{"year": "2015", "tier": "T1", "problem_label": "4", "problem_type": null, "exam": "RMM", "problem": "Let $A B C$ be a triangle, let $D$ be the touchpoint of the side $B C$ and the incircle of the triangle $A B C$, and let $J_{b}$ and $J_{c}$ be the incentres of the triangles $A B D$ and $A C D$, respectively. Prove that the circumcentre of the triangle $A J_{b} J_{c}$ lies on the bisectrix of the angle $B A C$.\n(Russia) Fedor Ivlev", "solution": "Let the incircle of the triangle $A B C$ meet $C A$ and $A B$ at points $E$ and $F$, respectively. Let the incircles of the triangles $A B D$ and $A C D$ meet $A D$ at points $X$ and $Y$, respectively. Then $2 D X=D A+D B-A B=D A+D B-B F-A F=D A-A F$; similarly, $2 D Y=D A-A E=2 D X$. Hence the points $X$ and $Y$ coincide, so $J_{b} J_{c} \\perp A D$.\n\nNow let $O$ be the circumcentre of the triangle $A J_{b} J_{c}$. Then $\\angle J_{b} A O=\\pi / 2-\\angle A O J_{b} / 2=$ $\\pi / 2-\\angle A J_{c} J_{b}=\\angle X A J_{c}=\\frac{1}{2} \\angle D A C$. Therefore, $\\angle B A O=\\angle B A J_{b}+\\angle J_{b} A O=\\frac{1}{2} \\angle B A D+$ $\\frac{1}{2} \\angle D A C=\\frac{1}{2} \\angle B A C$, and the conclusion follows.\n![](https://cdn.mathpix.com/cropped/2024_11_22_2a6f56d972086c73d081g-1.jpg?height=673&width=945&top_left_y=994&top_left_x=563)\n\nFig. 1", "metadata": {"resource_path": "RMM/segmented/en-2015-Solutions_RMM2015-2.jsonl", "problem_match": "\nProblem 4.", "solution_match": "\nSolution."}}
{"year": "2015", "tier": "T1", "problem_label": "5", "problem_type": null, "exam": "RMM", "problem": "Let $p \\geq 5$ be a prime number. For a positive integer $k$ we denote by $R(k)$ the remainder of $k$ when divided by $p$. Determine all positive integers $a<p$ such that\n\n$$\nm+R(m a)>a\n$$\n\nfor every $m=1,2, \\ldots, p-1$.\n(Bulgaria) Alexander Ivanov", "solution": "The required integers are $p-1$ along with all the numbers of the form $\\lfloor p / q\\rfloor, q=$ $2, \\ldots, p-1$. In other words, these are $p-1$, along with the numbers $1,2, \\ldots,\\lfloor\\sqrt{p}\\rfloor$, and also the (distinct) numbers $\\lfloor p / q\\rfloor, q=2, \\ldots,\\left\\lfloor\\sqrt{p}-\\frac{1}{2}\\right\\rfloor$.\n\nWe begin by showing that these numbers satisfy the conditions in the statement. It is readily checked that $p-1$ satisfies the required inequalities, since $m+R(m(p-1))=m+(p-m)=$ $p>p-1$ for all $m=1, \\ldots, p-1$.\n\nNow, consider any number $a$ of the form $a=\\lfloor p / q\\rfloor$, where $q$ is an integer greater than 1 but less than $p$; then $p=a q+r$ with $0<r<q$. Choose any integer $m \\in(0, p)$ and write $m=x q+y$ with $x, y \\in \\mathbb{Z}, 0<y \\leq q$ (notice that $x$ is nonnegative). Then\n\n$$\nR(m a)=R(a y+x a q)=R(a y+x p-x r)=R(a y-x r) .\n$$\n\nSince $a y-x r \\leq a y \\leq a q<p$, we obtain $R(a y-x r) \\geq a y-x r$ and hence\n\n$$\nm+R(m a) \\geq(x q+y)+(a y-x r)=x(q-r)+y(a+1) \\geq a+1\n$$\n\nby $q>r$ and $y \\geq 1$. Thus $a$ satisfies the required condition.\nFinally, we show that if an integer $a \\in(0, p-1)$ satisfies the required condition then $a$ is indeed of the form $a=\\lfloor p / q\\rfloor$ for some integer $q \\in(0, p)$. This is clear for $a=1$, so we may (and will) assume that $a \\geq 2$.\n\nWrite $p=a q+r$ with $q, r \\in \\mathbb{Z}$ and $0<r<a$; since $a \\geq 2$ we have $q<p / 2$. Choose $m=q+1<p$; we have $R(m a)=R(a q+a)=R(p+(a-r))=a-r$, so\n\n$$\na<m+R(m a)=q+1+a-r,\n$$\n\nwhich yields $r<q+1$. Moreover, if $r=q$, then $p=q(a+1)$ which is impossible by $1<a+1<p$. Thus $r<q$, and we have\n\n$$\n0 \\leq \\frac{p}{q}-a=\\frac{r}{q}<1,\n$$\n\nwhich proves $a=\\lfloor p / q\\rfloor$.", "metadata": {"resource_path": "RMM/segmented/en-2015-Solutions_RMM2015-2.jsonl", "problem_match": "\nProblem 5.", "solution_match": "\nSolution."}}
