{"year": "2023", "tier": "T1", "problem_label": "4", "problem_type": null, "exam": "RMM", "problem": "Given a triangle $A B C$, let $H$ and $O$ be its orthocentre and circumcentre, respectively. Let $K$ be the midpoint of the line segment $A H$. Let further $\\ell$ be a line through $O$, and let $P$ and $Q$ be the orthogonal projections of $B$ and $C$ onto $\\ell$, respectively. Prove that $K P+K Q \\geq B C$.\n\n# Russia, Vasily Mokin", "solution": ". Fix the origin at $O$ and the real axis along $\\ell$. A lower case letter denotes the complex coordinate of the corresponding point in the configuration. For convenience, let $|a|=|b|=|c|=1$.\n\nClearly, $k=a+\\frac{1}{2}(b+c), p=a+\\frac{1}{2}\\left(b+\\frac{1}{b}\\right)$ and $q=a+\\frac{1}{2}\\left(c+\\frac{1}{c}\\right)$.\n\nThen $|k-p|=\\left|a+\\frac{1}{2}\\left(c-\\frac{1}{b}\\right)\\right|=\\frac{1}{2}|2 a b+b c-1|$, since $|b|=1$.\n\nSimilarly, $|k-q|=\\frac{1}{2}|2 a c+b c-1|$, so, since $|a|=1$,\n$|k-p|+|k-q|=\\frac{1}{2}|2 a b+b c-1|+\\frac{1}{2}|2 a c+b c-1|$\n\n$$\n\\geq \\frac{1}{2}|2 a(b-c)|=|b-c|,\n$$\n\nas required.", "metadata": {"resource_path": "RMM/segmented/en-2023-RMM2023-Day2-English_Solutions.jsonl", "problem_match": "\nProblem 4.", "solution_match": "\nSolution 1"}}
{"year": "2023", "tier": "T1", "problem_label": "4", "problem_type": null, "exam": "RMM", "problem": "Given a triangle $A B C$, let $H$ and $O$ be its orthocentre and circumcentre, respectively. Let $K$ be the midpoint of the line segment $A H$. Let further $\\ell$ be a line through $O$, and let $P$ and $Q$ be the orthogonal projections of $B$ and $C$ onto $\\ell$, respectively. Prove that $K P+K Q \\geq B C$.\n\n# Russia, Vasily Mokin", "solution": ". Let $M$ be the midpoint of $B C$, and let $R$ be the projection of $M$ onto $\\ell$. In other words, $R$ is the midpoint of $P Q$. Since $\\angle B P O=\\angle B M O=$ $90^{\\circ}$, the points $B, P, O$, and $M$ are concyclic, so $\\angle(O M, O B)=\\angle(P M, P B)=\\angle(P M, M R)$, so the right triangles $M R P$ and $O M B$ are similar and have different orientation. Similarly, the triangles $M R Q$ and $O M C$ are similar and have different orientation, hence so are the triangles $O B C$ and $M P Q$.\n![](https://cdn.mathpix.com/cropped/2024_11_22_d8cbf77d76bc852500acg-1.jpg?height=716&width=846&top_left_y=1478&top_left_x=128)\n\nRecall that $\\overrightarrow{A H}=2 \\overrightarrow{O M}$, so $\\overrightarrow{O M}=\\overrightarrow{A K}$. Hence $A O M K$ is a parallelogram, so $M K=O A=O B=$ $O C$.\n\nConsider the rotation through $\\angle(\\overrightarrow{O C}, \\overrightarrow{O B})$ about $M$. It maps $P$ to $Q$; let it map $K$ to some point $L$. Then $M K=M L=O B=O C$ and $\\angle L M K=\\angle B O C$, so the triangles $O B C$ and $M K L$ are congruent. Hence $B C=K L \\leq K Q+L Q=$ $K Q+K P$, as required.", "metadata": {"resource_path": "RMM/segmented/en-2023-RMM2023-Day2-English_Solutions.jsonl", "problem_match": "\nProblem 4.", "solution_match": "\nSolution 2"}}
