problem: Determine all composite positive integers $n$ for which it is possible to arrange all divisors of $n$ that are greater than 1 in a circle so that no two adjacent divisors are relatively prime.

solution: The only bad ones are $n=p q$, products of two distinct primes. Clearly they can't be so arranged, so we show all others work.  - If $n$ is a power of a prime, the result is obvious; any arrangement works. - If $n=p_{1}^{e_{1}} \ldots p_{k}^{e_{k}}$ for some $k \geq 3$, then first situate $p_{1} p_{2}, p_{2} p_{3}, \ldots, p_{k} p_{1}$ on the circle. Then we can arbitrarily place any multiples of $p_{i}$ between $p_{i-1} p_{i}$ and $p_{i} p_{i+1}$. This finishes this case. - Finally suppose $n=p^{a} q^{b}$. If $a>1$, say, we can repeat the argument by first placing $p q$ and $p^{2} q$ and then placing multiples of $p$ in one arc and multiples of $q$ in the other arc. On the other hand the case $a=b=1$ is seen to be impossible.


problem: Prove that the system of equations  $$ \begin{aligned} x^{6}+x^{3}+x^{3} y+y & =147^{157} \\ x^{3}+x^{3} y+y^{2}+y+z^{9} & =157^{147} \end{aligned} $$  has no integer solutions.

solution: Therefore, there are no integer solutions.  Remark. In fact, $a^{2}+b^{3} \equiv 14(\bmod 19)$ has no solutions modulo 19 either.  Remark. It can also be checked that the original system has no equations modulo 13, although this requires using both equations rather than just their sum. (As in the modulo 19 situation, $z^{9}$ may be replaced by $z^{3}$ and this remark still holds.)


problem: Let $A B C$ be an acute-angled triangle, and let $P$ and $Q$ be two points on side $B C$. Construct a point $C_{1}$ in such a way that the convex quadrilateral $A P B C_{1}$ is cyclic, $\overline{Q C_{1}} \| \overline{C A}$, and $C_{1}$ and $Q$ lie on opposite sides of line $A B$. Construct a point $B_{1}$ in such a way that the convex quadrilateral $A P C B_{1}$ is cyclic, $\overline{Q B_{1}} \| \overline{B A}$, and $B_{1}$ and $Q$ lie on opposite sides of line $A C$. Prove that the points $B_{1}, C_{1}, P$, and $Q$ lie on a circle.

solution: It is enough to prove that $A, B_{1}$, and $C_{1}$ are collinear, since then $\measuredangle C_{1} Q P=\measuredangle A C P=$ $\measuredangle A B_{1} P=\measuredangle C_{1} B_{1} P$. ![](https://cdn.mathpix.com/cropped/2024_11_19_cb60a5d1ecd6117ce0eag-05.jpg?height=678&width=809&top_left_y=986&top_left_x=635) ब First solution. Let $T$ be the second intersection of $\overline{A C_{1}}$ with $(A P C)$. Then readily $\triangle P C_{1} T \sim \triangle A B C$. Consequently, $\overline{Q C_{1}} \| \overline{A C}$ implies $T C_{1} Q P$ cyclic. Finally, $\overline{T Q} \| \overline{A B}$ now follows from the cyclic condition, so $T=B_{1}$ as desired.


