Found problems: 433
In triangle $ABC$ let $A'$, $B'$, $C'$ respectively be the midpoints of the sides $BC$, $CA$, $AB$. Furthermore let $L$, $M$, $N$ be the projections of the orthocenter on the three sides $BC$, $CA$, $AB$, and let $k$ denote the nine-point circle. The lines $AA'$, $BB'$, $CC'$ intersect $k$ in the points $D$, $E$, $F$. The tangent lines on $k$ in $D$, $E$, $F$ intersect the lines $MN$, $LN$ and $LM$ in the points $P$, $Q$, $R$.
Prove that $P$, $Q$ and $R$ are collinear.
Let $I$ be the center of the incircle of non-isosceles triangle $ABC,A_{1}=AI\cap BC$ and $B_{1}=BI\cap AC.$ Let $l_{a}$ be the line through $A_{1}$ which is parallel to $AC$ and $l_{b}$ be the line through $B_{1}$ parallel to $BC.$ Let $l_{a}\cap CI=A_{2}$ and $l_{b}\cap CI=B_{2}.$ Also $N=AA_{2}\cap BB_{2}$ and $M$ is the midpoint of $AB.$ If $CN\parallel IM$ find $\frac{CN}{IM}$.
Points $H$ and $T$ are marked respectively on the sides $BC$ abd $AC$ of triangle $ABC$ so that $AH$ is the altitude and $BT$ is the bisectrix $ABC$. It is known that the gravity center of $ABC$ lies on the line $HT$.
a) Find $AC$ if $BC$=a nad $AB$=c.
b) Determine all possible values of $\frac{c}{b}$ for all triangles $ABC$ satisfying the given condition.
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 a triangle and $A',B',C'$ be the midpoints of $BC,CA,AB$ respectively. Let $P$ and $P'$ be points in plane such that $PA=P'A',PB=P'B',PC=P'C'$. Prove that all $PP'$ pass through a fixed point.
Let $ABC$ be a triangle. The incircle of $ABC$ touches the sides $AB$ and $AC$ at the points $Z$ and $Y$, respectively. Let $G$ be the point where the lines $BY$ and $CZ$ meet, and let $R$ and $S$ be points such that the two quadrilaterals $BCYR$ and $BCSZ$ are parallelogram.
Prove that $GR=GS$.
[i]Proposed by Hossein Karke Abadi, Iran[/i]
Triangle $ABC$ has $AB=27$, $AC=26$, and $BC=25$. Let $I$ denote the intersection of the internal angle bisectors of $\triangle ABC$. What is $BI$?
$ \textbf{(A)}\ 15\qquad\textbf{(B)}\ 5+\sqrt{26}+3\sqrt{3}\qquad\textbf{(C)}\ 3\sqrt{26}\qquad\textbf{(D)}\ \frac{2}{3}\sqrt{546}\qquad\textbf{(E)}\ 9\sqrt{3} $
Given a parallelogram $ABCD$. The line perpendicular to $AC$ passing through $C$ and the line perpendicular to $BD$ passing through $A$ intersect at point $P$. The circle centered at point $P$ and radius $PC$ intersects the line $BC$ at point $X$, ($X \ne C$) and the line $DC$ at point $Y$ , ($Y \ne C$). Prove that the line $AX$ passes through the point $Y$ .
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 $A$, $B$, $C$, $D$ be four points in space not all lying on the same plane. The segments $AB$, $BC$, $CD$, and $DA$ are tangent to the same sphere. Prove that their four points of tangency are coplanar.
Let $ABC$ be a triangle and let the tangent at $B{}$ to its circumcircle meet the internal bisector of the angle $A{}$ at $P{}$. The line through $P{}$ parallel to $AC$ meets $AB$ at $Q{}$. Assume that $Q{}$ lies in the interior of segment $AB$ and let the line through $Q{}$ parallel to $BC$ meet $AC$ at $X{}$ and $PC$ at $Y{}$. Prove that $PX$ is tangent to the circumcircle of the triangle $XYC$.
Let $ABC$ be a scalene triangle, $AM$ be the median through $A$, and $\omega$ be the incircle. Let $\omega$ touch $BC$ at point $T$ and segment $AT$ meet $\omega$ for the second time at point $S$. Let $\delta$ be the triangle formed by lines $AM$ and $BC$ and the tangent to $\omega$ at $S$. Prove that the incircle of triangle $\delta$ is tangent to the circumcircle of triangle $ABC$.
Let $ABC$ be a triangle with incenter $I$. Let $D$ be a point on side $BC$ and let $\omega_B$ and $\omega_C$ be the incircles of $\triangle ABD$ and $\triangle ACD$, respectively. Suppose that $\omega_B$ and $\omega_C$ are tangent to segment $BC$ at points $E$ and $F$, respectively. Let $P$ be the intersection of segment $AD$ with the line joining the centers of $\omega_B$ and $\omega_C$. Let $X$ be the intersection point of lines $BI$ and $CP$ and let $Y$ be the intersection point of lines $CI$ and $BP$. Prove that lines $EX$ and $FY$ meet on the incircle of $\triangle ABC$.
