John and Mary play the following game. First they choose integers $n > m > 0$ and put $n$ sweets on an empty table. Then they start to make moves alternately. A move consists of choosing a nonnegative integer $k\le m$ and taking $k$ sweets away from the table (if $k = 0$ , nothing happens in fact). In doing so no value for $k$ can be chosen more than once (by none of the players) or can be greater than the number of sweets at the table at the moment of choice. The game is over when one of the players can make no more moves. John and Mary decided that at the beginning Mary chooses the numbers $m$ and $n$ and then John determines whether the performer of the last move wins or looses. Can Mary choose $m$ and $n$ in such way that independently of John’s decision she will be able to win?

In triangle $ABC$, $AC=24$ inches, $BC=10$ inches, $AB=26$ inches. The radius of the inscribed circle is: $\textbf{(A)}\ 26\text{ in} \qquad \textbf{(B)}\ 4\text{ in} \qquad \textbf{(C)}\ 13\text{ in} \qquad \textbf{(D)}\ 8\text{ in} \qquad \textbf{(E)}\ \text{None of these}$

On a table there are $1999$ counters, red on one side and black on the other side, arranged arbitrarily. Two people alternately make moves, where each move is of one of the following two types: (1) Remove several counters which all have the same color up; (2) Reverse several counters which all have the same color up. The player who takes the last counter wins. Decide which of the two players (the one playing first or the other one) has a wining strategy.

Alex, Bertha, Cameron, Dylan, and Ellen each have a different toy. Each kid puts each of their own toys into a large bag. The toys are then randomly distributed such that each kid receives a toy. How many ways are there for exactly one kid to get the same toy that they put in? [i]2018 CCA Math Bonanza Lightning Round #2.4[/i]

For an integer $n \ge 1$ we consider sequences of $2n$ numbers, each equal to $0, -1$ or $1$. The [i]sum product value[/i] of such a sequence is calculated by first multiplying each pair of numbers from the sequence, and then adding all the results together. For example, if we take $n = 2$ and the sequence $0,1, 1, -1$, then we find the products $0\cdot 1, 0\cdot 1, 0\cdot -1, 1\cdot 1, 1\cdot -1, 1\cdot -1$. Adding these six results gives the sum product value of this sequence: $0+0+0+1+(-1)+(-1) = -1$. The sum product value of this sequence is therefore smaller than the sum product value of the sequence $0, 0, 0, 0$, which equals $0$. Determine for each integer $n \ge 1$ the smallest sum product value that such a sequence of $2n$ numbers could have. [i]Attention: you are required to prove that a smaller sum product value is impossible.[/i]

Suppose that $n$ is a natural number. we call the sequence $(x_1,y_1,z_1,t_1),(x_2,y_2,z_2,t_2),.....,(x_s,y_s,z_s,t_s)$ of $\mathbb Z^4$ [b]good[/b] if it satisfies these three conditions: [b]i)[/b] $x_1=y_1=z_1=t_1=0$. [b]ii)[/b] the sequences $x_i,y_i,z_i,t_i$ be strictly increasing. [b]iii)[/b] $x_s+y_s+z_s+t_s=n$. (note that $s$ may vary). Find the number of good sequences. [i]proposed by Mohammad Ghiasi[/i]

In the triangle $ABC$ let $B'$ and $C'$ be the midpoints of the sides $AC$ and $AB$ respectively and $H$ the foot of the altitude passing through the vertex $A$. Prove that the circumcircles of the triangles $AB'C'$,$BC'H$, and $B'CH$ have a common point $I$ and that the line $HI$ passes through the midpoint of the segment $B'C'.$

Let $\lfloor x \rfloor$ denote the greatest integer less than or equal to $x.$ Let $\lambda \geq 1$ be a real number and $n$ be a positive integer with the property that $\lfloor \lambda^{n+1}\rfloor, \lfloor \lambda^{n+2}\rfloor ,\cdots, \lfloor \lambda^{4n}\rfloor$ are all perfect squares$.$ Prove that $\lfloor \lambda \rfloor$ is a perfect square$.$

Let $S$ be a set consisting of all positive integers less than or equal to $100$. Let $P$ be a subset of $S$ such that there do not exist two elements $x,y\in P$ such that $x=2y$. Find the maximum possible number of elements of $P$. [i]Proposed by Nathan Ramesh

Let $O$ be the circumcenter and $I$ be the incenter of triangle $ABC.$ Prove that if $AI\perp OB$ and $BI\perp OC$ then $CI\parallel OA$.