{"year": "2015", "tier": "T1", "problem_label": "6", "problem_type": null, "exam": "RMM", "problem": "Given a positive integer $n$, determine the largest real number $\\mu$ satisfying the following condition: for every $4 n$-point configuration $C$ in an open unit square $U$, there exists an open rectangle in $U$, whose sides are parallel to those of $U$, which contains exactly one point of $C$, and has an area greater than or equal to $\\mu$.\n(Bulgaria) Nikolai Beluhov", "solution": "The required maximum is $\\frac{1}{2 n+2}$. To show that the condition in the statement is not met if $\\mu>\\frac{1}{2 n+2}$, let $U=(0,1) \\times(0,1)$, choose a small enough positive $\\epsilon$, and consider the configuration $C$ consisting of the $n$ four-element clusters of points $\\left(\\frac{i}{n+1} \\pm \\epsilon\\right) \\times\\left(\\frac{1}{2} \\pm \\epsilon\\right), i=1, \\ldots, n$, the four possible sign combinations being considered for each $i$. Clearly, every open rectangle in $U$, whose sides are parallel to those of $U$, which contains exactly one point of $C$, has area at $\\operatorname{most}\\left(\\frac{1}{n+1}+\\epsilon\\right) \\cdot\\left(\\frac{1}{2}+\\epsilon\\right)<\\mu$ if $\\epsilon$ is small enough.\n\nWe now show that, given a finite configuration $C$ of points in an open unit square $U$, there always exists an open rectangle in $U$, whose sides are parallel to those of $U$, which contains exactly one point of $C$, and has an area greater than or equal to $\\mu_{0}=\\frac{2}{|C|+4}$.\n\nTo prove this, usage will be made of the following two lemmas whose proofs are left at the end of the solution.\nLemma 1. Let $k$ be a positive integer, and let $\\lambda<\\frac{1}{\\lfloor k / 2\\rfloor+1}$ be a positive real number. If $t_{1}, \\ldots, t_{k}$ are pairwise distinct points in the open unit interval $(0,1)$, then some $t_{i}$ is isolated from the other $t_{j}$ by an open subinterval of $(0,1)$ whose length is greater than or equal to $\\lambda$.\nLemma 2. Given an integer $k \\geq 2$ and positive integers $m_{1}, \\ldots, m_{k}$,\n\n$$\n\\left\\lfloor\\frac{m_{1}}{2}\\right\\rfloor+\\sum_{i=1}^{k}\\left\\lfloor\\frac{m_{i}}{2}\\right\\rfloor+\\left\\lfloor\\frac{m_{k}}{2}\\right\\rfloor \\leq \\sum_{i=1}^{k} m_{i}-k+2\n$$\n\nBack to the problem, let $U=(0,1) \\times(0,1)$, project $C$ orthogonally on the $x$-axis to obtain the points $x_{1}<\\cdots<x_{k}$ in the open unit interval $(0,1)$, let $\\ell_{i}$ be the vertical through $x_{i}$, and let $m_{i}=\\left|C \\cap \\ell_{i}\\right|, i=1, \\ldots, k$.\n\nSetting $x_{0}=0$ and $x_{k+1}=1$, assume that $x_{i+1}-x_{i-1}>\\left(\\left\\lfloor m_{i} / 2\\right\\rfloor+1\\right) \\mu_{0}$ for some index $i$, and apply Lemma 1 to isolate one of the points in $C \\cap \\ell_{i}$ from the other ones by an open subinterval $x_{i} \\times J$ of $x_{i} \\times(0,1)$ whose length is greater than or equal to $\\mu_{0} /\\left(x_{i+1}-x_{i-1}\\right)$. Consequently, $\\left(x_{i-1}, x_{i+1}\\right) \\times J$ is an open rectangle in $U$, whose sides are parallel to those of $U$, which contains exactly one point of $C$ and has an area greater than or equal to $\\mu_{0}$.