Let $ABC$ be a non isosceles triangle with circumcircle $(O)$ and incircle $(I)$. Denote $(O_1)$ as the circle that external tangent to $(O)$ at $A'$ and also tangent to the lines $AB,AC$ at $A_b,A_c$ respectively. Define the circles $(O_2), (O_3)$ and the points $B', C', B_c , B_a, C_a, C_b$ similarly. 1. Denote J as the radical center of $(O_1), (O_2), (O_3) $and suppose that $JA'$ intersects $(O_1)$ at the second point $X, JB'$ intersects $(O_2)$ at the second point Y , JC' intersects $(O_3)$ at the second point $Z$. Prove that the circle $(X Y Z)$ is tangent to $(O_1), (O_2), (O_3)$. 2. Prove that $AA', BB', CC'$ are concurrent at the point $M$ and $3$ points $I,M,O$ are collinear.

p1. Circle $M$ is the incircle of ABC, while circle $N$ is the incircle of $ACD$. Circles $M$ and $N$ are tangent at point $E$. If side length $AD = x$ cm, $AB = y$ cm, $BC = z$ cm, find the length of side $DC$ (in terms of $x, y$, and $z$). [img]https://cdn.artofproblemsolving.com/attachments/d/5/66ddc8a27e20e5a3b27ab24ff1eba3abee49a6.png[/img] p2. The address of the house on Jalan Bahagia will be numbered with the following rules: $\bullet$ One side of the road is numbered with consecutive even numbers starting from number $2$. $\bullet$ The opposite side is numbered with an odd number starting from number $3$. $\bullet$ In a row of even numbered houses, there is some land vacant house that has not been built. $\bullet$ The first house numbered $2$ has a neighbor next door. When the RT management ordered the numbers of the house, it is known that the cost of making each digit is $12.000$ Rp. For that, the total cost to be incurred is $1.020.000$ Rp. It is also known that the cost of all even-sided house numbers is $132.000$ Rp. cheaper than the odd side. When the land is empty later a house has been built, the number of houses on the even and odd sides is the same. Determine the number of houses that are now on Jalan Bahagia . p3. Given the following problem: Each element in the set $A = \{10, 11, 12,...,2008\}$ multiplied by each element in the set $B = \{21, 22, 23,...,99\}$. The results are then added together to give value of $X$. Determine the value of $X$. Someone answers the question by multiplying $2016991$ with $4740$. How can you explain that how does that person make sense? p4. Let $P$ be the set of all positive integers between $0$ and $2008$ which can be expressed as the sum of two or more consecutive positive integers . (For example: $11 = 5 + 6$, $90 = 29 + 30 + 31$, $100 = 18 + 19 +20 + 21 + 22$. So $11, 90, 100$ are some members of $P$.) Find the sum of of all members of $P$. p5. A four-digit number will be formed from the numbers at $0, 1, 2, 3, 4, 5$ provided that the numbers in the number are not repeated, and the number formed is a multiple of $3$. What is the probability that the number formed has a value less than $3000$?

Represent the polynomial $P(X) = X^{100} + X^{20} + 1$ as the product of 4 polynomials with integer coefficients.

Each of the numbers in the set $N = \{1, 2, 3, \cdots, n - 1\}$, where $n \geq 3$, is colored with one of two colors, say red or black, so that: [i](i)[/i] $i$ and $n - i$ always receive the same color, and [i](ii)[/i] for some $j \in N$, relatively prime to $n$, $i$ and $|j - i|$ receive the same color for all $i \in N, i \neq j.$ Prove that all numbers in $N$ must receive the same color.

A $ 9 \times 9$ square is divided into unit squares. Is it possible to fill each unit square with a number $ 1, 2,..., 9$ in such a way that, whenever one places the tile so that it fully covers nine unit squares, the tile will cover nine different numbers?

In a triangle $ABC$, let $D$ be the midpoint of the side $BC$, and let $E$ be a point on the side $AC$. The lines $BE$ and $AD$ meet at a point $F$. Prove: If $\frac{BF}{FE}=\frac{BC}{AB}+1$, then the line $BE$ bisects the angle $ABC$.

Find all functions \(f: \mathbb{R} \rightarrow \mathbb{R}\) such that for any real numbers \(x\) and \(y\), the following inequality holds: \[ f\left(x^2+2y f(x)\right) + (f(y))^2 \leq f\left((x+y)^2\right) \] [i]Proposed by Anton Trygub[/i]

Each vertex of a convex polygon is projected to all nonadjacent sidelines. Can it happen that each of these projections lies outside the corresponding side?

Dürer's $n \times m$ garden is surgically divided into $n \times m$ unit squares, and in the middle of one of these squares, he planted his favourite petunia. Dürer's gardener struggles with a mole, trying to drive him out of the magnificent garden, so he builds an underground wall on the edge of the garden. The only problem is that the mole managed to stay inside the walls.. When the mole meets a wall, it changes it's direction as if it was "reflected", that is, proceeding his route in the direction that includes the same angle with the wall as his direction before. The mole starts beneath the petunia, in a direction that includes a $45^o$ angle with the walls. Is it possible for the mole to cross the petunia in a direction perpendicular to it's original direction? (Think in terms of $n,m$.)

