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

Tags were heavily modified to better represent problems.

AND
OR
NO

Found problems: 1065

Consider a triangle $ABC$ with $|AB|\neq |AC|$. The angle bisector of the angle $CAB$ intersects the circumcircle of $\triangle ABC$ at two points $A$ and $D$. The circle of center $D$ and radius $|DC|$ intersects the line $AC$ at two points $C$ and $B’$. The line $BB’$ intersects the circumcircle of $\triangle ABC$ at $B$ and $E$. Prove that $B’$ is the orthocenter of $\triangle AED$.
The incircle of triangle $ABC$ touches $BC$ at $D$ and $AC$ at $E$. Let $K$ be the point on $CB$ with $CK=BD$, and $L$ be the point on $CA$ with $AE=CL$. Lines $AK$ and $BL$ meet at $P$. If $Q$ is the midpoint of $BC$, $I$ the incenter, and $G$ the centroid of $\triangle ABC$, show that: $(a)$ $IQ$ and $AK$ are parallel, $(b)$ the triangles $AIG$ and $QPG$ have equal area.
In triangle $ ABC$, $ AB \equal{} 13$, $ BC \equal{} 14$, and $ AC \equal{} 15$. Let $ D$ denote the midpoint of $ \overline{BC}$ and let $ E$ denote the intersection of $ \overline{BC}$ with the bisector of angle $ BAC$. Which of the following is closest to the area of the triangle $ ADE$? $ \textbf{(A)}\ 2 \qquad \textbf{(B)}\ 2.5 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 3.5 \qquad \textbf{(E)}\ 4$
A line $r$ contains the points $A$, $B$, $C$, $D$ in that order. Let $P$ be a point not in $r$ such that $\angle{APB} = \angle{CPD}$. Prove that the angle bisector of $\angle{APD}$ intersects the line $r$ at a point $G$ such that: $\frac{1}{GA} + \frac{1}{GC} = \frac{1}{GB} + \frac{1}{GD}$
Let $ABC$ be a triangle with incenter $I$, centroid $G$, and $|AC|>|AB|$. If $IG\parallel BC$, $|BC|=2$, and $Area(ABC)=3\sqrt 5 / 8$, then what is $|AB|$? $ \textbf{(A)}\ \dfrac 98 \qquad\textbf{(B)}\ \dfrac {11}8 \qquad\textbf{(C)}\ \dfrac {13}8 \qquad\textbf{(D)}\ \dfrac {15}8 \qquad\textbf{(E)}\ \dfrac {17}8 $
$AD$ Is the angle bisector Of $\angle BAC$ Where $D$ lies on the The circumcircle of $\triangle ABC$. Show that $2AD>AB+AC$
Let $ABC$ be a triangle such that $BC>AB$, $L$ be the internal angle bisector of $\angle ABC$. Let $P,Q$ be the feet from $A,C$ to $L$, respectively. Suppose $M,N$ are the midpoints of $\overline{AC}$ and $\overline{BC}$, respectively. Let $O$ be the circumcenter of triangle $PQM$, and the circumcircle intersects $AC$ at point $H$. Prove that $O,M,N,H$ are concyclic.
Given a triangle $ \,ABC,\,$ let $ \,I\,$ be the center of its inscribed circle. The internal bisectors of the angles $ \,A,B,C\,$ meet the opposite sides in $ \,A^{\prime },B^{\prime },C^{\prime }\,$ respectively. Prove that \[ \frac {1}{4} < \frac {AI\cdot BI\cdot CI}{AA^{\prime }\cdot BB^{\prime }\cdot CC^{\prime }} \leq \frac {8}{27}. \]
Two circles $\Gamma_1, \Gamma_2$ with radii $r_i, r_2$, respectively, touch internally at the point $P$. A tangent parallel to the diameter through $P$ touches $ \Gamma_1$ at $R$ and intersects $\Gamma_2$ at $M$ and $N$. Prove that $PR$ bisects $\angle MPN$.
In triangle $ ABC$, side $ AC$ and the perpendicular bisector of $ BC$ meet in point $ D$, and $ BD$ bisects $ \angle ABC$. If $ AD \equal{} 9$ and $ DC \equal{} 7$, what is the area of triangle $ ABD$? $ \textbf{(A)}\ 14 \qquad \textbf{(B)}\ 21 \qquad \textbf{(C)}\ 28 \qquad \textbf{(D)}\ 14\sqrt5 \qquad \textbf{(E)}\ 28\sqrt5$
Triangle $ ABC$ has $ AC\equal{}3$, $ BC\equal{}4$, and $ AB\equal{}5$. Point $ D$ is on $ \overline{AB}$, and $ \overline{CD}$ bisects the right angle. The inscribed circles of $ \triangle ADC$ and $ \triangle BCD$ have radii $ r_a$ and $ r_b$, respectively. What is $ r_a/r_b$? $ \textbf{(A)}\ \frac{1}{28}\left(10\minus{}\sqrt{2}\right) \qquad \textbf{(B)}\ \frac{3}{56}\left(10\minus{}\sqrt{2}\right) \qquad \textbf{(C)}\ \frac{1}{14}\left(10\minus{}\sqrt{2}\right) \qquad \textbf{(D)}\ \frac{5}{56}\left(10\minus{}\sqrt{2}\right) \\ \textbf{(E)}\ \frac{3}{28}\left(10\minus{}\sqrt{2}\right)$
For every triangle $ABC$ inscribed in a circle $\Gamma$ , let $A',B',C'$ be the intersections of the bisectors of the angles at $A,B,C$ with $\Gamma$ . Consider the triangle $A'B'C'$ . (a) Do triangles $A'B'C'$ go over all possible triangles inscribed in $\Gamma$ as $\vartriangle ABC$ varies? If not, what are the constraints? (b) Prove that the angle bisectors of $\vartriangle ABC$ are the altitudes of $\vartriangle A',B',C'$ .
