A triangle $\triangle ABC$ satisfies $AB = 13$, $BC = 14$, and $AC = 15$. Inside $\triangle ABC$ are three points $X$, $Y$, and $Z$ such that: [list] [*] $Y$ is the centroid of $\triangle ABX$ [*] $Z$ is the centroid of $\triangle BCY$ [*] $X$ is the centroid of $\triangle CAZ$ [/list] What is the area of $\triangle XYZ$? [i]Proposed by Adam Bertelli[/i]

Supppose five points in a plane are situated so that no two of the straight lines joining them are parallel, perpendicular, or coincident. From each point perpendiculars are drawn to all the lines joining the other four points. Determine the maxium number of intersections that these perpendiculars can have.

On the side $ \overline{DC} $ of the rectangle $ ABCD $ in which $ \frac{AB}{AD} = \sqrt{2} $ a semicircle is built externally. Any point $ M $ of the semicircle is connected by segments with $ A $ and $ B $ to obtain points $ K $ and $ L $ on $ \overline{DC} $, respectively. Prove that $ DL^2 + KC^2 = AB^2 $.

Jeff the delivery driver starts at the point $(5, 0)$ and has to make deliveries at all the other lattice points with distance $5$ from the origin before returning to his starting point. The length of a shortest possible path he can make is $\sqrt{a}+\sqrt{b}$ for positive integers $a$ and $b.$ Find $a + b.$

Let $ABC$ be a triangle with $AB=AC$ and let $D$ be the midpoint of $AC$. The angle bisector of $\angle BAC$ intersects the circle through $D,B$ and $C$ at the point $E$ inside the triangle $ABC$. The line $BD$ intersects the circle through $A,E$ and $B$ in two points $B$ and $F$. The lines $AF$ and $BE$ meet at a point $I$, and the lines $CI$ and $BD$ meet at a point $K$. Show that $I$ is the incentre of triangle $KAB$. [i]Proposed by Jan Vonk, Belgium and Hojoo Lee, South Korea[/i]

Taotao wants to buy a bracelet. The bracelets have 7 different beads on them, arranged in a circle. Two bracelets are the same if one can be rotated or flipped to get the other. If she can choose the colors and placement of the beads, and the beads come in orange, white, and black, how many possible bracelets can she buy?

If $A$ and $B$ are fixed points on a given circle not collinear with centre $O$ of the circle, and if $XY$ is a variable diameter, find the locus of $P$ (the intersection of the line through $A$ and $X$ and the line through $B$ and $Y$).

In an acute non-isosceles triangle $ABC$, the altitude $AA'$ is drawn and point $H$ is the intersection point of the altitudes and and $O$ is the center of the circumscribed circle. Prove that the point symmetric to the circumcenter of triangle $HOA'$ wrt straight line $HO$, lies on a midline of triangle $ABC$.

Let $A_1A_2A_3$ be a triangle and, for $1 \leq i \leq 3$, let $B_i$ be an interior point of edge opposite $A_i$. Prove that the perpendicular bisectors of $A_iB_i$ for $1 \leq i \leq 3$ are not concurrent.

A rectangular box measures $a \times b \times c$, where $a,$ $b,$ and $c$ are integers and $1 \leq a \leq b \leq c$. The volume and surface area of the box are numerically equal. How many ordered triples $(a,b,c)$ are possible? $ \textbf{(A) }4\qquad\textbf{(B) }10\qquad\textbf{(C) }12\qquad\textbf{(D) }21\qquad\textbf{(E) }26 $

Let $ABCD$ a quadrilateral inscribed in a circumference, $l_1$ the parallel to $BC$ through $A$, and $l_2$ the parallel to $AD$ through $B$. The line $DC$ intersects $l_1$ and $l_2$ at $E$ and $F$, respectively. The perpendicular to $l_1$ through $A$ intersects $BC$ at $P$, and the perpendicular to $l_2$ through $B$ cuts $AD$ at $Q$. Let $\Gamma_1$ and $\Gamma_2$ be the circumferences that pass through the vertex of triangles $ADE$ and $BFC$, respectively. Prove that $\Gamma_1$ and $\Gamma_2$ are tangent to each other if and only if $DP$ is perpendicular to $CQ$.

Let $ABC$ be a triangle such that $|AC|>|AB|$. Let $X$ be on line $AB$ (closer to $A$) such that $|BX|=|AC|$ and let $Y$ be on the segment $AC$ such that $|CY|=|AB|$. Intersection of lines $XY$ and bisector of $BC$ is point $P$. Prove that $\angle BPC+\angle BAC = 180^\circ$.

