Let ABC be a triangle with $AB = 13, BC = 14, CA = 15$. The altitude from $A$ intersects $BC$ at $D$. Let $\omega_1$ and $\omega_2$ be the incircles of $ABD$ and $ACD$, and let the common external tangent of $\omega_1$ and $\omega_2$ (other than $BC$) intersect $AD$ at $E$. Compute the length of $AE$.

On a line $l$ there are three different points $A$, $B$ and $P$ in that order. Let $a$ be the line through $A$ perpendicular to $l$, and let $b$ be the line through $B$ perpendicular to $l$. A line through $P$, not coinciding with $l$, intersects $a$ in $Q$ and $b$ in $R$. The line through $A$ perpendicular to $BQ$ intersects $BQ$ in $L$ and $BR$ in $T$. The line through $B$ perpendicular to $AR$ intersects $AR$ in $K$ and $AQ$ in $S$. (a) Prove that $P$, $T$, $S$ are collinear. (b) Prove that $P$, $K$, $L$ are collinear. [i](2nd Benelux Mathematical Olympiad 2010, Problem 3)[/i]

Suppose that fixed circle $C_1$ with radius $a > 0$ is tangent to the fixed line $\ell$ at $A$. Variable circle $C_2$, with center $X$, is externally tangent to $C_1$ at $B \ne A$ and $\ell$ at $C$. Prove that the set of all $X$ is a parabola minus a point

Eight circles of radius $34$ can be placed tangent to side $\overline{BC}$ of $\triangle ABC$ such that the first circle is tangent to $\overline{AB}$, subsequent circles are externally tangent to each other, and the last is tangent to $\overline{AC}$. Similarly, $2024$ circles of radius $1$ can also be placed along $\overline{BC}$ in this manner. The inradius of $\triangle ABC$ is $\tfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

In a triangle $ABC$, the points $A_1, B_1, C_1$ are symmetric to $A, B,C$ with respect to $B,C, A$, respectively. Given the points $A_1, B_1,C_1$ reconstruct the triangle $ABC$.

Let $ABCD$ be a convex quadrilateral and draw regular triangles $ABM, CDP, BCN, ADQ$, the first two outward and the other two inward. Prove that $MN = AC$. What can be said about the quadrilateral $MNPQ$?