Point $O$ is the center of the circumscribed circle of triangle $ABC$. Ray $AO$ intersects the side $BC$ at point $T$. With $AT$ as a diameter, a circle is constructed. At the intersection with the sides of the triangle $ABC$, three arcs were formed outside it. Prove that the larger of these arcs is equal to the sum of the other two. (Oleksii Karliuchenko)

Let $a$, $b$, and $d$ be positive integers. It is known that $a+b$ is divisible by $d$ and $a\cdot b$ is divisible by $d^2$. Prove that both $a$ and $b$ are divisible by $d$.

Place the integers $1,2 , \ldots, n^{3}$ in the cells of a $n\times n \times n$ cube such that every number appears once. For any possible enumeration, write down the maximal difference between any two adjacent cells (adjacent means having a common vertex). What is the minimal number noted down?

Investigate topologically the Riemann surface of the function \[w=\arctan z\]

Evin and Jerry are playing a game with a pile of marbles. On each players' turn, they can remove $2$, $3$, $7$, or $8$ marbles. If they can’t make a move, because there's $0$ or $1$ marble left, they lose the game. Given that Evin goes first and both players play optimally, for how many values of $n$ from $1$ to $1434$ does Evin lose the game? [i]Proposed by Evin Liang[/i] [hide=Solution][i]Solution.[/i] $\boxed{573}$ Observe that no matter how many marbles a one of them removes, the next player can always remove marbles such that the total number of marbles removed is $10$. Thus, when the number of marbles is a multiple of $10$, the first player loses the game. We analyse this game based on the number of marbles modulo $10$: If the number of marbles is $0$ modulo $10$, the first player loses the game If the number of marbles is $2$, $3$, $7$, or $8$ modulo $10$, the first player wins the game by moving to $0$ modulo 10 If the number of marbles is $5$ modulo $10$, the first player loses the game because every move leads to $2$, $3$, $7$, or $8$ modulo $10$ In summary, the first player loses if it is $0$ mod 5, and wins if it is $2$ or $3$ mod $5$. Now we solve the remaining cases by induction. The first player loses when it is $1$ modulo $5$ and wins when it is $4$ modulo $5$. The base case is when there is $1$ marble, where the first player loses because there is no move. When it is $4$ modulo $5$, then the first player can always remove $3$ marbles and win by the inductive hypothesis. When it is $1$ modulo $5$, every move results in $3$ or $4$ modulo $5$, which allows the other player to win by the inductive hypothesis. Thus, Evin loses the game if n is $0$ or $1$ modulo $5$. There are $\boxed{573}$ such values of $n$ from $1$ to $1434$.[/hide]

Let $ \{a_n\}_{n=1}^\infty $ be an arithmetic progression with $ a_1 > 0 $ and $ 5\cdot a_{13} = 6\cdot a_{19} $ . What is the smallest integer $ n$ such that $ a_n<0 $?

Find all real solutions to the equation $(x^2 + 3x + 1)^{x^2-x-6} = 1$.

