Found problems: 847
In a nonisosceles triangle $ABC$, point $I$ is its incentre and $\Gamma$ is its circumcircle. Points $E$ and $D$ lie on $\Gamma$ and the circumcircle of triangle $BIC$ respectively such that $AE$ and $ID$ are both perpendicular to $BC$. Let $M$ be the midpoint of $BC$, $N$ be the midpoint of arc $BC$ on $\Gamma$ containing $A$, $F$ is the point of tangency of the $A-$excircle on $BC$, and $G$ is the intersection of line $DE$ with $\Gamma$. Prove that lines $GM$ and $NF$ intersect at a point located on $\Gamma$.
(Possibly proposed by Farras Faddila)
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 $\omega$.
The tangents to the circumcircle of $\triangle ABC$ at $B$ and $C$ meet at $D$. The circumcircle of $\triangle BCD$ meets sides $AC$ and $AB$ again at $E$ and $F$ respectively. Let $O$ be the circumcentre of $\triangle ABC$. Show that $AO$ is perpendicular to $EF$.
Let $ABC$ be an equilateral triangle and $D$ the midpoint of $BC$. Let $E$ and $F$ be points on $AC$ and $AB$ respectively such that $AF=CE$. $P=BE$ $\cap$ $CF$.
Show that $\angle$$APF=$ $\angle$$BPD$
Let $ABC$ be a triangle. The incircle of triangle $ABC$ touches the side $BC$ at $A^{\prime}$, and the line $AA^{\prime}$ meets the incircle again at a point $P$. Let the lines $CP$ and $BP$ meet the incircle of triangle $ABC$ again at $N$ and $M$, respectively. Prove that the lines $AA^{\prime}$, $BN$ and $CM$ are concurrent.
Let $ABCD$ be quadrilateral inscribed in a circle. Let $M$ be the midpoint of the segment $BD$. If the tangents of the circle at $ B$, and at $D$ are also concurrent with the extension of $AC$, prove that $\angle AMD = \angle CMD$.
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 $ABC$.
Triangle $ABC$ has circumcircle $(O,R)$, and orthocenter $H$. The symmedians through $A,B,C$ meet the perpendicular bisectors of $BC,CA,AB$ at $D,E, F$ respectively. Let $M,N, P$ be the perpendicular projections of H on the line $OD,OE,OF.$ Prove that $$\frac{OH^2}{R^2} =\frac{\overline{OM}}{\overline{OD}}+\frac{\overline{ON}}{\overline{OE}} +\frac{\overline{OP}}{\overline{OF}}$$
Đỗ Thanh Sơn
In a trapezium with bases $AD$ and $BC$, let $P$ and $Q$ be the middles of diagonals $AC$ and $BD$ respectively. Prove that if $\angle DAQ = \angle CAB$ then $\angle PBA = \angle DBC$.
Let $ABC$ be an acute triange with $B,C$ fixed. Let $D$ be the midpoint of $BC$ and $E,F$ be the foot of the perpendiculars from $D$ to $AB,AC$, respectively.
a) Let $O$ be the circumcenter of triangle $ABC$ and $M=EF\cap AO, N=EF\cap BC$. Prove that the circumcircle of triangle $AMN$ passes through a fixed point;
b) Assume that tangents of the circumcircle of triangle $AEF$ at $E,F$ are intersecting at $T$. Prove that $T$ is on a fixed line.
Let $\triangle ABC$ have side lengths $AB = 5, BC = 9,$ and $CA = 10.$ The tangents to the circumcircle of $\triangle ABC$ at $B$ and $C$ intersect at point $D,$ and $\overline{AD}$ intersects the circumcircle at $P \ne A.$ The length of $\overline{AP}$ is equal to $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$
Let a point $D$ lie on the median $AM$ of a triangle $ABC$. The tangents to the circumcircle of triangle $BDC$ at points $B$ and $C$ meet at point $K$. Prove that $DD'$ is parallel to $AK$, where $D'$ is isogonally conjugated to $D$ with respect to $ABC$.
Let $ABC $ be an acute triangle such that $AB\neq AC$ ,with circumcircle $ \Gamma$ and circumcenter $O$. Let $M$ be the midpoint of $BC$ and $D$ be a point on $ \Gamma$ such that $AD \perp BC$. let $T$ be a point such that $BDCT$ is a parallelogram and $Q$ a point on the same side of $BC$ as $A$ such that $\angle{BQM}=\angle{BCA}$ and $\angle{CQM}=\angle{CBA}$. Let the line $AO$ intersect $ \Gamma$ at $E$ $(E\neq A)$ and let the circumcircle of $\triangle ETQ$ intersect $ \Gamma$ at point $X\neq E$. Prove that the point $A,M$ and $X$ are collinear.
