{"year": "2024", "tier": "T1", "problem_label": "1", "problem_type": null, "exam": "USAMO", "problem": "Find all integers $n \\geq 3$ such that the following property holds: if we list the divisors of $n$ ! in increasing order as $1=d_{1}<d_{2}<\\cdots<d_{k}=n$ !, then we have  $$ d_{2}-d_{1} \\leq d_{3}-d_{2} \\leq \\cdots \\leq d_{k}-d_{k-1} $$", "solution": "  The answer is $n \\in\\{3,4\\}$. These can be checked by listing all the divisors:  - For $n=3$ we have $(1,2,3,6)$. - For $n=4$ we have $(1,2,3,4,6,8,12,24)$.  We prove these are the only ones. The numbers $5 \\leq n \\leq 12$ all fail because:  - For $n=5$ we have $20-15>24-20$. - For $n=6$ we have $18-15>20-18$. - For $7 \\leq n \\leq 12$ we have because $14-12>15-14$ (and $13 \\nmid n!$ ).  Now assume $n \\geq 13$. In that case, we have  $$ \\left\\lfloor\\frac{n}{2}\\right\\rfloor^{2}-1 \\geq 2 n $$  So by Bertrand postulate, we can find a prime $p$ such that  $$ n<p<\\left\\lfloor\\frac{n}{2}\\right\\rfloor^{2}-1 $$  However, note that  $$ \\left\\lfloor\\frac{n}{2}\\right\\rfloor^{2}-1=\\left(\\left\\lfloor\\frac{n}{2}\\right\\rfloor-1\\right)\\left(\\left\\lfloor\\frac{n}{2}\\right\\rfloor+1\\right), \\quad\\left\\lfloor\\frac{n}{2}\\right\\rfloor^{2} $$  are consecutive integers both dividing $n$ ! (the latter number divides $\\left\\lfloor\\frac{n}{2}\\right\\rfloor \\cdot\\left(2\\left\\lfloor\\frac{n}{2}\\right\\rfloor\\right)$ ). So the prime $p$ causes the desired contradiction.", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-2024-notes.jsonl"}}
{"year": "2024", "tier": "T1", "problem_label": "2", "problem_type": null, "exam": "USAMO", "problem": "Let $S_{1}, S_{2}, \\ldots, S_{100}$ be finite sets of integers whose intersection is not empty. For each non-empty $T \\subseteq\\left\\{S_{1}, S_{2}, \\ldots, S_{100}\\right\\}$, the size of the intersection of the sets in $T$ is a multiple of $|T|$. What is the smallest possible number of elements which are in at least 50 sets?", "solution": "  The answer is $50\\binom{100}{50}$. ब Rephrasing (cosmetic translation only, nothing happens yet). We encode with binary strings $v \\in \\mathbb{F}_{2}^{100}$ of length 100 . Write $v \\subseteq w$ if $w$ has 1 's in every component $v$ does, and let $|v|$ denote the number of 1 's in $v$. Then for each $v$, we let $f(v)$ denote the number of elements $x \\in \\bigcup S_{i}$ such that $x \\in S_{i} \\Longleftrightarrow v_{i}=1$. For example,  - $f(1 \\ldots 1)$ denotes $\\left|\\bigcap_{1}^{100} S_{i}\\right|$, so we know $f(1 \\ldots 1) \\equiv 0(\\bmod 100)$. - $f(1 \\ldots 10)$ denotes the number of elements in $S_{1}$ through $S_{99}$ but not $S_{100}$ so we know that $f(1 \\ldots 1)+f(1 \\ldots 10) \\equiv 0(\\bmod 99)$. - . . And so on.  So the problem condition means that $f(v)$ translates to the statement  $$ P(u): \\quad|u| \\text { divides } \\sum_{v \\supseteq u} f(v) $$  for any $u \\neq 0 \\ldots 0$, plus one extra condition $f(1 \\ldots 1)>0$. And the objective function is to minimize the quantity  $$ A:=\\sum_{|v| \\geq 50} f(v) . $$  So the problem is transformed into an system of equations over $\\mathbb{Z}_{\\geq 0}$ (it's clear any assignment of values of $f(v)$ can be translated to a sequence ( $S_{1}, \\ldots, S_{100}$ ) in the original notation).  Note already that: Claim - It suffices to assign $f(v)$ for $|v| \\geq 50$.   