A rectangle, with sides parallel to the $x-$axis and $y-$axis, has opposite vertices located at $(15, 3)$ and$(16, 5).$ A line is drawn through points $A(0, 0)$ and $B(3, 1).$ Another line is drawn through points $C(0, 10)$ and $D(2, 9).$ How many points on the rectangle lie on at least one of the two lines? [asy] size(9cm); draw((0,-.5)--(0,11),EndArrow(size=.15cm)); draw((1,0)--(1,11),mediumgray); draw((2,0)--(2,11),mediumgray); draw((3,0)--(3,11),mediumgray); draw((4,0)--(4,11),mediumgray); draw((5,0)--(5,11),mediumgray); draw((6,0)--(6,11),mediumgray); draw((7,0)--(7,11),mediumgray); draw((8,0)--(8,11),mediumgray); draw((9,0)--(9,11),mediumgray); draw((10,0)--(10,11),mediumgray); draw((11,0)--(11,11),mediumgray); draw((12,0)--(12,11),mediumgray); draw((13,0)--(13,11),mediumgray); draw((14,0)--(14,11),mediumgray); draw((15,0)--(15,11),mediumgray); draw((16,0)--(16,11),mediumgray); draw((-.5,0)--(17,0),EndArrow(size=.15cm)); draw((0,1)--(17,1),mediumgray); draw((0,2)--(17,2),mediumgray); draw((0,3)--(17,3),mediumgray); draw((0,4)--(17,4),mediumgray); draw((0,5)--(17,5),mediumgray); draw((0,6)--(17,6),mediumgray); draw((0,7)--(17,7),mediumgray); draw((0,8)--(17,8),mediumgray); draw((0,9)--(17,9),mediumgray); draw((0,10)--(17,10),mediumgray); draw((-.13,1)--(.13,1)); draw((-.13,2)--(.13,2)); draw((-.13,3)--(.13,3)); draw((-.13,4)--(.13,4)); draw((-.13,5)--(.13,5)); draw((-.13,6)--(.13,6)); draw((-.13,7)--(.13,7)); draw((-.13,8)--(.13,8)); draw((-.13,9)--(.13,9)); draw((-.13,10)--(.13,10)); draw((1,-.13)--(1,.13)); draw((2,-.13)--(2,.13)); draw((3,-.13)--(3,.13)); draw((4,-.13)--(4,.13)); draw((5,-.13)--(5,.13)); draw((6,-.13)--(6,.13)); draw((7,-.13)--(7,.13)); draw((8,-.13)--(8,.13)); draw((9,-.13)--(9,.13)); draw((10,-.13)--(10,.13)); draw((11,-.13)--(11,.13)); draw((12,-.13)--(12,.13)); draw((13,-.13)--(13,.13)); draw((14,-.13)--(14,.13)); draw((15,-.13)--(15,.13)); draw((16,-.13)--(16,.13)); label(scale(.7)*"$1$", (1,-.13), S); label(scale(.7)*"$2$", (2,-.13), S); label(scale(.7)*"$3$", (3,-.13), S); label(scale(.7)*"$4$", (4,-.13), S); label(scale(.7)*"$5$", (5,-.13), S); label(scale(.7)*"$6$", (6,-.13), S); label(scale(.7)*"$7$", (7,-.13), S); label(scale(.7)*"$8$", (8,-.13), S); label(scale(.7)*"$9$", (9,-.13), S); label(scale(.7)*"$10$", (10,-.13), S); label(scale(.7)*"$11$", (11,-.13), S); label(scale(.7)*"$12$", (12,-.13), S); label(scale(.7)*"$13$", (13,-.13), S); label(scale(.7)*"$14$", (14,-.13), S); label(scale(.7)*"$15$", (15,-.13), S); label(scale(.7)*"$16$", (16,-.13), S); label(scale(.7)*"$1$", (-.13,1), W); label(scale(.7)*"$2$", (-.13,2), W); label(scale(.7)*"$3$", (-.13,3), W); label(scale(.7)*"$4$", (-.13,4), W); label(scale(.7)*"$5$", (-.13,5), W); label(scale(.7)*"$6$", (-.13,6), W); label(scale(.7)*"$7$", (-.13,7), W); label(scale(.7)*"$8$", (-.13,8), W); label(scale(.7)*"$9$", (-.13,9), W); label(scale(.7)*"$10$", (-.13,10), W); dot((0,0)); label(scale(.65)*"$A$", (0,0), NE); dot((3,1)); label(scale(.65)*"$B$", (3,1), NE); dot((0,10)); label(scale(.65)*"$C$", (0,10), NE); dot((2,9)); label(scale(.65)*"$D$", (2,9), NE); draw((15,3)--(16,3)--(16,5)--(15,5)--cycle,linewidth(1.125)); dot((15,3)); dot((16,3)); dot((16,5)); dot((15,5)); [/asy] $\textbf{(A) } 0\qquad\textbf{(B) } 1\qquad\textbf{(C) } 2\qquad\textbf{(D) } 3\qquad\textbf{(E) } 4$

In a triangle the sum of squares of the sides is $96$. What is the maximum possible value of the sum of the medians?

Given a matrix $\{a_{ij}\}_{i,j=0}^{9}$, $a_{ij}=10i+j+1$. Andrei is going to cover its entries by $50$ rectangles $1\times 2$ (each such rectangle contains two adjacent entries) so that the sum of $50$ products in these rectangles is minimal possible. Help him. [i]A. Badzyan[/i]

Let $ A$ and $ B$ be the intersections of two circumferences $\Gamma_1$, and $\Gamma_2$. Let $C$ and $D$ points in $\Gamma_1$ and $\Gamma_2$ respectively such that $AC = AD$. Let $E$ and $F$ be points in $\Gamma_1$ and $\Gamma_2$, such that $\angle ABE = \angle ABF = 90^o$. Let $K_1$ and $K_2$ be circumferences with centers $E$ and $F$ and radii $EC$ and $FD$ respectively. Let $T$ be a point in the line $AB$, but outside the segment, with $T\ne A$ and $T \ne A'$, where $A'$ is the point symmetric to $A$ with respect to $ B$. Let $X$ be the point of tangency of a tangent to $K_1$ passing through $T$, such that there arc two points of intersection of the line $TX$ to $K_2$. Let $Y$ and $Z$ be such points. Prove that $$\frac{1}{XT}=\frac{1}{XY} + \frac{1}{XZ}.$$

Consider a cyclic quadrilateral $ABCD$, and let $S$ be the intersection of $AC$ and $BD$. Let $E$ and $F$ the orthogonal projections of $S$ on $AB$ and $CD$ respectively. Prove that the perpendicular bisector of segment $EF$ meets the segments $AD$ and $BC$ at their midpoints.

