Found problems: 847
Let $P$ be a point outside a circle $\Gamma$, and let the two tangent lines through $P$ touch $\Gamma$ at $A$ and $B$. Let $C$ be on the minor arc $AB$, and let ray $PC$ intersect $\Gamma$ again at $D$. Let $\ell$ be the line through $B$ and parallel to $PA$. $\ell$ intersects $AC$ and $AD$ at $E$ and $F$, respectively. Prove that $B$ is the midpoint of $EF$.
$ABC$ is a triangle
$E, F$ are points in $AB$, such that $AE = EF = FB$
$D$ is a point at the line $BC$ such that $ED$ is perpendiculat to $BC$
$AD$ is perpendicular to $CF$.
The angle CFA is the triple of angle BDF. ($3\angle BDF = \angle CFA$)
Determine the ratio $\frac{DB}{DC}$.
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Let $ABC$ be a triangle with $\angle A = 60^\circ$. The point $T$ lies inside the triangle in such a way that $\angle ATB = \angle BTC = \angle CTA = 120^\circ$. Let $M$ be the midpoint of $BC$. Prove that $TA + TB + TC = 2AM$.
Given acute triangle $ABC$ with $AB>AC$, let $M$ be the midpoint of $BC$. $P$ is a point in triangle $AMC$ such that $\angle MAB=\angle PAC$. Let $O,O_1,O_2$ be the circumcenters of $\triangle ABC,\triangle ABP,\triangle ACP$ respectively. Prove that line $AO$ passes through the midpoint of $O_1 O_2$.
Let $L$ be the common point of the symmedians of triangle $ABC$, and $BH$ be its altitude. It is known that $\angle ALH = 180^o -2\angle A$. Prove that $\angle CLH = 180^o - 2\angle C$.
Let $ABC$ be a triangle with incenter $I$ and circumcircle $\Omega$. A point $X$ on $\Omega$ which is different from $A$ satisfies $AI=XI$. The incircle touches $AC$ and $AB$ at $E, F$, respectively. Let $M_a, M_b, M_c$ be the midpoints of sides $BC, CA, AB$, respectively. Let $T$ be the intersection of the lines $M_bF$ and $M_cE$. Suppose that $AT$ intersects $\Omega$ again at a point $S$.
Prove that $X, M_a, S, T$ are concyclic.
[i]Proposed by ltf0501 and Li4[/i]
In $\triangle ABC$, a point $D$ lies on line $BC$. The circumcircle of $ABD$ meets $AC$ at $F$ (other than $A$), and the circumcircle of $ADC$ meets $AB$ at $E$ (other than $A$). Prove that as $D$ varies, the circumcircle of $AEF$ always passes through a fixed point other than $A$, and that this point lies on the median from $A$ to $BC$.
[i]Proposed by Allen Liu[/i]
Let $ABCD$ be a cyclic quadrilateral. Let $P$, $Q$, $R$ be the feet of the perpendiculars from $D$ to the lines $BC$, $CA$, $AB$, respectively. Show that $PQ=QR$ if and only if the bisectors of $\angle ABC$ and $\angle ADC$ are concurrent with $AC$.
Let $P$ and $Q$ be on segment $BC$ of an acute triangle $ABC$ such that $\angle PAB=\angle BCA$ and $\angle CAQ=\angle ABC$. Let $M$ and $N$ be the points on $AP$ and $AQ$, respectively, such that $P$ is the midpoint of $AM$ and $Q$ is the midpoint of $AN$. Prove that the intersection of $BM$ and $CN$ is on the circumference of triangle $ABC$.
[i]Proposed by Giorgi Arabidze, Georgia.[/i]
In a non-isosceles triangle $ABC$,$O$ and $I$ are circumcenter and incenter,respectively.$B^\prime$ is reflection of $B$ with respect to $OI$ and lies inside the angle $ABI$.Prove that the tangents to circumcirle of $\triangle BB^\prime I$ at $B^\prime$,$I$ intersect on $AC$. (A. Kuznetsov)
Let $ABC$ be an acute scalene triangle, $H$ its orthocenter and $G$ its geocenter. The circumference with diameter $AH$ cuts the circumcircle of $BHC$ in $A'$ ($A' \neq H$). Points $B'$ and $C'$ are defined similarly. Show that the points $A'$, $B'$, $C'$, and $G$ lie in one circumference.
Let $(O)$ be a fixed circle with the radius $R$. Let $A$ and $B$ be fixed points in $(O)$ such that $A,B,O$ are not collinear. Consider a variable point $C$ lying on $(O)$ ($C\neq A,B$). Construct two circles $(O_1),(O_2)$ passing through $A,B$ and tangent to $BC,AC$ at $C$, respectively. The circle $(O_1)$ intersects the circle $(O_2)$ in $D$ ($D\neq C$). Prove that:
a) \[ CD\leq R \]
b) The line $CD$ passes through a point independent of $C$ (i.e. there exists a fixed point on the line $CD$ when $C$ lies on $(O)$).