I Construction. Consider the construction  $$ f_{0}(v)=2|v|-100 $$  This construction is valid since if $|u|=100-k$ for $k \\leq 50$ then  $$ \\begin{aligned} \\sum_{v \\supseteq u} f_{0}(v) & =\\binom{k}{0} \\cdot 100+\\binom{k}{1} \\cdot 98+\\binom{k}{2} \\cdot 96+\\cdots+\\binom{k}{k} \\cdot(100-2 k) \\\\ & =(100-k) \\cdot 2^{k}=|u| \\cdot 2^{k} \\end{aligned} $$  is indeed a multiple of $|u|$, hence $P(u)$ is true. In that case, the objective function is  $$ A=\\sum_{i=50}^{100}\\binom{100}{i}(2 i-100)=50\\binom{100}{50} $$  as needed. Remark. This construction is the \"easy\" half of the problem because it coincides with what you get from a greedy algorithm by downwards induction on $|u|$ (equivalently, induction on $k=100-|u| \\geq 0)$. To spell out the first three steps,  - We know $f(1 \\ldots 1)$ is a nonzero multiple of 100 , so it makes sense to guess $f(1 \\ldots 1)=$ 100. - Then we have $f(1 \\ldots 10)+100 \\equiv 0(\\bmod 99)$, and the smallest multiple of 99 which is at least 100 is 198 . So it makes sense to guess $f(1 \\ldots 10)=98$, and similarly guess $f(v)=98$ whenever $|v|=99$. - Now when we consider, say $v=1 \\ldots 100$ with $|v|=98$, we get  $$ f(1 \\ldots 100)+\\underbrace{f(1 \\ldots 101)}_{=98}+\\underbrace{f(1 \\ldots 110)}_{=98}+\\underbrace{f(1 \\ldots 111)}_{=100} \\equiv 0 \\quad(\\bmod 98) $$  we obtain $f(1 \\ldots 100) \\equiv 96(\\bmod 98)$. That makes $f(1 \\ldots 100)=96$ a reasonable guess. Continuing in this way gives the construction above.   We define a push-down on $v$ as the following operation:  - Pick any $v$ such that $|v| \\geq 50$ and $f(v) \\geq|v|$. - Decrease $f(v)$ by $|v|$. - For every $w$ such that $w \\subseteq v$ and $|w|=|v|-1$, increase $f(w)$ by 1 .  Claim - Apply a push-down preserves the main divisibility condition. Moreover, it doesn't change $A$ unless $|v|=50$, where it decreases $A$ by 50 instead.   To see $A$ doesn't change for $|v|>50$, note $|v|$ terms increase by 1 while one term decreases by $-|v|$. When $|v|=50$, only $f(v)$ decreases by 50 .  Now, given a valid assignment, we can modify it as follows:  - First apply pushdowns on $1 \\ldots 1$ until $f(1 \\ldots 1)=100$; - Then we may apply pushdowns on each $v$ with $|v|=99$ until $f(v)<99$; - Then we may apply pushdowns on each $v$ with $|v|=98$ until $f(v)<98$; - . . .and so on, until we have $f(v)<50$ for $|v|=50$.  Hence we get $f(1 \\ldots 1)=100$ and $0 \\leq f(v)<|v|$ for all $50 \\leq|v| \\leq 100$. However, by downwards induction on $|v|=99,98, \\ldots, 50$, we also have  $$ f(v) \\equiv f_{0}(v) \\quad(\\bmod |v|) \\Longrightarrow f(v)=f_{0}(v) $$  since $f_{0}(v)$ and $f(v)$ are both strictly less than $|v|$. So in fact $f=f_{0}$, and we're done. Remark. The fact that push-downs actually don't change $A$ shows that the equality case we described is far from unique: in fact, we could have made nearly arbitrary sub-optimal decisions during the greedy algorithm and still ended up with an equality case. For a concrete example, the construction  $$ f(v)= \\begin{cases}500 & |v|=100 \\\\ 94 & |v|=99 \\\\ 100-2|v| & 50 \\leq|v| \\leq 98\\end{cases} $$  works fine as well (where we arbitrarily chose 500 at the start, then used the greedy algorithm thereafter).", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-2024-notes.jsonl"}}
