A triangle with vertices $(6,5)$, $(8,-3)$, and $(9,1)$ is reflected about the line $x=8$ to create a second triangle. What is the area of the union of the two triangles? $\textbf{(A) }9\qquad \textbf{(B) }\dfrac{28}{3}\qquad \textbf{(C) }10\qquad \textbf{(D) }\dfrac{31}{3}\qquad \textbf{(E) }\dfrac{32}{3}\qquad$

Consider the family of nonisosceles triangles $ABC$ satisfying the property $AC^2 + BC^2 = 2 AB^2$. Points $M$ and $D$ lie on side $AB$ such that $AM = BM$ and $\angle ACD = \angle BCD$. Point $E$ is in the plane such that $D$ is the incenter of triangle $CEM$. Prove that exactly one of the ratios \[ \frac{CE}{EM}, \quad \frac{EM}{MC}, \quad \frac{MC}{CE} \] is constant.

Let $A$ be a point on a circle with center $O$ and $B$ be the midpoint of $[OA]$. Let $C$ and $D$ be points on the circle such that they lie on the same side of the line $OA$ and $\widehat{CBO} = \widehat{DBA}$. Show that the reflection of the midpoint of $[CD]$ over $B$ lies on the circle.

Let $ABCD$ be a parallelogram and let $K$ and $L$ be points on the segments $BC$ and $CD$, respectively, such that $BK\cdot AD=DL\cdot AB$. Let the lines $DK$ and $BL$ intersect at $P$. Show that $\measuredangle DAP=\measuredangle BAC$.

Show that among any $n\geq 3$ vertices of a regular $(2n-1)$-gon we can find $3$ of them forming an isosceles triangle.

Let $ABC$ be a triangle with $AB \neq AC$ and circumcenter $O$. The bisector of $\angle BAC$ intersects $BC$ at $D$. Let $E$ be the reflection of $D$ with respect to the midpoint of $BC$. The lines through $D$ and $E$ perpendicular to $BC$ intersect the lines $AO$ and $AD$ at $X$ and $Y$ respectively. Prove that the quadrilateral $BXCY$ is cyclic.

Problem 4 Let $ABCD$ be a rectangle with diagonals $AC$ and $BD$. Let $E$ be the intersection of the bisector of $\angle CAD$ with segment $CD$, $F$ on $CD$ such that $E$ is midpoint of $DF$, and $G$ on $BC$ such that $BG = AC$ (with $C$ between $B$ and $G$). Prove that the circumference through $D$, $F$ and $G$ is tangent to $BG$.

in a quadrilateral $ABCD$ digonals are perpendicular to each other. let $S$ be the intersection of digonals. $K$,$L$,$M$ and $N$ are reflections of $S$ to $AB$,$BC$,$CD$ and $DA$. $BN$ cuts the circumcircle of $SKN$ in $E$ and $BM$ cuts the circumcircle of $SLM$ in $F$. prove that $EFLK$ is concyclic.(20 points)

Let $M$ be a point in the plane, distinct from the vertices of $\triangle ABC$. Consider $N,P,Q$ the reflections of $M$ with respect to lines $AB, BC$ and $CA$, in this order. a) Prove that $N, P ,Q$ are collinear if and only if $M$ lies on the circumcircle of $\triangle ABC$. b) If $M$ does not lie on the circumcircle of $\triangle ABC$ and the centroids of triangles $\triangle ABC$ and $\triangle NPQ$ coincide, prove that $\triangle ABC$ is equilateral.

In a triangle $ABC$, with $\widehat{A} > 90^\circ$, let $O$ and $H$ denote its circumcenter and orthocenter, respectively. Let $K$ be the reflection of $H$ with respect to $A$. Prove that $K, O$ and $C$ are collinear if and only if $\widehat{A} - \widehat{B} = 90^\circ$.

Line $PQ$ is tangent to the incircle of triangle $ABC$ in such a way that the points $P$ and $Q$ lie on the sides $AB$ and $AC$, respectively. On the sides $AB$ and $AC$ are selected points $M$ and $N$, respectively, so that $AM = BP$ and $AN = CQ$. Prove that all lines constructed in this manner $MN$ pass through one point

Let $ ABP, BCQ, CAR$ be three non-overlapping triangles erected outside of acute triangle $ ABC$. Let $ M$ be the midpoint of segment $ AP$. Given that $ \angle PAB \equal{} \angle CQB \equal{} 45^\circ$, $ \angle ABP \equal{} \angle QBC \equal{} 75^\circ$, $ \angle RAC \equal{} 105^\circ$, and $ RQ^2 \equal{} 6CM^2$, compute $ AC^2/AR^2$. [i]Zuming Feng.[/i]

Let $ABC$ be a triangle and $D$ be the midpoint of the segment $BC$. The circle that passes through $D$ and tangent to $AB$ at $B$, and the circle that passes through $D$ and tangent to $AC$ at $C$ intersect at $M\neq D$. Let $M'$ be the reflection of $M$ with respect to $BC$. Prove that $M'$ is on $AD$.

