{"year": "2013", "tier": "T1", "problem_label": "1", "problem_type": null, "exam": "RMM", "problem": "For a positive integer $a$, define a sequence of integers $x_{1}, x_{2}, \\ldots$ by letting $x_{1}=a$ and $x_{n+1}=2 x_{n}+1$ for $n \\geq 1$. Let $y_{n}=2^{x_{n}}-1$. Determine the largest possible $k$ such that, for some positive integer $a$, the numbers $y_{1}, \\ldots, y_{k}$ are all prime.\n(Russia) Valery Senderov", "solution": "The largest such is $k=2$. Notice first that if $y_{i}$ is prime, then $x_{i}$ is prime as well. Actually, if $x_{i}=1$ then $y_{i}=1$ which is not prime, and if $x_{i}=m n$ for integer $m, n>1$ then $2^{m}-1 \\mid 2^{x_{i}}-1=y_{i}$, so $y_{i}$ is composite. In particular, if $y_{1}, y_{2}, \\ldots, y_{k}$ are primes for some $k \\geq 1$ then $a=x_{1}$ is also prime.\n\nNow we claim that for every odd prime $a$ at least one of the numbers $y_{1}, y_{2}, y_{3}$ is composite (and thus $k<3$ ). Assume, to the contrary, that $y_{1}, y_{2}$, and $y_{3}$ are primes; then $x_{1}, x_{2}, x_{3}$ are primes as well. Since $x_{1} \\geq 3$ is odd, we have $x_{2}>3$ and $x_{2} \\equiv 3(\\bmod 4)$; consequently, $x_{3} \\equiv 7$ $(\\bmod 8)$. This implies that 2 is a quadratic residue modulo $p=x_{3}$, so $2 \\equiv s^{2}(\\bmod p)$ for some integer $s$, and hence $2^{x_{2}}=2^{(p-1) / 2} \\equiv s^{p-1} \\equiv 1(\\bmod p)$. This means that $p \\mid y_{2}$, thus $2^{x_{2}}-1=x_{3}=2 x_{2}+1$. But it is easy to show that $2^{t}-1>2 t+1$ for all integer $t>3$. A contradiction.\n\nFinally, if $a=2$, then the numbers $y_{1}=3$ and $y_{2}=31$ are primes, while $y_{3}=2^{11}-1$ is divisible by 23 ; in this case we may choose $k=2$ but not $k=3$.\n\nRemark. The fact that $23 \\mid 2^{11}-1$ can be shown along the lines in the solution, since 2 is a quadratic residue modulo $x_{4}=23$.", "metadata": {"resource_path": "RMM/segmented/en-2013-Solutions2013-1.jsonl", "problem_match": "\nProblem 1.", "solution_match": "\nSolution."}}
{"year": "2013", "tier": "T1", "problem_label": "2", "problem_type": null, "exam": "RMM", "problem": "Does there exist a pair $(g, h)$ of functions $g, h: \\mathbb{R} \\rightarrow \\mathbb{R}$ such that the only function $f: \\mathbb{R} \\rightarrow \\mathbb{R}$ satisfying $f(g(x))=g(f(x))$ and $f(h(x))=h(f(x))$ for all $x \\in \\mathbb{R}$ is the identity function $f(x) \\equiv x$ ?\n(United Kingdom) Alexander Betts", "solution": ". Such a tester pair exists. We may biject $\\mathbb{R}$ with the closed unit interval, so it suffices to find a tester pair for that instead. We give an explicit example: take some positive real numbers $\\alpha, \\beta$ (which we will specify further later). Take\n\n$$\ng(x)=\\max (x-\\alpha, 0) \\quad \\text { and } \\quad h(x)=\\min (x+\\beta, 1) .\n$$\n\nSay a set $S \\subseteq[0,1]$ is invariant if $f(S) \\subseteq S$ for all functions $f$ commuting with both $g$ and $h$. Note that intersections and unions of invariant sets are invariant. Preimages of invariant sets under $g$ and $h$ are also invariant; indeed, if $S$ is invariant and, say, $T=g^{-1}(S)$, then $g(f(T))=$ $f(g(T)) \\subseteq f(S) \\subseteq S$, thus $f(T) \\subseteq T$.\n\nWe claim that (if we choose $\\alpha+\\beta<1$ ) the intervals $[0, n \\alpha-m \\beta]$ are invariant where $n$ and $m$ are nonnegative integers with $0 \\leq n \\alpha-m \\beta \\leq 1$. We prove this by induction on $m+n$.