Let $ABCD$ a convex quadrilateral (this means that all interior angles are smaller than $180^o$), such that there exist a point $M$ on line segment $AB$ and a point $N$ on line segment $BC$ having the property that $AN$ cuts the quadrilateral in two parts of equal area, and such that the same property holds for $CM$. Prove that $MN$ cuts the diagonal $BD$ in two segments of equal length.

Point $ A$ lies at $ (0, 4)$ and point $ B$ lies at $ (3, 8)$. Find the $ x$-coordinate of the point $ X$ on the $ x$-axis maximizing $ \angle AXB$.

Let $ABC$ be an acute-angled triangle with $AB < AC$ and $\Gamma$ the circumference that passes through $A,\ B$ and $C$. Let $D$ be the point diametrically opposite $A$ on $\Gamma$ and $\ell$ the tangent through $D$ to $\Gamma$. Let $P, Q$ and $R$ be the intersection points of $B C$ with $\ell$, of $A P$ with $\Gamma$ such that $Q \neq A$ and of $Q D$ with the $A$-altitude of the triangle $ABC$, respectively. Define $S$ to be the intersection of $AB$ with $\ell$ and $T$ to be the intersection of $A C$ with $\ell$. Show that $S$ and $T$ lie on the circumference that passes through $A, Q$ and $R$.

As in the following diagram, square $ABCD$ and square $CEFG$ are placed side by side (i.e. $C$ is between $B$ and $E$ and $G$ is between $C$ and $D$). If $CE = 14$, $AB > 14$, compute the minimal area of $\triangle AEG$. [asy] size(120); defaultpen(linewidth(0.7)+fontsize(10)); pair D2(real x, real y) { pair P = (x,y); dot(P,linewidth(3)); return P; } int big = 30, small = 14; filldraw((0,big)--(big+small,0)--(big,small)--cycle, rgb(0.9,0.5,0.5)); draw(scale(big)*unitsquare); draw(shift(big,0)*scale(small)*unitsquare); label("$A$",D2(0,big),NW); label("$B$",D2(0,0),SW); label("$C$",D2(big,0),SW); label("$D$",D2(big,big),N); label("$E$",D2(big+small,0),SE); label("$F$",D2(big+small,small),NE); label("$G$",D2(big,small),NE); [/asy]

In acute triangle $\triangle {ABC}$, $\angle A > \angle B > \angle C$. $\triangle {AC_1B}$ and $\triangle {CB_1A}$ are isosceles triangles such that $\triangle {AC_1B} \stackrel{+}{\sim} \triangle {CB_1A}$. Let lines $BB_1, CC_1$ intersects at ${T}$. Prove that if all points mentioned above are distinct, $\angle ATC$ isn't a right angle.

A point $T$ is chosen inside a triangle $ABC$. Let $A_1$, $B_1$, and $C_1$ be the reflections of $T$ in $BC$, $CA$, and $AB$, respectively. Let $\Omega$ be the circumcircle of the triangle $A_1B_1C_1$. The lines $A_1T$, $B_1T$, and $C_1T$ meet $\Omega$ again at $A_2$, $B_2$, and $C_2$, respectively. Prove that the lines $AA_2$, $BB_2$, and $CC_2$ are concurrent on $\Omega$. [i]Proposed by Mongolia[/i]

Given is a parallelogram $ABCD$ with $\angle A < 90^o$ and $|AB| < |BC|$. The angular bisector of angle $A$ intersects side $BC$ in $M$ and intersects the extension of $DC$ in $N$. Point $O$ is the centre of the circle through $M, C$, and $N$. Prove that $\angle OBC = \angle ODC$. [asy] unitsize (1.2 cm); pair A, B, C, D, M, N, O; A = (0,0); B = (2,0); D = (1,3); C = B + D - A; M = extension(A, incenter(A,B,D), B, C); N = extension(A, incenter(A,B,D), D, C); O = circumcenter(C,M,N); draw(D--A--B--C); draw(interp(D,N,-0.1)--interp(D,N,1.1)); draw(A--interp(A,N,1.1)); draw(circumcircle(M,C,N)); label("$\circ$", A + (0.45,0.15)); label("$\circ$", A + (0.25,0.35)); dot("$A$", A, SW); dot("$B$", B, SE); dot("$C$", C, dir(90)); dot("$D$", D, dir(90)); dot("$M$", M, SE); dot("$N$", N, dir(90)); dot("$O$", O, SE); [/asy]

In triangle $ABC$, a point $M$ is the midpoint of $AB$, and a point $I$ is the incentre. Point $A_1$ is the reflection of $A$ in $BI$, and $B_1$ is the reflection of $B$ in $AI$. Let $N$ be the midpoint of $A_1B_1$. Prove that $IN > IM$.

