/en-2011-Sols2011D1.jsonl", "problem_match": "\nProblem 1.", "solution_match": "# Solution."}}
{"year": "2011", "tier": "T1", "problem_label": "2", "problem_type": null, "exam": "RMM", "problem": "Determine all positive integers $n$ for which there exists a polynomial $f(x)$ with real coefficients, with the following properties:\n(1) for each integer $k$, the number $f(k)$ is an integer if and only if $k$ is not divisible by $n$;\n(2) the degree of $f$ is less than $n$.\n(Hungary) GÉza Kós", "solution": "We will show that such polynomial exists if and only if $n=1$ or $n$ is a power of a prime.\n\nWe will use two known facts stated in Lemmata 1 and 2.\nLemma 1. If $p^{a}$ is a power of a prime and $k$ is an integer, then $\\frac{(k-1)(k-2) \\ldots\\left(k-p^{a}+1\\right)}{\\left(p^{a}-1\\right)!}$ is divisible by $p$ if and only if $k$ is not divisible by $p^{a}$.\n\nProof. First suppose that $p^{a} \\mid k$ and consider\n\n$$\n\\frac{(k-1)(k-2) \\cdots\\left(k-p^{a}+1\\right)}{\\left(p^{a}-1\\right)!}=\\frac{k-1}{p^{a}-1} \\cdot \\frac{k-2}{p^{a}-2} \\cdots \\frac{k-p^{a}+1}{1}\n$$\n\nIn every fraction on the right-hand side, $p$ has the same maximal exponent in the numerator as in the denominator.\n\nTherefore, the product (which is an integer) is not divisible by $p$.\n\nNow suppose that $p^{a} \\nmid k$. We have\n\n$$\n\\frac{(k-1)(k-2) \\cdots\\left(k-p^{a}+1\\right)}{\\left(p^{a}-1\\right)!}=\\frac{p^{a}}{k} \\cdot \\frac{k(k-1) \\cdots\\left(k-p^{a}+1\\right)}{\\left(p^{a}\\right)!} .\n$$\n\nThe last fraction is an integer. In the fraction $\\frac{p^{a}}{k}$, the denominator $k$ is not divisible by $p^{a}$.\nLemma 2. If $g(x)$ is a polynomial with degree less than $n$ then\n\n$$\n\\sum_{\\ell=0}^{n}(-1)^{\\ell}\\binom{n}{\\ell} g(x+n-\\ell)=0\n$$\n\nProof. Apply induction on $n$. For $n=1$ then $g(x)$ is a constant and\n\n$$\n\\binom{1}{0} g(x+1)-\\binom{1}{1} g(x)=g(x+1)-g(x)=0\n$$\n\nNow assume that $n>1$ and the Lemma holds for $n-1$. Let $h(x)=g(x+1)-g(x)$; the degree of $h$ is less than the degree of $g$, so the induction hypothesis applies for $g$ and $n-1$ :\n\n$$\n\\begin{gathered}\n\\sum_{\\ell=0}^{n-1}(-1)^{\\ell}\\binom{n-1}{\\ell} h(x+n-1-\\ell)=0 \\\\\n\\sum_{\\ell=0}^{n-1}(-1)^{\\ell}\\binom{n-1}{\\ell}(g(x+n-\\ell)-g(x+n-1-\\ell))=0 \\\\\n\\binom{n-1}{0} g(x+n)+\\sum_{\\ell=1}^{n-1}(-1)^{\\ell}\\left(\\binom{n-1}{\\ell-1}+\\right. \\\\\n\\left.\\binom{n-1}{\\ell}\\right) g(x+n-\\ell)-(-1)^{n-1}\\binom{n-1}{n-1} g(x)=0 \\\\\n\\sum_{\\ell=0}^{n}(-1)^{\\ell}\\binom{n}{\\ell} g(x+n-\\ell)=0\n\\end{gathered}\n$$\n\nLemma 3. If $n$ has at least two distinct prime divisors then the greatest common divisor of $\\binom{n}{1},\\binom{n}{2}, \\ldots,\\binom{n}{n-1}$ is 1 .\n\nProof. Suppose to the contrary that $p$ is a common prime divisor of $\\binom{n}{1}, \\ldots,\\binom{n}{n-1}$. In particular, $p \\left\\lvert\\,\\binom{ n}{1}=n\\right.