Found problems: 1065
Let $ABCD$ be a trapezoid such that $|AB|=9$, $|CD|=5$ and $BC\parallel AD$. Let the internal angle bisector of angle $D$ meet the internal angle bisectors of angles $A$ and $C$ at $M$ and $N$, respectively. Let the internal angle bisector of angle $B$ meet the internal angle bisectors of angles $A$ and $C$ at $L$ and $K$, respectively. If $K$ is on $[AD]$ and $\dfrac{|LM|}{|KN|} = \dfrac 37$, what is $\dfrac{|MN|}{|KL|}$?
$
\textbf{(A)}\ \dfrac{62}{63}
\qquad\textbf{(B)}\ \dfrac{27}{35}
\qquad\textbf{(C)}\ \dfrac{2}{3}
\qquad\textbf{(D)}\ \dfrac{5}{21}
\qquad\textbf{(E)}\ \dfrac{24}{63}
$
Let $ABC$ be a triangle, $H$ the orthocenter and $M$ the midpoint of $AC$. Let $\ell$ be the parallel through $M$ to the bisector of $\angle AHC$. Prove that $\ell$ divides the triangle in two parts of equal perimeters.
[i]Pedro Marrone, Panamá[/i]
A circle $\omega$ and a point $T$ outside the circle is given. Let a tangent from $T$ to $\omega$ touch $\omega$ at $A$, and take points $B,C$ lying on $\omega$ such that $T,B,C$ are colinear. The bisector of $\angle ATC$ intersects $AB$ and $AC$ at $P$ and $Q$,respectively. Prove that $PA=\sqrt{PB\cdot QC}$
It is possible to compose a triangle from the altitudes of a given triangle. Can we conclude that it is possible to compose a triangle from its bisectors?
Let $ABC$ be an isosceles triangle such that $\angle BAC = 100^{\circ}$. Let $D$ be an intersection point of angle bisector of $\angle ABC$ and side $AC$, prove that $AD+DB=BC$
A triangle $ABC$ is given in which $\angle BAC = 40^o$. and $\angle ABC = 20^o$. Find the length of the angle bisector drawn from the vertex $C$, if it is known that the sides $AB$ and $BC$ differ by $4$ centimeters.
Consider the set of 30 parabolas defined as follows: all parabolas have as focus the point (0,0) and the directrix lines have the form $y=ax+b$ with a and b integers such that $a\in \{-2,-1,0,1,2\}$ and $b\in \{-3,-2,-1,1,2,3\}$. No three of these parabolas have a common point. How many points in the plane are on two of these parabolas?
${ \textbf{(A)}\ 720\qquad\textbf{(B)}\ 760\qquad\textbf{(C)}\ 810\qquad\textbf{(D}}\ 840\qquad\textbf{(E)}\ 870 $
In triangle $ABC$, the angle bisectors $AD$, $BE$ and $CF$ are drawn, intersecting at point $I$. The perpendicular bisector of the segment $AD$ intersects lines $BE$ and $CF$ at points $M$ and $N$, respectively. Prove that points $A$, $I$, $M$ and $ N$ lie on the same circle.
In triangle $ABC$, angle $B$ is obtuse and equal to $a$. The bisectors of angles $A$ and $C$ intersect opposite sides at points $P$ and $M$, respectively. On the side $AC$, points $K$ and $L$ are taken so that $\angle ABK = \angle CBL = 2a - 180^o$. What is the angle between straight lines $KP$ and $LM$?
Point $O$ is the center of circumcircle $\omega$ of the isosceles triangle $ABC$ ($AB = AC$). Bisector of the angle $\angle C$ intersects $\omega$ at the point $W$. Point $Q$ is the center of the circumcircle of the triangle $OWB$. Construct the triangle $ABC$ given the points $Q,W, B$.
(Andrey Mostovy)
Let $\Gamma$ be the circumcircle of an acute triangle $ABC$. The perpendicular line to $AB$ passing through $C$ cuts $AB$ in $D$ and $\Gamma$ again in $E$. The bisector of the angle $C$ cuts $AB$ in $F$ and $\Gamma$ again in $G$. The line $GD$ meets $\Gamma$ again at $H$ and the line $HF$ meets $\Gamma$ again at $I$. Prove that $AI = EB$.
Say that a (nondegenerate) triangle is [i]funny[/i] if it satisfies the following condition: the altitude, median, and angle bisector drawn from one of the vertices divide the triangle into 4 non-overlapping triangles whose areas form (in some order) a 4-term arithmetic sequence. (One of these 4 triangles is allowed to be degenerate.) Find with proof all funny triangles.
Let $A_{1}$, $B_{1}$, $C_{1}$ be the feet of the altitudes of an acute-angled triangle $ABC$ issuing from the vertices $A$, $B$, $C$, respectively. Let $K$ and $M$ be points on the segments $A_{1}C_{1}$ and $B_{1}C_{1}$, respectively, such that $\measuredangle KAM = \measuredangle A_{1}AC$. Prove that the line $AK$ is the angle bisector of the angle $C_{1}KM$.