Let ABC be a scalene triangle. Let $K_a$, $L_a$ and $M_a$ be the respective intersections with BC of the internal angle bisector, external angle bisector, and the median from A. The circumcircle of $AK_aL_a$ intersects $AM_a$ a second time at point $X_a$ different from A. Define $X_b$ and $X_c$ analogously. Prove that the circumcenter of $X_aX_bX_c$ lies on the Euler line of ABC.
(The Euler line of ABC is the line passing through the circumcenter, centroid, and orthocenter of ABC.)
[i]Proposed by Ivan Borsenco[/i]
Let $ABC$ be a triangle with $AB=AC$, and let $M$ be the midpoint of $BC$. Let $P$ be a point such that $PB<PC$ and $PA$ is parallel to $BC$. Let $X$ and $Y$ be points on the lines $PB$ and $PC$, respectively, so that $B$ lies on the segment $PX$, $C$ lies on the segment $PY$, and $\angle PXM=\angle PYM$. Prove that the quadrilateral $APXY$ is cyclic.
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
Point $D$ is selected inside acute $\triangle ABC$ so that $\angle DAC = \angle ACB$ and $\angle BDC = 90^{\circ} + \angle BAC$. Point $E$ is chosen on ray $BD$ so that $AE = EC$. Let $M$ be the midpoint of $BC$.
Show that line $AB$ is tangent to the circumcircle of triangle $BEM$.
[i]Proposed by Anton Trygub[/i]
In a triangle $ABC$, with $AB \ne BC$, $E$ is a point on the line $AC$ such that $BE$ is perpendicular to $AC$. A circle passing through $A$ and touching the line $BE$ at a point $P \ne B$ intersects the line $AB$ for the second time at $X$. Let $Q$ be a point on the line $PB$ different from $P$ such that $BQ = BP$. Let $Y$ be the point of intersection of the lines $CP$ and $AQ$. Prove that the points $C, X, Y, A$ are concyclic if and only if $CX$ is perpendicular to $AB$.
In a triangle $ABC$ with $AB\not = AC$, the incircle touches the sides $BC, CA, AB$ at $D, E, F$ , respectively. Line $AD$ meets the incircle again at $P$ . The line $EF$ and the line through $P$ perpendicular to $AD$ meet at $Q$. Line $AQ$ intersects $DE$ at $X$ and $DF$ at $Y$ . Prove that $A$ is the midpoint of $XY$.
Restore a triangle $ABC$ by vertex $B$, the centroid and the common point of the symmedian from $B$ with the circumcircle.
Let $\omega$ be a circle with center $O,$ and let $A$ be a point with tangents $AP$ and $AQ$ to the circle. Denote by $K$ the intersection point of $AO$ and $PQ.$ $l_1$ and $l_2$ are two lines passing through $A$ that intersect $\omega.$ Call $B$ the intersection point of $l_1$ with $\omega,$ which is located nearer to $A$ on $l_1.$ Call $C$ the intersection point of $l_2$ with $\omega,$ which is located further to $A$ on $l_2.$ Prove that $\angle PAB = \angle QAC$ if and only if the points $B, K, C$ are on line.
Given is a triangle $ABC$ with barycenter $G$ and circumcenter $O$.The perpendicular bisectors of $GA,GB,GC$ intersect at $A_1,B_1,C_1$.Show that $O$ is the barycenter of $\triangle{A_1B_1C_1}$.
Let $ABC$ be a triangle such that $\angle A=90^{\circ }$ and $\angle B<\angle C$. The tangent at $A$ to the circumcircle $\omega$ of triangle $ABC$ meets the line $BC$ at $D$. Let $E$ be the reflection of $A$ in the line $BC$, let $X$ be the foot of the perpendicular from $A$ to $BE$, and let $Y$ be the midpoint of the segment $AX$. Let the line $BY$ intersect the circle $\omega$ again at $Z$.
Prove that the line $BD$ is tangent to the circumcircle of triangle $ADZ$.
[hide="comment"]
[i]Edited by Orl.[/i]
[/hide]
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 $\omega$.
We are given an acute triangle $ABC$. Let $D$ be the point on its circumcircle such that $AD$ is a diameter. Suppose that points $K$ and $L$ lie on segments $AB$ and $AC$, respectively, and that $DK$ and $DL$ are tangent to circle $AKL$.
Show that line $KL$ passes through the orthocenter of triangle $ABC$.