Found problems: 847
In a scalene triangle $ ABC, H$ and $ M$ are the orthocenter an centroid respectively. Consider the triangle formed by the lines through $ A,B$ and $ C$ perpendicular to $ AM,BM$ and $ CM$ respectively. Prove that the centroid of this triangle lies on the line $ MH$.
Denote by $\Omega$ the circumcircle of the acute triangle $ABC$. Point $D$ is the midpoint of the arc $BC$ of $\Omega$ not containing $A$. Circle $\omega$ centered at $D$ is tangent to the segment $BC$ at point $E$. Tangents to the circle $\omega$ passing through point $A$ intersect line $BC$ at points $K$ and $L$ such that points $B, K, L, C$ lie on the line $BC$ in that order. Circle $\gamma_1$ is tangent to the segments $AL$ and $BL$ and to the circle $\Omega$ at point $M$. Circle $\gamma_2$ is tangent to the segments $AK$ and $CK$ and to the circle $\Omega$ at point $N$. Lines $KN$ and $LM$ intersect at point $P$. Prove that $\sphericalangle KAP = \sphericalangle EAL$.
Let $ABC$ be an acute triangle with $AB > AC$. Let $\Gamma $ be its circumcircle, $H$ its orthocenter, and $F$ the foot of the altitude from $A$. Let $M$ be the midpoint of $BC$. Let $Q$ be the point on $\Gamma$ such that $\angle HQA = 90^{\circ}$ and let $K$ be the point on $\Gamma$ such that $\angle HKQ = 90^{\circ}$. Assume that the points $A$, $B$, $C$, $K$ and $Q$ are all different and lie on $\Gamma$ in this order.
Prove that the circumcircles of triangles $KQH$ and $FKM$ are tangent to each other.
Proposed by Ukraine
Let $ABCD$ be a cyclic quadrilateral. Let $P$, $Q$, $R$ be the feet of the perpendiculars from $D$ to the lines $BC$, $CA$, $AB$, respectively. Show that $PQ=QR$ if and only if the bisectors of $\angle ABC$ and $\angle ADC$ are concurrent with $AC$.
Let $ABC$ be an acute triangle.
Find the locus of the centers of the rectangles which have their vertices on the sides of $ABC$.
Let $I$ be the incenter of triangle $ABC$, and $\ell$ be the perpendicular bisector of $AI$. Suppose that $P$ is on the circumcircle of triangle $ABC$, and line $AP$ and $\ell$ intersect at point $Q$. Point $R$ is on $\ell$ such that $\angle IPR = 90^{\circ}$.Suppose that line $IQ$ and the midsegment of $ABC$ that is parallel to $BC$ intersect at $M$. Show that $\angle AMR = 90^{\circ}$
(Note: In a triangle, a line connecting two midpoints is called a midsegment.)
Let $M$ be the midpoint of $BC$ of triangle $ABC$. The circle with diameter $BC$, $\omega$, meets $AB,AC$ at $D,E$ respectively. $P$ lies inside $\triangle ABC$ such that $\angle PBA=\angle PAC, \angle PCA=\angle PAB$, and $2PM\cdot DE=BC^2$. Point $X$ lies outside $\omega$ such that $XM\parallel AP$, and $\frac{XB}{XC}=\frac{AB}{AC}$. Prove that $\angle BXC +\angle BAC=90^{\circ}$.
Let $ABC$ be a triangle with $\angle{C} = 90^{\circ}$, and let $H$ be the foot of the altitude from $C$. A point $D$ is chosen inside the triangle $CBH$ so that $CH$ bisects $AD$. Let $P$ be the intersection point of the lines $BD$ and $CH$. Let $\omega$ be the semicircle with diameter $BD$ that meets the segment $CB$ at an interior point. A line through $P$ is tangent to $\omega$ at $Q$. Prove that the lines $CQ$ and $AD$ meet on $\omega$.
Let $ABC$ be an isosceles triangle with $AB = AC$. Let P be a point in the interior of $ABC$ such that $PB > PC$ and $\angle PBA = \angle PCB$. Let $M$ be the midpoint of the side $BC$. Let $O$ be the circumcenter of the triangle $APM$. Prove that $\angle OAC=2 \angle BPM$ .
Let $O$ and $\Omega$ denote the circumcenter and circumcircle, respectively, of scalene triangle $\triangle ABC$. Furthermore, let $M$ be the midpoint of side $BC$. The tangent to $\Omega$ at $A$ intersects $BC$ and $OM$ at points $X$ and $Y$, respectively. If the circumcircle of triangle $\triangle OXY$ intersects $\Omega$ at two distinct points $P$ and $Q$, prove that $PQ$ bisects $\overline{AM}$.
[i]Proposed by Andrew Wen[/i]
Let $ABC$ be an acute-angled triangle such that $AB<AC$. Let $D$ be the point of intersection of the perpendicular bisector of the side $BC$ with the side $AC$. Let $P$ be a point on the shorter arc $AC$ of the circumcircle of the triangle $ABC$ such that $DP \parallel BC$. Finally, let $M$ be the midpoint of the side $AB$. Prove that $\angle APD=\angle MPB$.
[i]Proposed by Dominik Burek, Poland[/i]
Let $O$ be the circumcenter of an acute triangle $ABC$. Line $OA$ intersects the altitudes of $ABC$ through $B$ and $C$ at $P$ and $Q$, respectively. The altitudes meet at $H$. Prove that the circumcenter of triangle $PQH$ lies on a median of triangle $ABC$.
