For positive integer $k,$ we say that it is a [i]Taurus integer[/i] if we can delete one element from the set $M_k=\{1,2,\cdots,k\},$ such that the sum of remaining $k-1$ elements is a positive perfect square. For example, $7$ is a Taurus integer, because if we delete $3$ from $M_7=\{1,2,3,4,5,6,7\},$ the sum of remaining $6$ elements is $25,$ which is a positive perfect square. $(1)$ Determine whether $2021$ is a Taurus integer. $(2)$ For positive integer $n,$ determine the number of Taurus integers in $\{1,2,\cdots,n\}.$

Find the largest value of the area of the projection of the cylinder onto the plane if its radius is $r$ and its height is $h$ (orthogonal projection).

A cylinder is inscribed within a sphere of radius 10 such that its volume is [i]almost-half[/i] that of the sphere. If [i]almost-half[/i] is defined such that the cylinder has volume $\frac12+\frac{1}{250}$ times the sphere’s volume, find the sum of all possible heights for the cylinder.

$21$ bandits live in the city of Warmridge, each of them having some enemies among the others. Initially each bandit has $240$ bullets, and duels with all of his enemies. Every bandit distributes his bullets evenly between his enemies, this means that he takes the same number of bullets to each of his duels, and uses each of his bullets in only one duel. In case the number of his bullets is not divisible by the number of his enemies, he takes as many bullets to each duel as possible, but takes the same number of bullets to every duel, so it is possible that in the end the bandit will have some remaining bullets. Shooting is banned in the city, therefore a duel consists only of comparing the number of bullets in the guns of the opponents, and the winner is whoever has more bullets. After the duel the sheriff takes the bullets of the winner and as an act of protest the loser shoots all of his bullets into the air. What is the largest possible number of bullets the sheriff can have after all of the duels have ended? Being someones enemy is mutual. If two opponents have the same number of bullets in their guns during a duel, then the sheriff takes the bullets of the bandit who has the wider hat among them. Example: If a bandit has $13$ enemies then he takes $18$ bullets with himself to each duel, and they will have $6$ leftover bullets after finishing all their duels.

Suppose that $m$ and $n$ are positive integers such that $m<n$, the geometric mean of $m$ and $n$ is greater than $2007$, and the arithmetic mean of $m$ and $n$ is less than $2007$. How many pairs $(m,n)$ satisfy these conditions? $\textbf{(A) }0\hspace{14em}\textbf{(B) }1\hspace{14em}\textbf{(C) }2$ $\textbf{(D) }3\hspace{14em}\textbf{(E) }4\hspace{14em}\textbf{(F) }5$ $\textbf{(G) }6\hspace{14em}\textbf{(H) }7\hspace{14em}\textbf{(I) }2007$

In some tournament there were $8$ players. Every two players played exactly one match, each of them finished with the win of one of the players or with a draw. Winner of the match got $2$ points, his opponent $0$ points and in the case of draw every player got $1$ point. When all matches had finished it turned out that every player had the same number of points. Determine the minimal total numbers of draws.

Find all positive integers $n>2$ such that $$ n! \mid \prod_{ p<q\le n, p,q \, \text{primes}} (p+q)$$

We define a sequence $x_1 = \sqrt{3}, x_2 =-1, x_3 =2 - \sqrt{3},$ and for all $n \geq 4$ $$(x_n + x_{n-3})(1 - x^2_{n-1}x^2_{n-2}) = 2x_{n-1}(1 + x^2_{n-2}).$$ Suppose $m$ is the smallest positive integer for which $x_m$ is undefined. Compute $m.$

For integers $0 \le m,n \le 64$, let $\alpha(m,n)$ be the number of nonnegative integers $k$ for which $\left\lfloor m/2^k \right\rfloor$ and $\left\lfloor n/2^k \right\rfloor$ are both odd integers. Consider a $65 \times 65$ matrix $M$ whose $(i,j)$th entry (for $1 \le i, j \le 65$) is \[ (-1)^{\alpha(i-1, j-1)}. \] Compute the remainder when $\det M$ is divided by $1000$. [i] Proposed by Evan Chen [/i]

A ray originating at point $P$ intersects a circle with center $O$ at points $A$ and $B$, with $PB > PA$. Segment $\overline{OP}$ intersects the circle at point $C$. Given that $PA = 31$, $PC = 17$, and $\angle PBO = 60^o$, find the radius of the circle.

Five identical empty buckets of $2$-liter capacity stand at the vertices of a regular pentagon. Cinderella and her wicked Stepmother go through a sequence of rounds: At the beginning of every round, the Stepmother takes one liter of water from the nearby river and distributes it arbitrarily over the five buckets. Then Cinderella chooses a pair of neighbouring buckets, empties them to the river and puts them back. Then the next round begins. The Stepmother goal's is to make one of these buckets overflow. Cinderella's goal is to prevent this. Can the wicked Stepmother enforce a bucket overflow? [i]Proposed by Gerhard Woeginger, Netherlands[/i]

