Found problems: 3585
Isosceles triangle $ABC$ has $AB = AC = 3\sqrt6$, and a circle with radius $5\sqrt2$ is tangent to line $AB$ at $B$ and to line $AC$ at $C$. What is the area of the circle that passes through vertices $A$, $B$, and $C?$
$\textbf{(A) }24\pi\qquad\textbf{(B) }25\pi\qquad\textbf{(C) }26\pi\qquad\textbf{(D) }27\pi\qquad\textbf{(E) }28\pi$
Let $ABC$ be a triangle with $AB+AC=3BC$. The $B$-excircle touches side $AC$ and line $BC$ at $E$ and $D$, respectively. The $C$-excircle touches side $AB$ at $F$. Let lines $CF$ and $DE$ meet at $P$. Prove that $\angle PBC = 90^{\circ}$.
[i]Ray Li[/i]
Let $B$ be a point on a circle $S_1$, and let $A$ be a point distinct from $B$ on the tangent at $B$ to $S_1$. Let $C$ be a point not on $S_1$ such that the line segment $AC$ meets $S_1$ at two distinct points. Let $S_2$ be the circle touching $AC$ at $C$ and touching $S_1$ at a point $D$ on the opposite side of $AC$ from $B$. Prove that the circumcentre of triangle $BCD$ lies on the circumcircle of triangle $ABC$.
$D, \: E , \: F$ are points on the sides $AB, \: BC, \: CA,$ respectively, of a triangle $ABC$ such that $AD=AF, \: BD=BE,$ and $DE=DF.$ Let $I$ be the incenter of the triangle $ABC,$ and let $K$ be the point of intersection of the line $BI$ and the tangent line through $A$ to the circumcircle of the triangle $ABI.$ Show that $AK=EK$ if $AK=AD.$
In convex hexagon $ABCDEF$, we know that $\angle BCA = \angle DEC = \angle AFB = \angle CBD = \angle EDF.$ Prove that $AB = CD = EF.$
Let $P$ be a point inside triangle $ABC$. Let $AP$ meet $BC$ at $A_1$, let $BP$ meet $CA$ at $B_1$, and let $CP$ meet $AB$ at $C_1$. Let $A_2$ be the point such that $A_1$ is the midpoint of $PA_2$, let $B_2$ be the point such that $B_1$ is the midpoint of $PB_2$, and let $C_2$ be the point such that $C_1$ is the midpoint of $PC_2$. Prove that points $A_2, B_2$, and $C_2$ cannot all lie strictly inside the circumcircle of triangle $ABC$.
(Australia)
Let $ ABC$ be an acute triangle with $ \angle BAC\equal{}60^{\circ}$. Let $ P$ be a point in triangle $ ABC$ with $ \angle APB\equal{}\angle BPC\equal{}\angle CPA\equal{}120^{\circ}$. The foots of perpendicular from $ P$ to $ BC,CA,AB$ are $ X,Y,Z$, respectively. Let $ M$ be the midpoint of $ YZ$.
a) Prove that $ \angle YXZ\equal{}60^{\circ}$
b) Prove that $ X,P,M$ are collinear.
Convex quadrilateral $ABCD$ is inscribed in circle $w.$Rays $AB$ and $DC$ intersect at $K.\ L$ is chosen on the diagonal $BD$ so that $\angle BAC= \angle DAL.\ M$ is chosen on the segment $KL$ so that $CM \mid\mid BD.$ Prove that line $BM$ touches $w.$
[i](Kungozhin M.)[/i]
Let $ABC$ be a triangle with incenter $I$. Suppose the reflection of $AB$ across $CI$ and the reflection of $AC$ across $BI$ intersect at a point $X$. Prove that $XI$ is perpendicular to $BC$.
Let $ABC$ be triangle with incenter $I$. A point $P$ in the interior of the triangle satisfies \[\angle PBA+\angle PCA = \angle PBC+\angle PCB.\] Show that $AP \geq AI$, and that equality holds if and only if $P=I$.
Let $ABC$ be a triangle with $AB = AC$. The incircle touches $BC$, $AC$ and $AB$ at $D$, $E$ and $F$ respectively. Let $P$ be a point on the arc $\overarc{EF}$ that does not contain $D$. Let $Q$ be the second point of intersection of $BP$ and the incircle of $ABC$. The lines $EP$ and $EQ$ meet the line $BC$ at $M$ and $N$, respectively.
Prove that the four points $P, F, B, M$ lie on a circle and $\frac{EM}{EN} = \frac{BF}{BP}$.
Let $ABCD$ be a cyclic quadrilateral. Let $E$ be a point on side $BC,$ $F$ be a point on side $AE,$ $G$ be a point on the exterior angle bisector of $\angle BCD,$ such that $EG=FG,$ $\angle EAG=\dfrac12\angle BAD.$ Prove that $AB\cdot AF=AD\cdot AE.$
Let \( ABC \) be a scalene triangle. Let \( E \) and \( F \) be the midpoints of sides \( AC \) and \( AB \), respectively, and let \( D \) be any point on segment \( BC \). The circumcircles of triangles \( BDF \) and \( CDE \) intersect line \( EF \) at points \( K \neq F \), and \( L \neq E \), respectively, and intersect at points \( X \neq D \). The point \( Y \) is on line \( DX \) such that \( AY \) is parallel to \( BC \). Prove that points \( K \), \( L \), \( X \), and \( Y \) lie on the same circle.
