In a mathematical competition $n=10\,000$ contestants participate. During the final party, in sequence, the first one takes $1/n$ of the cake, the second one takes $2/n$ of the remaining cake, the third one takes $3/n$ of the cake that remains after the first and the second contestant, and so on until the last one, who takes all of the remaining cake. Determine which competitor takes the largest piece of cake.

Point $A$ lies inside an angle with vertex $M$. A ray issuing from point $A$ is reflected in one side of the angle at point $B$, then in the other side at point $C$ and then returns back to point $A$ (the ordinary rule of reflection holds). Prove that the center of the circle circumscribed about triangle $\triangle BCM$ lies on line $AM$.

An integer $x$ has the property that the sums of the digits of $x$ and of $3x$ are the same. Prove that $x$ is divisible by $9$.

Find the number of positive integer solutions $(a,b,c,d)$ to the equation \[(a^2+b^2)(c^2-d^2)=2020.\] Note: The solutions $(10,1,6,4)$ and $(1,10,6,4)$ are considered different.

$\triangle ABC$ is isosceles $AB = AC$. $P$ is a point inside $\triangle ABC$ such that $\angle BCP = 30$ and $\angle APB = 150$ and $\angle CAP = 39$. Find $\angle BAP$.

Is it possible to place the numbers $0,1,2,\dots,9$ on a circle so that the sum of any three consecutive numbers is a) 13, b) 14, c) 15?

The difference in the areas of two similar triangles is $18$ square feet, and the ratio of the larger area to the smaller is the square of an integer. The area of the smaller triange, in square feet, is an integer, and one of its sides is $3$ feet. The corresponding side of the larger triangle, in feet, is: $\textbf{(A)}\ 12\quad \textbf{(B)}\ 9\qquad \textbf{(C)}\ 6\sqrt{2}\qquad \textbf{(D)}\ 6\qquad \textbf{(E)}\ 3\sqrt{2}$

Let $P$ be the intersection point of the diagonals of the quadrangle $ABCD$, $M$ the intersection point of the lines connecting the midpoints of its opposite sides, $O$ the intersection point of the perpendicular bisectors of the diagonals, $H$ the intersection point of the lines connecting the orthocenters of the triangles $APD$ and $BCP$, $APB$ and $CPD$. Prove that $M$ is the midpoint of $OH$.

Let $\mathbb{N}$ be the set of positive integers. For every $n \in \mathbb{N}$, define $d(n)$ as the number of positive divisors of $n$. Find all functions $f : \mathbb{N} \rightarrow \mathbb{N}$ such that: a) $d(f(x)) = x$ for all $x \in \mathbb{N}$ b) $f(xy)$ divides $(x-1)y^{xy-1}f(x)$ for all $x,y \in \mathbb{N}$

Let $a_1,a_2,\ldots,a_k$ be real numbers and $a_1+a_2+\ldots+a_k=0$. Prove that \[ \max_{1\leq i \leq k} a_i^2 \leq \frac{k}{3} \left( (a_1-a_2)^2+(a_2-a_3)^2+\cdots +(a_{k-1}-a_k)^2\right). \]

The star shown is symmetric with respect to each of the six diagonals shown. All segments connecting the points $A_1, A_2, . . . , A_6$ with the centre of the star have the length $1$, and all the angles at $B_1, B_2, . . . , B_6$ indicated in the figure are right angles. Calculate the area of the star. [img]https://1.bp.blogspot.com/-Rso2aWGUq_k/XzcAm4BkAvI/AAAAAAAAMW0/277afcqTfCgZOHshf_6ce2XpinWWR4SZACLcBGAsYHQ/s0/2006%2BMohr%2Bp1.png[/img]

Integers $n$ and $k$ satisfy $n > 2023k^3$. Kingdom Kitty has $n$ cities, with at most one road between each pair of cities. It is known that the total number of roads in the kingdom is at least $2n^{3/2}$. Prove that we can choose $3k + 1$ cities such that the total number of roads with both ends being a chosen city is at least $4k$.

Let $B=\{2^1,2^2,2^3,\dots,2^{21}\}.$ Find the remainder when \[ \sum_{m, n \in B: \ m<n}\gcd(m,n) \] is divided by $1000,$ where the sum is taken over all pairs of elements $(m,n)$ of $B$ such that $m<n.$

In what case does the system of equations $\begin{matrix} x + y + mz = a \\ x + my + z = b \\ mx + y + z = c \end{matrix}$ have a solution? Find conditions under which the unique solution of the above system is an arithmetic progression.

