Found problems: 847
Let $ABC$ be an acute-angled triangle with circumcenter $O$. Let $I$ be a circle with center on the altitude from $A$ in $ABC$, passing through vertex $A$ and points $P$ and $Q$ on sides $AB$ and $AC$. Assume that
\[BP\cdot CQ = AP\cdot AQ.\] Prove that $I$ is tangent to the circumcircle of triangle $BOC$.
Let $ABC$ be an acute-angled triangle, $\Gamma$ its circumcircle and $M$ the midpoint of $BC$. Let $N$ be a point in the arc $BC$ of $\Gamma$ not containing $A$ such that $\angle NAC= \angle BAM$. Let $R$ be the midpoint of $AM$, $S$ the midpoint of $AN$ and $T$ the foot of the altitude through $A$. Prove that $R$, $S$ and $T$ are collinear.
Bisector of side $BC$ intersects circumcircle of triangle $ABC$ in points $P$ and $Q$. Points $A$ and $P$ lie on the same side of line $BC$. Point $R$ is an orthogonal projection of point $P$ on line $AC$. Point $S$ is middle of line segment $AQ$. Show that points $A, B, R, S$ lie on one circle.
Let point $H$ be the orthocenter of a scalene triangle $ABC$. Line $AH$ intersects with the circumcircle $\Omega$ of triangle $ABC$ again at point $P$. Line $BH, CH$ meets with $AC,AB$ at point $E$ and $F$, respectively. Let $PE, PF$ meet $\Omega$ again at point $Q,R$, respectively. Point $Y$ lies on $\Omega$ so that lines $AY,QR$ and $EF$ are concurrent. Prove that $PY$ bisects $EF$.
In the Quadrilateral $ABCD$ we have $ \measuredangle B=\measuredangle D = 60^\circ $.$M$ is midpoint of side $AD$.The line through $M$ parallel to $CD$ meets $BC$ at $P$.Point $X$ lying on $CD$ such that $BX=MX$.Prove that $AB=BP$ if and only if $\measuredangle MXB=60^\circ$.
Author: Davoud Vakili, Iran
Let $I$ be the incentre of acute triangle $ABC$ with $AB\neq AC$. The incircle $\omega$ of $ABC$ is tangent to sides $BC, CA$, and $AB$ at $D, E,$ and $F$, respectively. The line through $D$ perpendicular to $EF$ meets $\omega$ at $R$. Line $AR$ meets $\omega$ again at $P$. The circumcircles of triangle $PCE$ and $PBF$ meet again at $Q$.
Prove that lines $DI$ and $PQ$ meet on the line through $A$ perpendicular to $AI$.
[i]Proposed by Anant Mudgal, India[/i]
Let $\triangle ABC$ be an acute triangle of circumcentre $O$. Let the tangents to the circumcircle of $\triangle ABC$ in points $B$ and $C$ meet at point $P$. The circle of centre $P$ and radius $PB=PC$ meets the internal angle bisector of $\angle BAC$ inside $\triangle ABC$ at point $S$, and $OS \cap BC = D$. The projections of $S$ on $AC$ and $AB$ respectively are $E$ and $F$. Prove that $AD$, $BE$ and $CF$ are concurrent.
[i]Author: Cosmin Pohoata[/i]
Let $ABC$ be a triangle. Let $P_1$ and $P_2$ be points on the side $AB$ such that $P_2$ lies on the segment $BP_1$ and $AP_1 = BP_2$; similarly, let $Q_1$ and $Q_2$ be points on the side $BC$ such that $Q_2$ lies on the segment $BQ_1$ and $BQ_1 = CQ_2$. The segments $P_1Q_2$ and $P_2Q_1$ meet at $R$, and the circles $P_1P_2R$ and $Q_1Q_2R$ meet again at $S$, situated inside triangle $P_1Q_1R$. Finally, let $M$ be the midpoint of the side $AC$. Prove that the angles $P_1RS$ and $Q_1RM$ are equal.
Let $ABCD$ be a cyclic quadrilateral. Prove that there exists a point $X$ on segment $\overline{BD}$ such that $\angle BAC=\angle XAD$ and $\angle BCA=\angle XCD$ if and only if there exists a point $Y$ on segment $\overline{AC}$ such that $\angle CBD=\angle YBA$ and $\angle CDB=\angle YDA$.