{"year": "2024", "tier": "T1", "problem_label": "3", "problem_type": null, "exam": "USAMO", "problem": "Let $(m, n)$ be positive integers with $n \\geq 3$ and draw a regular $n$-gon. We wish to triangulate this $n$-gon into $n-2$ triangles, each colored one of $m$ colors, so that each color has an equal sum of areas. For which $(m, n)$ is such a triangulation and coloring possible?", "solution": "  The answer is if and only if $m$ is a proper divisor of $n$.  ## Lemma  The triangle with vertices $\\omega^{k}, \\omega^{k+a}, \\omega^{k+b}$ has signed area  $$ T(a, b):=\\frac{\\left(\\omega^{a}-1\\right)\\left(\\omega^{b}-1\\right)\\left(\\omega^{-a}-\\omega^{-b}\\right)}{2 i} . $$   $$ \\frac{1}{2 i} \\operatorname{det}\\left[\\begin{array}{ccc} 1 & 1 & 1 \\\\ \\omega^{a} & \\omega^{-a} & 1 \\\\ \\omega^{b} & \\omega^{-b} & 1 \\end{array}\\right]=\\frac{1}{2 i} \\operatorname{det}\\left[\\begin{array}{ccc} 0 & 0 & 1 \\\\ \\omega^{a}-1 & \\omega^{-a}-1 & 1 \\\\ \\omega^{b}-1 & \\omega^{-b}-1 & 1 \\end{array}\\right] $$  which equals the above. 【 Construction. It suffices to actually just take all the diagonals from the vertex 1 , and then color the triangles with the $m$ colors in cyclic order. For example, when $n=9$ and $m=3$, a coloring with red, green, blue would be: ![](https://cdn.mathpix.com/cropped/2024_11_19_bff0e8c9a0bc19d4709dg-07.jpg?height=617&width=615&top_left_y=1873&top_left_x=726)  To see this works one can just do the shoelace calculation: for a given residue $r \\bmod m$, we get an area  $$ \\begin{aligned} \\sum_{j \\equiv r \\bmod m} \\operatorname{Area}\\left(\\omega^{j}, \\omega^{0}, \\omega^{j+1}\\right) & =\\sum_{j \\equiv r \\bmod m} T(-j, 1) \\\\ & =\\sum_{j \\equiv r \\bmod m} \\frac{\\left(\\omega^{-j}-1\\right)\\left(\\omega^{1}-1\\right)\\left(\\omega^{j}-\\omega^{-1}\\right)}{2 i} \\\\ & =\\frac{\\omega-1}{2 i} \\sum_{j \\equiv r \\bmod m}\\left(\\omega^{-j}-1\\right)\\left(\\omega^{j}-\\omega^{-1}\\right) \\\\ & =\\frac{\\omega-1}{2 i}\\left(\\frac{n}{m}\\left(1+\\omega^{-1}\\right)+\\sum_{j \\equiv r \\bmod m}\\left(\\omega^{-j}-\\omega^{j}\\right)\\right) . \\end{aligned} $$  (We allow degenerate triangles where $j \\in\\{-1,0\\}$ with area zero.) However, if $m$ is a proper divisor of $m$, then $\\sum_{j \\equiv r \\bmod m} \\omega^{j}=\\omega^{r}\\left(1+\\omega^{m}+\\omega^{2 m}+\\cdots+\\omega^{n-m}\\right)=0$. Similarly, $\\sum_{j \\equiv r \\bmod m} \\omega^{-j}=0$. So the inner sum vanishes, and the total area of the $m$ th color equals  $$ \\frac{n}{m} \\frac{(\\omega-1)\\left(\\omega^{-1}+1\\right)}{2 i} $$  which does not depend on the residue $r$, proving the coloring works.  Repeating the same calculation as above, we find that if there was a valid triangulation and coloring, the total area of each color would equal  $$ S:=\\frac{n}{m} \\frac{(\\omega-1)\\left(\\omega^{-1}+1\\right)}{2 i} . $$  ## However:  Claim — The number $2 i \\cdot S$ is not an algebraic integer when $m \\nmid n$.  However, each of the quantities $T(a, b)$ is $\\frac{1}{2 i}$ times an algebraic integer. Since a finite sum of algebraic integers is also an algebraic integer, such areas can never sum to $S$.  Remark. If one wants to avoid citing the fact that $\\mathcal{O}_{K}=\\mathbb{Z}[\\omega]$, then one can instead note that $T(a, b)$ is actually always divisible by $(\\omega-1)\\left(\\omega^{-1}+1\\right)$ over the algebraic integers (at least one of $\\left\\{\\omega^{a}-1, \\omega^{b}-1, \\omega^{-a}-\\omega^{-b}\\right\\}$ is a multiple of $\\omega+1$, by casework on $a, b \\bmod 2$ ). Then one using $\\frac{2 i}{(\\omega-1)\\left(\\omega^{-1}+1\\right)}$ as the scaling factor instead of $2 i$, one sees that we actually need $\\frac{n}{m}$ to be an algebraic integer, which happens only when $m$ divides $n$.", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-2024-notes.jsonl"}}