The function $f(x)$ satisfies $f(2+x) = f(2-x)$ for all real numbers $x$. If the equation $f(x) = 0$ has exactly four distinct real roots, then the sum of these roots is A. 0 B. 2 C. 4 D. 6 E. 8

In equilateral triangle $ABC$, $AB=2$ and $M$ is the midpoint of $AB$. A laser is shot from $M$ in a certain direction, and whenever it collides with a side of $ABC$ it will reflect off the side such that the acute angle formed by the incident ray and the side is equal to the acute angle formed by the reflected ray and the side. Once the laser coincides with a vertex, it stops. Find the sum of the smallest three possible integer distances that the laser could have traveled. [i]Proposed by Jerry Xu[/i] [hide=Solution] [i]Solution.[/i] $\boxed{21}$ Whenever the laser hits a side of the triangle, reflect the laser's path over that side so that the path of the laser forms a straight line. We want the path of the laser to coincide with a vertex of one of the reflected triangles. Thus, we can restate the problem as follows: Tessellate the plane with equilateral triangles of side length $3$. Consider one of these equilateral triangles $ABC$ with $M$ being the midpoint of $AB=2$. Find the sum of the three minimum integer distances from $M$ to any vertex in the plane. [asy] import geometry; size(8cm); pair A = (0,sqrt(3)); pair B = (-1,0); pair C = (1,0); pair M = (0,0); for (int i = -1; i <= 2; ++i) { draw((i-3,i*sqrt(3))--(-i+3,i*sqrt(3))); draw(((i-1)*2,-sqrt(3))--(i+1,(2-i)*sqrt(3))); draw((-i-1,(2-i)*sqrt(3))--((1-i)*2,-sqrt(3))); } draw(A--B--C--A, red); dot(M); label("$A$",A+(0,0.25),N); label("$B$",B-(0.25,0),SW); label("$C$",C+(0.25,0),SE); label("$M$",M,S); [/asy] It is trivial to see that the vertical distance between $M$ and a given vertex is $n\sqrt{3}$ for $n \in \mathbb{N}^{0}$. If $n$ is even, the horizontal distance between $O$ and a given vertex is $1+2m$ for $m \in \mathbb{N}^{0}$. If $n$ is odd, the horizontal distance is $2m$ for $m \in \mathbb{N}^{0}$. We consider two separate cases: $1.$ $n$ is even. We thus want to find $l \in \mathbb{N}$ such that $$\left(n\sqrt{3}\right)^2+(1+2m)^2=l^2.$$Make the substitution $1+2m=k$ to get that $$3n^2+k^2=l^2.$$Notice that these equations form a family of generalized Pell equations $y^2-3x^2=N$ with $N=k^2$. We can find some set of roots to these equations using the multiplicative principle: we will use this idea to find three small $l$ values, and that gives us an upper bound on what the three $l$ values can be. From there, a simple bash of lower $l$ values to see if solutions to each generalized Pell equation not given by the multiplicative principle exist finishes this case. By the multiplicative principle some set of solutions $(x_n,y_n)$ to the above equation with sufficiently small $x_n$ follow the formula$$x_n\sqrt{3}+y_n=\left(x_0\sqrt{3}+y_0\right)\left(u_n\sqrt{3}+v_n\right),$$where $\left(x_0,y_0\right)$ is a solution to the generalized Pell equation and $\left(u_n,v_n\right)$ are solutions to the Pell equation $y^2-3x^2=1$. Remember that the solutions to this last Pell equation satisfy$$u_n\sqrt{3}+v_n=\left(u_0\sqrt{3}+v_0\right)^k$$where the trivial positive integer solution $$\left(u_0, v_0\right)=(1,2)$$(this can easily be found by inspection or by taking the convergents of the continued fraction expansion of $\sqrt{3}$). We thus get that$$\left(u_1,v_1\right)=(4,7),\left(u_2,v_2\right)=(15,26),\left(u_2,v_2\right)=(56,97)\dots$$(also don't forget that $(u,v)=(0,1)$ is another solution). From here, note that $k$ must be odd since $k=1+2m$ for $m \in \mathbb{N}^{0}$. For $k=1$, the smallest three solutions to the Pell equation with $n$ even are \begin{align*} (x,y)&=(0,1),(4,7),(56,97) \\ \longrightarrow (n,m,l)&=(0,0,1),(4,0,7),(56,0,97) \end{align*}Our current smallest three values of $l$ are thus $1,7,97$. A quick check confirms that all of these solutions are not extraneous (extraneous solutions appear when the path taken by the laser prematurely hits a vertex). For $k=3$, using the multiplicative principle we get two new smaller solutions \begin{align*} (x,y)&=(0,3),(12,21) \\ \longrightarrow (n,m,l)&=(0,1,3),(12,1,21) \end{align*}However, note that $(n,m,l)=(0,1,3)$ is extraneous since is equivalent to the path that is traced out by the solution $(n,m,l)=(0,0,1)$ found previously and will thus hit a vertex prematurely. Thus, our new three smallest values of $l$ are $1,7,21$. For $k \ge 5$, it is evident that there are no more smaller integral values of $l$ that can be found using the multiplicative principle: the solution set $(n,m,l)=\left(0,\dfrac{k-1}{2},k\right)$ is always extraneous for $k > 1$ since it is equivalent to the path traced out by $(0,0,1)$ as described above, and any other solutions will give larger values of $l$. Thus, we now only need to consider solutions to each generalized Pell equation not found by the multiplicative principle. A quick bash shows that $l=3,5,9,11$ gives no solutions for any odd $k$ and even $n$, however $n=13$ gives $k=11$ and $n=4$, a non-extraneous solution smaller than one of the three we currently have. Thus, our new three smallest $l$ values are $1,7,13$. $2$. $n$ is odd. We thus want to find $l \in \mathbb{N}$ such that $$\left(n\sqrt{3}\right)^2+(2m)^2=l^2.$$Make the substitution $2m=k$ to get that $$3n^2+k^2=l^2.$$This is once again a family of generalized Pell equations with $N=k^2$, however this time we must have $k$ even instead of $k$ odd. However, note that there are no solutions to this family of Pell equation with $n$ odd: $k^2 \equiv 0 \text{ (mod }4)$ since $k$ is even, and $3n^2 \equiv 3 \text{ (mod }4)$ since $n$ is odd, however $0+3 \equiv 3 \text{ (mod }4)$ is not a possible quadratic residue mod $4$. Thus, this case gives no solutions. Our final answer is thus $1+7+13=\boxed{21}$. [/hide]