$. Let $a$ be the exponent of $p$ in the prime factorization of $n$. Since $n$ has at least two prime divisors, we have $1<p^{a}<n$. Hence, $\\binom{n}{p^{a}-1}$ and $\\binom{n}{p^{a}}$ are listed among $\\binom{n}{1}, \\ldots,\\binom{n}{n-1}$ and thus $p \\left\\lvert\\,\\binom{ n}{p^{a}}\\right.$ and $p \\left\\lvert\\,\\binom{ n}{p^{a}-1}\\right.$. But then $p$ divides $\\binom{n}{p^{a}}-\\binom{n}{p^{a}-1}=\\binom{n-1}{p^{a}-1}$, which contradicts Lemma 1.\n\nNext we construct the polynomial $f(x)$ when $n=1$ or $n$ is a power of a prime.\n\nFor $n=1, f(x)=\\frac{1}{2}$ is such a polynomial.\nIf $n=p^{a}$ where $p$ is a prime and $a$ is a positive integer then let\n\n$$\nf(x)=\\frac{1}{p}\\binom{x-1}{p^{a}-1}=\\frac{1}{p} \\cdot \\frac{(x-1)(x-2) \\cdots\\left(x-p^{a}+1\\right)}{\\left(p^{a}-1\\right)!} .\n$$\n\nThe degree of this polynomial is $p^{a}-1=n-1$.\nThe number $\\frac{(k-1)(k-2) \\cdots\\left(k-p^{a}+1\\right)}{\\left(p^{a}-1\\right)!}$ is an integer for any integer $k$, and, by Lemma 1 , it is divisible by $p$ if and only if $k$ is not divisible by $p^{a}=n$.\n\nFinally we prove that if $n$ has at least two prime divisors then no polynomial $f(x)$ satisfies $(1,2)$. Suppose that some polynomial $f(x)$ satisfies (1,2), and apply Lemma 2 for $g=f$ and $x=-k$ where $1 \\leq k \\leq n-1$. We get that\n\n$$\n\\binom{n}{k} f(0)=\\sum_{0 \\leq \\ell \\leq n, \\ell \\neq k}(-1)^{k-\\ell}\\binom{n}{\\ell} f(-k+\\ell)\n$$\n\nSince $f(-k), \\ldots, f(-1)$ and $f(1), \\ldots, f(n-k)$ are all integers, we conclude that $\\binom{n}{k} f(0)$ is an integer for every $1 \\leq k \\leq n-1$.\nBy dint of Lemma 3, the greatest common divisor of $\\binom{n}{1},\\binom{n}{2}, \\ldots,\\binom{n}{n-1}$ is 1 . Hence, there will exist some integers $u_{1}, u_{2}, \\ldots, u_{n-1}$ for which $u_{1}\\binom{n}{1}+\\cdots+u_{n-1}\\binom{n}{n-1}=1$. Then\n\n$$\nf(0)=\\left(\\sum_{k=1}^{n-1} u_{k}\\binom{n}{k}\\right) f(0)=\\sum_{k=1}^{n-1} u_{k}\\binom{n}{k} f(0)\n$$\n\nis a sum of integers. This contradicts the fact that $f(0)$ is not an integer. So such polynomial $f(x)$ does not exist.\n\nAlternative Solution. (I. Bogdanov) We claim the answer is $n=p^{\\alpha}$ for some prime $p$ and nonnegative $\\alpha$.\n\nLemma. For every integers $a_{1}, \\ldots, a_{n}$ there exists an integervalued polynomial $P(x)$ of degree $<n$ such that $P(k)=a_{k}$ for all $1 \\leq k \\leq n$.\n\nProof. Induction on $n$. For the base case $n=1$ one may set $P(x)=a_{1}$. For the induction step, suppose that the polynomial $P_{1}(x)$ satisfies the desired property for all $1 \\leq k \\leq n-1$. Then set $P(x)=P_{1}(x)+\\left(a_{n}-P_{1}(n)\\right)\\binom{x-1}{n-1}$; since $\\binom{k-1}{n-1}=0$ for $1 \\leq k \\leq n-1$ and $\\binom{n-1}{n-1}=1$, the polynomial $P(x)$ is a sought one.