This website contains problems from math contests. Problems and corresponding tags were obtained from the Art of Problem Solving website.

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Found problems: 1065

Consider the triangle $\triangle ABC$ with $AB \not = AC$. Let point $O$ be the circumcenter of $\triangle ABC$. Let the angle bisector of $\angle BAC$ intersect $BC$ at point $D$. Let $E$ be the reflection of point $D$ across the midpoint of the segment $BC$. The lines perpendicular to $BC$ in points $D,E$ intersect the lines $AO,AD$ at the points $X,Y$ respectively. Prove that the quadrilateral $B,X,C,Y$ is cyclic.
$ABC$ is a triangle such that $BC = 10$, $CA = 12$. Let $M$ be the midpoint of side $AC$. Given that $BM$ is parallel to the external bisector of $\angle A$, find area of triangle $ABC$. (Lines $AB$ and $AC$ form two angles, one of which is $\angle BAC$. The external angle bisector of $\angle A$ is the line that bisects the other angle.
In the acute triangle $ABC$, $\angle BAC$ is less than $\angle ACB $. Let $AD$ be a diameter of $\omega$, the circle circumscribed to said triangle. Let $E$ be the point of intersection of the ray $AC$ and the tangent to $\omega$ passing through $B$. The perpendicular to $AD$ that passes through $E$ intersects the circle circumscribed to the triangle $BCE$, again, at the point $F$. Show that $CD$ is an angle bisector of $\angle BCF$.
Let $ABCD$ be a parallelogram. Let $M$ be a point on the side $AB$ and $N$ be a point on the side $BC$ such that the segments $AM$ and $CN$ have equal lengths and are non-zero. The lines $AN$ and $CM$ meet at $Q$. Prove that the line $DQ$ is the bisector of the angle $\measuredangle ADC$. [i]Alternative formulation.[/i] Let $ABCD$ be a parallelogram. Let $M$ and $N$ be points on the sides $AB$ and $BC$, respectively, such that $AM=CN\neq 0$. The lines $AN$ and $CM$ intersect at a point $Q$. Prove that the point $Q$ lies on the bisector of the angle $\measuredangle ADC$.
$ABC$ is a triangle with $|AB|=6$, $|BC|=7$, and $|AC|=8$. Let the angle bisector of $\angle A$ intersect $BC$ at $D$. If $E$ is a point on $[AC]$ such that $|CE|=2$, what is $|DE|$? $ \textbf{(A)}\ 3 \qquad\textbf{(B)}\ \frac {17}5 \qquad\textbf{(C)}\ \frac 72 \qquad\textbf{(D)}\ 2\sqrt 3 \qquad\textbf{(E)}\ 3\sqrt 2 $
In the isosceles triangle $ABC$, with $AB=AC$, the angle bisector of $\angle B$ meets the side $AC$ at $B'$. Suppose that $BB'+B'A=BC$. Find the angles of the triangle $ABC$. [i]Dan Nedeianu[/i]
Let $ABC$ be an acute angled triangle (with $AB<AC<BC$) inscribed in circle $c(O,R)$ (with center $O$ and radius $R$). Circle $c_1(A,AB)$ (with center $A$ and radius $AB$) intersects side $BC$ at point $D$ and the circumcircle $c(O,R)$ at point $E$. Prove that side $AC$ bisects angle $\angle DAE$.
The lengths of sides $CB$ and $CA$ of $\vartriangle ABC$ are $a$ and $b$, and the angle between them is $\gamma = 120^o$. Express the length of the bisector of $\gamma$ in terms of $a$ and $b$.
Given a triangle $ABC$ with side $AB$ greater than $BC$, let $M$ be the midpoint of $AC$ and $L$ be the point at which the bisector of angle $\angle B$ intersects side $AC$. The line parallel to $AB$, which intersects the bisector $BL$ at $D$, is drawn by $M$, and the line parallel to the side $BC$ that intersects the median $BM$ at $E$ is drawn by $L$. Show that $ED$ is perpendicular to $BL$.
In triangle $ABC$, sides $CB$ and $CA$ are equal to $a$ and $b$, respectively. The bisector of the angle $ACB$ intersects the side $AB$ at the point $K$ and the circumscribed circle of the triangle at point $M$. The circumscribed circle of the triangle $AMK$ intersects for second time the straight line $CA$ at the point $P$. Find the length of the segment $AP$.
Let $ABC$ be a triangle, and $C^{\prime}$ and $A^{\prime}$ the midpoints of its sides $AB$ and $BC$. Consider two lines $g$ and $g^{\prime}$ which both pass through the vertex $A$ and are symmetric to each other with respect to the angle bisector of the angle $CAB$. Further, let $Y$ and $Y^{\prime}$ be the orthogonal projections of the point $B$ on these lines $g$ and $g^{\prime}$. Show that the points $Y$ and $Y^{\prime}$ are symmetric to each other with respect to the line $C^{\prime}A^{\prime}$.