\n\nThe set $\\{0\\}$ is invariant, as for any $f$ commuting with $g$ we have $g(f(0))=f(g(0))=f(0)$, so $f(0)$ is a fixed point of $g$. This gives that $f(0)=0$, thus the induction base is established.\n\nSuppose now we have some $m, n$ such that $\\left[0, n^{\\prime} \\alpha-m^{\\prime} \\beta\\right]$ is invariant whenever $m^{\\prime}+n^{\\prime}<$ $m+n$. At least one of the numbers $(n-1) \\alpha-m \\beta$ and $n \\alpha-(m-1) \\beta$ lies in $(0,1)$. Note however that in the first case $[0, n \\alpha-m \\beta]=g^{-1}([0,(n-1) \\alpha-m \\beta])$, so $[0, n \\alpha-m \\beta]$ is invariant. In the second case $[0, n \\alpha-m \\beta]=h^{-1}([0, n \\alpha-(m-1) \\beta])$, so again $[0, n \\alpha-m \\beta]$ is invariant. This completes the induction.\n\nWe claim that if we choose $\\alpha+\\beta<1$, where $0<\\alpha \\notin \\mathbb{Q}$ and $\\beta=1 / k$ for some integer $k>1$, then all intervals $[0, \\delta]$ are invariant for $0 \\leq \\delta<1$. This occurs, as by the previous claim, for all nonnegative integers $n$ we have $[0,(n \\alpha \\bmod 1)]$ is invariant. The set of $n \\alpha \\bmod 1$ is dense in $[0,1]$, so in particular\n\n$$\n[0, \\delta]=\\bigcap_{(n \\alpha \\bmod 1)>\\delta}[0,(n \\alpha \\bmod 1)]\n$$\n\nis invariant.\nA similar argument establishes that $[\\delta, 1]$ is invariant, so by intersecting these $\\{\\delta\\}$ is invariant for $0<\\delta<1$. Yet we also have $\\{0\\},\\{1\\}$ both invariant, which proves $f$ to be the identity.", "metadata": {"resource_path": "RMM/segmented/en-2013-Solutions2013-1.jsonl", "problem_match": "\nProblem 2.", "solution_match": "\nSolution 1"}}
{"year": "2013", "tier": "T1", "problem_label": "2", "problem_type": null, "exam": "RMM", "problem": "Does there exist a pair $(g, h)$ of functions $g, h: \\mathbb{R} \\rightarrow \\mathbb{R}$ such that the only function $f: \\mathbb{R} \\rightarrow \\mathbb{R}$ satisfying $f(g(x))=g(f(x))$ and $f(h(x))=h(f(x))$ for all $x \\in \\mathbb{R}$ is the identity function $f(x) \\equiv x$ ?\n(United Kingdom) Alexander Betts", "solution": ". Let us agree that a sequence $\\mathbf{x}=\\left(x_{n}\\right)_{n=1,2, \\ldots}$ is cofinally non-constant if for every index $m$ there exists an index $n>m$ such that $x_{m} \\neq x_{n}$.\n\nBiject $\\mathbb{R}$ with the set of cofinally non-constant sequences of 0 's and 1 's, and define $g$ and $h$ by\n\n$$\ng(\\epsilon, \\mathbf{x})=\\left\\{\\begin{array}{ll}\n\\epsilon, \\mathbf{x} & \\text { if } \\epsilon=0 \\\\\n\\mathbf{x} & \\text { else }\n\\end{array} \\quad \\text { and } \\quad h(\\epsilon, \\mathbf{x})= \\begin{cases}\\epsilon, \\mathbf{x} & \\text { if } \\epsilon=1 \\\\\n\\mathbf{x} & \\text { else }\\end{cases}\\right.\n$$\n\nwhere $\\epsilon, \\mathbf{x}$ denotes the sequence formed by appending $\\mathbf{x}$ to the single-element sequence $\\epsilon$. Note that $g$ fixes precisely those sequences beginning with 0 , and $h$ fixes precisely those beginning with 1.\n\nNow assume that $f$ commutes with both $f$ and $g$. To prove that $f(\\mathbf{x})=\\mathbf{x}$ for all $\\mathbf{x}$ we show that $\\mathbf{x}$ and $f(\\mathbf{x})$ share the same first $n$ terms, by induction on $n$.\n\nThe base case $n=1$ is simple, as we have noticed above that the set of sequences beginning with a 0 is precisely the set of $g$-fixed points, so is preserved by $f$, and similarly for the set of sequences starting with 1.