Let $ABC$ be an acute triangle such that $CA \neq CB$ with circumcircle $\omega$ and circumcentre $O$. Let $t_A$ and $t_B$ be the tangents to $\omega$ at $A$ and $B$ respectively, which meet at $X$. Let $Y$ be the foot of the perpendicular from $O$ onto the line segment $CX$. The line through $C$ parallel to line $AB$ meets $t_A$ at $Z$. Prove that the line $YZ$ passes through the midpoint of the line segment $AC$.
[i]Proposed by Dominic Yeo, United Kingdom[/i]
In a non-isosceles triangle $ABC$ the bisectors of angles $A$ and $B$ are inversely proportional to the respective sidelengths. Find angle $C$.
Let $\gamma,\Gamma$ be two concentric circles with radii $r,R$ with $r<R$. Let $ABCD$ be a cyclic quadrilateral inscribed in $\gamma$. If $\overrightarrow{AB}$ denotes the Ray starting from $A$ and extending indefinitely in $B's$ direction then Let $\overrightarrow{AB}, \overrightarrow{BC}, \overrightarrow{CD} , \overrightarrow{DA}$ meet $\Gamma$ at the points $C_1,D_1,A_1,B_1$ respectively. Prove that
\[\frac{[A_1B_1C_1D_1]}{[ABCD]} \ge \frac{R^2}{r^2}\]
where $[.]$ denotes area.
Let $ABC$ be an acute scalene triangle with $O$ as its circumcenter. Point $P$ lies inside triangle $ABC$ with $\angle PAB = \angle PBC$ and $\angle PAC = \angle PCB$. Point $Q$ lies on line $BC$ with $QA = QP$. Prove that $\angle AQP = 2\angle OQB$.
Given a $\Delta ABC$ where $\angle C = 90^{\circ}$, $D$ is a point in the interior of $\Delta ABC$ and lines $AD$ $,$ $BD$ and $CD$ intersect $BC$, $CA$ and $AB$ at points $P$ ,$Q$ and $R$ ,respectively. Let $M$ be the midpoint of $\overline{PQ}$. Prove that, if $\angle BRP$ $ =$ $ \angle PRC$ then $MR=MC$.
The diagonals of the inscribed quadrilateral $ABCD$ meet at the point $M$, $\angle AMB = 60^o$. Equilateral triangles $ADK$ and $BCL$ are built outward on sides $AD$ and $BC$. Line $KL$ meets the circle circumscribed ariound $ABCD$ at points $P$ and $Q$. Prove that $PK = LQ$.
In a convex quadrilateral $ABCD$, $\angle ABC = \angle ADC = 90^\circ$. A point $P$ is chosen from the diagonal $BD$ such that $\angle APB = 2\angle CPD$, points $X$, $Y$ is chosen from the segment $AP$ such that $\angle AXB = 2\angle ADB$, $\angle AYD = 2\angle ABD$. Prove that: $BD = 2XY$.
The inscribed circle of triangle $ABC$, with centre $I$, touches sides $BC$, $CA$ and $AB$ at $D$, $E$ and
$F$, respectively. Let $P$ be a point, on the same side of $FE$ as $A$, for which $\angle PFE = \angle BCA$
and $\angle PEF = \angle ABC$. Prove that $P$, $I$ and $D$ lie on a straight line.
In acute scalene triangle $ABC$ the external angle bisector of $\angle BAC$ meet $BC$ at point $X$.Lines $l_b$ and $l_c$ which tangents of $B$ and $C$ with respect to $(ABC)$.The line pass through $X$ intersects $l_b$ and $l_c$ at points $Y$ and $Z$ respectively. Suppose $(AYB)\cap(AZC)=N$ and $l_b\cap l_c=D$. Show that $ND$ is angle bisector of $\angle YNZ$.
Proposed by [i]Alireza Haghi[/i]
Let $ ABCDE$ be a convex pentagon such that
\[ \angle BAC \equal{} \angle CAD \equal{} \angle DAE\qquad \text{and}\qquad \angle ABC \equal{} \angle ACD \equal{} \angle ADE.
\]The diagonals $BD$ and $CE$ meet at $P$. Prove that the line $AP$ bisects the side $CD$.
[i]Proposed by Zuming Feng, USA[/i]
Let $ABC$ be a triangle, satisfying $2AC=AB+BC$. If $O$ and $I$ are its circumcenter and incenter, show that $\angle OIB=90^{\circ}$.
Points $P$ and $Q$ on side $AB$ of a convex quadrilateral $ABCD$ are given such that $AP = BQ.$ The circumcircles of triangles $APD$ and $BQD$ meet again at $K$ and those of $APC$ and $BQC$ meet again at $L$. Show that the points $D,C,K,L$ lie on a circle.