Let $a$ be a positive real number. Collinear points $Z_1, Z_2, Z_3, Z_4$ (in that order) are plotted on the $(x, y)$ Cartesian plane. Suppose that the graph of the equation \[ x^2 + (y+a)^2 + x^2 + (y-a)^2 = 4a^2 + \sqrt{(x^2 + (y+a)^2)(x^2 + (y-a)^2)} \] passes through points $Z_1$ and $Z_4$, and the graph of the equation \[ x^2 + (y+a)^2 + x^2 + (y-a)^2 = 4a^2 - \sqrt{(x^2 + (y+a)^2)(x^2 + (y-a)^2)} \] passes through points $Z_2$ and $Z_3$. If $Z_1Z_2 = 5$, $Z_2Z_3 = 1$, and $Z_3Z_4 = 3$, then $a^2$ can be written as $\frac{m + n\sqrt{p}}{q}$, where $m$, $n$, $p$, and $q$ are positive integers, $m$, $n$, and $q$ are relatively prime, and $p$ is squarefree. Find $m + n + p + q$.

Eight identical cubes with of size $1 \times 1 \times 1$ each have the numbers $1$ through $6$ written on their faces with the number $1$ written on the face opposite number $2$, number $3$ written on the face opposite number $5$, and number $4$ written on the face opposite number $6$. The eight cubes are stacked into a single $2 \times 2 \times 2$ cube. Add all of the numbers appearing on the outer surface of the new cube. Let $M$ be the maximum possible value for this sum, and $N$ be the minimum possible value for this sum. Find $M - N$.

Let $P_1$ and $P_2$ be regular polygons of 1985 sides and perimeters $x$ and $y$ respectively. Each side of $P_1$ is tangent to a given circle of circumference $c$ and this circle passes through each vertex of $P_2$. Prove $x + y \ge 2c$. (You may assume that $\tan \theta \ge \theta$ for $0 \le \theta < \frac{\pi}{2}$.)

Let $ABCDE$ be a convex pentagon with all five sides equal in length. The diagonals $AD$ and $EC$ meet in $S$ with $\angle ASE = 60^\circ$. Prove that $ABCDE$ has a pair of parallel sides.

Given any $60$ points on a circle of radius $1$, prove that there is a point on the circle the sum of whose distances to these $60$ points is at most $80$.

The $A,B,C$ and $D$ points (from left to right) belong to the straight line. Prove that every point $E$, that doesn't belong to the line satisfy: $$|AE| + |ED| + | |AB| - |CD| | > |BE| + |CE|$$

Let $ABC$ be a triangle such that $|AB|=8$ and $|AC|=2|BC|$. What is the largest value of altitude from side $[AB]$? $ \textbf{(A)}\ 3\sqrt 2 \qquad\textbf{(B)}\ 3\sqrt 3 \qquad\textbf{(C)}\ 5 \qquad\textbf{(D)}\ \dfrac {16}3 \qquad\textbf{(E)}\ 6 $

Let $ABC$ be an acute triangle, and let $M$ and $N$ be two points on the line $AC$ such that the vectors $MN$ and $AC$ are identical. Let $X$ be the orthogonal projection of $M$ on $BC$, and let $Y$ be the orthogonal projection of $N$ on $AB$. Finally, let $H$ be the orthocenter of triangle $ABC$. Show that the points $B$, $X$, $H$, $Y$ lie on one circle.

$\triangle ABC$ is a triangle such that $\angle A = 60^{\circ}$. The incenter of $\triangle ABC$ is $I$. $AI$ intersects with $BC$ at $D$, $BI$ intersects with $CA$ at $E$, and $CI$ intersects with $AB$ at $F$, respectively. Also, the circumcircle of $\triangle DEF$ is $\omega$. The tangential line of $\omega$ at $E$ and $F$ intersects at $T$. Show that $\angle BTC \ge 60^{\circ}$

Let $ABC$ be an acute triangle, $D$ be the midpoint of $BC$. Bisectors of angles $ADB$ and $ADC$ intersect the circles circumscribed around the triangles $ADB$ and $ADC$ at points $E$ and $F$, respectively. Prove that $EF\perp AD$.