solution: It is enough to prove that $A, B_{1}$, and $C_{1}$ are collinear, since then $\measuredangle C_{1} Q P=\measuredangle A C P=$ $\measuredangle A B_{1} P=\measuredangle C_{1} B_{1} P$. ![](https://cdn.mathpix.com/cropped/2024_11_19_cb60a5d1ecd6117ce0eag-05.jpg?height=678&width=809&top_left_y=986&top_left_x=635) \ Second solution. One may also use barycentric coordinates. Let $P=(0, m, n)$ and $Q=(0, r, s)$ with $m+n=r+s=1$. Once again,  $$ (A P B):-a^{2} y z-b^{2} z x-c^{2} x y+(x+y+z)\left(a^{2} m \cdot z\right)=0 $$  Set $C_{1}=(s-z, r, z)$, where $C_{1} Q \| A C$ follows by $(s-z)+r+z=1$. We solve for this $z$.  $$ \begin{aligned} 0 & =-a^{2} r z+(s-z)\left(-b^{2} z-c^{2} r\right)+a^{2} m z \\ & =b^{2} z^{2}+\left(-s b^{2}+r c^{2}\right) z-a^{2} r z+a^{2} m z-c^{2} r s \\ & =b^{2} z^{2}+\left(-s b^{2}+r c^{2}+a^{2}(m-r)\right) z-c^{2} r s \\ \Longrightarrow 0 & =r b^{2}\left(\frac{z}{r}\right)^{2}+\left(-s b^{2}+r c^{2}+a^{2}(m-r)\right)\left(\frac{z}{r}\right)-c^{2} s \end{aligned} $$  So the quotient of the $z$ and $y$ coordinates of $C_{1}$ satisfies this quadratic. Similarly, if $B_{1}=(r-y, y, s)$ we obtain that  $$ 0=s c^{2}\left(\frac{y}{s}\right)^{2}+\left(-r c^{2}+s b^{2}+a^{2}(n-s)\right)\left(\frac{y}{s}\right)-b^{2} r $$  Since these two quadratics are the same when one is written backwards (and negated), it follows that their roots are reciprocals. But the roots of the quadratics represent $\frac{z}{y}$ and $\frac{y}{z}$ for the points $C_{1}$ and $B_{1}$, respectively. This implies (with some configuration blah) that the points $B_{1}$ and $C_{1}$ are collinear with $A=(1,0,0)$ (in some line of the form $\frac{y}{z}=k$ ), as desired.


problem: Legs $L_{1}, L_{2}, L_{3}, L_{4}$ of a square table each have length $n$, where $n$ is a positive integer. For how many ordered 4 -tuples ( $k_{1}, k_{2}, k_{3}, k_{4}$ ) of nonnegative integers can we cut a piece of length $k_{i}$ from the end of leg $L_{i}$ and still have a stable table? (The table is stable if it can be placed so that all four of the leg ends touch the floor. Note that a cut leg of length 0 is permitted.)

solution: Flip the table upside-down so that that the table's surface rests on the floor. Then, we see that we want the truncated legs to have endpoints $A, B, C, D$ which are coplanar (say).  Claim - This occurs if and only if $A B C D$ is a parallelogram.   In still other words, we are counting the number of solutions to  $$ \left(n-k_{1}\right)+\left(n-k_{3}\right)=\left(n-k_{2}\right)+\left(n-k_{4}\right) \Longleftrightarrow k_{1}+k_{3}=k_{2}+k_{4} . $$  Define  $$ a_{r}=\#\{(a, b) \mid a+b=r, 0 \leq a, b \leq n\} $$   $$ \begin{aligned} \sum_{r=0}^{2 n} a_{r}^{2} & =1^{2}+2^{2}+\cdots+n^{2}+(n+1)^{2}+n^{2}+\cdots+1^{2} \\ & =\frac{1}{3}(n+1)\left(2 n^{2}+4 n+3\right) \end{aligned} $$


problem: Let $n>1$ be an integer. Suppose $2 n$ points are given in the plane, no three of which are collinear. Suppose $n$ of the given $2 n$ points are colored blue and the other $n$ colored red. A line in the plane is called a balancing line if it passes through one blue and one red point and, for each side of the line, the number of blue points on that side is equal to the number of red points on the same side. Prove that there exist at least two balancing lines.

