Let $(1+x+x^2)^n=a_0+a_1x+a_2x^2+...+a_{2n}x^{2n}$ be an identity in $x$. If we lt $s=a_0+a_2+a_4+...+a_{2n}$, then $s$ equals: $\text{(A)}\ 2^n\qquad \text{(B)}\ 2^n+1\qquad \text{(C)}\ \dfrac{3^n-1}{2}\qquad \text{(D)}\ \dfrac{3^n}{2}\qquad \text{(E)}\ \dfrac{3^n+1}{2}$

The gure below shows a large circle with area $120$ containing a circle with half of the radius of the large circle and six circles with a quarter of the radius of the large circle. Find the area of the shaded region. [img]https://cdn.artofproblemsolving.com/attachments/7/9/064a05feb9bd67896c079a5141bf7556d7165b.png[/img]

Let $K=(V, E)$ be a finite, simple, complete graph. Let $\phi: E \to \mathbb{R}^2$ be a map from the edge set to the plane, such that the preimage of any point in the range defines a connected graph on the entire vertex set $V$, and the points assigned to the edges of any triangle are collinear. Show that the range of $\phi$ is contained in a line.

Compute $\sum^{999}_{x=1}\gcd (x, 10x + 9)$.

We are given an acute triangle $ABC$ and points $D$ on $BC$ and $E$ on $AC$ such that $AD$ is perpendicular to $BC$ and $BE$ is perpendicular to $AC$. The intersection of $AD$ and $BE$ is called $H$. A line through $H$ intersects line segment $BC$ in $P$, and intersects line segment $AC$ in $Q$. Furthermore, $K$ is a point on $BE$ such that $PK$ is perpendicular to $BE$, and $L$ is a point on $AD$ such that $QL$ is perpendicular to $AD$. Prove that $DK$ and $EL$ are parallel. [asy] unitsize(1 cm); pair A, B, C, D, E, H, K, L, P, Q; A = (0,0); B = (6,0); C = (3.5,4); D = (A + reflect(B,C)*(A))/2; E = (B + reflect(A,C)*(B))/2; H = extension(A, D, B, E); P = extension(H, H + dir(-10), B, C); Q = extension(H, H + dir(-10), A, C); K = (P + reflect(B,E)*(P))/2; L = (Q + reflect(A,D)*(Q))/2; draw(A--B--C--cycle); draw(A--D); draw(B--E); draw(K--P--Q--L); draw(rightanglemark(B,D,A,5)); draw(rightanglemark(B,E,A,5)); draw(rightanglemark(P,K,B,5)); draw(rightanglemark(A,L,Q,5)); dot("$A$", A, SW); dot("$B$", B, SE); dot("$C$", C, N); dot("$D$", D, NE); dot("$E$", E, NW); dot("$H$", H, N); dot("$K$", K, SW); dot("$L$", L, SE); dot("$P$", P, NE); dot("$Q$", Q, NW); [/asy]

Chords $AB$ and $CD$ of a circle $\omega$ meet at point $E$ in such a way that $AD = AE = EB$. Let $F$ be a point of segment $CE$ such that $ED = CF$. The bisector of angle $AFC$ meets an arc $DAC$ at point $P$. Prove that $A$, $E$, $F$, and $P$ are concyclic.

Lucy starts by writing $s$ integer-valued $2022$-tuples on a blackboard. After doing that, she can take any two (not necessarily distinct) tuples $\mathbf{v}=(v_1,\ldots,v_{2022})$ and $\mathbf{w}=(w_1,\ldots,w_{2022})$ that she has already written, and apply one of the following operations to obtain a new tuple: \begin{align*} \mathbf{v}+\mathbf{w}&=(v_1+w_1,\ldots,v_{2022}+w_{2022}) \\ \mathbf{v} \lor \mathbf{w}&=(\max(v_1,w_1),\ldots,\max(v_{2022},w_{2022})) \end{align*} and then write this tuple on the blackboard. It turns out that, in this way, Lucy can write any integer-valued $2022$-tuple on the blackboard after finitely many steps. What is the smallest possible number $s$ of tuples that she initially wrote?