{"year": "2023", "tier": "T1", "problem_label": "4", "problem_type": null, "exam": "RMM", "problem": "Given a triangle $A B C$, let $H$ and $O$ be its orthocentre and circumcentre, respectively. Let $K$ be the midpoint of the line segment $A H$. Let further $\\ell$ be a line through $O$, and let $P$ and $Q$ be the orthogonal projections of $B$ and $C$ onto $\\ell$, respectively. Prove that $K P+K Q \\geq B C$.\n\n# Russia, Vasily Mokin", "solution": ". Let $\\alpha=\\angle(P B, B C)=\\angle(Q C, B C)$. Since $P$ lies on the circle of diameter $O B$, $\\angle(O P, O M)=\\alpha$. Since also $Q$ lies on the circle of diameter $O C$, it immediately follows that\n$M P=M Q=R \\sin \\alpha$ by sine theorem in triangles $\\triangle O P M$ and $\\triangle O Q M$.\n\nBecause $P Q$ is the projection of $B C$ on line $\\ell$, it follows that $P Q=B C \\sin \\alpha$. Just like in the first solution, $K M=A O=R$ (the circumradius of triangle $\\triangle A B C)$.\n\nNow apply Ptolemy's inequality for the quadrilateral $K P M Q$ : $K P \\cdot M Q+K Q \\cdot M P \\geq P Q \\cdot K M$, and now substitute the relations from above, leading to\n\n$$\nR \\sin \\alpha(K P+K Q) \\geq R \\sin \\alpha \\cdot B C\n$$\n\nwhich is precisely the conclusion whenever $\\sin \\alpha \\neq 0$. The case when $\\sin \\alpha=0$ can be treated either directly, or via a limit argument.", "metadata": {"resource_path": "RMM/segmented/en-2023-RMM2023-Day2-English_Solutions.jsonl", "problem_match": "\nProblem 4.", "solution_match": "\nSolution 3"}}
{"year": "2023", "tier": "T1", "problem_label": "4", "problem_type": null, "exam": "RMM", "problem": "Given a triangle $A B C$, let $H$ and $O$ be its orthocentre and circumcentre, respectively. Let $K$ be the midpoint of the line segment $A H$. Let further $\\ell$ be a line through $O$, and let $P$ and $Q$ be the orthogonal projections of $B$ and $C$ onto $\\ell$, respectively. Prove that $K P+K Q \\geq B C$.\n\n# Russia, Vasily Mokin", "solution": ". Denote by $R$ and $O$ the circumradius and the circumcentre of triangle $A B C$, respectively. As in Solution 1, we see that $M K=R$.\n\nAssume now that $\\ell$ is fixed, while $A$ moves along the fixed circle $(A B C)$. Then $K$ will move along a cricle centred at $M$ with radius $R$. We must show that for each point $K$ on this circle we have $B C \\leq K P+K Q$. In doing so, we prove that the afore-mentioned circle contains an ellipse with foci at $Q$ and $P$ with distance $B C$.\n\nLet $S$ be the foot of the perpendicular from $M$ to $P Q$, it is easy to verify that $S$ is the center of the ellipse. We shall then consider it as the origin. Let $u=\\frac{B C}{2}$ and $t=\\frac{P Q}{2}$; notice that $u$ is the major semi-axis of the ellipse and $\\sqrt{u^{2}-t^{2}}$ is the minor one. Assume $X(x, y)$ is a point on this ellipse. We now need to prove $M X \\leq R$.