\n\nNow, if for some $n$ there exists some polynomial $f(x)$ satisfying the problem conditions, one may choose some integer-valued polynomial $P(x)$ (of degree $<n-1$ ) coinciding with $f(x)$ at points $1, \\ldots, n-1$. The difference $f_{1}(x)=$ $f(x)-P(x)$ also satisfies the problem conditions, therefore we may restrict ourselves to the polynomials vanishing at points $1, \\ldots, n-1-$ that are, the polynomials of the form $f(x)=c \\prod_{i=1}^{n-1}(x-i)$ for some (surely rational) constant $c$.\n\nLet $c=p / q$ be its irreducible form, and $q=\\prod_{j=1}^{d} p_{j}^{\\alpha_{j}}$ be the prime decomposition of the denominator.\n\n1. Assume that a desired polynomial $f(x)$ exists. Since $f(0)$ is not an integer, we have $9 \\nmid(-1)^{n-1}(n-1)$ ! and hence $p_{j}^{\\alpha_{j}} \\nmid(-1)^{n-1}(n-1)$ ! for some $j$. Hence\n\n$$\n\\prod_{i=1}^{n-1}\\left(p_{j}^{\\alpha_{j}}-i\\right) \\equiv(-1)^{n-1}(n-1)!\\not \\equiv 0 \\quad\\left(\\bmod p_{j}^{\\alpha_{j}}\\right)\n$$\n\ntherefore $f\\left(p_{i}^{\\alpha_{i}}\\right)$ is not integer, too. By the condition (i), this means that $n \\mid p_{i}^{\\alpha_{i}}$, and hence $n$ should be a power of a prime.\n2. Now let us construct a desired polynomial $f(x)$ for any power of a prime $n=p^{\\alpha}$. We claim that the polynomial\n\n$$\nf(x)=\\frac{1}{p}\\binom{x-1}{n-1}=\\frac{n}{p x}\\binom{x}{n}\n$$\n\nfits. Actually, consider some integer $x$. From the first representation, the denominator of the irreducible form of $f(x)$ may be 1 or $p$ only. If $p^{\\alpha} \\nmid x$, then the prime decomposition of the fraction $n /(p x)$ contains $p$ with a nonnegative exponent; hence $f(x)$ is integer. On the other hand, if $n=p^{\\alpha} \\mid x$, then the numbers $x-1, x-2, \\ldots, x-(n-1)$ contain the same exponents of primes as the numbers $n-1, n-2, \\ldots, 1$ respectively; hence the number\n\n$$\n\\binom{x-1}{n-1}=\\frac{\\prod_{i=1}^{n-1}(x-i)}{\\prod_{i=1}^{n-1}(n-i)}\n$$\n\nis not divisible by $p$. Thus $f(x)$ is not an integer.", "metadata": {"resource_path": "RMM/segmented/en-2011-Sols2011D1.jsonl", "problem_match": "\nProblem 2.", "solution_match": "\nSolution."}}
{"year": "2011", "tier": "T1", "problem_label": "3", "problem_type": null, "exam": "RMM", "problem": "A triangle $A B C$ is inscribed in a circle $\\omega$. A variable line $\\ell$ chosen parallel to $B C$ meets segments $A B$, $A C$ at points $D, E$ respectively, and meets $\\omega$ at points $K, L$ (where $D$ lies between $K$ and $E$ ). Circle $\\gamma_{1}$ is tangent to the segments $K D$ and $B D$ and also tangent to $\\omega$, while circle $\\gamma_{2}$ is tangent to the segments $L E$ and $C E$ and also tangent to $\\omega$. Determine the locus, as $\\ell$ varies, of the meeting point of the common inner tangents to $\\gamma_{1}$ and $\\gamma_{2}$.