Found problems: 847
An acute triangle $ABC$ ($AB\ne AC$) is inscribed in the circle $\omega$ with center at $O$. The point $M$ is the midpoint of the side $BC$. The tangent line to $\omega$ at point $A$ intersects the line $BC$ at point $D$. A circle with center at point $M$ with radius $MA$ intersects the extensions of sides $AB$ and $AC$ at points $K$ and $L$, respectively. Let $X$ be such a point that $BX\parallel KM$ and $CX\parallel LM$. Prove that the points $X$, $D$, $O$ are collinear.
In triangle $ABC$ with $AB<AC$ and $AB+AC=2BC$, let $M$ be the midpoint of $\overline{BC}$. Choose point $P$ on the extension of $\overline{BA}$ past $A$ and point $Q$ on segment $\overline{AC}$ such that $M$ lies on $\overline{PQ}$. Let $X$ be on the opposite side of $\overline{AB}$ from $C$ such that $\overline{AX} \parallel \overline{BC}$ and $AX=AP=AQ$. Let $\overline{BX}$ intersect the circumcircle of $BMQ$ again at $Y \neq B$, and let $\overline{CX}$ intersect the circumcircle of $CMP$ again at $Z \neq C$. Prove that $A$, $Y$, and $Z$ are collinear.
[i]Tiger Zhang[/i]
(A.Zaslavsky, 9--10) The reflections of diagonal $ BD$ of a quadrilateral $ ABCD$ in the bisectors of angles $ B$ and $ D$ pass
through the midpoint of diagonal $ AC$. Prove that the reflections of diagonal $ AC$ in the bisectors of angles $ A$ and $ C$ pass
through the midpoint of diagonal $ BD$ (There was an error in published condition of this problem).
Let $O,H$ be the circumcenter and orthocenter of acute triangle $ABC$ with $AB\neq AC$, respectively. Let $M$ be the midpoint of $BC$ and $K$ be the intersection of $AM$ and the circumcircle of $\triangle BHC$, such that $M$ lies between $A$ and $K$. Let $N$ be the intersection of $HK$ and $BC$. Show that if $\angle BAM=\angle CAN$, then $AN\perp OH$.
Let $ABCD$ be a convex quadrilateral with $A,B,C,D$ concyclic. Assume $\angle ADC$ is acute and $\frac{AB}{BC}=\frac{DA}{CD}$. Let $\Gamma$ be a circle through $A$ and $D$, tangent to $AB$, and let $E$ be a point on $\Gamma$ and inside $ABCD$.
Prove that $AE\perp EC$ if and only if $\frac{AE}{AB}-\frac{ED}{AD}=1$.
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]
A scalene triangle $ABC$ is inscribed within circle $\omega$. The tangent to the circle at point $C$ intersects line $AB$ at point $D$. Let $I$ be the center of the circle inscribed within $\triangle ABC$. Lines $AI$ and $BI$ intersect the bisector of $\angle CDB$ in points $Q$ and $P$, respectively. Let $M$ be the midpoint of $QP$. Prove that $MI$ passes through the middle of arc $ACB$ of circle $\omega$.
Let $ABC$ be a triangle and let $D\in (BC)$ be the foot of the $A$- altitude. The circle $w$ with the diameter $[AD]$
meet again the lines $AB$ , $AC$ in the points $K\in (AB)$ , $L\in (AC)$ respectively. Denote the meetpoint $M$
of the tangents to the circle $w$ in the points $K$ , $L$ . Prove that the ray $[AM$ is the $A$-median in $\triangle ABC$ ([b][u]Serbia[/u][/b]).
Given a cyclic quadrilateral $ABCD$, let the diagonals $AC$ and $BD$ meet at $E$ and the lines $AD$ and $BC$ meet at $F$. The midpoints of $AB$ and $CD$ are $G$ and $H$, respectively. Show that $EF$ is tangent at $E$ to the circle through the points $E$, $G$ and $H$.
[i]Proposed by David Monk, United Kingdom[/i]
Let $\Gamma_1$ and $\Gamma_2$ be two circles intersecting at $P$ and $Q$. The common tangent, closer to $P$, of $\Gamma_1$ and $\Gamma_2$ touches $\Gamma_1$ at $A$ and $\Gamma_2$ at $B$. The tangent of $\Gamma_1$ at $P$ meets $\Gamma_2$ at $C$, which is different from $P$, and the extension of $AP$ meets $BC$ at $R$.
Prove that the circumcircle of triangle $PQR$ is tangent to $BP$ and $BR$.
Let $ABC$ be an acute triangle. Tangents on the circumscribed circle of triangle $ABC$ at points $B$ and $C$ intersect at point $T$. Let $D$ and $E$ be a foot of the altitudes from $T$ onto $AB$ and $AC$ and let $M$ be the midpoint of $BC$. Prove:
A) Prove that $M$ is the orthocenter of the triangle $ADE$.
B) Prove that $TM$ cuts $DE$ in half.
Let $ABC$ be a triangle with $CA \neq CB$. Let $D$, $F$, and $G$ be the midpoints of the sides $AB$, $AC$, and $BC$ respectively. A circle $\Gamma$ passing through $C$ and tangent to $AB$ at $D$ meets the segments $AF$ and $BG$ at $H$ and $I$, respectively. The points $H'$ and $I'$ are symmetric to $H$ and $I$ about $F$ and $G$, respectively. The line $H'I'$ meets $CD$ and $FG$ at $Q$ and $M$, respectively. The line $CM$ meets $\Gamma$ again at $P$. Prove that $CQ = QP$.
[i]Proposed by El Salvador[/i]
An acute triangle $\bigtriangleup ABC$ is given which $BC>CA>AB$.
