Found problems: 429
Let $ABC$ be a triangle with $AB = AC \neq BC$ and let $I$ be its incentre. The line $BI$ meets $AC$ at $D$, and the line through $D$ perpendicular to $AC$ meets $AI$ at $E$. Prove that the reflection of $I$ in $AC$ lies on the circumcircle of triangle $BDE$.
Chords $A_1A_2, A_3A_4, A_5A_6$ of a circle $\Omega$ concur at point $O$. Let $B_i$ be the second common point of $\Omega$ and the circle with diameter $OA_i$ . Prove that chords $B_1B_2, B_3B_4, B_5B_6$ concur.
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]
Let $ABC$ be an isosceles triangle with $AC=BC$, whose incentre is $I$. Let $P$ be a point on the circumcircle of the triangle $AIB$ lying inside the triangle $ABC$. The lines through $P$ parallel to $CA$ and $CB$ meet $AB$ at $D$ and $E$, respectively. The line through $P$ parallel to $AB$ meets $CA$ and $CB$ at $F$ and $G$, respectively. Prove that the lines $DF$ and $EG$ intersect on the circumcircle of the triangle $ABC$.
[i]Proposed by Hojoo Lee, Korea[/i]
Let $ A_1A_2A_3$ be a non-isosceles triangle with incenter $ I.$ Let $ C_i,$ $ i \equal{} 1, 2, 3,$ be the smaller circle through $ I$ tangent to $ A_iA_{i\plus{}1}$ and $ A_iA_{i\plus{}2}$ (the addition of indices being mod 3). Let $ B_i, i \equal{} 1, 2, 3,$ be the second point of intersection of $ C_{i\plus{}1}$ and $ C_{i\plus{}2}.$ Prove that the circumcentres of the triangles $ A_1 B_1I,A_2B_2I,A_3B_3I$ are collinear.
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 $ABC$ 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 $ABC$ be a triangle with $\angle B > \angle C$. Let $P$ and $Q$ be two different points on line $AC$ such that $\angle PBA = \angle QBA = \angle ACB $ and $A$ is located between $P$ and $C$. Suppose that there exists an interior point $D$ of segment $BQ$ for which $PD=PB$. Let the ray $AD$ intersect the circle $ABC$ at $R \neq A$. Prove that $QB = QR$.
Let $ABC$ be a triangle with incenter $I$ and circumcircle $\Gamma$. Circles $\omega_{B}$ passing through $B$ and $\omega_{C}$ passing through $C$ are tangent at $I$. Let $\omega_{B}$ meet minor arc $AB$ of $\Gamma$ at $P$ and $AB$ at $M\neq B$, and let $\omega_{C}$ meet minor arc $AC$ of $\Gamma$ at $Q$ and $AC$ at $N\neq C$. Rays $PM$ and $QN$ meet at $X$. Let $Y$ be a point such that $YB$ is tangent to $\omega_{B}$ and $YC$ is tangent to $\omega_{C}$.
Show that $A,X,Y$ are collinear.
Two circles are internally tangent at a point $A$. Let $C$ be a point on the smaller circle other than $A$. The tangent line to the smaller circle at $C$ meets the bigger circle at $D$ and $E$; and the line $AC$ meets the bigger circle at $A$ and $P$. Show that the line $PE$ is tangent to the circle through $A$, $C$, and $E$.
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.
Circles $\omega$ and $\Omega$ meet at points $A$ and $B$. Let $M$ be the midpoint of the arc $AB$ of circle $\omega$ ($M$ lies inside $\Omega$). A chord $MP$ of circle $\omega$ intersects $\Omega$ at $Q$ ($Q$ lies inside $\omega$). Let $\ell_P$ be the tangent line to $\omega$ at $P$, and let $\ell_Q$ be the tangent line to $\Omega$ at $Q$. Prove that the circumcircle of the triangle formed by the lines $\ell_P$, $\ell_Q$ and $AB$ is tangent to $\Omega$.
[i]Ilya Bogdanov, Russia and Medeubek Kungozhin, Kazakhstan[/i]
Given a triangle $ABC$, let $I$ be the incenter, and $J$ be the $A$-excenter. A line $\ell$ through $A$ perpendicular to $BC$ intersect the lines $BI$, $CI$, $BJ$, $CJ$ at $P$, $Q$, $R$, $S$ respectively. Suppose the angle bisector of $\angle BAC$ meet $BC$ at $K$, and $L$ is a point such that $AL$ is a diameter in $(ABC)$.
Prove that the line $KL$, $\ell$, and the line through the centers of circles $(IPQ)$ and $(JRS)$, are concurrent.
[i]Proposed by Chuah Jia Herng & Ivan Chan Kai Chin[/i]
Let scalene triangle $ABC$ have circumcenter $O$ and incenter $I$. Its incircle $\omega$ is tangent to sides $BC,CA,$ and $AB$ at $D,E,$ and $F$, respectively. Let $P$ be the foot of the altitude from $D$ to $EF$, and let line $DP$ intersect $\omega$ again at $Q \ne D$. The line $OI$ intersects the altitude from $A$ to$ BC$ at $T$. Given that $OI \|BC,$ show that $PQ=PT$.