\n\nSince $X$ is on the ellipse, we can write $(x, y)=$ $\\left(u \\cos \\theta, \\sqrt{u^{2}-t^{2}} \\sin \\theta\\right)$, for some $\\theta \\in(0,2 \\pi)$. Since $M X^{2}=x^{2}+(y+M S)^{2}$, we can expand and obtain $M X^{2}=u^{2}+M S^{2}-t^{2} \\cdot \\sin ^{2} \\theta+2 M S \\cdot \\sqrt{u^{2}-t^{2}} \\cdot \\sin \\theta$.\n\nAdd and subtract $M S^{2}\\left(u^{2}-t^{2}\\right) / t^{2}$ in order to obtain a square on the right hand side: $M X^{2}=u^{2}+$ $M S^{2}+\\frac{M S^{2}\\left(u^{2}-t^{2}\\right)}{t^{2}}-\\left(t \\sin \\theta-\\frac{M S \\sqrt{u^{2}-t^{2}}}{t}\\right)^{2}$. It now suffices to show that $u^{2}+M S^{2}+\\frac{M S^{2}\\left(u^{2}-t^{2}\\right)}{t^{2}}=$ $R^{2}$, since then it would immediately follow that $M X^{2} \\leq R^{2}$.\n\nApplying Pythagorean theorem in triangles $O B M$ and $O S M$, we obtain $R^{2}=u^{2}+O M^{2}$ and $O M^{2}=M S^{2}+O S^{2}$, so it remains to prove that $O S^{2}=\\frac{M S^{2}\\left(u^{2}-t^{2}\\right)}{t^{2}} . \\quad$ Let $\\alpha=\\angle(O P, B M)$, then $O S / M S=\\tan ^{t^{2}} \\alpha$ and $t / u=\\cos \\alpha$, so $O S^{2}=$ $M S^{2} \\tan ^{2} \\alpha=M S^{2}\\left(\\frac{1-\\cos ^{2} \\alpha}{\\cos ^{2} \\alpha}\\right)=M S^{2} \\cdot \\frac{u^{2}-t^{2}}{t^{2}}$, which is the desired result.", "metadata": {"resource_path": "RMM/segmented/en-2023-RMM2023-Day2-English_Solutions.jsonl", "problem_match": "\nProblem 4.", "solution_match": "\nSolution 4"}}
{"year": "2023", "tier": "T1", "problem_label": "5", "problem_type": null, "exam": "RMM", "problem": "Let $P(x), Q(x), R(x)$ and $S(x)$ be non-constant polynomials with real coefficients such that $P(Q(x))=R(S(x))$. Suppose that the degree of $P(x)$ is divisible by the degree of $R(x)$.\nProve that there is a polynomial $T(x)$ with real coefficients such that $P(x)=R(T(x))$.\n\n## Iran, NaVid Safaei", "solution": ". Degree comparison of $P(Q(x))$ and $R(S(x))$ implies that $q=\\operatorname{deg} Q \\mid \\operatorname{deg} S=s$. We will show that $S(x)=T(Q(x))$ for some polynomial $T$. Then $P(Q(x))=R(S(x))=R(T(Q(x)))$, so the polynomial $P(t)-R(T(t))$ vanishes upon substitution $t=S(x)$; it therefore vanishes identically, as desired.\n\nChoose the polynomials $T(x)$ and $M(x)$ such that\n\n$$\nS(x)=T(Q(x))+M(x)\n$$\n\nwhere $\\operatorname{deg} M$ is minimised; if $M=0$, then we get the desired result. For the sake of contradiction, suppose $M \\neq 0$. Then $q \\nmid m=\\operatorname{deg} M$; otherwise, $M(x)=\\beta Q(x)^{m / q}+M_{1}(x)$, where $\\beta$ is some number and $\\operatorname{deg} M_{1}<\\operatorname{deg} M$, contradicting the choice of $M$. In particular, $0<m<s$ and hence $\\operatorname{deg} T(Q(x))=s$.\n\nSubstitute now (*) into $R(S(x))-P(Q(x))=$ 0 ; let $\\alpha$ be the leading coefficient of $R(x)$ and let $r=\\operatorname{deg} R(x)$. Expand the brackets to get a sum of powers of $Q(x)$ and other terms including powers of $M(x)$ as well. Amongst the latter, the unique term of highest degree is $\\operatorname{arM}(x) T(Q(x))^{r-1}$. So, for some polyno$\\operatorname{mial} N(x), \\quad N(Q(x))=\\alpha r M(x) T(Q(x))^{r-1}+$ a polynomial of lower degree.\n\nThis is impossible, since $q$ divides the degree of the left-hand member, but not that of the righthand member.", "metadata": {"resource_path": "RMM/segmented/en-2023-RMM2023-Day2-English_Solutions.jsonl", "problem_match": "\nProblem 5.", "solution_match": "\nSolution 1"}}