One day while Tony plays in the back yard of the Kubik's home, he wonders about the width of the back yard, which is in the shape of a rectangle. A row of trees spans the width of the back of the yard by the fence, and Tony realizes that all the trees have almost exactly the same diameter, and the trees look equally spaced. Tony fetches a tape measure from the garage and measures a distance of almost exactly $12$ feet between a consecutive pair of trees. Tony realizes the need to include the width of the trees in his measurements. Unsure as to how to do this, he measures the distance between the centers of the trees, which comes out to be around $15$ feet. He then measures $2$ feet to either side of the first and last trees in the row before the ends of the yard. Tony uses these measurements to estimate the width of the yard. If there are six trees in the row of trees, what is Tony's estimate in feet? [asy] size(400); defaultpen(linewidth(0.8)); draw((0,-3)--(0,3)); int d=8; for(int i=0;i<=5;i=i+1) { draw(circle(7/2+d*i,3/2)); } draw((5*d+7,-3)--(5*d+7,3)); draw((0,0)--(2,0),Arrows(size=7)); draw((5,0)--(2+d,0),Arrows(size=7)); draw((7/2+d,0)--(7/2+2*d,0),Arrows(size=7)); label("$2$",(1,0),S); label("$12$",((7+d)/2,0),S); label("$15$",((7+3*d)/2,0),S); [/asy]

Let $ABC$ be an isosceles triangle with vertex $A$ and $AB> BC$. Let $k$ be the circle with center $A$ passsing through $B$ and $C$. Let $H$ be the second intersection of $k$ with the altitude of the triangle $ABC$ through $B$. Further let $G$ be the second intersection of $k$ with the median through $B$ in triangle $ABC$. Let $X$ be the intersection of the lines $AC$ and $GH$. Show that $C$ is the midpoint of $AX$.

Let $f(x) = 15x - 2016$. If $f(f(f(f(f(x))))) = f(x)$, find the sum of all possible values of $x$.

All the high schools in a large school district are involved in a fundraiser selling T-shirts. Which of the choices below is logically equivalent to the statement “No school bigger than Euclid HS sold more T-shirts than Euclid HS”? $(\textbf{A})$ All schools smaller than Euclid HS sold fewer T-shirts than Euclid HS. $(\textbf{B})$ No school that sold more T-shirts than Euclid HS is bigger than Euclid HS. $(\textbf{C})$ All schools bigger than Euclid HS sold fewer T-shirts than Euclid HS. $(\textbf{D})$ All schools that sold fewer T-shirts than Euclid HS are smaller than Euclid HS. $(\textbf{E})$ All schools smaller than Euclid HS sold more T-shirts than Euclid HS.

Walking down Jane Street, Ralph passed four houses in a row, each painted a different color. He passed the orange house before the red house, and he passed the blue house before the yellow house. The blue house was not next to the yellow house. How many orderings of the colored houses are possible? ${ \textbf{(A)}\ 2\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}}\ 5\qquad\textbf{(E)}\ 6$

In parallelogram $ABCD$ with acute angle $A$ a point $N$ is chosen on the segment $AD$, and a point $M$ on the segment $CN$ so that $AB = BM = CM$. Point $K$ is the reflection of $N$ in line $MD$. The line $MK$ meets the segment $AD$ at point $L$. Let $P$ be the common point of the circumcircles of $AMD$ and $CNK$ such that $A$ and $P$ share the same side of the line $MK$. Prove that $\angle CPM = \angle DPL$.

Here $G_{n}$ denotes a simple undirected graph with $n$ vertices, $K_{n}$ denotes the complete graph with $n$ vertices, $K_{n,m}$ the complete bipartite graph whose components have $m$ and $n$ vertices, and $C_{n}$ a circuit with $n$ vertices. The number of edges in the graph $G_{n}$ is denoted $e(G_{n})$. (a) If $G_{n}$ does not contain $K_{2,3}$ , prove that $e(G_{n}) \leq\frac{n\sqrt{n}}{\sqrt{2}}+n$. (b) Given $n \geq 16$ distinct points $P_{1}, . . . , P_{n}$ in the plane, prove that at most $n\sqrt{n}$ of the segments $P_{i}P_{j}$ have unit length.

In a quadrilateral $ABCD$, the diagonals are perpendicular and intersect at the point $S$. Let $K, L, M$, and $N$ be points symmetric to $S$ with respect to the lines $AB, BC, CD$, and $DA$, respectively, $BN$ intersects the circumcircle of the triangle $SKN$ at point $E$, and $BM$ intersects circumscribed the circle of the triangle $SLM$ at the point $F$. Prove that the quadrilateral $EFLK$ is cyclic .

Determine all possible values of positive integer $n$, such that there are $n$ different 3-element subsets $A_1,A_2,...,A_n$ of the set $\{1,2,...,n\}$, with $|A_i \cap A_j| \not= 1$ for all $i \not= j$.

find all polynomials with integer coefficients that $P(\mathbb{Z})= ${$p(a):a\in \mathbb{Z}$} has a Geometric progression.

Let $ABCD$ be a quadrilateral with $AD = 20$ and $BC = 13$. The area of $\triangle ABC$ is $338$ and the area of $\triangle DBC$ is $212$. Compute the smallest possible perimeter of $ABCD$. [i]Proposed by Evan Chen[/i]

Prove that this triangle cut out of paper can be folded so that the surface of a regular unit tetradragon (i.e., a triangular pyramid, all edges of which are equal to $1$) is obtained if: a) this triangle is isosceles, the lateral sides are equal to $2$ , the angle between them is $120^o$, b) two sides of this triangle are equal to $2$ and $2\sqrt3$, the angle between them is $150^o$.

Given are $8$ weights weighing $1, 2, . . . , 8$ grams, but it is not known which one how much does it weigh. Baron Munchausen claims that he remembers which of the weights weighs how much, and to prove that he is right he is ready to conduct one weighing, as a result of which the weight of at least one of the weights will be unambiguously established. Is he cheating?

Positive numbers $x$ and $y$ satisfy $xy=2^{15}$ and $\log_2{x} \cdot \log_2{y} = 60$. Find $\sqrt[3]{(\log_2{x})^3+(\log_2{y})^3}$