Let $I$ be the incentre of acute-angled triangle $ABC$. Let the incircle meet $BC, CA$, and $AB$ at $D, E$, and $F,$ respectively. Let line $EF$ intersect the circumcircle of the triangle at $P$ and $Q$, such that $F$ lies between $E$ and $P$. Prove that $\angle DPA + \angle AQD =\angle QIP$.
(Slovakia)
Let $I_a$ be the $A$-excenter of a scalene triangle $ABC$. And let $M$ be the point symmetric to $I_a$ about line $BC$.
Prove that line $AM$ is parallel to the line through the circumcenter and the orthocenter of triangle $I_aCB$.
Let $ABC$ be a triangle and let $D$ be a point on the segment $BC, D\neq B$ and $D\neq C$. The circle $ABD$ meets the segment $AC$ again at an interior point $E$. The circle $ACD$ meets the segment $AB$ again at an interior point $F$. Let $A'$ be the reflection of $A$ in the line $BC$. The lines $A'C$ and $DE$ meet at $P$, and the lines $A'B$ and $DF$ meet at $Q$. Prove that the lines $AD, BP$ and $CQ$ are concurrent (or all parallel).
In isosceles triangle $ABC(AC=BC)$ the point $M$ is in the segment $AB$ such that $AM=2MB,$ $F$ is the midpoint of $BC$ and $H$ is the orthogonal projection of $M$ in $AF.$ Prove that $\angle BHF=\angle ABC.$
Let $ABC$ be an acute triangle with circumcenter $O$ and orthocenter $H$. Rays $AO$, $CO$ intersect sides $BC, BA$ in points $A_1, C_1$ respectively, $K$ is the projection of $O$ onto the segment $A_1C_1$, $M$ is the midpoint of $AC$. Prove that $\angle HMA = \angle BKC_1$.
[i]Proposed by Anton Trygub[/i]
Let $ABC$ be a triangle with circumcircle $k$. The tangents at $A$ and $C$ intersect at $T$. The circumcircle of triangle $ABT$ intersects the line $CT$ at $X$ and $Y$ is the midpoint of $CX$. Prove that the lines $AX$ and $BY$ intersect on $k$.
We say a finite set $S$ of points in the plane is [i]very[/i] if for every point $X$ in $S$, there exists an inversion with center $X$ mapping every point in $S$ other than $X$ to another point in $S$ (possibly the same point).
(a) Fix an integer $n$. Prove that if $n \ge 2$, then any line segment $\overline{AB}$ contains a unique very set $S$ of size $n$ such that $A, B \in S$.
(b) Find the largest possible size of a very set not contained in any line.
(Here, an [i]inversion[/i] with center $O$ and radius $r$ sends every point $P$ other than $O$ to the point $P'$ along ray $OP$ such that $OP\cdot OP' = r^2$.)
[i]Proposed by Sammy Luo[/i]
Let $ABC$ be a triangle with $AC>AB$ , and denote its circumcircle by $\Omega$ and incentre by $I$. Let its incircle meet sides $BC,CA,AB$ at $D,E,F$ respectively. Let $X$ and $Y$ be two points on minor arcs $\widehat{DF}$ and $\widehat{DE}$ of the incircle, respectively, such that $\angle BXD = \angle DYC$. Let line $XY$ meet line $BC$ at $K$. Let $T$ be the point on $\Omega$ such that $KT$ is tangent to $\Omega$ and $T$ is on the same side of line $BC$ as $A$. Prove that lines $TD$ and $AI$ meet on $\Omega$.
[right][i]Tommy Walker Mackay, United Kingdom[/i][/right]
Let $A$ be a point in the plane, and $\ell$ a line not passing through $A$. Evan does not have a straightedge, but instead has a special compass which has the ability to draw a circle through three distinct noncollinear points. (The center of the circle is [i]not[/i] marked in this process.) Additionally, Evan can mark the intersections between two objects drawn, and can mark an arbitrary point on a given object or on the plane.
(i) Can Evan construct* the reflection of $A$ over $\ell$?
(ii) Can Evan construct the foot of the altitude from $A$ to $\ell$?
*To construct a point, Evan must have an algorithm which marks the point in finitely many steps.
[i]Proposed by Zack Chroman[/i]
Let $O$ be the circumcenter of $\triangle ABC$($AB<AC$), and $D$ be a point on the internal angle bisector of $\angle BAC$. Point $E$ lies on $BC$, satisfying $OE\parallel AD$, $DE\perp BC$. Point $K$ lies on $EB$ extended such that $EK=EA$. The circumcircle of $\triangle ADK$ meets $BC$ at $P\neq K$, and meets the circumcircle of $\triangle ABC$ at $Q\neq A$. Prove that $PQ$ is tangent to the circumcircle of $\triangle ABC$.
Let $ABC$ be a triangle with $AB < AC$, incenter $I$, and $A$ excenter $I_{A}$. The incircle meets $BC$ at $D$. Define $E = AD\cap BI_{A}$, $F = AD\cap CI_{A}$. Show that the circumcircle of $\triangle AID$ and $\triangle I_{A}EF$ are tangent to each other