{"year": "2024", "tier": "T1", "problem_label": "4", "problem_type": null, "exam": "USAMO", "problem": "Let $m$ and $n$ be positive integers. A circular necklace contains $m n$ beads, each either red or blue. It turned out that no matter how the necklace was cut into $m$ blocks of $n$ consecutive beads, each block had a distinct number of red beads. Determine, with proof, all possible values of the ordered pair $(m, n)$.", "solution": "  The answer is $m \\leq n+1$ only.  Construction when $m=n+1$. For concreteness, here is the construction for $n=4$, which obviously generalizes. The beads are listed in reading order as an array with $n+1$ rows and $n$ columns. Four of the blue beads have been labeled $B_{1}, \\ldots, B_{n}$ to make them easier to track as they move.  $$ T_{0}=\\left[\\begin{array}{llll} R & R & R & R \\\\ R & R & R & B_{1} \\\\ R & R & B & B_{2} \\\\ R & B & B & B_{3} \\\\ B & B & B & B_{4} \\end{array}\\right] $$  To prove this construction works, it suffices to consider the $n$ cuts $T_{0}, T_{1}, T_{2}, \\ldots, T_{n-1}$ made where $T_{i}$ differs from $T_{i-1}$ by having the cuts one bead later also have the property each row has a distinct red count:  $$ T_{1}=\\left[\\begin{array}{llll} R & R & R & R \\\\ R & R & B_{1} & R \\\\ R & B & B_{2} & R \\\\ B & B & B_{3} & B \\\\ B & B & B_{4} & R \\end{array}\\right] \\quad T_{2}=\\left[\\begin{array}{cccc} R & R & R & R \\\\ R & B_{1} & R & R \\\\ B & B_{2} & R & B \\\\ B & B_{3} & B & B \\\\ B & B_{4} & R & R \\end{array}\\right] \\quad T_{3}=\\left[\\begin{array}{cccc} R & R & R & R \\\\ B_{1} & R & R & B \\\\ B_{2} & R & B & B \\\\ B_{3} & B & B & B \\\\ B_{4} & R & R & R \\end{array}\\right] $$  We can construct a table showing for each $1 \\leq k \\leq n+1$ the number of red beads which are in the $(k+1)$ st row of $T_{i}$ from the bottom:  | $k$ | $T_{0}$ | $T_{1}$ | $T_{2}$ | $T_{3}$ | | :---: | :---: | :---: | :---: | :---: | | $k=4$ | 4 | 4 | 4 | 4 | | $k=3$ | 3 | 3 | 3 | 2 | | $k=2$ | 2 | 2 | 1 | 1 | | $k=1$ | 1 | 0 | 0 | 0 | | $k=0$ | 0 | 1 | 2 | 3 |.  This suggests following explicit formula for the entry of the $(i, k)$ th cell which can then be checked straightforwardly:  $$ \\#\\left(\\text { red cells in } k \\text { th row of } T_{i}\\right)= \\begin{cases}k & k>i \\\\ k-1 & i \\geq k>0 \\\\ i & k=0\\end{cases} $$  And one can see for each $i$, the counts are all distinct (they are ( $i, 0,1, \\ldots, k-1, k+1, \\ldots, k)$ from bottom to top). This completes the construction.  Construction when $m<n+1$. Fix $m$. Take the construction for $(m, m-1)$ and add $n+1-m$ cyan beads to the start of each row; for example, if $n=7$ and $m=5$ then the new construction is  $$ T=\\left[\\begin{array}{lllllll} C & C & C & R & R & R & R \\\\ C & C & C & R & R & R & B_{1} \\\\ C & C & C & R & R & B & B_{2} \\\\ C & C & C & R & B & B & B_{3} \\\\ C & C & C & B & B & B & B_{4} \\end{array}\\right] $$  This construction still works for the same reason (the cyan beads do nothing for the first $n+1-m$ shifts, then one reduces to the previous case). If we treat cyan as a shade of blue, this finishes the problem.", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-2024-notes.jsonl"}}