Suppose that $ p_1,p_2,p_3,q_1,q_2,q_3$ are six points in the plane and that the distance between $ p_i$ and $ q_j$ ($ i,j \equal{} 1,2,3$) is $ i \plus{} j$. Show that the six points are collinear.

For how many positive integers $n\le100$ is it true that $10n$ has exactly three times as many positive divisors as $n$ has?

Players $A$ and $B$ play a "paintful" game on the real line. Player $A$ has a pot of paint with four units of black ink. A quantity $p$ of this ink suffices to blacken a (closed) real interval of length $p$. In every round, player $A$ picks some positive integer $m$ and provides $1/2^m $ units of ink from the pot. Player $B$ then picks an integer $k$ and blackens the interval from $k/2^m$ to $(k+1)/2^m$ (some parts of this interval may have been blackened before). The goal of player $A$ is to reach a situation where the pot is empty and the interval $[0,1]$ is not completely blackened. Decide whether there exists a strategy for player $A$ to win in a finite number of moves.

$p$ is a prime. Find the largest integer $d$ such that $p^d$ divides $p^4!$.

Given a triangle $ ABC$ with angle $ C \geq 60^{\circ}$. Prove that: $ \left(a \plus{} b\right) \cdot \left(\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c} \right) \geq 4 \plus{} \frac {1}{\sin\left(\frac {C}{2}\right)}.$

Find a way of arranging $n$ girls and $n$ boys around a round table for which $d_n-c_n$ is maximum, where dn is the number of girls sitting between two boys and $c_n$ is the number of boys sitting between two girls.

Circles $\omega_1$ and $\omega_2$ meet at points $A$ and $B$. Let points $K_1$ and $K_2 $ of $\omega_1$ and $\omega_2$ respectively be such that $K_1A$ touches $\omega_2$, and $K_2A$ touches $\omega_1$. The circumcircle of triangle $K_1BK_2$ meets lines $AK_1$ and $AK_2$ for the second time at points $L_1$ and $L_2$ respectively. Prove that $L_1$ and $L_2$ are equidistant from line $AB$.

Two triangles $ABC$ and $DEF$ have the same incircle. If a circle passes through $A,B,C,D,E$ prove that it also passes through $F$.

$ P(x)$ is a quadratic trinomial. What maximum number of terms equal to the sum of the two preceding terms can occur in the sequence $ P(1)$, $ P(2)$, $ P(3)$, $ \dots?$ [i]Proposed by A. Golovanov[/i]

On the legs $AC$ and $BC$ of an isosceles right-angled triangle with a right angle $C$, points $D$ and $E$ are taken, respectively, so that $CD = CE$. Perpendiculars on line $AE$ from points $C$ and $D$ intersect segment $AB$ at points $P$ and $Q$, respectively. Prove that $BP = PQ$.

Simon expands factored polynomials with his favorite AI, ChatSFFT. However, he has not paid for a premium ChatSFFT account, so when he goes to expand $(m - a)(n - b),$ where $a, b, m, n$ are integers, ChatSFFT returns the sum of the two factors instead of the product. However, when Simon plugs in certain pairs of integer values for $m$ and $n,$ he realizes that the value of ChatSFFT’s result is the same as the real result in terms of $a$ and $b$. How many such pairs are there?

In the interior of a square $ABCD$ we construct the equilateral triangles $ABK, BCL, CDM, DAN.$ Prove that the midpoints of the four segments $KL, LM, MN, NK$ and the midpoints of the eight segments $AK, BK, BL, CL, CM, DM, DN, AN$ are the 12 vertices of a regular dodecagon.

Find all integers $ k$ such that for infinitely many integers $ n \ge 3$ the polynomial \[ P(x) =x^{n+ 1}+ kx^n - 870x^2 + 1945x + 1995\] can be reduced into two polynomials with integer coefficients.

Let $k$ be a fixed positive integer. The $n$th derivative of $\tfrac{1}{x^k-1}$ has the form $\tfrac{P_n(x)}{(x^k-1)^{n+1}}$, where $P_n(x)$ is a polynomial. Find $P_n(1)$.

For an integer $n>2$, the tuple $(1, 2, \ldots, n)$ is written on a blackboard. On each turn, one can choose two numbers from the tuple such that their sum is a perfect square and swap them to obtain a new tuple. Find all integers $n > 2$ for which all permutations of $\{1, 2,\ldots, n\}$ can appear on the blackboard in this way.

A lattice point in the coordinate plane with origin $O$ is called invisible if the segment $OA$ contains a lattice point other than $O,A$. Let $L$ be a positive integer. Show that there exists a square with side length $L$ and sides parallel to the coordinate axes, such that all points in the square are invisible.