Found problems: 87
Let triangle \( ABC \) be an acute-angled triangle. Square \( AEFB \) and \( ADGC \) lie outside triangle \( ABC \). \( BD \) intersects \( CE \) at point \( H \), and \( BG \) intersects \( CF \) at point \( I \). The circumcircle of triangle \( BFI \) intersects the circumcircle of triangle \( CGI \) again at point \( K \). Prove that line segment \( HK \) bisects \( BC \).
Given a triangle $ABC$, with $I$ as its incenter and $\Gamma$ as its circumcircle, $AI$ intersects $\Gamma$ again at $D$. Let $E$ be a point on the arc $BDC$, and $F$ a point on the segment $BC$, such that $\angle BAF=\angle CAE < \dfrac12\angle BAC$. If $G$ is the midpoint of $IF$, prove that the meeting point of the lines $EI$ and $DG$ lies on $\Gamma$.
[i]Proposed by Tai Wai Ming and Wang Chongli, Hong Kong[/i]
In triangle $ABC$ with incenter $I$ and circumcenter $O$, let $A',B',C'$ be the points of tangency of its circumcircle with its $A,B,C$-mixtilinear circles, respectively. Let $\omega_A$ be the circle through $A'$ that is tangent to $AI$ at $I$, and define $\omega_B, \omega_C$ similarly. Prove that $\omega_A,\omega_B,\omega_C$ have a common point $X$ other than $I$, and that $\angle AXO = \angle OXA'$.
[i]Proposed by Sammy Luo[/i]
On $\triangle ABC$ if there existed a point $D$ in $AC$ such that $\angle CBD=\angle ABD+60$ and $\angle BDC=30$ and $AB \cdot BC=BD^2$, then find the angles inside the triangle $\triangle ABC$
An equilateral triangle $ABC$ with center $O$ and radius $OA = R$ is given, and consider the seven regions that the lines of the sides determine on the plane. It is asked to draw and describe the region of the plane transformed from the two shaded regions in the attached figure, by the inversion of center $O$ and power $R^2$.
[img]https://cdn.artofproblemsolving.com/attachments/e/c/bf1cb12c961467d216d54885f3387b328ce744.png[/img]
Given a triangle $ABC$, with $I$ as its incenter and $\Gamma$ as its circumcircle, $AI$ intersects $\Gamma$ again at $D$. Let $E$ be a point on the arc $BDC$, and $F$ a point on the segment $BC$, such that $\angle BAF=\angle CAE < \dfrac12\angle BAC$. If $G$ is the midpoint of $IF$, prove that the meeting point of the lines $EI$ and $DG$ lies on $\Gamma$.
[i]Proposed by Tai Wai Ming and Wang Chongli, Hong Kong[/i]
Let $ ABCD$ be a convex quadrilateral and let $ P$ and $ Q$ be points in $ ABCD$ such that $ PQDA$ and $ QPBC$ are cyclic quadrilaterals. Suppose that there exists a point $ E$ on the line segment $ PQ$ such that $ \angle PAE \equal{} \angle QDE$ and $ \angle PBE \equal{} \angle QCE$. Show that the quadrilateral $ ABCD$ is cyclic.
[i]Proposed by John Cuya, Peru[/i]
Let $ABC$ be a triangle with circumcircle $\omega$, incenter $I$, and $A$-excenter $I_A$. Let the incircle and the $A$-excircle hit $BC$ at $D$ and $E$, respectively, and let $M$ be the midpoint of arc $BC$ without $A$. Consider the circle tangent to $BC$ at $D$ and arc $BAC$ at $T$. If $TI$ intersects $\omega$ again at $S$, prove that $SI_A$ and $ME$ meet on $\omega$.
[i]Amol Aggarwal.[/i]
Triangle $ABC$ has incircle $\omega$ and $A$-excircle $\omega_A.$ Circle $\gamma_B$ passes through $B$ and is externally tangent to $\omega$ and $\omega_A.$ Circle $\gamma_C$ passes through $C$ and is externally tangent to $\omega$ and $\omega_A.$ If $\gamma_B$ intersects line $BC$ again at $D,$ and $\gamma_C$ intersects line $BC$ again at $E,$ prove that $BD=EC.$
Let $ABCD$ be an isosceles trapezoid with $AB\parallel CD$. Let $E$ be the midpoint of $AC$. Denote by $\omega$ and $\Omega$ the circumcircles of the triangles $ABE$ and $CDE$, respectively. Let $P$ be the crossing point of the tangent to $\omega$ at $A$ with the tangent to $\Omega$ at $D$. Prove that $PE$ is tangent to $\Omega$.
[i]Jakob Jurij Snoj, Slovenia[/i]
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 $BDE$.
Let $ ABCD$ be a convex quadrilateral and let $ P$ and $ Q$ be points in $ ABCD$ such that $ PQDA$ and $ QPBC$ are cyclic quadrilaterals. Suppose that there exists a point $ E$ on the line segment $ PQ$ such that $ \angle PAE \equal{} \angle QDE$ and $ \angle PBE \equal{} \angle QCE$. Show that the quadrilateral $ ABCD$ is cyclic.