Given a cube with side length $ 1$, we perform six cuts as follows: one cut parallel to the $xy$-plane, two cuts parallel to the $yz$-plane, and three cuts parallel to the $xz$-plane, where the cuts are made uniformly independent of each other. What is the expected value of the volume of the largest piece?

In triangle $ABC, \angle A = 90^o$. $M$ is the midpoint of $BC$ and $H$ is the foot of the altitude from $A$ to $BC$. The line passing through $M$ and perpendicular to $AC$ meets the circumcircle of triangle $AMC$ again at $P$. If $BP$ intersects $AH$ at $K$, prove that $AK = KH$.

Inside an $11\times 11$ square, Pablo drew a rectangle and extending its sides divided the square into $5$ rectangles, as shown in the figure. [img]https://cdn.artofproblemsolving.com/attachments/5/a/7774da7085f283b3aae74fb5ff472572571827.gif[/img] Sofía did the same, but she also managed to make the lengths of the sides of the $5$ rectangles be whole numbers between $1$ and $10$, all different. Show a figure like the one Sofia made.

[b]p1.[/b] $17$ rooks are placed on an $8\times 8$ chess board. Prove that there must be at least one rook that is attacking at least $2$ other rooks. [b]p2.[/b] In New Scotland there are three kinds of coins: $1$ cent, $6$ cent, and $36$ cent coins. Josh has $99$ of the $36$-cent coins (and no other coins). He is allowed to exchange a $36$ cent coin for $6$ coins of $6$ cents, and to exchange a $6$ cent coin for $6$ coins of $1$ cent. Is it possible that after several exchanges Josh will have $500$ coins? [b]p3.[/b] Find all solutions $a, b, c, d, e, f, g, h, i$ if these letters represent distinct digits and the following multiplication is correct: $\begin{tabular}{ccccc} & & a & b & c \\ x & & & d & e \\ \hline & f & a & c & c \\ + & g & h & i & \\ \hline f & f & f & c & c \\ \end{tabular}$ [b]p4.[/b] Pinocchio rode a bicycle for $3.5$ hours. During every $1$-hour period he went exactly $5$ km. Is it true that his average speed for the trip was $5$ km/h? Explain your reasoning. [b]p5.[/b] Let $a, b, c$ be odd integers. Prove that the equation $ax^2 + bx + c = 0$ cannot have a rational solution. PS. You should use hide for answers. Collected [url=https://artofproblemsolving.com/community/c5h2760506p24143309]here[/url].

In the triangle $ABC$ it is known that $\angle ACB> 90 {} ^ \circ$, $\angle CBA> 45 {} ^ \circ$. On the sides $AC$ and $AB$, respectively, there are points $P$ and $T$ such that $ABC$ and $PT = BC$. The points ${{P} _ {1}}$ and ${{T} _ {1}}$ on the sides $AC$ and $AB$ are such that $AP = C {{P} _ {1}}$ and $AT = B {{T} _ {1}}$. Prove that $\angle CBA- \angle {{P} _ {1}} {{T} _ {1}} A = 45 {} ^ \circ$. (Anton Trygub)

In a trapezoid $ABCD$ with $AB // CD$, the diagonals $AC$ and $BD$ intersect at point $E$. Let $F$ and $G$ be the orthocenters of the triangles $EBC$ and $EAD$. Prove that the midpoint of $GF$ lies on the perpendicular from $E$ to $AB$.

In an acute-angled triangle $ABC$, the point $H{}$ is the orthocenter, $M{}$ is the midpoint of the side $BC$ and $\omega$ is the circumcircle. The lines $AH, BH$ and $CH{}$ intersect $\omega$ a second time at points $D, E$ and $F{}$ respectively. The ray $MH$ intersects $\omega$ at point $J{}$. The points $K{}$ and $L{}$ are the centers of the inscribed circles of the triangles $DEJ$ and $DFJ$ respectively. Prove that $KL\parallel BC$.