A circle $\omega$ with center $I$ is inscribed into a segment of the disk, formed by an arc and a chord $AB$. Point $M$ is the midpoint of this arc $AB$, and point $N$ is the midpoint of the complementary arc. The tangents from $N$ touch $\omega$ in points $C$ and $D$. The opposite sidelines $AC$ and $BD$ of quadrilateral $ABCD$ meet in point $X$, and the diagonals of $ABCD$ meet in point $Y$. Prove that points $X, Y, I$ and $M$ are collinear.
Prove: In an acute triangle $ ABC$ angle bisector $ w_{\alpha},$ median $ s_b$ and the altitude $ h_c$ intersect in one point if $ w_{\alpha},$ side $ BC$ and the circle around foot of the altitude $ h_c$ have vertex $ A$ as a common point.
Prove that the orthogonal projections of the vertex $ D $ of the tetrahedron $ ABCD $ onto the bisectors of the internal and external dihedral angles at the edges $ \overline{AB} $, $ \overline{BC} $ and $ \overline{CA} $ belong to one plane .
Let $E$ and $F$ be the intersections of opposite sides of a convex quadrilateral $ABCD$. The two diagonals meet at $P$. Let $O$ be the foot of the perpendicular from $P$ to $EF$. Show that $\angle BOC=\angle AOD$.
In a triangle $ABC$, $AB > AC$, the external bisector of angle $A$ meets the circumcircle of triangle $ABC$ at $E$, and $F$ is the foot of the perpendicular from $E$ onto $AB$. Prove that $2AF = AB - AC$
Let \( AL \) be the bisector of triangle \( ABC \), \( O \) the center of its circumcircle, and \( D \) and \( E \) the midpoints of \( BL \) and \( CL \), respectively. Points \( P \) and \( Q \) are chosen on segments \( AD \) and \( AE \) such that \( APLQ \) is a parallelogram. Prove that \( PQ \perp AO \). [i]Proposed by Mykhailo Plotnikov[/i]
Let $D$ be the foot of the altitude towards $BC$ in the triangle $ABC$. Let $E$ be the intersection of $AB$ with the bisector of angle $\angle C$. Assume that the angle $\angle AEC = 45^o$ . Determine the angle $\angle EDB$.
Two circles $\gamma_1, \gamma_2$ in a plane, with centers $A$ and $B$ respectively, intersect at $C$ and $D$. Suppose that the circumcircle of $ABC$ intersects $\gamma_1$ in $E$ and $\gamma_2$ in $F$, where the arc $EF$ not containing $C$ lies outside $\gamma_1$ and $\gamma_2$. Prove that this arc $EF$ is bisected by the line $CD$.
In $\triangle ABC$, $AB = 40$, $BC = 60$, and $CA = 50$. The angle bisector of $\angle A$ intersects the circumcircle of $\triangle ABC$ at $A$ and $P$. Find $BP$. [i]Proposed by Eugene Chen[/i]
The sides $ [AB]$ and $ [AC]$ of the triangle $ ABC$ are tangent to the incircle with center $ I$ of the $ \triangle ABC$ at the points $ M$ and $ N$, respectively. The internal bisectors of the $ \triangle ABC$ drawn form $ B$ and $ C$ intersect the line $ MN$ at the points $ P$ and $ Q$, respectively. Suppose that $ F$ is the intersection point of the lines $ CP$ and $ BQ$. Prove that $ FI\perp BC$.
Median $AM$ and the angle bisector $CD$ of a right triangle $ABC$ ($\angle B=90^o$) intersect at the point $O$. Find the area of the triangle $ABC$ if $CO=9, OD=5$.
In an acute triangle $ABC$ the angle $C$ is greater than the angle $A$. Let $AE$ be a diameter of the circumcircle of the triangle. Let the intersection point of the ray $AC$ and the tangent of the circumcircle through the vertex $B$ be $K$. The perpendicular to $AE$ through $K$ intersects the circumcircle of the triangle $BCK$ for the second time at point $D$. Prove that $CE$ bisects the angle $BCD$.
In the figure shown below, $ABCDE$ is a regular pentagon and $AG=1$. What is $FG+JH+CD$? [asy] import cse5;pathpen=black;pointpen=black; size(2inch); pair A=dir(90), B=dir(18), C=dir(306), D=dir(234), E=dir(162); D(MP("A",A,A)--MP("B",B,B)--MP("C",C,C)--MP("D",D,D)--MP("E",E,E)--cycle,linewidth(1.5)); D(A--C--E--B--D--cycle); pair F=IP(A--D,B--E), G=IP(B--E,C--A), H=IP(C--A,B--D), I=IP(D--B,E--C), J=IP(C--E,D--A); D(MP("F",F,dir(126))--MP("I",I,dir(270))--MP("G",G,dir(54))--MP("J",J,dir(198))--MP("H",H,dir(342))--cycle); [/asy] $\textbf{(A) } 3 \qquad\textbf{(B) } 12-4\sqrt5 \qquad\textbf{(C) } \dfrac{5+2\sqrt5}{3} \qquad\textbf{(D) } 1+\sqrt5 \qquad\textbf{(E) } \dfrac{11+11\sqrt5}{10} $