{"year": "2024", "tier": "T1", "problem_label": "5", "problem_type": null, "exam": "USAMO", "problem": "Point $D$ is selected inside acute triangle $A B C$ so that $\\angle D A C=\\angle A C B$ and $\\angle B D C=90^{\\circ}+\\angle B A C$. Point $E$ is chosen on ray $B D$ so that $A E=E C$. Let $M$ be the midpoint of $B C$. Show that line $A B$ is tangent to the circumcircle of triangle $B E M$.", "solution": "  ![](https://cdn.mathpix.com/cropped/2024_11_19_bff0e8c9a0bc19d4709dg-11.jpg?height=661&width=712&top_left_y=960&top_left_x=678)  Claim - We have DQCF is cyclic.  $$ \\begin{aligned} \\measuredangle F D C & =-\\measuredangle C D B=180^{\\circ}-\\left(90^{\\circ}+\\measuredangle C A B\\right)=90^{\\circ}-\\measuredangle C A B \\\\ & =90^{\\circ}-\\measuredangle Q C A=\\measuredangle F Q C . \\end{aligned} $$  To conclude, note that  $$ \\measuredangle B E M=\\measuredangle B F C=\\measuredangle D F C=\\measuredangle D Q C=\\measuredangle A Q C=\\measuredangle A B C=\\measuredangle A B M . $$  Remark (Motivation). Here is one possible way to come up with the construction of point $F$ (at least this is what led Evan to find it). If one directs all the angles in the obvious way, there are really two points $D$ and $D^{\\prime}$ that are possible, although one is outside the triangle; they give corresponding points $E$ and $E^{\\prime}$. The circles $B E M$ and $B E^{\\prime} M$ must then actually coincide since they are both alleged to be tangent to line $A B$. See the figure below. ![](https://cdn.mathpix.com/cropped/2024_11_19_bff0e8c9a0bc19d4709dg-12.jpg?height=1112&width=1041&top_left_y=238&top_left_x=513)  One can already prove using angle chasing that $\\overline{A B}$ is tangent to $\\left(B E E^{\\prime}\\right)$. So the point of the problem is to show that $M$ lies on this circle too. However, from looking at the diagram, one may realize that in fact it seems  $$ \\triangle M E E^{\\prime} \\stackrel{\\triangle}{\\sim} \\triangle C D D^{\\prime} $$  is going to be true from just those marked in the figure (and this would certainly imply the desired concyclic conclusion). Since $M$ is a midpoint, it makes sense to dilate $\\triangle E M E^{\\prime}$ from $B$ by a factor of 2 to get $\\triangle F C F^{\\prime}$ so that the desired similarity is actually a spiral similarity at $C$. Then the spiral similarity lemma says that the desired similarity is equivalent to requiring $\\overline{D D^{\\prime}} \\cap \\overline{F F^{\\prime}}=Q$ to lie on both $(C D F)$ and $\\left(C D^{\\prime} F^{\\prime}\\right)$. Hence the key construction and claim from the solution are both discovered naturally, and we find the solution above. (The points $D^{\\prime}, E^{\\prime}, F^{\\prime}$ can then be deleted to hide the motivation.) Another short solution. Let $Z$ be on line $B D E$ such that $\\angle B A Z=90^{\\circ}$. This lets us interpret the angle condition as follows:  Claim - Points $A, D, Z, C$ are cyclic.  ![](https://cdn.mathpix.com/cropped/2024_11_19_bff0e8c9a0bc19d4709dg-13.jpg?height=804&width=700&top_left_y=249&top_left_x=678)  Define $W$ as the midpoint of $\\overline{B Z}$, so $\\overline{M W} \\| \\overline{C Z}$. And let $O$ denote the center of $(A B C)$.  