Let \( AL \) be the bisector of triangle \( ABC \), \( O \) the center of its circumcircle, and \( D \) and \( E \) the midpoints of \( BL \) and \( CL \), respectively. Points \( P \) and \( Q \) are chosen on segments \( AD \) and \( AE \) such that \( APLQ \) is a parallelogram. Prove that \( PQ \perp AO \). [i]Proposed by Mykhailo Plotnikov[/i]

The base of isosceles $\triangle{ABC}$ is $24$ and its area is $60$. What is the length of one of the congruent sides? $\textbf{(A)}\ 5 \qquad \textbf{(B)}\ 8 \qquad \textbf{(C)}\ 13 \qquad \textbf{(D)}\ 14 \qquad \textbf{(E)}\ 18$

in a convex quadrilateral $ABCD$ , $M,N$ are midpoints of $BC,AD$ respectively. If $AM=BN$ and $DM=CN$ then prove that $AC=BD$. S. Berlov

Given three tetrahedrons $A_iB_i C_i D_i$ ($i=1,2,3$), planes $\alpha _i,\beta _i,\gamma _i$ ($i=1,2,3$) are drawn through $B_i ,C_i ,D_i$ respectively, and they are perpendicular to edges $A_i B_i, A_i C_i, A_i D_i$ ($i=1,2,3$) respectively. Suppose that all nine planes $\alpha _i,\beta _i,\gamma _i$ ($i=1,2,3$) meet at a point $E$, and points $A_1,A_2,A_3$ lie on line $l$. Determine the intersection (shape and position) of the circumscribed spheres of the three tetrahedrons.

Let $ABC$ be a right triangle, right at $B$, and let $M$ be the midpoint of the side $BC$. Let $P$ be the point in bisector of the angle $ \angle BAC$ such that $PM$ is perpendicular to $BC (P$ is outside the triangle $ABC$). Determine the triangle area $ABC$ if $PM = 1$ and $MC = 5$.

Hassan has a piece of paper in the shape of a hexagon. The interior angles are all $120^o$, and the side lengths are $1$, $2$, $3$, $4$, $5$, $6$, although not in that order. Initially, the paper is in the shape of an equilateral triangle, then Hassan has cut off three smaller equilateral triangle shapes, one at each corner of the paper. What is the minimum possible side length of the original triangle?

Let $O$ be the circumcenter and $H$ the orthocenter of an acute triangle $ABC$. Show that there exist points $D$, $E$, and $F$ on sides $BC$, $CA$, and $AB$ respectively such that \[ OD + DH = OE + EH = OF + FH\] and the lines $AD$, $BE$, and $CF$ are concurrent.

In triangle $ABC$(with incenter $I$) let the line parallel to $BC$ from $A$ intersect circumcircle of $\triangle ABC$ at $A_1$ let $AI\cap BC=D$ and $E$ is tangency point of incircle with $BC$ let $ EA_1\cap \odot (\triangle ADE)=T$ prove that $AI=TI$.

Let $ABC$ be a triangle with a right angle at $C$. Let $I$ be the incentre of triangle $ABC$, and let $D$ be the foot of the altitude from $C$ to $AB$. The incircle $\omega$ of triangle $ABC$ is tangent to sides $BC$, $CA$, and $AB$ at $A_1$, $B_1$, and $C_1$, respectively. Let $E$ and $F$ be the reflections of $C$ in lines $C_1A_1$ and $C_1B_1$, respectively. Let $K$ and $L$ be the reflections of $D$ in lines $C_1A_1$ and $C_1B_1$, respectively. Prove that the circumcircles of triangles $A_1EI$, $B_1FI$, and $C_1KL$ have a common point.

A piece of paper in the shape of a square $FBHD$ with side $a$ is given. Points $G,A$ on $FB$ and $E,C$ on $BH$ are marked so that $FG=GA=AB$ and $BE=EC=CH$. The paper is folded along $DG,DA,DC$ and $AC$ so that $G$ overlaps with $B$, and $F$ and $H$ overlap with $E$. Compute the volume of the obtained tetrahedron $ABCD$.

On a circle $O$ with radius $\overline{OA}$, points $B$ and $C$ are drawn such that $\angle AOC = \angle BOA = 30^\circ$, as shown. A second circle passing through $B$, $C$, and the midpoint of $\overline{OA}$ is drawn. The ratio of the radius of this new circle to the radius of circle $O$ can be expressed in the form $\tfrac{a \sqrt 3 - b}{c}$ where $a$, $b$, and $c$ are positive integers and $c$ is as small as possible. What is $a + b + c$? [asy] size(100); pair O,A,B,C; O = (0,0); label("$O$",O,W); A = (2,0); label("$A$",A,E); B = (sqrt(3),1); label("$B$",B,N*1.8); C = (sqrt(3),-1); label("$C$",C,S*1.8); draw(Circle(O,2)); dot((1,0)^^A^^B^^C^^O); draw(O--B); draw(O--C); draw(O--A); draw(Circle((2.04904,0),1.04904),dashed); [/asy] [center]Note: In the diagram, $A$ is not necessarily the center of the second circle.[/center] $\textbf{(A) }10\qquad\textbf{(B) }12\qquad\textbf{(C) }15\qquad\textbf{(D) }21\qquad\textbf{(E) }27$

$AB,AC$ are tangent to $\Omega$ at $B$ and $C$, respectively. $D,E,F$ lie on segments $BC,CA,AB$ such that $AF<AE$ and $\angle FDB= \angle EDC$. The circumcircle of $\triangle FEC$ intersects $\Omega$ at $G$ and $C$. Show that $ \angle AEF= \angle BGD$

Let $ABC$ be a triangle in which $\angle BAC = 60^{\circ}$ . Let $P$ (similarly $Q$) be the point of intersection of the bisector of $\angle ABC$(similarly of $\angle ACB$) and the side $AC$(similarly $AB$). Let $r_1$ and $r_2$ be the in-radii of the triangles $ABC$ and $AP Q$, respectively. Determine the circum-radius of $APQ$ in terms of $r_1$ and $r_2$.