problem: For each positive integer $n$, find the number of $n$-digit positive integers for which no two consecutive digits are equal, and the last digit is a prime.

solution: Almost trivial. Let $a_{n}$ be the desired answer. We have  $$ a_{n}+a_{n-1}=4 \cdot 9^{n-1} $$  for all $n$, by padding the $(n-1)$ digit numbers with a leading zero. Since $a_{0}=0, a_{1}=4$, solving the recursion gives  $$ a_{n}=\frac{2}{5}\left(9^{n}-(-1)^{n}\right) $$  The end. Remark. For concreteness, the first few terms are $0,4,32,292, \ldots$.


problem: Let $a, b, c$ be positive real numbers such that $a+b+c=4 \sqrt[3]{a b c}$. Prove that  $$ 2(a b+b c+c a)+4 \min \left(a^{2}, b^{2}, c^{2}\right) \geq a^{2}+b^{2}+c^{2} $$

solution: WLOG let $c=\min (a, b, c)=1$ by scaling. The given inequality becomes equivalent to  $$ 4 a b+2 a+2 b+3 \geq(a+b)^{2} \quad \forall a+b=4(a b)^{1 / 3}-1 $$  Now, let $t=(a b)^{1 / 3}$ and eliminate $a+b$ using the condition, to get  $$ 4 t^{3}+2(4 t-1)+3 \geq(4 t-1)^{2} \Longleftrightarrow 0 \leq 4 t^{3}-16 t^{2}+16 t=4 t(t-2)^{2} $$  which solves the problem. Equality occurs only if $t=2$, meaning $a b=8$ and $a+b=7$, which gives  $$ \{a, b\}=\left\{\frac{7 \pm \sqrt{17}}{2}\right\} $$  with the assumption $c=1$. Scaling gives the curve of equality cases.


problem: Let $A B C D$ be a quadrilateral inscribed in circle $\omega$ with $\overline{A C} \perp \overline{B D}$. Let $E$ and $F$ be the reflections of $D$ over $\overline{B A}$ and $\overline{B C}$, respectively, and let $P$ be the intersection of $\overline{B D}$ and $\overline{E F}$. Suppose that the circumcircles of $E P D$ and $F P D$ meet $\omega$ at $Q$ and $R$ different from $D$. Show that $E Q=F R$.

solution: Most of this problem is about realizing where the points $P, Q, R$ are. 【 First solution (Evan Chen). Let $X, Y$, be the feet from $D$ to $\overline{B A}, \overline{B C}$, and let $Z=\overline{B D} \cap \overline{A C}$. By Simson theorem, the points $X, Y, Z$ are collinear. Consequently, the point $P$ is the reflection of $D$ over $Z$, and so we conclude $P$ is the orthocenter of $\triangle A B C$. ![](https://cdn.mathpix.com/cropped/2024_11_19_c3196aba41526ceabc46g-5.jpg?height=738&width=1007&top_left_y=1070&top_left_x=533)  Suppose now we extend ray $C P$ to meet $\omega$ again at $Q^{\prime}$. Then $\overline{B A}$ is the perpendicular bisector of both $\overline{P Q^{\prime}}$ and $\overline{D E}$; consequently, $P Q^{\prime} E D$ is an isosceles trapezoid. In particular, it is cyclic, and so $Q^{\prime}=Q$. In the same way $R$ is the second intersection of ray $\overline{A P}$ with $\omega$.  Now, because of the two isosceles trapezoids we have found, we conclude  $$ E Q=P D=F R $$  as desired. Remark. Alternatively, after identifying $P$, one can note $\overline{B Q E}$ and $\overline{B R F}$ are collinear. Since $B E=B D=B F$, upon noticing $B Q=B P=B R$ we are also done.


solution: Most of this problem is about realizing where the points $P, Q, R$ are. 【I Second solution (Danielle Wang). Here is a solution which does not identify the point $P$ at all. We know that $B E=B D=B F$, by construction. ![](https://cdn.mathpix.com/cropped/2024_11_19_c3196aba41526ceabc46g-6.jpg?height=669&width=1172&top_left_y=242&top_left_x=419)  Claim - The points $B, Q, E$ are collinear. Similarly the points $B, R, F$ are collinear.   Observe that $B E=B D=B F$, so $B$ is the circumcenter of $\triangle D E F$. Thus, $\measuredangle D E P=$ $\measuredangle D E F=90^{\circ}-\beta$. Then  $$ \begin{aligned} \measuredangle D P E & =\measuredangle D E P+\measuredangle P D E=\left(90^{\circ}-\beta\right)+\alpha \\ & =\alpha-\beta+90^{\circ} \\ \measuredangle D Q^{\prime} B & =\measuredangle D C B=\measuredangle D C A+\measuredangle A C B \\ & =\measuredangle D B A-\left(90^{\circ}-\measuredangle D B C\right)=-\left(90^{\circ}-\alpha\right)-\left(90^{\circ}-\left(90^{\circ}-\beta\right)\right) \\ & =\alpha-\beta+90^{\circ} . \end{aligned} $$  Thus $Q^{\prime}$ lies on the desired circle, so $Q^{\prime}=Q$. Now, by power of a point we have $B Q \cdot B E=B P \cdot B D=B R \cdot B F$, so $B Q=B P=B R$. Hence $E Q=P D=F R$.