A polygon is decomposed into triangles by drawing some non-intersecting interior diagonals in such a way that for every pair of triangles of the triangulation sharing a common side, the sum of the angles opposite to this common side is greater than $180^o$. a) Prove that this polygon is convex. b) Prove that the circumcircle of every triangle used in the decomposition contains the entire polygon. [i]Proposed by Morteza Saghafian - Iran[/i]

Let $\Delta ABC$ be a triangle inscribed in the circumference $\omega$ of center $O$. Let $T$ be the symmetric of $C$ respect to $O$ and $T'$ be the reflection of $T$ respect to line $AB$. Line $BT'$ intersects $\omega$ again at $R$. The perpendicular to $CT$ through $O$ intersects line $AC$ at $L$. Let $N$ be the intersection of lines $TR$ and $AC$. Prove that $\overline{CN}=2\overline{AL}$.

Given a triangle $ABC$, with right $\angle A$, we know: the point $T$ of tangency of the circumference inscribed in $ABC$ with the hypotenuse $BC$, the point $D$ of intersection of the angle bisector of $\angle B$ with side AC and the point E of intersection of the angle bisector of $\angle C$ with side $AB$ . Describe a construction with ruler and compass for points $A$, $B$, and $C$. Justify.

Draw a circle that has a given radius $R$ and is tangent to a given line and a given circle. How many solutions does this problem have?

Let $ABCD$ be an isosceles trapezoid with $AB = 1$, $BC = DA = 5$, $CD = 7$. Let $P$ be the intersection of diagonals $AC$ and $BD$, and let $Q$ be the foot of the altitude from $D$ to $BC$. Let $PQ$ intersect $AB$ at $R$. Compute $\sin \angle RP D$

Let $ABC$ be a non-isosceles triangle with incenter $I$ whose incircle is tangent to $\overline{BC}$, $\overline{CA}$, $\overline{AB}$ at $D$, $E$, $F$, respectively. Denote by $M$ the midpoint of $\overline{BC}$. Let $Q$ be a point on the incircle such that $\angle AQD = 90^{\circ}$. Let $P$ be the point inside the triangle on line $AI$ for which $MD = MP$. Prove that either $\angle PQE = 90^{\circ}$ or $\angle PQF = 90^{\circ}$. [i]Proposed by Evan Chen[/i]