\n\nSuppose that $f(\\mathbf{x})$ and $\\mathbf{x}$ agree for the first $n$ terms, whatever $\\mathbf{x}$. Consider any sequence, and write it as $\\mathbf{x}=\\epsilon, \\mathbf{y}$. Without loss of generality, we may (and will) assume that $\\epsilon=0$, so $f(\\mathbf{x})=0, \\mathbf{y}^{\\prime}$ by the base case. Yet then $f(\\mathbf{y})=f(h(\\mathbf{x}))=h(f(\\mathbf{x}))=h\\left(0, \\mathbf{y}^{\\prime}\\right)=\\mathbf{y}^{\\prime}$. Consequently, $f(\\mathbf{x})=0, f(\\mathbf{y})$, so $f(\\mathbf{x})$ and $\\mathbf{x}$ agree for the first $n+1$ terms by the inductive hypothesis.\n\nThus $f$ fixes all of cofinally non-constant sequences, and the conclusion follows.", "metadata": {"resource_path": "RMM/segmented/en-2013-Solutions2013-1.jsonl", "problem_match": "\nProblem 2.", "solution_match": "\nSolution 2"}}
{"year": "2013", "tier": "T1", "problem_label": "2", "problem_type": null, "exam": "RMM", "problem": "Does there exist a pair $(g, h)$ of functions $g, h: \\mathbb{R} \\rightarrow \\mathbb{R}$ such that the only function $f: \\mathbb{R} \\rightarrow \\mathbb{R}$ satisfying $f(g(x))=g(f(x))$ and $f(h(x))=h(f(x))$ for all $x \\in \\mathbb{R}$ is the identity function $f(x) \\equiv x$ ?\n(United Kingdom) Alexander Betts", "solution": ". (Ilya Bogdanov) We will show that there exists a tester pair of bijective functions $g$ and $h$.\n\nFirst of all, let us find out when a pair of functions is a tester pair. Let $g, h: \\mathbb{R} \\rightarrow \\mathbb{R}$ be arbitrary functions. We construct a directed graph $G_{g, h}$ with $\\mathbb{R}$ as the set of vertices, its edges being painted with two colors: for every vertex $x \\in \\mathbb{R}$, we introduce a red edge $x \\rightarrow g(x)$ and a blue edge $x \\rightarrow h(x)$.\n\nNow, assume that the function $f: \\mathbb{R} \\rightarrow \\mathbb{R}$ satisfies $f(g(x))=g(f(x))$ and $f(h(x))=h(f(x))$ for all $x \\in \\mathbb{R}$. This means exactly that if there exists an edge $x \\rightarrow y$, then there also exists an edge $f(x) \\rightarrow f(y)$ of the same color; that is $-f$ is an endomorphism of $G_{g, h}$.\n\nThus, a pair $(g, h)$ is a tester pair if and only if the graph $G_{g, h}$ admits no nontrivial endomorphisms. Notice that each endomorphism maps a component into a component. Thus, to construct a tester pair, it suffices to construct a continuum of components with no nontrivial endomorphisms and no homomorphisms from one to another. It can be done in many ways; below we present one of them.\n\nLet $g(x)=x+1$; the construction of $h$ is more involved. For every $x \\in[0,1)$ we define the set $S_{x}=x+\\mathbb{Z}$; the sets $S_{x}$ will be exactly the components of $G_{g, h}$. Now we will construct these components.\n\nLet us fix any $x \\in[0,1)$; let $x=0 . x_{1} x_{2} \\ldots$ be the binary representation of $x$. Define $h(x-n)=x-n+1$ for every $n>3$. Next, let $h(x-3)=x, h(x)=x-2, h(x-2)=x-1$, and $h(x-1)=x+1$ (that would be a \"marker\" which fixes a point in our component).\n\nNext, for every $i=1,2, \\ldots$, we define\n(1) $h(x+3 i-2)=x+3 i-1, h(x+3 i-1)=x+3 i$, and $h(x+3 i)=x+3 i+1$, if $x_{i}=0$;\n(2) $h(x+3 i-2)=x+3 i, h(x+3 i)=3 i-1$, and $h(x+3 i-1)=x+3 i+1$, if $x_{i}=1$.\n\nClearly, $h$ is a bijection mapping each $S_{x}$ to itself. Now we claim that the graph $G_{g, h}$ satisfies the desired conditions.