Given a right triangle $ABC$ with the hypotenuse $AB$. Let $K$ be any interior point of triangle $ABC$ and points $L, M$ are symmetric of point $K$ wrt lines $BC, AC$ respectively. Specify all possible values for $S_{ABLM} / S_{ABC}$, where $S_{XY ... Z}$ indicates the area of the polygon $XY...Z$ .

The points $D,E,F$ belong to the sides $(AB), (BC)$ and $(CA)$ of the triangle $ABC$ respectively (but they are not vertices). Let us denote with $d_0, d_1, d_2$, and $d_3$ the maximal side length of the triangles $DEF$, $DEA$, $DBF$, $CEF$, respectively. Prove that $$d_0 \ge \frac{\sqrt3}{2} min\{d_1, d_2, d_3\}$$ When the equality takes place?

In a softball league, after each team has played every other team 4 times, the total accumulated points are: Lions 22, Tigers 19, Mounties 14, and Royals 12. If each team received 3 points for a win, 1 point for a tie and no points for a loss, how many games ended in a tie? $\textbf{(A)}\ 3 \qquad \textbf{(B)}\ 4 \qquad \textbf{(C)}\ 5 \qquad \textbf{(D)}\ 7 \qquad \textbf{(E)}\ 10$

Let $ a, b, c, r,$ and $ s$ be real numbers. Show that if $ r$ is a root of $ ax^2\plus{}bx\plus{}c \equal{} 0$ and s is a root of $ \minus{}ax^2\plus{}bx\plus{}c \equal{} 0,$ then \[ \frac{a}{2} x^2 \plus{} bx \plus{} c \equal{} 0\] has a root between $ r$ and $ s.$

In the sequence of digits $2,0,2,9,3,...$ any digit it equal to the last digit in the decimal representation of the sum of four previous digits. Do the four numbers $2,0,1,5$ in that order occur in the sequence? Folklore

Let $n$ be a given positive integer. We say that a positive integer $m$ is [i]$n$-good[/i] if and only if there are at most $2n$ distinct primes $p$ satisfying $p^2\mid m$. (a) Show that if two positive integers $a,b$ are coprime, then there exist positive integers $x,y$ so that $ax^n+by^n$ is $n$-good. (b) Show that for any $k$ positive integers $a_1,\ldots,a_k$ satisfying $\gcd(a_1,\ldots,a_k)=1$, there exist positive integers $x_1,\ldots,x_k$ so that $a_1x_1^n+a_2x_2^n+\cdots+a_kx_k^n$ is $n$-good. (Remark: $a_1,\ldots,a_k$ are not necessarily pairwise distinct) [i]Proposed by usjl.[/i]

Show that if $\lambda > \frac{1}{2}$ there does not exist a real-valued function $u(x)$ such that for all $x$ in the closed interval $[0,1]$ the following holds: $$u(x)= 1+ \lambda \int_{x}^{1} u(y) u(y-x) \; dy.$$

Can a $5\times 7$ checkerboard be covered by L's (figures formed from a $2\times2$ square by removing one of its four $1\times1$ corners), not crossing its borders, in several layers so that each square of the board is covered by the same number of L's? [i]M. Evdokimov[/i]

Define a permutation of the set $\{1,2,3,...,n\}$ to be $\textit{sortable}$ if upon cancelling an appropriate term of such permutation, the remaining $n-1$ terms are in increasing order. If $f(n)$ is the number of sortable permutations of $\{1,2,3,...,n\}$, find the remainder when $$\sum\limits_{i=1}^{2018} (-1)^i \cdot f(i)$$ is divided by $1000$. Note that the empty set is considered sortable. [i]Proposed by [b]FedeX333X[/b][/i]