Let $ABC$ be an acute triangle with $AB<AC$, and let $D$ and $E$ be points on side $BC$ such that $BD=CE$ and $D$ lies between $B$ and $E$. Suppose there exists a point $P$ inside $ABC$ such that $PD\parallel AE$ and $\angle PAB=\angle EAC$. Prove that $\angle PBA=\angle PCA$. [i]Calvin Deng.[/i]

Given that circle $ \omega$ is tangent internally to circle $ \Gamma$ at $ S.$ $ \omega$ touches the chord $ AB$ of $ \Gamma$ at $ T$. Let $ O$ be the center of $ \omega.$ Point $ P$ lies on the line $ AO.$ Show that $ PB\perp AB$ if and only if $ PS\perp TS.$

The centre of the circumcircle of quadrilateral $ABCD$ is $O$ and $O$ is not on any of the sides of $ABCD$. $P=AC \cap BD$. The circumecentres of $\triangle{OAB}$, $\triangle{OBC}$, $\triangle{OCD}$ and $\triangle{ODA}$ are $O_1$, $O_2$, $O_3$ and $O_4$ respectively. Prove that $O_1O_3$, $O_2O_4$ and $OP$ are concurrent.

Prove that for any convex pentagon $A_1A_2A_3A_4A_5$, there exists a unique pair of points $\{P,Q\}$ (possibly with $P=Q$) such that $\measuredangle{PA_i A_{i-1}} = \measuredangle{A_{i+1}A_iQ}$ for $1\le i\le 5$, where indices are taken $\pmod5$ and angles are directed $\pmod\pi$. [i]Calvin Deng.[/i]

Let $ABC$ be an isosceles triangle with $AB=AC$ . Let also $\omega$ be a circle of center $K$ tangent to the line $AC$ at $C$ which intersects the segment $BC$ again at $H$ . Prove that $HK \bot AB $.

Points $ A_{1}$, $ B_{1}$, $ C_{1}$ are chosen on the sides $ BC$, $ CA$, $ AB$ of a triangle $ ABC$ respectively. The circumcircles of triangles $ AB_{1}C_{1}$, $ BC_{1}A_{1}$, $ CA_{1}B_{1}$ intersect the circumcircle of triangle $ ABC$ again at points $ A_{2}$, $ B_{2}$, $ C_{2}$ respectively ($ A_{2}\neq A, B_{2}\neq B, C_{2}\neq C$). Points $ A_{3}$, $ B_{3}$, $ C_{3}$ are symmetric to $ A_{1}$, $ B_{1}$, $ C_{1}$ with respect to the midpoints of the sides $ BC$, $ CA$, $ AB$ respectively. Prove that the triangles $ A_{2}B_{2}C_{2}$ and $ A_{3}B_{3}C_{3}$ are similar.

Let $ABC$, $AC \not= BC$, be an acute triangle with orthocenter $H$ and incenter $I$. The lines $CH$ and $CI$ meet the circumcircle of $\bigtriangleup ABC$ at points $D$ and $L$, respectively. Prove that $\angle CIH = 90^{\circ}$ if and only if $\angle IDL = 90^{\circ}$

In scalene $\triangle ABC$, $I$ is the incenter, $I_a$ is the $A$-excenter, $D$ is the midpoint of arc $BC$ of the circumcircle of $ABC$ not containing $A$, and $M$ is the midpoint of side $BC$. Extend ray $IM$ past $M$ to point $P$ such that $IM = MP$. Let $Q$ be the intersection of $DP$ and $MI_a$, and $R$ be the point on the line $MI_a$ such that $AR\parallel DP$. Given that $\frac{AI_a}{AI}=9$, the ratio $\frac{QM} {RI_a}$ can be expressed in the form $\frac{m}{n}$ for two relatively prime positive integers $m,n$. Compute $m+n$. [i]Ray Li.[/i] [hide="Clarifications"][list=1][*]"Arc $BC$ of the circumcircle" means "the arc with endpoints $B$ and $C$ not containing $A$".[/list][/hide]

Let $ ABC$ be a triangle with circumcentre $ O$. The points $ P$ and $ Q$ are interior points of the sides $ CA$ and $ AB$ respectively. Let $ K,L$ and $ M$ be the midpoints of the segments $ BP,CQ$ and $ PQ$. respectively, and let $ \Gamma$ be the circle passing through $ K,L$ and $ M$. Suppose that the line $ PQ$ is tangent to the circle $ \Gamma$. Prove that $ OP \equal{} OQ.$ [i]Proposed by Sergei Berlov, Russia [/i]

Let $ABC$ be a triangle where $AC\neq BC$. Let $P$ be the foot of the altitude taken from $C$ to $AB$; and let $V$ be the orthocentre, $O$ the circumcentre of $ABC$, and $D$ the point of intersection between the radius $OC$ and the side $AB$. The midpoint of $CD$ is $E$. a) Prove that the reflection $V'$ of $V$ in $AB$ is on the circumcircle of the triangle $ABC$. b) In what ratio does the segment $EP$ divide the segment $OV$?

Do there exist such a triangle $T$, that for any coloring of a plane in two colors one may found a triangle $T'$, equal to $T$, such that all vertices of $T'$ have the same color. [b] proposed by S. Berlov[/b]

Let $ABC$ be a nondegenerate acute triangle with circumcircle $\omega$ and let its incircle $\gamma$ touch $AB, AC, BC$ at $X, Y, Z$ respectively. Let $XY$ hit arcs $AB, AC$ of $\omega$ at $M, N$ respectively, and let $P \neq X, Q \neq Y$ be the points on $\gamma$ such that $MP=MX, NQ=NY$. If $I$ is the center of $\gamma$, prove that $P, I, Q$ are collinear if and only if $\angle BAC=90^\circ$. [i]Proposed by David Stoner[/i]