\n(Russia) VASily Mokin \\& Fedor IvLev", "solution": "Let $P$ be the meeting point of the common inner tangents to $\\gamma_{1}$ and $\\gamma_{2}$. Also, let $b$ be the angle bisector of $\\angle B A C$. Since $K L \\| B C, b$ is also the angle bisector of $\\angle K A L$.\n\nLet $\\mathfrak{H}$ be the composition of the symmetry $\\mathfrak{S}$ with respect to $b$ and the inversion $\\mathfrak{I}$ of centre $A$ and ratio $\\sqrt{A K \\cdot A L}$ (it is readily seen that $\\mathfrak{S}$ and $\\mathfrak{I}$ commute, so since $\\mathfrak{S}^{2}=\\mathfrak{I}^{2}=$ id, then also $\\mathfrak{H}^{2}=\\mathrm{id}$, the identical transformation). The elements of the configuration interchanged by $\\mathfrak{H}$ are summarized in Table I.\n\nLet $O_{1}$ and $O_{2}$ be the centres of circles $\\gamma_{1}$ and $\\gamma_{2}$. Since the circles $\\gamma_{1}$ and $\\gamma_{2}$ are determined by their construction (in a unique way), they are interchanged by $\\mathfrak{H}$, therefore the rays $A O_{1}$ and $A O_{2}$ are symmetrical with respect\nto $b$. Denote by $\\rho_{1}$ and $\\rho_{2}$ the radii of $\\gamma_{1}$ and $\\gamma_{2}$. Since $\\angle O_{1} A B=\\angle O_{2} A C$, we have $\\rho_{1} / \\rho_{2}=A O_{1} / A O_{2}$. On the other hand, from the definition of $P$ we have $O_{1} P / O_{2} P=$ $\\rho_{1} / \\rho_{2}=A O_{1} / A O_{2}$; this means that $A P$ is the angle bisector of $\\angle O_{1} A O_{2}$ and therefore of $\\angle B A C$.\n\nThe limiting, degenerated, cases are when the parallel line passes through $A$ - when $P$ coincides with $A$; respectively when the parallel line is $B C$ - when $P$ coincides with the foot $A^{\\prime} \\in B C$ of the angle bisector of $\\angle B A C$ (or any other point on $B C$ ). By continuity, any point $P$ on the open segment $A A^{\\prime}$ is obtained for some position of the parallel, therefore the locus is the open segment $A A^{\\prime}$ of the angle bisector $b$ of $\\angle B A C$.\n\n| point $K$ | $\\longleftrightarrow$ | point $L$ |\n| :---: | :---: | :---: |\n| line $K L$ | $\\longleftrightarrow$ | circle $\\omega$ |\n| ray $A B$ | $\\longleftrightarrow$ | ray $A C$ |\n| point $B$ | $\\longleftrightarrow$ | point $E$ |\n| point $C$ | $\\longleftrightarrow$ | point $D$ |\n| segment $B D$ | $\\longleftrightarrow$ | segment $E C$ |\n| $\\operatorname{arc} B K$ | $\\longleftrightarrow$ | segment $E L$ |\n| $\\operatorname{arc} C L$ | $\\longleftrightarrow$ | segment $D K$ |\n\n![](https://cdn.mathpix.com/cropped/2024_11_22_04b99d8d12f731235412g-3.jpg?height=655&width=729&top_left_y=438&top_left_x=1229)\n\nTABLE I: Elements interchanged by $\\mathfrak{H}$.", "metadata": {"resource_path": "RMM/segmented/en-2011-Sols2011D1.jsonl", "problem_match": "\nProblem 3.", "solution_match": "\nSolution."}}