[i]Proposed by John Cuya, Peru[/i]
Given two circles ${\omega _1},{\omega _2}$, $S$ denotes all $\Delta ABC$ satisfies that ${\omega _1}$ is the circumcircle of $\Delta ABC$, ${\omega _2}$ is the $A$- excircle of $\Delta ABC$ , ${\omega _2}$ touches $BC,CA,AB$ at $D,E,F$.
$S$ is not empty, prove that the centroid of $\Delta DEF$ is a fixed point.
In triangle $ABC$ with incenter $I$ and circumcenter $O$, let $A',B',C'$ be the points of tangency of its circumcircle with its $A,B,C$-mixtilinear circles, respectively. Let $\omega_A$ be the circle through $A'$ that is tangent to $AI$ at $I$, and define $\omega_B, \omega_C$ similarly. Prove that $\omega_A,\omega_B,\omega_C$ have a common point $X$ other than $I$, and that $\angle AXO = \angle OXA'$.
[i]Proposed by Sammy Luo[/i]
In $\triangle ABC$ $\angle B$ is obtuse and $AB \ne BC$. Let $O$ is the circumcenter and $\omega$ is the circumcircle of this triangle. $N$ is the midpoint of arc $ABC$. The circumcircle of $\triangle BON$ intersects $AC$ on points $X$ and $Y$. Let $BX \cap \omega = P \ne B$ and $BY \cap \omega = Q \ne B$. Prove that $P, Q$ and reflection of $N$ with respect to line $AC$ are collinear.
We are given a $\Delta ABC$. Point $D$ on the circumscribed circle k is such that $CD$ is a symmedian in $\Delta ABC$. Let $X$ and $Y$ be on the rays $\overrightarrow{CB}$ and $\overrightarrow{CA}$, so that $CX=2CA$ and $CY=2CB$. Prove that the circle, tangent externally to $k$ and to the lines $CA$ and $CB$, is tangent to the circumscribed circle of $\Delta XDY$.
Let $D$ be a point in the interior of $\triangle{ABC}$ such that $AB=ab$, $AC=ac$, $BC=bc$, $AD=ad$, $BD=bd$, $CD=cd$. Show that $\angle{ABD}+\angle{ACD}=60^{\circ}$.
Source: 2016 Singapore Mathematical Olympiad (Open) Round 2, Problem 1
Let $D$ and $E$ be points in the interiors of sides $AB$ and $AC$, respectively, of a triangle $ABC$, such that $DB = BC = CE$. Let the lines $CD$ and $BE$ meet at $F$. Prove that the incentre $I$ of triangle $ABC$, the orthocentre $H$ of triangle $DEF$ and the midpoint $M$ of the arc $BAC$ of the circumcircle of triangle $ABC$ are collinear.
Given two circles ${\omega _1},{\omega _2}$, $S$ denotes all $\Delta ABC$ satisfies that ${\omega _1}$ is the circumcircle of $\Delta ABC$, ${\omega _2}$ is the $A$- excircle of $\Delta ABC$ , ${\omega _2}$ touches $BC,CA,AB$ at $D,E,F$.
$S$ is not empty, prove that the centroid of $\Delta DEF$ is a fixed point.
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 $BDE$.
Trapezoid $ ABCD$, with $ \overline{AB}\parallel{}\overline{CD}$, is inscribed in circle $ \omega$ and point $ G$ lies inside triangle $ BCD$. Rays $ AG$ and $ BG$ meet $ \omega$ again at points $ P$ and $ Q$, respectively. Let the line through $ G$ parallel to $ \overline{AB}$ intersects $ \overline{BD}$ and $ \overline{BC}$ at points $ R$ and $ S$, respectively. Prove that quadrilateral $ PQRS$ is cyclic if and only if $ \overline{BG}$ bisects $ \angle CBD$.
Let $I$ be the incenter of the $ABC$ triangle. The circumference that passes through $I$ and has center
in $A$ intersects the circumscribed circumference of the $ABC$ triangle at points $M$ and
$N$. Prove that the line $MN$ is tangent to the inscribed circle of the $ABC$ triangle.
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 $BDE$.
Let $ABCD$ be an isosceles trapezoid with $AB\parallel CD$. Let $E$ be the midpoint of $AC$. Denote by $\omega$ and $\Omega$ the circumcircles of the triangles $ABE$ and $CDE$, respectively. Let $P$ be the crossing point of the tangent to $\omega$ at $A$ with the tangent to $\Omega$ at $D$. Prove that $PE$ is tangent to $\Omega$.
[i]Jakob Jurij Snoj, Slovenia[/i]
Let $ABC$ be a triangle with circumcircle $\omega$, incenter $I$, and $A$-excenter $I_A$. Let the incircle and the $A$-excircle hit $BC$ at $D$ and $E$, respectively, and let $M$ be the midpoint of arc $BC$ without $A$. Consider the circle tangent to $BC$ at $D$ and arc $BAC$ at $T$. If $TI$ intersects $\omega$ again at $S$, prove that $SI_A$ and $ME$ meet on $\omega$.
[i]Amol Aggarwal.[/i]