Let $ABC$ be a triangle with $\angle{C} = 90^{\circ}$, and let $H$ be the foot of the altitude from $C$. A point $D$ is chosen inside the triangle $CBH$ so that $CH$ bisects $AD$. Let $P$ be the intersection point of the lines $BD$ and $CH$. Let $\omega$ be the semicircle with diameter $BD$ that meets the segment $CB$ at an interior point. A line through $P$ is tangent to $\omega$ at $Q$. Prove that the lines $CQ$ and $AD$ meet on $\omega$.
Let $ABC$ be an acute-angled triangle with $AB<AC$ and let $O$ be the centre of its circumcircle $\omega$. Let $D$ be a point on the line segment $BC$ such that $\angle BAD = \angle CAO$. Let $E$ be the second point of intersection of $\omega$ and the line $AD$. If $M$, $N$ and $P$ are the midpoints of the line segments $BE$, $OD$ and $AC$, respectively, show that the points $M$, $N$ and $P$ are collinear.
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]
Prove that the parallelogram $ABCD$ with relation $\angle ABD + \angle DAC = 90^o$, is either a rectangle or a rhombus.
Let $R$ and $S$ be different points on a circle $\Omega$ such that $RS$ is not a diameter. Let $\ell$ be the tangent line to $\Omega$ at $R$. Point $T$ is such that $S$ is the midpoint of the line segment $RT$. Point $J$ is chosen on the shorter arc $RS$ of $\Omega$ so that the circumcircle $\Gamma$ of triangle $JST$ intersects $\ell$ at two distinct points. Let $A$ be the common point of $\Gamma$ and $\ell$ that is closer to $R$. Line $AJ$ meets $\Omega$ again at $K$. Prove that the line $KT$ is tangent to $\Gamma$.
[i]Proposed by Charles Leytem, Luxembourg[/i]
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]
Let $ABCDE$ be a convex pentagon such that $BC \parallel AE,$ $AB = BC + AE,$ and $\angle ABC = \angle CDE.$ Let $M$ be the midpoint of $CE,$ and let $O$ be the circumcenter of triangle $BCD.$ Given that $\angle DMO = 90^{\circ},$ prove that $2 \angle BDA = \angle CDE.$
[i]Proposed by Nazar Serdyuk, Ukraine[/i]
Let $ABC$ be a triangle with $\angle C=90^\circ$. A line joining the midpoint of its altitude $CH$ and the vertex $A$ meets $CB$ at point $K$. Let $L$ be the midpoint of $BC$ ,and $T$ be a point of segment $AB$ such that $\angle ATK=\angle LTB$. It is known that $BC=1$. Find the perimeter of triangle $KTL$.
Let $\triangle {ABC} $ be isosceles triangle with $AC=BC$ . The point $D$ lies on the extension of $AC$ beyond $C$ and is that $AC>CD$. The angular bisector of $ \angle BCD $ intersects $BD$ at point $N$ and let $M$ be the midpoint of $BD$. The tangent at $M$ to the circumcircle of triangle $AMD$ intersects the side $BC$ at point $P$. Prove that points $A,P,M$ and $N$ lie on a circle.
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]
In triangle $ABC$ let $F$ denote the midpoint of side $BC$. Let the circle passing through point $A$ and tangent to side $BC$ at point $F$ intersect sides $AB$ and $AC$ at points $M$ and $N$, respectively. Let the line segments $CM$ and $BN$ intersect in point $X$. Let $P$ be the second point of intersection of the circumcircles of triangles $BMX$ and $CNX$. Prove that points $A, F$ and $P$ are collinear.
Proposed by Imolay András, Budapest
In acute triangle $ABC$ ($AB<AC$) point $O$ is center of its circumcircle $\Omega$. Let the tangent to $\Omega$ drawn at point $A$ intersect the line $BC$ at point $D$. Let the line $DO$ intersects the segments $AB$ and $AC$ at points $E$ and $F$, respectively. Point $G$ is constructed such that $AEGF$ is a parallelogram. Let $K$ and $H$ be points of intersection of segment $BC$ with segments $EG$ and $FG$, respectively. Prove that the circle $(GKH)$ touches the circle $\Omega$.
[i] Proposed by Dong Luu [/i]
Let $ABC$ ($BC > AC$) be an acute triangle with circumcircle $k$ centered at $O$. The tangent to $k$ at $C$ intersects the line $AB$ at the point $D$. The circumcircles of triangles $BCD, OCD$ and $AOB$ intersect the ray $CA$ (beyond $A$) at the points $Q, P$ and $K$, respectively, such that $P \in (AK)$ and $K \in (PQ)$. The line $PD$ intersects the circumcircle of triangle $BKQ$ at the point $T$, so that $P$ and $T$ are in different halfplanes with respect to $BQ$. Prove that $TB = TQ$.