$I$ is the interior and the incircle of $\bigtriangleup ABC$ meets $BC, CA, AB$ at $D,E,F$. $AD$ and $BE$ meet at $P$. Let $l_{1}$ be a tangent from D to the circumcircle of $\bigtriangleup DIP$, and define $l_{2}$ and $l_{3}$ on $E$ and $F$, respectively.
Prove $l_{1},l_{2},l_{3}$ meet at one point.
Let $ABC$ be an isosceles right triangle with $\angle A=90^o$. Point $D$ is the midpoint of the side $[AC]$, and point $E \in [AC]$ is so that $EC = 2AE$. Calculate $\angle AEB + \angle ADB$ .
Given an isosceles triangle $ABC$ with base $AB$. Point $K$ is taken on the extension of the side $AC$ (beyond the point $C$ ) so that $\angle KBC = \angle ABC$. Denote $S$ the intersection point of angle - bisectors of $\angle BKC$ and $\angle ACB$. Lines $AB$ and $KS$ intersect at point $L$, lines $BS$ and $CL$ intersect at point $M$ . Prove that line $KM$ passes through the midpoint of the segment $BC$.
Let $ ABC$ be a triangle with incenter $ I$ and let $ \Gamma$ be a circle centered at $ I$, whose radius is greater than the inradius and does not pass through any vertex. Let $ X_{1}$ be the intersection point of $ \Gamma$ and line $ AB$, closer to $ B$; $ X_{2}$, $ X_{3}$ the points of intersection of $ \Gamma$ and line $ BC$, with $ X_{2}$ closer to $ B$; and let $ X_{4}$ be the point of intersection of $ \Gamma$ with line $ CA$ closer to $ C$. Let $ K$ be the intersection point of lines $ X_{1}X_{2}$ and $ X_{3}X_{4}$. Prove that $ AK$ bisects segment $ X_{2}X_{3}$.
Let $ABC$ be an acute triangle with altitudes $AA'$ and $BB'$ and orthocenter $H$. Let $C_0$ be the midpoint of the segment $AB$. Let $g$ be the line symmetric to the line $CC_0$ with respect to the angular bisector of $\angle ACB$. Let $h$ be the line symmetric to the line $HC_0$ with respect to the angular bisector of $\angle AHB$.
Show that the lines $g$ and $h$ intersect on the line $A'B'$.
The sides of a convex $200$-gon $A_1 A_2 \dots A_{200}$ are colored red and blue in an alternating fashion.
Suppose the extensions of the red sides determine a regular $100$-gon, as do the extensions of the blue sides.
Prove that the $50$ diagonals $\overline{A_1A_{101}},\ \overline{A_3A_{103}},\ \dots,
\ \overline{A_{99}A_{199}}$ are concurrent.
[i]Proposed by: [b]Ankan Bhattacharya[/b][/i]
Let $ABC$ be an acute triangle with orthocenter $H$. Let $A_1$, $B_1$ and $C_1$ be the feet of the altitudes of triangle $ABC$ opposite to vertices $A$, $B$, and $C$ respectively. Let $B_2$ and $C_2$ be the midpoints of $BB_1$ and $CC_1$, respectively. Let $O$ be the intersection of lines $BC_2$ and $CB_2$. Prove that $O$ is the circumcenter of triangle $ABC$ if and only if $H$ is the midpoint of $AA_1$.
[i]Proposed by Dorlir Ahmeti[/i]
Given a triangle $ABC$, the line passing through the vertex $A$ symmetric to the median $AM$ wrt the line containing the bisector of the angle $\angle BAC$ intersects the circle circumscribed around the triangle $ABC$ at points $A$ and $K$. Let $L$ be the midpoint of the segment $AK$. Prove that $\angle BLC=2\angle BAC$.
Let ${I}$ be the incenter of $\triangle {ABC}$ and ${BX}$, ${CY}$ are its two angle bisectors. ${M}$ is the midpoint of arc $\overset{\frown}{BAC}$. It is known that $MXIY$ are concyclic. Prove that the area of quadrilateral $MBIC$ is equal to that of pentagon $BXIYC$.
[i]Proposed by Dominik Burek - Poland[/i]
Let be given a triangle $ ABC$ and its internal angle bisector $ BD$ $ (D\in BC)$. The line $ BD$ intersects the circumcircle $ \Omega$ of triangle $ ABC$ at $ B$ and $ E$. Circle $ \omega$ with diameter $ DE$ cuts $ \Omega$ again at $ F$. Prove that $ BF$ is the symmedian line of triangle $ ABC$.
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 $ADZ$.
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[i]Edited by Orl.[/i]
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Let $P$ be a point on the median $AM$ of a triangle $ABC$. Suppose that the tangents to the circumcircles of $ABP$ and $ACP$ at $B$ and $C$ respectively meet at $Q$. Show that $\angle PAB = \angle CAQ$.
Canmoo is trying to do constructions, but doesn't have a ruler or compass. Instead, Canmoo has a device that, given four distinct points $A$, $B$, $C$, $P$ in the plane, will mark the isogonal conjugate of $P$ with respect to triangle $ABC$, if it exists. Show that if two points are marked on the plane, then Canmoo can construct their midpoint using this device, a pencil for marking additional points, and no other tools.
(Recall that the [i]isogonal conjugate[/i] of $P$ with respect to triangle $ABC$ is the point $Q$ such that lines $AP$ and $AQ$ are reflections around the bisector of $\angle BAC$, lines $BP$ and $BQ$ are reflections around the bisector of $\angle CBA$, lines $CP$ and $CQ$ are reflections around the bisector of $\angle ACB$. Additional points marked by the pencil can be assumed to be in general position, meaning they don't lie on any line through two existing points or any circle through three existing points.)
[i]Maxim Li[/i]