[i]Proposed by Ray Li[/i]
Let $ABCD$ be a cyclic quadrilateral. Assume that the points $Q, A, B, P$ are collinear in this order, in such a way that the line $AC$ is tangent to the circle $ADQ$, and the line $BD$ is tangent to the circle $BCP$. Let $M$ and $N$ be the midpoints of segments $BC$ and $AD$, respectively. Prove that the following three lines are concurrent: line $CD$, the tangent of circle $ANQ$ at point $A$, and the tangent to circle $BMP$ at point $B$.
Let $ABC$ be an isosceles triangle, in which $AB=AC$ , and let $M$ and $N$ be two points on the sides $BC$ and $AC$, respectively such that $\angle BAM = \angle MNC$. Suppose that the lines $MN$ and $AB$ intersects at $P$. Prove that the bisectors of the angles $\angle BAM$ and $\angle BPM$ intersects at a point lying on the line $BC$
Let $ABC$ be a triangle with circumcircle $\Gamma$. Let $\omega_0$ be a circle tangent to chord $AB$ and arc $ACB$. For each $i = 1, 2$, let $\omega_i$ be a circle tangent to $AB$ at $T_i$ , to $\omega_0$ at $S_i$ , and to arc $ACB$. Suppose $\omega_1 \ne \omega_2$. Prove that there is a circle passing through $S_1, S_2, T_1$, and $T_2$, and tangent to $\Gamma$ if and only if $\angle ACB = 90^o$.
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The diagonals $AC$ and $CE$ of the regular hexagon $ABCDEF$ are divided by inner points $M$ and $N$ respectively, so that \[ {AM\over AC}={CN\over CE}=r. \] Determine $r$ if $B,M$ and $N$ are collinear.
Let $ ABCD$ be a convex quadrangle, $ P$ the intersection of lines $ AB$ and $ CD$, $ Q$ the intersection of lines $ AD$ and $ BC$ and $ O$ the intersection of diagonals $ AC$ and $ BD$. Show that if $ \angle POQ\equal{} 90^\circ$ then $ PO$ is the bisector of $ \angle AOD$ and $ OQ$ is the bisector of $ \angle AOB$.
Let $ABC$ be an acute triangle and $H$ its orthocenter. Let $D$ be the intersection of the altitude from $A$ to $BC$. Let $M$ and $N$ be the midpoints of $BH$ and $CH$, respectively. Let the lines $DM$ and $DN$ intersect $AB$ and $AC$ at points $X$ and $Y$ respectively. If $P$ is the intersection of $XY$ with $BH$ and $Q$ the intersection of $XY$ with $CH$, show that $H, P, D, Q$ lie on a circumference.
Points $A_1$, $A_2$, $B_1$, $B_2$ lie on the circumcircle of a triangle $ABC$ in such a way that $A_1B_1 \parallel AB$, $A_1A_2 \parallel BC$, $B_1B_2 \parallel AC$. The line $AA_2$ and $CA_1$ meet at point $A'$, and the lines $BB_2$ and $CB_1$ meet at point $B'$. Prove that all lines $A'B'$ concur.
Let $\displaystyle ABC$ and $\displaystyle DBC$ be isosceles triangle with the base $\displaystyle BC$. We know that $\displaystyle \measuredangle ABD = \frac{\pi}{2}$. Let $\displaystyle M$ be the midpoint of $\displaystyle BC$. The points $\displaystyle E,F,P$ are chosen such that $\displaystyle E \in (AB)$, $\displaystyle P \in (MC)$, $\displaystyle C \in (AF)$, and $\displaystyle \measuredangle BDE = \measuredangle ADP = \measuredangle CDF$. Prove that $\displaystyle P$ is the midpoint of $\displaystyle EF$ and $\displaystyle DP \perp EF$.
In the triangle $ABC, AC=BC, \angle C=90^o, D$ is the midpoint of $BC, E$ is the point on $AB$ such that $AD$ is perpendicular to $CE$. Prove that $AE=2EB$.
Let $ABC$ be a right triangle with hypotenuse $BC$. The tangent to the circumcircle of triangle $ABC$ at $A$ intersects the line $BC$ at $T$. The points $D$ and $E$ are chosen so that $AD = BD, AE = CE,$ and $\angle CBD = \angle BCE < 90^{\circ}$. Prove that $D, E,$ and $T$ are collinear.
Proposed by [i]Nikola Velov, Macedonia[/i]
Let a line $\ell$ intersect the line $AB$ at $F$, the sides $AC$ and $BC$ of a triangle $ABC$ at $D$ and $E$, respectively and the internal bisector of the angle $BAC$ at $P$. Suppose that $F$ is at the opposite side of $A$ with respect to the line $BC$, $CD = CE$ and $P$ is in the interior the triangle $ABC$. Prove that
\[FB \cdot FA+CP^2 = CF^2 \iff AD \cdot BE = PD^2.\]
Let $ABCD$ be a cyclic quadrilateral. Assume that the points $Q, A, B, P$ are collinear in this order, in such a way that the line $AC$ is tangent to the circle $ADQ$, and the line $BD$ is tangent to the circle $BCP$. Let $M$ and $N$ be the midpoints of segments $BC$ and $AD$, respectively. Prove that the following three lines are concurrent: line $CD$, the tangent of circle $ANQ$ at point $A$, and the tangent to circle $BMP$ at point $B$.