{"year": "2023", "tier": "T1", "problem_label": "5", "problem_type": null, "exam": "RMM", "problem": "Let $P(x), Q(x), R(x)$ and $S(x)$ be non-constant polynomials with real coefficients such that $P(Q(x))=R(S(x))$. Suppose that the degree of $P(x)$ is divisible by the degree of $R(x)$.\nProve that there is a polynomial $T(x)$ with real coefficients such that $P(x)=R(T(x))$.\n\n## Iran, NaVid Safaei", "solution": ". All polynomials in the solution have real coefficients. As usual, the degree of a polynomial $f(x)$ is denoted $\\operatorname{deg} f(x)$.\n\nOf all pairs of polynomials $P(x), R(x)$, satisfying the conditions in the statement, choose one, say, $P_{0}(x), R_{0}(x)$, so that $P_{0}(Q(x))=R_{0}(S(x))$ has a minimal (positive) degree. We will show that $\\operatorname{deg} R_{0}(x)=1$, say, $R_{0}(x)=\\alpha x+\\beta$ for some real numbers $\\alpha \\neq 0$ and $\\beta$, so $P_{0}(Q(x))=\\alpha S(x)+\\beta$. Hence $S(x)=T(Q(x))$ for some polynomial $T(x)$.\n\nNow, if $P(x)$ and $R(x)$ are polynomials satisfying $P(Q(x))=R(S(x))$, then $P(Q(x))=$ $R(T(Q(x)))$. Since $Q(x)$ is not constant, it takes infinitely many values, so $P(x)$ and $R(T(x))$ agree at infinitely many points, implying that $P(x)=$ $R(T(x))$, as required.\n\nIt is therefore sufficient to solve the problem in the particular case where $F(x)=P(Q(x))=$ $R(S(x))$ has a minimal degree. Let $d=$ $\\operatorname{gcd}(\\operatorname{deg} Q(x), \\operatorname{deg} S(x))$ to write $\\operatorname{deg} Q(x)=a d$ and $\\operatorname{deg} S(x)=b d$, where $\\operatorname{gcd}(a, b)=1$. Then\n$\\operatorname{deg} P(x)=b c, \\operatorname{deg} R(x)=a c$ and $\\operatorname{deg} F(x)=a b c d$ for some positive integer $c$. We will show that minimality of $\\operatorname{deg} F(x)$ forces $c=1$, so $\\operatorname{deg} P(x)=b$, $\\operatorname{deg} R(x)=a$ and $\\operatorname{deg} F(x)=a b d$. The conditions $a=\\operatorname{deg} R(x) \\mid \\operatorname{deg} P(x)=b$ and $\\operatorname{gcd}(a, b)=1$ then force $a=1$, as stated above.\n\nConsequently, the only thing we are left with is the proof of the fact that $c=1$. For convenience, we may and will assume that $P(x), Q(x), R(x), S(x)$ are all monic; hence so is $F(x)$. The argument hinges on the lemma below.\nLemma. If $f(x)$ is a monic polynomial of degree $m n$, then there exists a degree $n$ monic polynomial $g(x)$ such that $\\operatorname{deg}\\left(f(x)-g(x)^{m}\\right)<(m-1) n$. (If $m=0$ or 1 , or $n=0$, the conclusion is still consistent with the usual convention that the identically zero polynomial has degree $-\\infty$.)