Let $ABC$ be a triangle with circumcircle $\Gamma$ and incenter $I$ and let $M$ be the midpoint of $\overline{BC}$. The points $D$, $E$, $F$ are selected on sides $\overline{BC}$, $\overline{CA}$, $\overline{AB}$ such that $\overline{ID} \perp \overline{BC}$, $\overline{IE}\perp \overline{AI}$, and $\overline{IF}\perp \overline{AI}$. Suppose that the circumcircle of $\triangle AEF$ intersects $\Gamma$ at a point $X$ other than $A$. Prove that lines $XD$ and $AM$ meet on $\Gamma$.
[i]Proposed by Evan Chen, Taiwan[/i]
In triangle $ ABC$, $ AC \equal{} 13, BC \equal{} 14,$ and $ AB\equal{}15$. Points $ M$ and $ D$ lie on $ AC$ with $ AM\equal{}MC$ and $ \angle ABD \equal{} \angle DBC$. Points $ N$ and $ E$ lie on $ AB$ with $ AN\equal{}NB$ and $ \angle ACE \equal{} \angle ECB$. Let $ P$ be the point, other than $ A$, of intersection of the circumcircles of $ \triangle AMN$ and $ \triangle ADE$. Ray $ AP$ meets $ BC$ at $ Q$. The ratio $ \frac{BQ}{CQ}$ can be written in the form $ \frac{m}{n}$, where $ m$ and $ n$ are relatively prime positive integers. Find $ m\minus{}n$.
$ABC$ is a triangle with incenter $I$. The foot of the perpendicular from $I$ to $BC$ is $D$, and the foot of the perpendicular from $I$ to $AD$ is $P$. Prove that $\angle BPD = \angle DPC$.
[i]Alex Zhu.[/i]
Let $\Gamma$ be a circle centered at $O$ with chord $AB$. The tangents to $\Gamma$ at $A$ and $B$ meet at $C$. A secant from $C$ intersects chord $AB$ at $D$ and $\Gamma$ at $E$ such that $D$ lies on segment $CE$. Given that $\angle BOD + \angle EAD = 180^\circ$, $AE = 1$, and $BE = 2$, find $CE$.
Let $ABC$ be an acute-angled scalene triangle and $\Gamma$ be its circumcircle. $K$ is the foot of the internal bisector of $\angle BAC$ on $BC$. Let $M$ be the midpoint of the arc $BC$ containing $A$. $MK$ intersect $\Gamma$ again at $A'$. $T$ is the intersection of the tangents at $A$ and $A'$. $R$ is the intersection of the perpendicular to $AK$ at $A$ and perpendicular to $A'K$ at $A'$. Show that $T, R$ and $K$ are collinear.
Let $X$ and $Y$ be the diameter's extremes of a circunference $\Gamma$ and $N$ be the midpoint of one of the arcs $XY$ of $\Gamma$. Let $A$ and $B$ be two points on the segment $XY$. The lines $NA$ and $NB$ cuts $\Gamma$ again in $C$ and $D$, respectively. The tangents to $\Gamma$ at $C$ and at $D$ meets in $P$. Let $M$ the the intersection point between $XY$ and $NP$. Prove that $M$ is the midpoint of the segment $AB$.
In acute triangle $ABC$ ($\triangle ABC$ is not an isosceles triangle), $I$ is its incentre, and circle $ \omega$ is its inscribed circle. $\odot\omega$ touches $BC, CA, AB$ at $D, E, F$ respectively. $AD$ intersects with $\odot\omega$ at $J$ ($J\neq D$), and the circumcircle of $\triangle BCJ$ intersects $\odot\omega$ at $K$ ($K\neq J$). The circumcircle of $\triangle BFK$ and $\triangle CEK$ meet at $L$ ($L\neq K$). Let $M$ be the midpoint of the major arc $BAC$.