solution: Consider the convex hull $\mathcal{H}$ of the polygon. There are two cases. 【 Easy case: the convex hull has both colors. If the convex hull $\mathcal{H}$ is not all the same color, there exist two edges of $\mathcal{H}$ (at least) which have differently colored endpoints. The extensions of those sides form balancing lines; indeed given any such line $\ell$ one side of $\ell$ has no points, the other has $n-1$ red and $n-1$ blue points.  【 Harder case: the convex hull is all one color. Now assume $\mathcal{H}$ is all blue (WLOG). We will prove there are at least $|\mathcal{H}|$ balancing lines in the following way.  Claim - For any vertex $B$ of $\mathcal{H}$ there is a balancing line through it. ![](https://cdn.mathpix.com/cropped/2024_11_19_cb60a5d1ecd6117ce0eag-08.jpg?height=504&width=498&top_left_y=1621&top_left_x=779)  During this process, we consider the set of points on the same side of $\ell$ as $C$, and let $x$ be the number of such red points minus the number of such blue points. Note that:  - Every time $\ell$ touches a blue point, $x$ increases by 1 . - Every time $\ell$ touches a red point, $x$ decreases by 1 . - Initially, $x=+1$. - Just before reaching the end we have $x=-1$.  So at the moment where $x$ first equals zero, we have found our balancing line.


problem: For a positive integer $m$, let $s(m)$ denote the sum of the decimal digits of $m$. A set $S$ positive integers is $k$-stable if $s\left(\sum_{x \in X} x\right)=k$ for any nonempty subset $X \subseteq S$. For each integer $n \geq 2$ let $f(n)$ be the minimal $k$ for which there exists a $k$-stable set with $n$ integers. Prove that there are constants $0<C_{1}<C_{2}$ with  $$ C_{1} \log _{10} n \leq f(n) \leq C_{2} \log _{10} n . $$

solution: 【 Construction showing $f(n) \leq 9\left\lceil\log _{10}\binom{n+1}{2}\right\rceil$. Let $n \geq 1$ and $e \geq 1$ be integers satisfying $1+2+\cdots+n<10^{e}$. Consider the set  $$ S=\left\{10^{e}-1,2\left(10^{e}-1\right), \ldots, n\left(10^{e}-1\right)\right\} $$  For example, if $n=6$ and $e=3$, we have $S=\{999,1998,2997,3996,4995,5994\}$. The set $S$ here is easily seen to be $9 e$-stable. Thus $f(n) \leq 9\left\lceil\log _{10}\binom{n+1}{2}\right\rceil$, proving one direction.  Remark. I think the problem is actually more natural with a multiset $S$ rather than a vanilla set, in which case $S=\left\{10^{e}-1,10^{e}-1, \ldots, 10^{e}-1\right\}$ works fine, and is easier to think of. In some sense the actual construction is obtained by starting with this one, and then pushing together the terms together in order to get the terms to be distinct, hence the $1+2+\cdots+n$ appearance.   Claim - Let $k$ be a positive integer. In any (multi)set $S$ of more than $12^{k}$ integers, there exists a subset whose sum of decimal digits exceeds $k$.   - If the $i$ th digit from the right of $\Sigma$ is nonzero, then arbitrarily partition the numbers on the board into groups of 10 , erasing any leftover numbers. Within each group of 10 , we can find a nonempty subset with sum $0 \bmod 10^{i}$; we then erase each group and replace it with that sum. - If the $i$ th digit from the right of $\Sigma$ is zero, but some entry on the board is not divisible by $10^{i}$, then we erase that entry and add it to $\Sigma$. Then we do the grouping as in the previous step. - If the $i$ th digit from the right of $\Sigma$ is zero, and all entries on the board are divisible by $10^{i}$, we do nothing and move on to the next step.  This process ends when no numbers remain on the blackboard. The first and second cases occur at least $k+1$ times (the number of entries decreases by a factor of at most 12 each step), and each time $\Sigma$ gets some nonzero digit, which is never changed at later steps. Therefore $\Sigma$ has sum of digits at least $k+1$ as needed.