problem: Find all real numbers $x$ for which there exists a triangle $A B C$ with circumradius 2 , such that $\angle A B C \geq 90^{\circ}$, and  $$ x^{4}+a x^{3}+b x^{2}+c x+1=0 $$  where $a=B C, b=C A, c=A B$.

solution: The answer is $x=-\frac{1}{2}(\sqrt{6} \pm \sqrt{2})$. We prove this the only possible answer. Evidently $x<0$. Now, note that  $$ a^{2}+c^{2} \leq b^{2} \leq 4 b $$  since $b \leq 4$ (the diameter of its circumcircle). Then,  $$ \begin{aligned} 0 & =x^{4}+a x^{3}+b x^{2}+c x+1 \\ & =x^{2}\left[\left(x+\frac{1}{2} a\right)^{2}+\left(\frac{1}{x}+\frac{1}{2} c\right)^{2}+\left(b-\frac{a^{2}+c^{2}}{4}\right)\right] \\ & \geq 0+0+0=0 \end{aligned} $$  In order for equality to hold, we must have $x=-\frac{1}{2} a, 1 / x=-\frac{1}{2} c$, and $a^{2}+c^{2}=b^{2}=4 b$. This gives us $b=4, a c=4, a^{2}+c^{2}=16$. Solving for $a, c>0$ implies  $$ \{a, c\}=\{\sqrt{6} \pm \sqrt{2}\} $$  This gives the $x$ values claimed above; by taking $a, b, c$ as deduced here, we find they work too.  Remark. Note that by perturbing $\triangle A B C$ slightly, we see $a$ priori that the set of possible $x$ should consist of unions of intervals (possibly trivial). So it makes sense to try inequalities no matter what.


problem: Let $p$ be a prime, and let $a_{1}, \ldots, a_{p}$ be integers. Show that there exists an integer $k$ such that the numbers  $$ a_{1}+k, a_{2}+2 k, \ldots, a_{p}+p k $$  produce at least $\frac{1}{2} p$ distinct remainders upon division by $p$.

solution: For each $k=0, \ldots, p-1$ let $G_{k}$ be the graph on $\{1, \ldots, p\}$ where we join $\{i, j\}$ if and only if  $$ a_{i}+i k \equiv a_{j}+j k \quad(\bmod p) \Longleftrightarrow k \equiv-\frac{a_{i}-a_{j}}{i-j} \quad(\bmod p) $$  So we want a graph $G_{k}$ with at least $\frac{1}{2} p$ connected components. However, each $\{i, j\}$ appears in exactly one graph $G_{k}$, so some graph has at most $\frac{1}{p}\binom{p}{2}=\frac{1}{2}(p-1)$ edges (by "pigeonhole"). This graph has at least $\frac{1}{2}(p+1)$ connected components, as desired.  Remark. Here is an example for $p=5$ showing equality can occur:  $$ \left[\begin{array}{lllll} 0 & 0 & 3 & 4 & 3 \\ 0 & 1 & 0 & 2 & 2 \\ 0 & 2 & 2 & 0 & 1 \\ 0 & 3 & 4 & 3 & 0 \\ 0 & 4 & 1 & 1 & 4 \end{array}\right] . $$  Ankan Bhattacharya points out more generally that $a_{i}=i^{2}$ is sharp in general.


problem: Karl starts with $n$ cards labeled $1,2, \ldots, n$ lined up in random order on his desk. He calls a pair $(a, b)$ of cards swapped if $a>b$ and the card labeled $a$ is to the left of the card labeled $b$. Karl picks up the card labeled 1 and inserts it back into the sequence in the opposite position: if the card labeled 1 had $i$ cards to its left, then it now has $i$ cards to its right. He then picks up the card labeled 2 and reinserts it in the same manner, and so on, until he has picked up and put back each of the cards $1, \ldots, n$ exactly once in that order. For example, if $n=4$, then one example of a process is  $$ 3142 \longrightarrow 3412 \longrightarrow 2341 \longrightarrow 2431 \longrightarrow 2341 $$  which has three swapped pairs both before and after. Show that, no matter what lineup of cards Karl started with, his final lineup has the same number of swapped pairs as the starting lineup.

solution: We define a new process $P^{\prime}$ where, when re-inserting card $i$, we additionally change its label from $i$ to $n+i$. An example of $P^{\prime}$ also starting with 3142 is:  $$ 3142 \longrightarrow 3452 \longrightarrow 6345 \longrightarrow 6475 \longrightarrow 6785 . $$  Note that now, each step of $P^{\prime}$ preserves the number of inversions. Moreover, the final configuration of $P^{\prime}$ is the same as the final configuration of $P$ with all cards incremented by $n$, and of course thus has the same number of inversions. Boom.