Claim - Points $M, E, O, W$ are cyclic.   $$ \\begin{aligned} \\measuredangle M O E & =\\measuredangle(\\overline{O M}, \\overline{B C})+\\measuredangle(\\overline{B C}, \\overline{A C})+\\measuredangle(\\overline{A C}, \\overline{O E}) \\\\ & =90^{\\circ}+\\measuredangle B C A+90^{\\circ} \\\\ & =\\measuredangle B C A=\\measuredangle C A D=\\measuredangle C Z D=\\measuredangle M W D=\\measuredangle M W E . \\end{aligned} $$  To finish, note  $$ \\begin{aligned} \\measuredangle M E B & =\\measuredangle M E W=\\measuredangle M O W \\\\ & =\\measuredangle(\\overline{M O}, \\overline{B C})+\\measuredangle(\\overline{B C}, \\overline{A B})+\\measuredangle(\\overline{A B}, \\overline{O W}) \\\\ & =90^{\\circ}+\\measuredangle C B A+90^{\\circ}=\\measuredangle C B A=\\measuredangle M B A . \\end{aligned} $$  This implies the desired tangency.", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-2024-notes.jsonl"}}
{"year": "2024", "tier": "T1", "problem_label": "5", "problem_type": null, "exam": "USAMO", "problem": "Point $D$ is selected inside acute triangle $A B C$ so that $\\angle D A C=\\angle A C B$ and $\\angle B D C=90^{\\circ}+\\angle B A C$. Point $E$ is chosen on ray $B D$ so that $A E=E C$. Let $M$ be the midpoint of $B C$. Show that line $A B$ is tangent to the circumcircle of triangle $B E M$.", "solution": "  ![](https://cdn.mathpix.com/cropped/2024_11_19_bff0e8c9a0bc19d4709dg-11.jpg?height=661&width=712&top_left_y=960&top_left_x=678)  Claim - We have DQCF is cyclic.  $$ \\begin{aligned} \\measuredangle F D C & =-\\measuredangle C D B=180^{\\circ}-\\left(90^{\\circ}+\\measuredangle C A B\\right)=90^{\\circ}-\\measuredangle C A B \\\\ & =90^{\\circ}-\\measuredangle Q C A=\\measuredangle F Q C . \\end{aligned} $$  To conclude, note that  $$ \\measuredangle B E M=\\measuredangle B F C=\\measuredangle D F C=\\measuredangle D Q C=\\measuredangle A Q C=\\measuredangle A B C=\\measuredangle A B M . $$  Remark (Motivation). Here is one possible way to come up with the construction of point $F$ (at least this is what led Evan to find it). If one directs all the angles in the obvious way, there are really two points $D$ and $D^{\\prime}$ that are possible, although one is outside the triangle; they give corresponding points $E$ and $E^{\\prime}$. The circles $B E M$ and $B E^{\\prime} M$ must then actually coincide since they are both alleged to be tangent to line $A B$. See the figure below. ![](https://cdn.mathpix.com/cropped/2024_11_19_bff0e8c9a0bc19d4709dg-12.jpg?height=1112&width=1041&top_left_y=238&top_left_x=513)  One can already prove using angle chasing that $\\overline{A B}$ is tangent to $\\left(B E E^{\\prime}\\right)$. So the point of the problem is to show that $M$ lies on this circle too. However, from looking at the diagram, one may realize that in fact it seems  $$ \\triangle M E E^{\\prime} \\stackrel{\\triangle}{\\sim} \\triangle C D D^{\\prime} $$  is going to be true from just those marked in the figure (and this would certainly imply the desired concyclic conclusion). Since $M$ is a midpoint, it makes sense to dilate $\\triangle E M E^{\\prime}$ from $B$ by a factor of 2 to get $\\triangle F C F^{\\prime}$ so that the desired similarity is actually a spiral similarity at $C$. Then the spiral similarity lemma says that the desired