\n\nConsider any homomorphism $f_{x}: S_{x} \\rightarrow S_{y}$ ( $x$ and $y$ may coincide). Since $g$ is a bijection, consideration of the red edges shows that $f_{x}(x+n)=x+n+k$ for a fixed real $k$. Next, there exists a blue edge $(x-3) \\rightarrow x$, and the only blue edge of the form $(y+m-3) \\rightarrow(y+m)$ is $(y-3) \\rightarrow y$; thus $f_{x}(x)=y$, and $k=0$.\n\nNext, if $x_{i}=0$ then there exists a blue edge $(x+3 i-2) \\rightarrow(x+3 i-1)$; then the edge $(y+3 i-2) \\rightarrow(y+3 i-1)$ also should exist, so $y_{i}=0$. Analogously, if $x_{i}=1$ then there exists a blue edge $(x+3 i-2) \\rightarrow(x+3 i)$; then the edge $(y+3 i-2) \\rightarrow(y+3 i)$ also should exist, so $y_{i}=1$. We conclude that $x=y$, and $f_{x}$ is the identity mapping, as required.\n\nRemark. If $g$ and $h$ are injections, then the components of $G_{g, h}$ are at most countable. So the set of possible pairwise non-isomorphic such components is continual; hence there is no bijective tester pair for a hyper-continual set instead of $\\mathbb{R}$.", "metadata": {"resource_path": "RMM/segmented/en-2013-Solutions2013-1.jsonl", "problem_match": "\nProblem 2.", "solution_match": "\nSolution 3"}}
{"year": "2013", "tier": "T1", "problem_label": "3", "problem_type": null, "exam": "RMM", "problem": "Let $A B C D$ be a quadrilateral inscribed in a circle $\\omega$. The lines $A B$ and $C D$ meet at $P$, the lines $A D$ and $B C$ meet at $Q$, and the diagonals $A C$ and $B D$ meet at $R$. Let $M$ be the midpoint of the segment $P Q$, and let $K$ be the common point of the segment $M R$ and the circle $\\omega$. Prove that the circumcircle of the triangle $K P Q$ and $\\omega$ are tangent to one another.\n(Russia) MedeubeK Kungozhin", "solution": "Let $O$ be the centre of $\\omega$. Notice that the points $P, Q$, and $R$ are the poles (with respect to $\\omega$ ) of the lines $Q R, R P$, and $P Q$, respectively. Hence we have $O P \\perp Q R, O Q \\perp R P$, and $O R \\perp P Q$, thus $R$ is the orthocentre of the triangle $O P Q$. Now, if $M R \\perp P Q$, then the points $P$ and $Q$ are the reflections of one another in the line $M R=M O$, and the triangle $P Q K$ is symmetrical with respect to this line. In this case the statement of the problem is trivial.\n\nOtherwise, let $V$ be the foot of the perpendicular from $O$ to $M R$, and let $U$ be the common point of the lines $O V$ and $P Q$. Since $U$ lies on the polar line of $R$ and $O U \\perp M R$, we obtain that $U$ is the pole of $M R$. Therefore, the line $U K$ is tangent to $\\omega$. Hence it is enough to prove that $U K^{2}=U P \\cdot U Q$, since this relation implies that $U K$ is also tangent to the circle $K P Q$.\n\nFrom the rectangular triangle $O K U$, we get $U K^{2}=U V \\cdot U O$. Let $\\Omega$ be the circumcircle of triangle $O P Q$, and let $R^{\\prime}$ be the reflection of its orthocentre $R$ in the midpoint $M$ of the side $P Q$. It is well known that $R^{\\prime}$ is the point of $\\Omega$ opposite to $O$, hence $O R^{\\prime}$ is the diameter of $\\Omega$. Finally, since $\\angle O V R^{\\prime}=90^{\\circ}$, the point $V$ also lies on $\\Omega$, hence $U P \\cdot U Q=U V \\cdot U O=U K^{2}$, as required.\n![](https://cdn.mathpix.com/cropped/2024_11_22_9d27bd20b70215e0f78cg-4.jpg?height=974&width=1220&top_left_y=1128&top_left_x=420)\n\nRemark. The statement of the problem is still true if $K$ is the other common point of the line $M R$ and $\\omega$.", "metadata": {"resource_path": "RMM/segmented/en-2013-Solutions2013-1.jsonl", "problem_match": "\nProblem 3.", "solution_match": "\nSolution."}}