\nProof. Write $f(x)=\\sum_{k=0}^{m n} \\alpha_{k} x^{k}, \\alpha_{m n}=1$, and seek $g(x)=\\sum_{k=0}^{n} \\beta_{k} x^{k}, \\beta_{n}=1$, so as to fit the bill. To this end, notice that, for each positive integer $k \\leq n$, the coefficient of $x^{m n-k}$ in the expansion of $g(x)^{m}$ is of the form $m \\beta_{n-k}+\\varphi_{k}\\left(\\beta_{n}, \\ldots, \\beta_{n-k+1}\\right)$, where $\\varphi_{k}\\left(\\beta_{n}, \\ldots, \\beta_{n-k+1}\\right)$ is an algebraic expression in $\\beta_{n}, \\ldots, \\beta_{n-k+1}$. Recall that $\\beta_{n}=1$ to determine the $\\beta_{n-k}$ recursively by requiring $\\beta_{n-k}=$ $\\frac{1}{m}\\left(a_{m n-k}-\\varphi_{k}\\left(\\beta_{n}, \\ldots, \\beta_{n-k+1}\\right)\\right), k=1, \\ldots, n$.\n\nThe outcome is then the desired polynomial $g(x)$.\n\nWe are now in a position to prove that $c=$ 1. Suppose, if possible, that $c>1$. By the lemma, there exist monic polynomials $U(x)$ and $V(x)$ of degree $b$ and $a$, respectively, such that $\\operatorname{deg}\\left(P(x)-U(x)^{c}\\right)<(c-1) b$ and $\\operatorname{deg}(R(x)-$ $\\left.V(x)^{c}\\right)<(c-1) a$. Then $\\operatorname{deg}\\left(F(x)-U(Q(x))^{c}\\right)=$ $\\operatorname{deg}\\left(P(Q(x))-U(Q(x))^{c}\\right)<(c-1) a b d, \\operatorname{deg}(F(x)-$ $\\left.V(S(x))^{c}\\right)=\\operatorname{deg}\\left(R(S(x))-V(S(x))^{c}\\right)<(c-1) a b d$, so $\\operatorname{deg}\\left(U(Q(x))^{c}-V(S(x))^{c}\\right)=\\operatorname{deg}((F(x)-$ $\\left.\\left.V(S(x))^{c}\\right)-\\left(F(x)-U(Q(x))^{c}\\right)\\right)<(c-1) a b d$.\n\nOn the other hand, $U(Q(x))^{c}-V(S(x))^{c}=$ $(U(Q(x))-V(S(x)))\\left(U(Q(x))^{c-1}+\\cdots+\\right.$ $\\left.V(S(x))^{c-1}\\right)$.\n\nBy the preceding, the degree of the left-hand member is (strictly) less than $(c-1) a b d$ which is precisely the degree of the second factor in the righthand member. This forces $U(Q(x))=V(S(x))$, so $U(Q(x))=V(S(x))$ has degree $a b d<a b c d=$ $\\operatorname{deg} F(x)$ - a contradiction. Consequently, $c=1$. This completes the argument and concludes the proof.", "metadata": {"resource_path": "RMM/segmented/en-2023-RMM2023-Day2-English_Solutions.jsonl", "problem_match": "\nProblem 5.", "solution_match": "\nSolution 2"}}
{"year": "2023", "tier": "T1", "problem_label": "6", "problem_type": null, "exam": "RMM", "problem": "Let $r, g, b$ be non-negative integers. Let $\\Gamma$ be a connected graph on $r+g+b+1$ vertices. The edges of $\\Gamma$ are each coloured red, green or blue. It turns out that $\\Gamma$ has\n\n- a spanning tree in which exactly $r$ of the edges are red,\n- a spanning tree in which exactly $g$ of the edges are green and\n- a spanning tree in which exactly $b$ of the edges are blue.\n\nProve that $\\Gamma$ has a spanning tree in which exactly $r$ of the edges are red, exactly $g$ of the edges are green and exactly $b$ of the edges are blue.