similarity is equivalent to requiring $\\overline{D D^{\\prime}} \\cap \\overline{F F^{\\prime}}=Q$ to lie on both $(C D F)$ and $\\left(C D^{\\prime} F^{\\prime}\\right)$. Hence the key construction and claim from the solution are both discovered naturally, and we find the solution above. (The points $D^{\\prime}, E^{\\prime}, F^{\\prime}$ can then be deleted to hide the motivation.) I A Menelaus-based approach (Kevin Ren). Let $P$ be on $\\overline{B C}$ with $A P=P C$. Let $Y$ be the point on line $A B$ such that $\\angle A C Y=90^{\\circ}$; as $\\angle A Y C=90^{\\circ}-A$ it follows $B D Y C$ is cyclic. Let $K=\\overline{A P} \\cap \\overline{C Y}$, so $\\triangle A C K$ is a right triangle with $P$ the midpoint of its hypotenuse. ![](https://cdn.mathpix.com/cropped/2024_11_19_bff0e8c9a0bc19d4709dg-14.jpg?height=1198&width=1080&top_left_y=246&top_left_x=491)  Claim - Triangles $B P E$ and $D Y K$ are similar.   Claim - Triangles $B E M$ and $Y D C$ are similar.   $$ \\frac{B P}{B C} \\frac{Y C}{Y K} \\frac{A K}{A P}=1 $$  Since $A K / A P=2$ (note that $P$ is the midpoint of the hypotenuse of right triangle $A C K)$ and $B C=2 B M$, this simplifies to  $$ \\frac{B P}{B M}=\\frac{Y K}{Y C} $$  To finish, note that  $$ \\measuredangle D B A=\\measuredangle D B Y=\\measuredangle D C Y=\\measuredangle B M E $$  implying the desired tangency.  ![](https://cdn.mathpix.com/cropped/2024_11_19_bff0e8c9a0bc19d4709dg-15.jpg?height=1097&width=1021&top_left_y=431&top_left_x=523)  Denote by $S$ the second intersection of $\\Gamma$ and $\\Omega$. The main idea behind is to consider the spiral similarity  $$ \\Psi: \\Omega \\rightarrow \\Gamma \\quad C \\mapsto M \\text { and } Y \\mapsto B $$  centered at $S$ (due to the spiral similarity lemma), and show that $\\Psi(D)=E$. The spiral similarity lemma already promises $\\Psi(D)$ lies on line $B D$.  Claim - We have $\\Psi(A)=O$, the circumcenter of $A B C$.   Claim - $\\Psi$ maps line $A D$ to line $O P$.   $$ \\measuredangle(\\overline{A D}, \\overline{O P})=\\measuredangle A P O=\\measuredangle O P C=\\measuredangle Y C P=\\measuredangle(\\overline{Y C}, \\overline{B M}) $$  As $\\Psi$ maps line $Y C$ to line $B M$ and $\\Psi(A)=O$, we're done. Hence $\\Psi(D)$ should not only lie on $B D$ but also line $O P$. This proves $\\Psi(D)=E$, so $E \\in \\Gamma$ as needed.", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-2024-notes.jsonl"}}
{"year": "2024", "tier": "T1", "problem_label": "6", "problem_type": null, "exam": "USAMO", "problem": "Let $n>2$ be an integer and let $\\ell \\in\\{1,2, \\ldots, n\\}$. A collection $A_{1}, \\ldots, A_{k}$ of (not necessarily distinct) subsets of $\\{1,2, \\ldots, n\\}$ is called $\\ell$-large if $\\left|A_{i}\\right| \\geq \\ell$ for all $1 \\leq i \\leq k$. Find, in terms of $n$ and $\\ell$, the largest real number $c$ such that the inequality  $$ \\sum_{i=1}^{k} \\sum_{j=1}^{k} x_{i} x_{j} \\frac{\\left|A_{i} \\cap A_{j}\\right|^{2}}{\\left|A_{i}\\right| \\cdot\\left|A_{j}\\right|} \\geq c\\left(\\sum_{i=1}^{k} x_{i}\\right)^{2} $$  holds for all positive integer $k$, all nonnegative real numbers $x_{1}, x_{2}, \\ldots, x_{k}$, and all $\\ell$-large collections $A_{1}, A_{2}, \\ldots, A_{k}$ of subsets of $\\{1,2, \\ldots, n\\}$.", "solution": "  The answer turns out to be  $$ c=\\frac{n+\\ell^{2}-2 \\ell}{n(n-1)} $$   【 Rewriting as a dot