Triangle $WSE$ has side lengths $WS=13$, $SE=15$, and $WE=14$. Points $J$ and $H$ lie on $\overline{SE}$ such that $SJ=JH=HE=5$. Let the angle bisector of $\angle{WES}$ intersect $\overline{WH}$ and $\overline{WJ}$ at points $M$ and $T$, respectively. Find the area of quadrilateral $JHMT$.
Let $ABC$ be a triangle with $AB \neq BC$ and let $BD$ the interior bisectrix of $ \angle ABC$ with $D \in AC$ . Let $M$ be the midpoint of the arc $AC$ that contains the point $B$ in the circumcircle of the triangle $ABC$ .The circumcircle of the triangle $BDM$ intersects the segment $AB$ in $K \neq B$ . Denote by $J$ the symmetric of $A$ with respect to $K$ .If $DJ$ intersects $AM$ in $O$ then prove that $J,B,M,O$ are concyclic.
Let $ ABC$ be a triangle, and $ I$ its incenter. Let the incircle of $ ABC$ touch side $ BC$ at $ D$, and let lines $ BI$ and $ CI$ meet the circle with diameter $ AI$ at points $ P$ and $ Q$, respectively. Given $ BI \equal{} 6, CI \equal{} 5, DI \equal{} 3$, determine the value of $ \left( DP / DQ \right)^2$.
The bisectors $BB_1$ and $CC_1$ of the acute triangle $ABC$ intersect in point $I$. On the extensions of the segments $BB_1$ and $CC_1$, the points $B'$ and $C'$ are marked, respectively So, the quadrilateral $AB'IC'$ is a parallelogram. Prove that if $\angle BAC = 60^o$, then the straight line $B'C'$ passes through the intersection point of the circumscribed circles of the triangles $BC_1B'$ and $CB_1C'$.
Given an equilateral $\triangle ABC$, find the locus of points $P$ such that $\angle APB=\angle BPC$.
$\triangle ABC$, $\angle A=23^{\circ},\angle B=46^{\circ}$. Let $\Gamma$ be a circle with center $C$, radius $AC$. Let the external angle bisector of $\angle B$ intersects $\Gamma$ at $M,N$. Find $\angle MAN$.
Let $ABC$ be a triangle with $\angle A = 60^\circ$, and $AA', BB', CC'$ be its internal angle bisectors. Prove that $\angle B'A'C' \le 60^\circ$.
Consider $ n$ points in a plane which are vertices of a convex polygon. Prove that the set of the lengths of the sides and the diagonals of the polygon has at least $ \lfloor n/2\rfloor$ elements.
Given $\triangle ABC$ with circumcircle $\ell$. Point $M$ in $\triangle ABC$ such that $AM$ is the angle bisector of $\angle BAC$. Circle with center $M$ and radius $MB$ intersects $\ell$ and $BC$ at $D$ and $E$ respectively, $(B \not= D, B \not= E)$. Let $P$ be the midpoint of arc $BC$ in $\ell$ that didn't have $A$. Prove that $AP$ angle bisector of $\angle DPE$ if and only if $\angle B = 90^{\circ}$.
Let $ABC$ be a triangle with circumcircle $\omega$. Let the bisector of $\angle ABC$ meet segment $AC$ at $D$ and circle $\omega$ at $M\ne B$. The circumcircle of $\triangle BDC$ meets line $AB$ at $E\ne B$, and $CE$ meets $\omega$ at $P\ne C$. The bisector of $\angle PMC$ meets segment $AC$ at $Q\ne C$. Given that $PQ = MC$, determine the degree measure of $\angle ABC$.
[i]Ray Li.[/i]
In the figure below, $E$ is the midpoint of the arc $ABEC$ and the segment $ED$ is perpendicular to the chord $BC$ at $D$. If the length of the chord $AB$ is $l_1$, and that of the segment $BD$ is $l_2$, determine the length of $DC$ in terms of $l_1, l_2$.
[asy]
unitsize(1 cm);
pair A=2dir(240),B=2dir(190),C=2dir(30),E=2dir(135),D=foot(E,B,C);
draw(circle((0,0),2)); draw(A--B--C); draw(E--D); draw(rightanglemark(C,D,E,8));
label("$A$",A,.5A); label("$B$",B,.5B); label("$C$",C,.5C); label("$E$",E,.5E); label("$D$",D,dir(-60));
[/asy]
Let $ABC$ a scalene triangle and $AD, BE, CF$ your angle bisectors, with $D$ in the segment $BC, E$ in the segment $AC$ and $F$ in the segment $AB$. If $\angle AFE = \angle ADC$.
Determine $\angle BCA$.