\n\nRussia, Vasily Mokin", "solution": ". Induct on $n=r+g+b$. The base case, $n=1$, is clear.\n\nLet now $n>1$. Let $V$ denote the vertex set of $\\Gamma$, and let $T_{r}, T_{g}$, and $T_{b}$ be the trees with exactly $r$ red edges, $g$ green edges, and $b$ blue edges, respectively. Consider two cases.\nCase 1: There exists a partition $V=A \\sqcup B$ of the vertex set into two non-empty parts such that the edges joining the parts all bear the same colour, say, blue.\n\nSince $\\Gamma$ is connected, it has a (necessarily blue) edge connecting $A$ and $B$. Let $e$ be one such.\n\nAssume that $T$, one of the three trees, does not contain $e$. Then the graph $T \\cup\\{e\\}$ has a cycle $C$ through $e$. The cycle $C$ should contain another edge $e^{\\prime}$ connecting $A$ and $B$; the edge $e^{\\prime}$ is also blue. Replace $e^{\\prime}$ by $e$ in $T$ to get another tree $T^{\\prime}$ with the same number of edges of each colour as in $T$, but containing $e$.\n\nPerforming such an operation to all three trees, we arrive at the situation where the three trees $T_{r}^{\\prime}$, $T_{g}^{\\prime}$, and $T_{b}^{\\prime}$ all contain $e$. Now shrink $e$ by identifying its endpoints to obtain a graph $\\Gamma^{*}$, and set $r^{*}=r, g^{*}=g$, and $b^{*}=b-1$. The new graph satisfies the conditions in the statement for those new values - indeed, under the shrinking, each of the trees $T_{r}^{\\prime}, T_{g}^{\\prime}$, and $T_{b}^{\\prime}$ loses a blue edge. So $\\Gamma^{*}$ has a spanning tree with exactly $r$ red, exactly $g$\ngreen, and exactly $b-1$ blue edges. Finally, pass back to $\\Gamma$ by restoring $e$, to obtain the a desired spanning tree in $\\Gamma$.\n\nCase 2: There is no such a partition.\nConsider all possible collections $(R, G, B)$, where $R, G$ and $B$ are acyclic sets consisting of $r$ red edges, $g$ green edges, and $b$ blue edges, respectively. By the problem assumptions, there is at least one such collection. Amongst all such collections, consider one such that the graph on $V$ with edge set $R \\cup G \\cup B$ has the smallest number $k$ of components. If $k=1$, then the collection provides the edges of a desired tree (the number of edges is one less than the number of vertices).\n\nAssume now that $k \\geq 2$; then in the resulting graph some component $K$ contains a cycle $C$. Since $R, G$, and $B$ are acyclic, $C$ contains edges of at least two colours, say, red and green. By assumption, the edges joining $V(K)$ to $V \\backslash V(K)$ bear at least two colours; so one of these edges is either red or green. Without loss of generality, consider a red such edge $e$.\n\nLet $e^{\\prime}$ be a red edge in $C$ and set $R^{\\prime}=R \\backslash\\left\\{e^{\\prime}\\right\\} \\cup$ $\\{e\\}$. Then $\\left(R^{\\prime}, G, B\\right)$ is a valid collection providing a smaller number of components. This contradicts minimality of the choice above and concludes the proof.", "metadata": {"resource_path": "RMM/segmented/en-2023-RMM2023-Day2-English_Solutions.jsonl", "problem_match": "\nProblem 6.", "solution_match": "\nSolution 1"}}