Right triangle $ ABC,$ with $ \angle A\equal{}90^{\circ},$ is inscribed in circle $ \Gamma.$ Point $ E$ lies on the interior of arc $ {BC}$ (not containing $ A$) with $ EA>EC.$ Point $ F$ lies on ray $ EC$ with $ \angle EAC \equal{} \angle CAF.$ Segment $ BF$ meets $ \Gamma$ again at $ D$ (other than $ B$). Let $ O$ denote the circumcenter of triangle $ DEF.$ Prove that $ A,C,O$ are collinear.
In an acute triangle $ABC$, altitudes at vertices $A$ and $B$ and bisector line at angle $C$ intersect the circumcircle again at points $A_1, B_1$ and $C_0$. Using the straightedge and compass, reconstruct the triangle by points $A_1, B_1$ and $C_0$.
Let $ABC$ be a triangle with $|AB|> |BC|$. Let $D$ be the midpoint of $AC$. Let $E$ be the intersection of the angular bisector of $\angle ABC$ and the line $AC$. Let $F$ be the point on $BE$ such that $CF$ is perpendicular to $BE$. Finally, let $G$ be the intersection of $CF$ and $BD$. Prove that $DF$ divides the line segment $EG$ into two equal parts.
Let $ABC$ be a triangle with $AB < AC$ and incentre $I$. Let $E$ be the point on the side $AC$ such that $AE = AB$. Let $G$ be the point on the line $EI$ such that $\angle IBG = \angle CBA$ and such that $E$ and $G$ lie on opposite sides of $I$. Prove that the line $AI$, the line perpendicular to $AE$ at $E$, and the bisector of the angle $\angle BGI$ are concurrent.
Let $\beta$ be the length of the bisector of angle $B$, and $a', c'$ be the lengths of the segments into which this bisector divides the side $AC$ of the triangle $ABC$. Prove the relation $\beta^2 = ac-a'c'$ and derive from this the formula $\beta^2=ac-\frac{b^2ac}{(a+c)^2}$.
The circle inscribed in triangle $ABC$ touches $AC$ at point $F$. The perpendicular from point $F$ on $BC$ intersects the bisector of angle $C$ at point $N$. Prove that segment $FN$ is equal to the radius of the circle inscribed in triangle $ABC$. (Oleksii Karliuchenko)
Let $ABC$ ne a right triangle with $\angle ACB=90^o$. Let $E, F$ be respecitvely the midpoints of the $BC, AC$ and $CD$ be it's altitude. Next, let $P$ be the intersection of the internal angle bisector from $A$ and the line $EF$. Prove that $P$ is the center of the circle inscribed in the triangle $CDE$ .
Let $ABCD$ be a cyclic quadrilaterla and let $E$ be the point of intersection of its diagonals $AC$ and $BD$. Suppose $AD$ and $BC$ meet in $F$. Let the midpoints of $AB$ and $CD$ be $G$ and $H$ respectively. If $\Gamma $ is the circumcircle of triangle $EGH$, prove that $FE$ is tangent to $\Gamma $.
Let $ABC$ be a triangle with $\angle BAC = 60^{\circ}$. Let $AP$ bisect $\angle BAC$ and let $BQ$ bisect $\angle ABC$, with $P$ on $BC$ and $Q$ on $AC$. If $AB + BP = AQ + QB$, what are the angles of the triangle?
The angle bisectors at $A$ and $C$ in a non-isosceles triangle $ABC$ with incenter $I$ intersect its circumcircle $k$ at $A_0$ and $C_0$, respectively. The line through $I$, parallel to $AC$, intersects $A_0C_0$ at $P$. Prove that $PB$ is tangent to $k$.
Let $ k$ be the inscribed circle of non-isosceles triangle $ \triangle ABC$, which center is $ S$. Circle $ k$ touches sides $ BC,CA,AB$ in points $ P,Q,R$ respectively. Line $ QR$ intersects $ BC$ in point $ M$. Let a circle which contains points $ B$ and $ C$ touch $ k$ in point $ N$. Circumscribed circle of $ \triangle MNP$ intersects line $ AP$ in point $ L$, different from $ P$. Prove that points $ S,L$ and $ M$ are collinear.
Let $ABCD$ be a cyclic quadrilateral (a quadrilateral which can be inscribed in a circle). Let $E$ and $F$ be variable points on the sides $AB$ and $CD$, respectively, such that $\frac{AE}{EB} = \frac{C}{FD}$. Let $P$ be the point on the segment $EF$ such that $\frac{PE}{PF} = \frac{AB}{CD}$. Prove that the ratio between the areas of triangle $APD$ and $BPC$ does not depend on the choice of $E$ and $F$.
In the acute-angled triangle $ ABC$, $ D$ is the intersection of the altitude passing through $ A$ with $ BC$ and $ I_a$ is the excenter of the triangle with respect to $ A$. $ K$ is a point on the extension of $ AB$ from $ B$, for which $ \angle AKI_a\equal{}90^\circ\plus{}\frac 34\angle C$. $ I_aK$ intersects the extension of $ AD$ at $ L$. Prove that $ DI_a$ bisects the angle $ \angle AI_aB$ iff $ AL\equal{}2R$. ($ R$ is the circumradius of $ ABC$)
Characterize all triangles $ABC$ s.t. \[ AI_a : BI_b : CI_c = BC: CA : AB \] where $I_a$ etc. are the corresponding excentres to the vertices $A, B , C$