Mike leaves home and drives slowly east through city traffic. When he reaches the highway he drives east more rapidly until he reaches the shopping mall where he stops. He shops at the mall for an hour. Mike returns home by the same route as he came, driving west rapidly along the highway and then slowly through city traffic. Each graph shows the distance from home on the vertical axis versus the time elapsed since leaving home on the horizontal axis. Which graph is the best representation of Mike's trip? [asy] import graph; unitsize(12); real a(real x) {return ((x-15)^2)/2;} real b(real x) {return ((x-25)^2)/2;} real c(real x) {return ((x-30)^2 * (x-40)^2) * 8/625;} real d(real x) {return ((x-15)^2)*8/25-15;} real e(real x) {return ((x-25)^2)*8/25-15;} draw((0,9)--(0,0)--(11,0)); draw((15,9)--(15,0)--(26,0)); draw((30,9)--(30,0)--(41,0)); draw((0,-6)--(0,-15)--(11,-15)); draw((15,-6)--(15,-15)--(26,-15)); draw((0,0)--(3,8)--(7,8)--(10,0)); draw(graph(a,15,17)); draw((17,2)--(18,8)--(22,8)--(23,2)); draw(graph(b,23,25)); draw(graph(c,30,40)); draw((0,-15)--(5,-7)--(10,-15)); draw(graph(d,15,20)); draw(graph(e,20,25)); for (int k=0; k<3; ++k) { label("d",(15*k-1,8),N); label("i",(15*k-1,7),N); label("s",(15*k-1,6),N); label("t",(15*k-1,5),N); label("a",(15*k-1,4),N); label("n",(15*k-1,3),N); label("c",(15*k-1,2),N); label("e",(15*k-1,1),N); label("time",(15*k+8,0),S); } for (int k=0; k<2; ++k) { label("d",(15*k-1,8-15),N); label("i",(15*k-1,7-15),N); label("s",(15*k-1,6-15),N); label("t",(15*k-1,5-15),N); label("a",(15*k-1,4-15),N); label("n",(15*k-1,3-15),N); label("c",(15*k-1,2-15),N); label("e",(15*k-1,1-15),N); label("time",(15*k+8,0-15),S); } label("(A)",(5,9),N); label("(B)",(20,9),N); label("(C)",(35,9),N); label("(D)",(5,-6),N); label("(E)",(20,-6),N); [/asy]

There is a unique degree-$10$ monic polynomial with integer coefficients $f(x)$ such that $$f \left( \sum^9_{j=0}\sqrt[10]{2021^j}\right)= 0.$$ Find the remainder when $f(1)$ is divided by $1000$.

The set Φ consists of a finite number of points on the plane. The distance between any two points from Φ is at least $\sqrt{2}$. It is known that a regular triangle with side lenght $3$ cut out of paper can cover all points of Φ. What is the greatest number of points that Φ can consist of?

Let $ABC$ be an acute angled isosceles triangle with $AB=AC$ and circumcentre $O$. Lines $BO$ and $CO$ intersect $AC, AB$ respectively at $B', C'$. A straight line $l$ is drawn through $C'$ parallel to $AC$. Prove that the line $l$ is tangent to the circumcircle of $\triangle B'OC$.

Several positive integers are written in a row. Iteratively, Alice chooses two adjacent numbers $x$ and $y$ such that $x>y$ and $x$ is to the left of $y$, and replaces the pair $(x,y)$ by either $(y+1,x)$ or $(x-1,x)$. Prove that she can perform only finitely many such iterations. [i]Proposed by Warut Suksompong, Thailand[/i]

In a triangle $ABC$, $BC=a$, $AC=b$, $\angle B=\beta$ and $\angle C=\gamma$. Prove that the bisector of the angle at $A$ is equal to the altitude from $B$ if and only if $b=a\cos\frac{\beta-\gamma}2$.

Two congruent rectangles are located as shown in the figure. Find the area of the shaded part. [img]https://cdn.artofproblemsolving.com/attachments/2/e/10b164535ab5b3a3b98ce1a0b84892cd11d76f.png[/img]

Let $x,y,z $ be positive reals such that $\sqrt{a}=x(y-z)^2$, $\sqrt{b}=y(z-x)^2$ and $\sqrt{c}=z(x-y)^2$. Prove that \[a^2+b^2+c^2 \geq 2(ab+bc+ca)\]