{"year": "2023", "tier": "T1", "problem_label": "6", "problem_type": null, "exam": "RMM", "problem": "Let $r, g, b$ be non-negative integers. Let $\\Gamma$ be a connected graph on $r+g+b+1$ vertices. The edges of $\\Gamma$ are each coloured red, green or blue. It turns out that $\\Gamma$ has\n\n- a spanning tree in which exactly $r$ of the edges are red,\n- a spanning tree in which exactly $g$ of the edges are green and\n- a spanning tree in which exactly $b$ of the edges are blue.\n\nProve that $\\Gamma$ has a spanning tree in which exactly $r$ of the edges are red, exactly $g$ of the edges are green and exactly $b$ of the edges are blue.", "solution": ". For a spanning tree $T$ in $\\Gamma$, denote by $r(T), g(T)$, and $b(T)$ the number of red, green, and blue edges in $T$, respectively.\n\nAssume that $\\mathcal{C}$ is some collection of spanning trees in $\\Gamma$. Write\n\n$$\n\\begin{array}{rlrl}\nr(\\mathcal{C}) & =\\min _{T \\in \\mathcal{C}} r(T), & & g(\\mathcal{C})=\\min _{T \\in \\mathcal{C}} g(T) \\\\\nb(\\mathcal{C}) & =\\min _{T \\in \\mathcal{C}} b(T), & & R(\\mathcal{C})=\\max _{T \\in \\mathcal{C}}(T) \\\\\nG(\\mathcal{C}) & =\\max _{T \\in \\mathcal{C}} g(T), & B(\\mathcal{C})=\\max _{T \\in \\mathcal{C}} b(T)\n\\end{array}\n$$\n\nSay that a collection $\\mathcal{C}$ is good if $r \\in[r(\\mathcal{C}, R(\\mathcal{C})]$, $g \\in[g(\\mathcal{C}, G(\\mathcal{C})]$, and $b \\in[b(\\mathcal{C}, B(\\mathcal{C})]$. By the problem conditions, the collection of all spanning trees in $\\Gamma$ is good.\n\nFor a good collection $\\mathcal{C}$, say that an edge $e$ of $\\Gamma$ is suspicious if $e$ belongs to some tree in $\\mathcal{C}$ but not to all trees in $\\mathcal{C}$. Choose now a good collection $\\mathcal{C}$ minimizing the number of suspicious edges. If $\\mathcal{C}$ contains a desired tree, we are done. Otherwise, without loss of generality, $r(\\mathcal{C})<r$ and $G(\\mathcal{C})>g$.\n\nWe now distinguish two cases.\nCase 1: $B(\\mathcal{C})=b$ 。\nLet $T^{0}$ be a tree in $\\mathcal{C}$ with $g\\left(T^{0}\\right)=g(\\mathcal{C}) \\leq g$. Since $G(\\mathcal{C})>g$, there exists a green edge $e$ contained in some tree in $\\mathcal{C}$ but not in $T^{0}$; clearly, $e$ is suspicious. Fix one such green edge $e$.\n\nNow, for every $T$ in $\\mathcal{C}$, define a spanning tree $T_{1}$ of $\\Gamma$ as follows. If $T$ does not contain $e$, then $T_{1}=T$; in particular, $\\left(T^{0}\\right)_{1}=T^{0}$. Otherwise, the graph $T \\backslash\\{e\\}$ falls into two components. The tree $T_{0}$ contains some edge $e^{\\prime}$ joining those components; this edge is necessarily suspicious. Choose one such edge and define $T_{1}=T \\backslash\\{e\\} \\cup\\left\\{e^{\\prime}\\right\\}$.