Two equal circles $S_1$ and $S_2$ meet at two different points. The line $\ell$ intersects $S_1$ at points $A,C$ and $S_2$ at points $B,D$ respectively (the order on $\ell$: $A,B,C,D$) . Define circles $\Gamma_1$ and $\Gamma_2$ as follows: both $\Gamma_1$ and $\Gamma_2$ touch $S_1$ internally and $S_2$ externally, both $\Gamma_1$ and $\Gamma_2$ line $\ell$, $\Gamma_1$ and $\Gamma_2$ lie in the different halfplanes relatively to line $\ell$. Suppose that $\Gamma_1$ and $\Gamma_2$ touch each other. Prove that $AB=CD$. I. Voronovich

A teacher writes the positive integers from $1$ to $12$ on a blackboard. Every minute, they choose a number $k$ uniformly at random from the written numbers, subtract $k$ from each number $n \geq k$ on the blackboard (without touching the numbers $n<k$), and erase every $0$ on the board. Estimate the expected number of minutes that pass before the board is empty. An estimate of $E$ earns $2^{1-0.5|E-A|}$ points, where $A$ is the actual answer. [i]2020 CCA Math Bonanza Lightning Round #5.2[/i]

Let us denote by $<a>$ the distance from $a$ to the nearest integer. (For example, $<1,2> = 0.2$, $<\sqrt3> = 2-\sqrt3$) How many solutions does the system of equations have $$\begin{cases} <19x>=y \\ <96y>=x \end{cases} \,\,\, ?$$

Find all possible values of $a\in \mathbb{R}$ and $n\in \mathbb{N^*}$ such that $f(x)=(x-1)^n+(x-2)^{2n+1}+(1-x^2)^{2n+1}+a$ is divisible by $\phi (x)=x^2-x+1$

Santa Claus wants to wrap presents. These are available in $n$ sizes $A_1 <A_2 <...<A_n$, and analogously, there are $n$ packaging sizes $B_1 <B_2 <...<B_n$, where $B_i$ is enough to all gift sizes $A_j$ can be grouped with $j\le i$, but too small for those with $j> i$. On the shelf to the right of Santa Claus are the gifts sorted by size, where the smallest are on the right, of course there can be several gifts of the same size, or none of a size at all. To his left is a shelf with packaging, and also these are sorted from small to large in the same direction. He's brooding in what way he should wrap the gifts and sees two methods for doing this, which depend on his thinking and laziness of movement have been optimized: a) He takes the present closest to him and puts it in the closest packaging, in which it fits in. b) He takes the packaging closest to him and packs in it the closest thing to him gift. In both cases he then does the same again, although of course the one he was using the gift and its packaging are missing, and so on. Once it is not large enough if the packaging or the present is not small enough, he / she will provide the present or the packaging back to its place on the shelf and takes the next-closest. Prove that both methods lead to the same result in the end, they are considered to be exactly the same gifts packed in the same packaging.

In Simoland there are $2017n$ cities arranged in a $2017\times n$ lattice grid. There are $2016$ MRT (train) tracks and each track can only go north and east, or south and east. Suppose that all the tracks together pass through all the cities. Determine the largest possible value of $n$.

At a prom, there are $4$ boys and $3$ girls. Each boy picks a girl to dance with, and each girl picks a boy to dance with. Assuming that each choice is uniformly random, the probability that at least one boy and one girl choose each other as dance partners is $\tfrac{p}{q},$ where $p$ and $q$ are relatively prime positive integers. Compute $p+q.$

Let $p, q$, and $r$ be prime numbers such that $2pqr + p + q + r = 2020$. Find $pq + qr + rp$.

Prove that the number $9^n+8^n+7^n+6^n-4^n-3^n-2^n-1^n$ is divisible by $10$ for all non-negative $n$.

Show that if $a \ge b \ge c \ge 0$ then $$a^2b(a - b) + b^2c(b - c) + c^2a(c - a) \ge 0.$$

Let $a$ be a real number. Find the number of functions $f:\mathbb{R}\rightarrow \mathbb{R}$ depending on $a$, such that $f(xy+f(y))=f(x)y+a$ holds for every $x, y\in \mathbb{R}$.

Given a triangle $ABC$, let $D$ be the intersection of the line through $A$ perpendicular to $AB$, and the line through $B$ perpendicular to $BC$. Let $P$ be a point inside the triangle. Show that $DAPB$ is cyclic if and only if $\angle BAP = \angle CBP$.

What is $20 - 2^1$? $\textbf{(A) }1\qquad\textbf{(B) }18\qquad\textbf{(C) }19\qquad\textbf{(D) }20\qquad\textbf{(E) }21$

Given positive integers $n$, $P_1$, $P_2$, …$P_n$ and two sets \[B=\{ (a_1,a_2,…,a_n)|a_i=0 \vee 1,\ \forall i \in \mathbb{N} \}, S=\{ (x_1,x_2,…,x_n)|1 \leq x_i \leq P_i \wedge x_i \in \mathbb{N} ,\ \forall i \in \mathbb{N} \}\] A function $f:S \to \mathbb{Z}$ is called [b]Real[/b], if and only if for any positive integers $(y_1,y_2,…,y_n)$ and positive integer $a$ which satisfied $ 1 \leq y_i \leq P_i-a$ $\forall i \in \mathbb{N}$, we always have: \begin{align*} \sum_{(a_1,a_2,…,a_n) \in B \wedge 2| \sum_{i=1}^na_i}f(y+a \times a_1,y+a \times a_2,……,y+a \times a_n)&>\\ \sum_{(a_1,a_2,…,a_n) \in B \wedge 2 \nmid \sum_{i=1}^na_i}f(y+a \times a_1,y+a \times a_2,……,y+a \times a_n)&. \end{align*} Find the minimum of $\sum_{i_1=1}^{P_1}\sum_{i_2=1}^{P_2}....\sum_{i_n=1}^{P_n}|f(i_1,i_2,...,i_n)|$, where $f$ is a [b]Real[/b] function. [i]Proposed by tob8y[/i]

For every nonempty set $M{}$ of integers denote $S(M)$ the sum of all its elements. Let $A=\{a_1,a_2,\ldots,a_{11}\}$ be a set of positive integers with the properties: 1) $a_1<a_2<\ldots<a_{11};$ 2) for every positive integer $n\leq 1500$ there is a subset $M{}$ of $A{}$ for which $S(M)=n.$ Find the smallest possible value of $a_{10}.$

Find positive integers $a_1 < a_2<... <a_{2010}$ such that $$a_1(1!)^{2010} + a_2(2!)^{2010} + ... + a_{2010}(2010!)^{2010} = (2011 !)^{2010}. $$

Let triangle $ABC$ with incenter $I$ satisfy $AB = 10$, $BC = 21$, and $CA = 17$. Points $D$ and E lie on side $BC$ such that $BD = 4$, $DE = 6$, and $EC = 11$. The circumcircles of triangles $BIE$ and $CID$ meet again at point $P$, and line $IP$ meets the altitude from $A$ to $BC$ at $X$. Find $(DX \cdot EX)^2$.