Prove that $M, I, L$ are collinear.
let the incircle of a triangle ABC touch BC,AC,AB at A1,B1,C1 respectively. M and N are the midpoints of AB1 and AC1 respectively. MN meets A1C1 at T . draw two tangents TP and TQ through T to incircle. PQ meets MN at L and B1C1 meets PQ at K . assume I is the center of the incircle .
prove IK is parallel to AL
Let $ ABC $ be an acute triangle. Denote by $ D $ the foot of the perpendicular line drawn from the point $ A $ to the side $ BC $, by $M$ the midpoint of $ BC $, and by $ H $ the orthocenter of $ ABC $. Let $ E $ be the point of intersection of the circumcircle $ \Gamma $ of the triangle $ ABC $ and the half line $ MH $, and $ F $ be the point of intersection (other than $E$) of the line $ ED $ and the circle $ \Gamma $. Prove that $ \tfrac{BF}{CF} = \tfrac{AB}{AC} $ must hold.
(Here we denote $XY$ the length of the line segment $XY$.)
In $\triangle {ABC}$, ${D}$ is on the internal angle bisector of $\angle BAC$ and $\angle ADB=\angle ACD$. $E, F$ is on the external angle bisector of $\angle BAC$, such that $AE=BE$ and $AF=CF$. The circumcircles of $\triangle ACE$ and $\triangle ABF$ intersects at ${A}$ and ${K}$ and $A'$ is the reflection of ${A}$ with respect to $BC$. Prove that: if $AD=BC$, then the circumcenter of $\triangle AKA'$ is on line $AD$.
Let $P$ be a point in the interior of an acute triangle $ABC$, and let $Q$ be its isogonal conjugate. Denote by $\omega_P$ and $\omega_Q$ the circumcircles of triangles $BPC$ and $BQC$, respectively. Suppose the circle with diameter $\overline{AP}$ intersects $\omega_P$ again at $M$, and line $AM$ intersects $\omega_P$ again at $X$. Similarly, suppose the circle with diameter $\overline{AQ}$ intersects $\omega_Q$ again at $N$, and line $AN$ intersects $\omega_Q$ again at $Y$.
Prove that lines $MN$ and $XY$ are parallel.
(Here, the points $P$ and $Q$ are [i]isogonal conjugates[/i] with respect to $\triangle ABC$ if the internal angle bisectors of $\angle BAC$, $\angle CBA$, and $\angle ACB$ also bisect the angles $\angle PAQ$, $\angle PBQ$, and $\angle PCQ$, respectively. For example, the orthocenter is the isogonal conjugate of the circumcenter.)
[i]Proposed by Sammy Luo[/i]
Let $ABC$ be a triangle with $\angle A=60^o$ and $T$ be a point such that $\angle ATB=\angle BTC=\angle ATC$. A circle passing through $B,C$ and $T$ meets $AB$ and $AC$ for the second time at points $K$ and $L$.Prove that the distances from $K$ and $L$ to $AT$ are equal.
In triangle $ABC$ with incenter $I$ and circumcircle $\omega$, the tangent through $C$ to $\omega$ intersects $AB$ at point $D$. The angle bisector of $\angle CDB$ intersects $AI$ and $BI$ at $E$ and $F$, respectively. Let $M$ be the midpoint of $[EF]$. Prove that line $MI$ passes through the midpoint of arc $ACB$ of $w$ .
The tangents at $B$ and $C$ to the circumcircle of the acute-angled triangle $ABC$ meet at $X$. Let $M$ be the midpoint of $BC$. Prove that
[i](a)[/i] $\angle BAM = \angle CAX$, and
[i](b)[/i] $\frac{AM}{AX} = \cos\angle BAC.$
Let $A_1$, $B_1$ and $C_1$ be the midpoints of sides $BC$, $CA$ and $AB$ of triangle $ABC$, respectively. Points $B_2$ and $C_2$ are the midpoints of segments $BA_1$ and $CA_1$ respectively. Point $B_3$ is symmetric to $C_1$ wrt $B$, and $C_3$ is symmetric to $B_1$ wrt $C$.
Prove that one of common points of circles $BB_2B_3$ and $CC_2C_3$ lies on the circumcircle of triangle $ABC$.
Let $ABC$ be an acute triangle with $AB<BC$. Let $I$ be the incenter of $ABC$, and let $\omega$ be the circumcircle of $ABC$. The incircle of $ABC$ is tangent to the side $BC$ at $K$. The line $AK$ meets $\omega$ again at $T$. Let $M$ be the midpoint of the side $BC$, and let $N$ be the midpoint of the arc $BAC$ of $\omega$. The segment $NT$ intersects the circumcircle of $BIC$ at $P$. Prove that $PM\parallel AK$.