product. For $i=1, \\ldots, n$ define $\\mathbf{v}_{i}$ by  $$ \\mathbf{v}_{i}[p, q]:=\\left\\{\\begin{array}{ll} \\frac{1}{\\left|A_{i}\\right|} & p \\in A_{i} \\text { and } q \\in A_{i} \\\\ 0 & \\text { otherwise; } \\end{array} \\quad \\mathbf{v}:=\\sum_{i} x_{i} \\mathbf{v}_{i}\\right. $$  Then  $$ \\begin{aligned} \\sum_{i} \\sum_{j} x_{i} x_{j} \\frac{\\left|A_{i} \\cap A_{j}\\right|^{2}}{\\left|A_{i}\\right|\\left|A_{j}\\right|} & =\\sum_{i} \\sum_{j} x_{i} x_{j}\\left\\langle\\mathbf{v}_{i}, \\mathbf{v}_{j}\\right\\rangle \\\\ & =\\left\\langle\\sum_{i} x_{i} \\mathbf{v}_{i}, \\sum_{j} x_{i} \\mathbf{v}_{i}\\right\\rangle=\\left\\|\\sum_{i} x_{i} \\mathbf{v}_{i}\\right\\|^{2}=\\|\\mathbf{v}\\|^{2} . \\end{aligned} $$   $$ \\begin{aligned} & \\langle\\mathbf{e}, \\mathbf{v}\\rangle=\\sum_{i} x_{i}\\left\\langle\\mathbf{e}, \\mathbf{v}_{i}\\right\\rangle=\\sum_{i} x_{i} \\\\ & \\langle\\mathbf{1}, \\mathbf{v}\\rangle=\\sum_{i} x_{i}\\left\\langle\\mathbf{1}, \\mathbf{v}_{i}\\right\\rangle=\\sum_{i} x_{i}\\left|A_{i}\\right| . \\end{aligned} $$  That means for any positive real constants $\\alpha$ and $\\beta$, by Cauchy-Schwarz for vectors, we should have  $$ \\begin{aligned} \\|\\alpha \\mathbf{e}+\\beta \\mathbf{1}\\|\\|\\mathbf{v}\\| & \\geq\\langle\\alpha \\mathbf{e}+\\beta \\mathbf{1}, \\mathbf{v}\\rangle=\\alpha\\langle\\mathbf{e}, \\mathbf{v}\\rangle+\\beta\\langle\\mathbf{1}, \\mathbf{v}\\rangle \\\\ & =\\alpha \\cdot \\sum x_{i}+\\beta \\cdot \\sum x_{i}\\left|A_{i}\\right| \\\\ & \\geq(\\alpha+\\ell \\beta) \\sum x_{i} . \\end{aligned} $$  Set $\\mathbf{w}:=\\alpha \\mathbf{e}+\\beta \\mathbf{1}$ for brevity. Then  $$ \\mathbf{w}[p, q]= \\begin{cases}\\alpha+\\beta & \\text { if } p=q \\\\ \\beta & \\text { if } p \\neq q\\end{cases} $$  SO  $$ \\|\\mathbf{w}\\|=\\sqrt{n \\cdot(\\alpha+\\beta)^{2}+\\left(n^{2}-n\\right) \\cdot \\beta^{2}} $$  Therefore, we get an lower bound  $$ \\frac{\\|\\mathbf{v}\\|}{\\sum x_{i}} \\geq \\frac{\\alpha+\\ell \\beta}{\\sqrt{n \\cdot(\\alpha+\\beta)^{2}+\\left(n^{2}-n\\right) \\cdot \\beta^{2}}} $$  Letting $\\alpha=n-\\ell$ and $\\beta=\\ell-1$ gives a proof that the constant  $$ c=\\frac{((n-\\ell)+\\ell(\\ell-1))^{2}}{n \\cdot(n-1)^{2}+\\left(n^{2}-n\\right) \\cdot(\\ell-1)^{2}}=\\frac{\\left(n+\\ell^{2}-2 \\ell\\right)^{2}}{n(n-1)\\left(n+\\ell^{2}-2 \\ell\\right)}=\\frac{n+\\ell^{2}-2 \\ell}{n(n-1)} $$  makes the original inequality always true. (The choice of $\\alpha: \\beta$ is suggested by the example below.)  【 Example showing this $c$ is best possible. Let $k=\\binom{n}{\\ell}$, let $A_{i}$ run over all $\\binom{n}{\\ell}$ subsets of $\\{1, \\ldots, n\\}$ of size $\\ell$, and let $x_{i}=1$ for all $i$. We claim this construction works.  To verify this, it would be sufficient to show that $\\mathbf{w}$ and $\\mathbf{v}$ are scalar multiples, so that the above Cauchy-Schwarz is equality. However, we can compute  $$ \\mathbf{w}[p, q]=\\left\\{\\begin{array}{ll} n-1 & \\text { if } p=q \\\\ \\ell-1 & \\text { if } p \\neq q \\end{array}, \\quad \\mathbf{v}[p, q]= \\begin{cases}\\binom{n-1}{\\ell-1} \\cdot \\frac{1}{\\ell} & \\text { if } p=q \\\\ \\binom{n-2}{\\ell-2} \\cdot \\frac{1}{\\ell} & \\text { if } p \\neq q\\end{cases}\\right. $$  which are indeed scalar multiples, finishing the proof.", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-2024-notes.jsonl"}}