The sum of the base-$ 10$ logarithms of the divisors of $ 10^n$ is $ 792$. What is $ n$? $ \textbf{(A)}\ 11\qquad \textbf{(B)}\ 12\qquad \textbf{(C)}\ 13\qquad \textbf{(D)}\ 14\qquad \textbf{(E)}\ 15$

In our number system the base is ten. If the base were changed to four you would count as follows: $ 1,2,3,10,11,12,13,20,21,22,23,30,\ldots$ The twentieth number would be: $ \textbf{(A)}\ 20 \qquad\textbf{(B)}\ 38 \qquad\textbf{(C)}\ 44 \qquad\textbf{(D)}\ 104 \qquad\textbf{(E)}\ 110$

Let $ABC$ be a scalene triangle with orthocenter $H$ and circumcenter $O$. Let $P$ be the midpoint of $\overline{AH}$ and let $T$ be on line $BC$ with $\angle TAO=90^{\circ}$. Let $X$ be the foot of the altitude from $O$ onto line $PT$. Prove that the midpoint of $\overline{PX}$ lies on the nine-point circle* of $\triangle ABC$. *The nine-point circle of $\triangle ABC$ is the unique circle passing through the following nine points: the midpoint of the sides, the feet of the altitudes, and the midpoints of $\overline{AH}$, $\overline{BH}$, and $\overline{CH}$. [i]Proposed by Zack Chroman[/i]

Alice and Bob are playing a game, starting with a binary string$ b$ of length $2022$. In each step, the rightmost digit of the string is deleted. If the deleted digit was $1$, Alice gets to choose which digit she wants to append on the left. Otherwise, Bob gets to choose the digit to append on the left of the string. Alice would like to turn the string $b$ into the all-zero string $\underbrace{00 . . . 0}_{2022}$, in the least number of steps possible, while Bob would like to maximize the number of steps necessary, or prevent Alice from doing this at all. a) Is there a string $b$ for which Bob can prevent Alice in her goal, if both players play optimally? b) If the answer to part a is yes, find all such strings $b$. If the answer is no, find the maximal game time and find the set of strings $b$ for which the game time is maximal.

Let $D$ be the interior of the circle $C$ and let $A \in C$. Show that the function $f : D \to \mathbb R, f(M)=\frac{|MA|}{|MM'|}$ where $M' = AM \cap C$, is strictly convex; i.e., $f(P) <\frac{f(M_1)+f(M_2)}{2}, \forall M_1,M_2 \in D, M_1 \neq M_2$ where $P$ is the midpoint of the segment $M_1M_2.$

In triangle $ABC$ , $\angle A = 45^o$ and $M$ is the midpoint of $\overline{BC}$. $\overline{AM}$ intersects the circumcircle of $ABC$ for the second time at $D$, and $AM = 2MD$. Find $cos\angle AOD$, where $O$ is the circumcenter of $ABC$.

If numbers $x_1$, $x_2$,...,$x_n$ are from interval $\left( \frac{1}{4},1 \right)$ prove the inequality: $\log _{x_1} {\left(x_2-\frac{1}{4} \right)} + \log _{x_2} {\left(x_3-\frac{1}{4} \right)}+ ... + \log _{x_{n-1}} {\left(x_n-\frac{1}{4} \right)} + \log _{x_n} {\left(x_1-\frac{1}{4} \right)} \geq 2n$

Let $ G$ be a finite non-commutative group of order $ t \equal{} 2^nm$, where $ n, m$ are positive and $ m$ is odd. Prove, that if the group contains an element of order $ 2^n$, then (i) $ G$ is not simple; (ii) $ G$ contains a normal subgroup of order $ m$.

In triangle $ABC$ let $N, M, P$ be the midpoints of the sides $BC, CA, AB$ and $G$ be the centroid of this triangle. Let the circle circumscribed to $BGP$ intersect the line $MP$ in point $K$, $P \neq K$, and the circle circumscribed to $CGN$ intersect the line $MN$ in point $L$, $N \neq L$. Prove that $ \angle BAK = \angle CAL $.