A conical surface $C$ is cut by a plane $T$ as shown in the figure on the back of this sheet. Show that $C \cap T$ is an ellipse. You can use as an aid the fact that if you consider the two spheres tangent to $C$ and $T$ as shown in the figure, they intersect $T$ in the bulbs. [asy] // calculate intersection of line and plane // p = point on line // d = direction of line // q = point in plane // n = normal to plane triple lineintersectplan(triple p, triple d, triple q, triple n) { return (p + dot(n,q - p)/dot(n,d)*d); } // projection of point A onto line BC triple projectionofpointontoline(triple A, triple B, triple C) { return lineintersectplan(B, B - C, A, B - C); } // calculate area of space triangle with vertices A, B, and C real trianglearea(triple A, triple B, triple C) { return abs(cross(A - C, B - C)/2); } // calculate incentre of space triangle ABC triple triangleincentre(triple A, triple B, triple C) { return (abs(B - C) * A + abs(C - A) * B + abs(A - B) * C)/(abs(B - C) + abs(C - A) + abs(A - B)); } // calculate inradius of space triangle ABC real triangleinradius(triple A, triple B, triple C) { return 2*trianglearea(A,B,C)/(abs(B - C) + abs(C - A) + abs(A - B)); } // calculate excentre of space triangle ABC triple triangleexcentre(triple A, triple B, triple C) { return (-abs(B - C) * A + abs(C - A) * B + abs(A - B) * C)/(-abs(B - C) + abs(C - A) + abs(A - B)); } // calculate exradius of space triangle ABC real triangleexradius(triple A, triple B, triple C) { return 2*trianglearea(A,B,C)/(-abs(B - C) + abs(C - A) + abs(A - B)); } unitsize(2 cm); pair project (triple A, real t) { return((A.x, A.y*Sin(t) + A.z*Cos(t))); } real alpha, beta, theta, t; real coneradius = 1, coneheight = 3; real a, b, c; real[] m, r; triple A, B, V; triple ellipsecenter, ellipsex, ellipsey; triple[] F, O, P, R, W; path[] ellipse, spherering; theta = 15; V = (0,0,-coneheight); m[1] = sqrt(Cos(theta)^2*coneheight^2 - Sin(theta)^2*coneradius^2)/coneradius; m[2] = -m[1]; alpha = -aTan(Sin(theta)/m[1]); beta = -aTan(Sin(theta)/m[2]) + 180; A = (coneradius*Cos(alpha), coneradius*Sin(alpha), 0); B = (coneradius*Cos(beta), coneradius*Sin(beta), 0); W[1] = interp(V,(coneradius,0,0),0.6); W[2] = interp(V,(-coneradius,0,0),0.4); O[1] = triangleexcentre(V,W[1],W[2]); O[2] = triangleincentre(V,W[1],W[2]); r[1] = triangleexradius(V,W[1],W[2]); r[2] = triangleinradius(V,W[1],W[2]); F[1] = projectionofpointontoline(O[1],W[1],W[2]); F[2] = projectionofpointontoline(O[2],W[1],W[2]); P[1] = O[1] - (0,0,r[1]*coneradius/sqrt(coneradius^2 + coneheight^2)); P[2] = O[2] - (0,0,r[2]*coneradius/sqrt(coneradius^2 + coneheight^2)); spherering[11] = shift(project(P[1],theta))*yscale(Sin(theta))*arc((0,0),r[1]*coneheight/sqrt(coneradius^2 + coneheight^2),alpha,beta); spherering[12] = shift(project(P[1],theta))*yscale(Sin(theta))*arc((0,0),r[1]*coneheight/sqrt(coneradius^2 + coneheight^2),beta,alpha + 360); spherering[21] = shift(project(P[2],theta))*yscale(Sin(theta))*arc((0,0),r[2]*coneheight/sqrt(coneradius^2 + coneheight^2),alpha,beta); spherering[22] = shift(project(P[2],theta))*yscale(Sin(theta))*arc((0,0),r[2]*coneheight/sqrt(coneradius^2 + coneheight^2),beta,alpha + 360); ellipsecenter = (W[1] + W[2])/2; a = abs(W[1] - ellipsecenter); c = abs(F[1] - ellipsecenter); b = sqrt(a^2 - c^2); ellipsex = (W[1] - W[2])/abs(W[1] - W[2]); ellipsey = (0,1,0); ellipse[1] = project(ellipsecenter + a*ellipsex, theta); for (t = 0; t <= 180; t = t + 5) { ellipse[1] = ellipse[1]--project(ellipsecenter + a*Cos(t)*ellipsex + b*Sin(t)*ellipsey, theta); } ellipse[2] = project(ellipsecenter - a*ellipsex, theta); for (t = 180; t <= 360; t = t + 5) { ellipse[2] = ellipse[2]--project(ellipsecenter + a*Cos(t)*ellipsex + b*Sin(t)*ellipsey, theta); } R[1] = ellipsecenter + 1*ellipsex + ellipsey; R[2] = ellipsecenter - 1.2*ellipsex + ellipsey; R[3] = ellipsecenter - 1*ellipsex - ellipsey; R[4] = ellipsecenter + 1.2*ellipsex - ellipsey; fill(ellipse[1]--ellipse[2]--cycle, gray(0.9)); draw(yscale(Sin(theta))*Circle((0,0),coneradius)); draw(project(V,theta)--project(A,theta)); draw(project(V,theta)--project(B,theta)); draw(Circle(project(O[1],theta),r[1])); draw(Circle(project(O[2],theta),r[2])); draw(spherering[11], dashed); draw(spherering[12]); draw(spherering[21], dashed); draw(spherering[22]); draw(ellipse[1], dashed); draw(ellipse[2]); draw(project(R[1],theta)--interp(project(R[1],theta),project(R[2],theta),0.13)); draw(interp(project(R[1],theta),project(R[2],theta),0.13)--interp(project(R[1],theta),project(R[2],theta),0.76), dashed); draw(interp(project(R[1],theta),project(R[2],theta),0.76)--project(R[2],theta)); draw(project(R[2],theta)--project(R[3],theta)--project(R[4],theta)--project(R[1],theta)); label("$C$", (-1,0.3)); label("$T$", (1.2,-0.8)); dot(project(F[1],theta)); dot(project(F[2],theta)); //dot("$F_1$", project(F[1],theta)); //dot("$F_2$", project(F[2],theta)); //dot("$O_1$", project(O[1],theta)); //dot("$O_2$", project(O[2],theta)); //dot("$P_1$", project(P[1],theta)); //dot("$V$", project(V,theta)); //dot("$W_1$", project(W[1],theta)); //dot("$W_2$", project(W[2],theta)); [/asy]