\n\nLet $\\mathcal{C}_{1}=\\left\\{T_{1}: T \\in \\mathcal{C}\\right\\}$. All edges suspicious for $\\mathcal{C}_{1}$ are also suspicious for $\\mathcal{C}$, but no tree in $\\mathcal{C}_{1}$ con-\n\n## Russia, Vasily Mokin\n\ntains $e$. So the number of suspicious edges for $\\mathcal{C}_{1}$ is strictly smaller than that for $\\mathcal{C}$.\n\nWe now show that $\\mathcal{C}_{1}$ is good, reaching thereby a contradiction with the choice of $\\mathcal{C}$. For every $T$ in $\\mathcal{C}$, the tree $T_{1}$ either coincides with $T$ or is obtained from it by removing a green edge and adding an edge of some colour. This already shows that $g\\left(\\mathcal{C}_{1}\\right) \\leq g(\\mathcal{C}) \\leq g, G\\left(\\mathcal{C}_{1}\\right) \\geq G(\\mathcal{C})-1 \\geq g$, $R\\left(\\mathcal{C}_{1}\\right) \\geq R(\\mathcal{C}) \\geq r, r\\left(\\mathcal{C}_{1}\\right) \\leq r(\\mathcal{C})+1 \\leq r$, and $B\\left(\\mathcal{C}_{1}\\right) \\geq B(\\mathcal{C}) \\geq b$. Finally, we get $b\\left(T^{0}\\right) \\leq$ $B(\\mathcal{C})=b$; since $\\mathcal{C}_{1}$ contains $T^{0}$, it follows that $b\\left(\\mathcal{C}_{1}\\right) \\leq b\\left(T^{0}\\right) \\leq b$, which concludes the proof.\nCase 2: $B(\\mathcal{C})>b$.\nConsider a tree $T^{0}$ in $\\mathcal{C}$ satisfying $r\\left(T^{0}\\right)=$ $R(\\mathcal{C}) \\geq r$. Since $r(\\mathcal{C})<r$, the tree $T^{0}$ contains a suspicious red edge. Fix one such edge $e$.\n\nNow, for every $T$ in $\\mathcal{C}$, define a spanning tree $T_{2}$ of $\\Gamma$ as follows. If $T$ contains $e$, then $T_{2}=T$; in particular, $\\left(T^{0}\\right)_{2}=T^{0}$. Otherwise, the graph $T \\cup\\{e\\}$ contains a cycle $C$ through $e$. This cycle contains an edge $e^{\\prime}$ absent from $T^{0}$ (otherwise $T^{0}$ would contain the cycle $C$ ), so $e^{\\prime}$ is suspicious. Choose one such edge and define $T_{2}=T \\backslash\\left\\{e^{\\prime}\\right\\} \\cup\\{e\\}$.\n\nLet $\\mathcal{C}_{2}=\\left\\{T_{2}: T \\in \\mathcal{C}\\right\\}$. All edges suspicious for $\\mathcal{C}_{2}$ are also suspicious for $\\mathcal{C}$, but all trees in $\\mathcal{C}_{2}$ contain $e$. So the number of suspicious edges for $\\mathcal{C}_{2}$ is strictly smaller than that for $\\mathcal{C}$.\n\nWe now show that $\\mathcal{C}_{2}$ is good, reaching again a contradiction. For every $T$ in $\\mathcal{C}$, the tree $T_{2}$ either coincides with $T$ or is obtained from it by removing some edge and adding a red edge. This shows that $r\\left(\\mathcal{C}_{2}\\right) \\leq r(\\mathcal{C})+1 \\leq r, R\\left(\\mathcal{C}_{2}\\right) \\geq R(\\mathcal{C}) \\geq r$, $G\\left(\\mathcal{C}_{2}\\right) \\geq G(\\mathcal{C})-1 \\geq g, g\\left(\\mathcal{C}_{2}\\right) \\leq g(\\mathcal{C}) \\leq g$, $b\\left(\\mathcal{C}_{1}\\right) \\leq b(\\mathcal{C}) \\leq b$ and $B\\left(\\mathcal{C}_{2}\\right) \\geq B(\\mathcal{C})-1 \\geq b$. This concludes the proof.", "metadata": {"resource_path": "RMM/segmented/en-2023-RMM2023-Day2-English_Solutions.jsonl", "problem_match": "\nProblem 6.", "solution_match": "\nSolution 2"}}