Consider a triangle $ABC$ with altitudes $AD, BE$, and $CF$, and orthocenter $H$. Let the perpendicular line from $H$ to $EF$ intersects $EF, AB$ and $AC$ at $P, T$ and $L$, respectively. Point $K$ lies on the side $BC$ such that $BD=KC$. Let $\omega$ be a circle that passes through $H$ and $P$, that is tangent to $AH$. Prove that circumcircle of triangle $ATL$ and $\omega$ are tangent, and $KH$ passes through the tangency point.
Consider a circle $(O)$ and two fixed points $B,C$ on $(O)$ such that $BC$ is not the diameter of $(O)$. $A$ is an arbitrary point on $(O)$, distinct from $B,C$. Let $D,J,K$ be the midpoints of $BC,CA,AB$, respectively, $E,M,N$ be the feet of perpendiculars from $A$ to $BC$, $B$ to $DJ$, $C$ to $DK$, respectively. The two tangents at $M,N$ to the circumcircle of triangle $EMN$ meet at $T$. Prove that $T$ is a fixed point (as $A$ moves on $(O)$).
In unit square $ABCD,$ the inscribed circle $\omega$ intersects $\overline{CD}$ at $M,$ and $\overline{AM}$ intersects $\omega$ at a point $P$ different from $M.$ What is $AP?$
$\textbf{(A) } \frac{\sqrt5}{12} \qquad \textbf{(B) } \frac{\sqrt5}{10} \qquad \textbf{(C) } \frac{\sqrt5}{9} \qquad \textbf{(D) } \frac{\sqrt5}{8} \qquad \textbf{(E) } \frac{2\sqrt5}{15}$
The incircle of a triangle $ABC$ touches the sides $BC$ and $AC$ at points $D$ and $E$, respectively. Suppose $P$ is the point on the shorter arc $DE$ of the incircle such that $\angle APE = \angle DPB$. The segments $AP$ and $BP$ meet the segment $DE$ at points $K$ and $L$, respectively. Prove that $2KL = DE$.
[i]Dušan Djukić[/i]
Let $ABCD$ be a parallelogram. The tangent to the circumcircle of triangle $BCD$ at $C$ intersects $AB$ at $P$ and intersects $AD$ at $Q$. The tangents to the circumcircle of triangle $APQ$ at $P$ and $Q$ meet at $R$. Show that points $A$, $C$, and $R$ are collinear.
Let $ABC$ be an acute triangle with orthocenter $H$ and circumcircle $\Gamma$. Let $BH$ intersect $AC$ at $E$, and let $CH$ intersect $AB$ at $F$. Let $AH$ intersect $\Gamma$ again at $P \neq A$. Let $PE$ intersect $\Gamma$ again at $Q \neq P$. Prove that $BQ$ bisects segment $\overline{EF}$.
[i]Proposed by Luke Robitaille[/i]
In the isosceles triangle $ABC$ where $AB = AC$, the point $I{}$ is the center of the inscribed circle. Through the point $A{}$ all the rays lying inside the angle $BAC$ are drawn. For each such ray, we denote by $X{}$ and $Y{}$ the points of intersection with the arc $BIC$ and the straight line $BC$ respectively. The circle $\gamma$ passing through $X{}$ and $Y{}$, which touches the arc $BIC$ at the point $X{}$ is considered. Prove that all the circles $\gamma$ pass through a fixed point.
Let$ AB$ and $AC$ be the tangents from a point $A$ to a circle $ \Omega$. Let $M$ be the midpoint of $BC$ and $P$ be an arbitrary point on this segment. A line $AP$ meets $ \Omega$ at points $D$ and $E$. Prove that the common external tangents to circles $MDP$ and $MPE$ meet on the midline of triangle $ABC$.
The altitudes $AA_1, BB_1$ and $CC_1$ were drawn in the triangle $ABC$. Point $K$ is a projection of point $B$ on $A_1C_1$. Prove that the symmmedian $\vartriangle ABC$ from the vertex $B$ divides the segment $B_1K$ in half.
(Anton Trygub)
$ABC$ is a triangle, and $E$ and $F$ are points on the segments $BC$ and $CA$ respectively, such that $\frac{CE}{CB}+\frac{CF}{CA}=1$ and $\angle CEF=\angle CAB$. Suppose that $M$ is the midpoint of $EF$ and $G$ is the point of intersection between $CM$ and $AB$. Prove that triangle $FEG$ is similar to triangle $ABC$.
Let $ABC$ be a triangle and $M$ the midpoint of $AB$. Let circumcircles of triangles $CMO$ and $ABC$ intersect at $K$ where $O$ is the circumcenter of $ABC$. Let $P$ be the intersection of lines $OM$ and $CK$. Prove that $\angle{PAK} = \angle{MCB}$.
Incircle of $\triangle ABC$ touch $AC$ at $D$. $BD$ intersect incircle at $E$. Points $F,G$ on incircle are such points, that $FE \parallel BC,GE \parallel AB$. $I_1,I_2$ are incenters of $DEF,DEG$. Prove that angle bisector of $\angle GDF$ passes though the midpoint of $I_1I_2 $.
Acute-angled triangle $ABC$ is inscribed into circle $\Omega$. Lines tangent to $\Omega$ at $B$ and $C$ intersect at $P$. Points $D$ and $E$ are on $AB$ and $AC$ such that $PD$ and $PE$ are perpendicular to $AB$ and $AC$ respectively. Prove that the orthocentre of triangle $ADE$ is the midpoint of $BC$.