In a triangle scalene $ABC$, let $I$ be its incenter and $D$ the intersection of $AI$ and $BC$. Let $M$ and $N$ points where the incircle touches $AB$ and $AC$, respectively. Let $F$ be the second intersection of the circumcircle $(AMN)$ with the circumcircle $(ABC)$. Let $T$ the intersection of $AF$ and $BC$. Let $J$ be the intersection of $TI$ with the line parallel of $FI$ that passes through $D$. Prove that the line $AJ$ is perpendicular to $BC$.

In triangle $ABC$ let $M$ and $N$ be midpoints of $BC$ and $AC,$ respectively. Point $P$ is inside $ABC$ such that $\angle BAP = \angle PBC = \angle PCA .$ Prove that if $\angle PNA = \angle AMB,$ then $ABC$ is isosceles triangle.

Let ${ABC}$ be an acute angled triangle, and ${H}$ a point in its interior. Let the reflections of ${H}$ through the sides ${AB}$ and ${AC}$ be called ${H_{c} }$ and ${H_{b} }$ , respectively, and let the reflections of H through the midpoints of these same sidesbe called ${H_{c}^{'} }$ and ${H_{b}^{'} }$, respectively. Show that the four points ${H_{b}, H_{b}^{'} , H_{c}}$, and ${H_{c}^{'} }$ are concyclic if and only if at least two of them coincide or ${H}$ lies on the altitude from ${A}$ in triangle ${ABC}$.

A park is in the shape of a regular hexagon $2$ km on a side. Starting at a corner, Alice walks along the perimeter of the park for a distance of $5$ km. How many kilometers is she from her starting point? $ \textbf{(A)}\ \sqrt{13}\qquad\textbf{(B)}\ \sqrt{14}\qquad\textbf{(C)}\ \sqrt{15}\qquad\textbf{(D)}\ \sqrt{16}\qquad\textbf{(E)}\ \sqrt{17}$

Let $ABCD$ be an orthodiagonal trapezoid such that $\measuredangle A = 90^{\circ}$ and $AB$ is the larger base. The diagonals intersect at $O$, $\left( OE \right.$ is the bisector of $\measuredangle AOD$, $E \in \left( AD \right)$ and $EF \| AB$, $F \in \left( BC \right)$. Let $P,Q$ the intersections of the segment $EF$ with $AC,BD$. Prove that: (a) $EP=QF$; (b) $EF=AD$. [i]Claudiu-Stefan Popa[/i]

Let $\vartriangle ABC$ be a triangle. The angle bisector of $\angle CAB$ intersects$ BC$ at $L$. On the interior of line segments $AC$ and $AB$, two points, $M$ and $N$, respectively, are chosen in such a way that the lines $AL, BM$ and $CN$ are concurrent, and such that $\angle AMN = \angle ALB$. Prove that $\angle NML = 90^o$.

Let $ABC$ be a triangle with $\angle BAC=60$. Let $B_1$ and $C_1$ be the feet of the bisectors from $B$ and $C$. Let $A_1$ be the symmetrical of $A$ according to line $B_1C_1$. Prove that $A_1, B, C$ are colinear.