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.


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).


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 $n$.


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.


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.


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.


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.