\n\nNext, we rule out the case $x_{i+1}-x_{i-1} \\leq\\left(\\left\\lfloor m_{i} / 2\\right\\rfloor+1\\right) \\mu_{0}$ for all indices $i$. If this were the case, notice that necessarily $k>1$; also, $x_{1}-x_{0}<x_{2}-x_{0} \\leq\\left(\\left\\lfloor m_{1} / 2\\right\\rfloor+1\\right) \\mu_{0}$ and $x_{k+1}-x_{k}<$ $x_{k+1}-x_{k-1} \\leq\\left(\\left\\lfloor m_{k} / 2\\right\\rfloor+1\\right) \\mu_{0}$. With reference to Lemma 2, write\n\n$$\n\\begin{aligned}\n2=2\\left(x_{k+1}-x_{0}\\right) & =\\left(x_{1}-x_{0}\\right)+\\sum_{i=1}^{k}\\left(x_{i+1}-x_{i-1}\\right)+\\left(x_{k+1}-x_{k}\\right) \\\\\n& <\\left(\\left(\\left\\lfloor\\frac{m_{1}}{2}\\right\\rfloor+1\\right)+\\sum_{i=1}^{k}\\left(\\left\\lfloor\\frac{m_{i}}{2}\\right\\rfloor+1\\right)+\\left(\\left\\lfloor\\frac{m_{k}}{2}\\right\\rfloor+1\\right)\\right) \\cdot \\mu_{0} \\\\\n& \\leq\\left(\\sum_{i=1}^{k} m_{i}+4\\right) \\mu_{0}=(|C|+4) \\mu_{0}=2\n\\end{aligned}\n$$\n\nand thereby reach a contradiction.\n\nFinally, we prove the two lemmas.\nProof of Lemma 1. Suppose, if possible, that no $t_{i}$ is isolated from the other $t_{j}$ by an open subinterval of $(0,1)$ whose length is greater than or equal to $\\lambda$. Without loss of generality, we may (and will) assume that $0=t_{0}<t_{1}<\\cdots<t_{k}<t_{k+1}=1$. Since the open interval $\\left(t_{i-1}, t_{i+1}\\right)$ isolates $t_{i}$ from the other $t_{j}$, its length, $t_{i+1}-t_{i-1}$, is less than $\\lambda$. Consequently, if $k$ is odd we have $1=\\sum_{i=0}^{(k-1) / 2}\\left(t_{2 i+2}-t_{2 i}\\right)<\\lambda\\left(1+\\frac{k-1}{2}\\right)<1$; if $k$ is even, we have $1<1+t_{k}-t_{k-1}=$ $\\sum_{i=0}^{k / 2-1}\\left(t_{2 i+2}-t_{2 i}\\right)+\\left(t_{k+1}-t_{k-1}\\right)<\\lambda\\left(1+\\frac{k}{2}\\right)<1$. A contradiction in either case.\nProof of Lemma 2. Let $I_{0}$, respectively $I_{1}$, be the set of all indices $i$ in the range $2, \\ldots, k-1$ such that $m_{i}$ is even, respectively odd. Clearly, $I_{0}$ and $I_{1}$ form a partition of that range. Since $m_{i} \\geq 2$ if $i$ is in $I_{0}$, and $m_{i} \\geq 1$ if $i$ is in $I_{1}$ (recall that the $m_{i}$ are positive integers),\n\n$$\n\\sum_{i=2}^{k-1} m_{i}=\\sum_{i \\in I_{0}} m_{i}+\\sum_{i \\in I_{1}} m_{i} \\geq 2\\left|I_{0}\\right|+\\left|I_{1}\\right|=2(k-2)-\\left|I_{1}\\right|, \\quad \\text { or } \\quad\\left|I_{1}\\right| \\geq 2(k-2)-\\sum_{i=2}^{k-1} m_{i}\n$$\n\nTherefore,\n\n$$\n\\begin{aligned}\n\\left\\lfloor\\frac{m_{1}}{2}\\right\\rfloor+\\sum_{i=1}^{k}\\left\\lfloor\\frac{m_{i}}{2}\\right\\rfloor+\\left\\lfloor\\frac{m_{k}}{2}\\right\\rfloor & \\leq m_{1}+\\left(\\sum_{i=2}^{k-1} \\frac{m_{i}}{2}-\\frac{\\left|I_{1}\\right|}{2}\\right)+m_{k} \\\\\n& \\leq m_{1}+\\left(\\frac{1}{2} \\sum_{i=2}^{k-1} m_{i}-(k-2)+\\frac{1}{2} \\sum_{i=2}^{k-1} m_{i}\\right)+m_{k} \\\\\n& =\\sum_{i=1}^{k} m_{i}-k+2\n\\end{aligned}\n$$\n\nRemark. In case $4 n$ is replaced by a positive integer $k$ not divisible by 4 , we do not yet know the maximal $\\mu$ satisfying the corresponding condition.", "metadata": {"resource_path": "RMM/segmented/en-2015-Solutions_RMM2015-2.jsonl", "problem_match": "\nProblem 6.", "solution_match": "\nSolution."}}