Found problems: 239
The points $D, E$ on the side $AB$ of the triangle $ABC$ are such that $\frac{AD}{DB}\frac{AE}{EB} = \left(\frac{AC}{CB}\right)^2$. Show that $\angle ACD = \angle BCE$.
A hexagon inscribed in a circle has three consecutive sides each of length $3$ and three consecutive sides each of length $5$. The chord of the circle that divides the hexagon into two trapezoids, one with three sides each of length $3$ and the other with three sides each of length $5$, has length equal to $\frac mn$, where $m$ and $n$ are relatively prime positive integers. Find $m + n$.
$\text{(A)}\ 309 \qquad \text{(B)}\ 349 \qquad \text{(C)}\ 369 \qquad \text{(D)}\ 389\qquad \text{(E)}\ 409$
Triangle $ABC$ is isosceles with $AC = BC$ and $\angle ACB = 106^\circ$. Point $M$ is in the interior of the triangle so that $\angle MAC = 7^\circ$ and $\angle MCA = 23^\circ$. Find the number of degrees in $\angle CMB$.
A segment through the focus $F$ of a parabola with vertex $V$ is perpendicular to $\overline{FV}$ and intersects the parabola in points $A$ and $B$. What is $\cos(\angle AVB)$?
$ \textbf{(A)}\ -\frac{3\sqrt{5}}{7} \qquad
\textbf{(B)}\ -\frac{2\sqrt{5}}{5} \qquad
\textbf{(C)}\ -\frac{4}{5} \qquad
\textbf{(D)}\ -\frac{3}{5} \qquad
\textbf{(E)}\ -\frac{1}{2} $
Let $ABC$ be a triangle with $AB = 42$, $AC = 39$, $BC = 45$. Let $E$, $F$ be on the sides $\overline{AC}$ and $\overline{AB}$ such that $AF = 21, AE = 13$. Let $\overline{CF}$ and $\overline{BE}$ intersect at $P$, and let ray $AP$ meet $\overline{BC}$ at $D$. Let $O$ denote the circumcenter of $\triangle DEF$, and $R$ its circumradius. Compute $CO^2-R^2$.
[i]Proposed by Yang Liu[/i]
Let $ABC$ be an equilateral triangle with side length $1$. This triangle is rotated by some angle about its center to form triangle $DEF.$ The intersection of $ABC$ and $DEF$ is an equilateral hexagon with an area that is $\frac{4} {5}$ the area of $ABC.$ The side length of this hexagon can be expressed in the form $\frac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. What is $m+n$?
[i]Author: Ray Li[/i]
Triangle $ ABC$ has side lengths $ AB \equal{} 5$, $ BC \equal{} 6$, and $ AC \equal{} 7$. Two bugs start simultaneously from $ A$ and crawl along the sides of the triangle in opposite directions at the same speed. They meet at point $ D$. What is $ BD$?
$ \textbf{(A)}\ 1 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 5$
See all the problems from 5-th Kyiv math festival
[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=506789#p506789]here[/url]
Let $O$ be the circumcenter and $H$ be the intersection point of the altitudes of acute triangle $ABC.$ The straight lines $BH$ and $CH$ intersect the segments $CO$ and $BO$ at points $D$ and $E$ respectively. Prove that if triangles $ODH$ and $OEH$ are isosceles then triangle $ABC$ is isosceles too.
Let $ABC$ be an isosceles triangle with $\angle{ABC} = \angle{ACB} = 80^\circ$. Let $D$ be a point on $AB$ such that $\angle{DCB} = 60^\circ$ and $E$ be a point on $AC$ such that $\angle{ABE} = 30^\circ$. Find $\angle{CDE}$ in degrees.
In the triangle $A,B,C$, let $D$ be the middle of $BC$ and $E$ the projection of $C$ on $AD$. Suppose $\angle ACE = \angle ABC$. Show that the triangle $ABC$ is isosceles or rectangle.
Convex quadrilateral $ABCD$ has sides $AB=BC=7$, $CD=5$, and $AD=3$. Given additionally that $m\angle ABC=60^\circ$, find $BD$.
Prove that $\cos{\frac{\pi}{7}}-\cos{\frac{2\pi}{7}}+\cos{\frac{3\pi}{7}}=\frac{1}{2}$
Let $O$ be the center of a circle of radius $26$, and let $A$, $B$ be two distinct points on the circle, with $M$ being the midpoint of $AB$. Consider point $C$ for which $CO=34$ and $\angle COM=15^\circ$. Let $N$ be the midpoint of $CO$. Suppose that $\angle ACB=90^\circ$. Find $MN$.
Let $ABC$ be equilateral, and $D, E,$ and $F$ be the midpoints of $\overline{BC}, \overline{CA},$ and $\overline{AB},$ respectively. There exist points $P, Q,$ and $R$ on $\overline{DE}, \overline{EF},$ and $\overline{FD},$ respectively, with the property that $P$ is on $\overline{CQ}, Q$ is on $\overline{AR},$ and $R$ is on $\overline{BP}.$ The ratio of the area of triangle $ABC$ to the area of triangle $PQR$ is $a+b\sqrt{c},$ where $a, b$ and $c$ are integers, and $c$ is not divisible by the square of any prime. What is $a^{2}+b^{2}+c^{2}$?
In convex quadrilateral $ABCD$, $\angle A \cong \angle C$, $AB = CD = 180$, and $AD \neq BC$. The perimeter of $ABCD$ is 640. Find $\lfloor 1000 \cos A \rfloor$. (The notation $\lfloor x \rfloor$ means the greatest integer that is less than or equal to $x$.)
Given any convex polygon, show that there are three consecutive vertices such that the polygon lies inside the circle through them.
Let $ ABC$ be a triangle with incenter $ I$ and let $ \Gamma$ be a circle centered at $ I$, whose radius is greater than the inradius and does not pass through any vertex. Let $ X_{1}$ be the intersection point of $ \Gamma$ and line $ AB$, closer to $ B$; $ X_{2}$, $ X_{3}$ the points of intersection of $ \Gamma$ and line $ BC$, with $ X_{2}$ closer to $ B$; and let $ X_{4}$ be the point of intersection of $ \Gamma$ with line $ CA$ closer to $ C$. Let $ K$ be the intersection point of lines $ X_{1}X_{2}$ and $ X_{3}X_{4}$. Prove that $ AK$ bisects segment $ X_{2}X_{3}$.
Triangle $ ABC$ has $ AB \equal{} 2 \cdot AC$. Let $ D$ and $ E$ be on $ \overline{AB}$ and $ \overline{BC}$, respectively, such that $ \angle{BAE} \equal{} \angle{ACD}.$ Let $ F$ be the intersection of segments $ AE$ and $ CD$, and suppose that $ \triangle{CFE}$ is equilateral. What is $ \angle{ACB}$?
$ \textbf{(A)}\ 60^{\circ}\qquad \textbf{(B)}\ 75^{\circ}\qquad \textbf{(C)}\ 90^{\circ}\qquad \textbf{(D)}\ 105^{\circ}\qquad \textbf{(E)}\ 120^{\circ}$
Given an acute triangle $ABC$ with altituties AD and BE. O circumcinter of $ABC$.If o lies on the segment DE then find the value of $sinAsinBcosC$
Let $\triangle{ABC}$ be an acute-angled triangle and let $D$, $E$, $F$ be points on $BC$, $CA$, $AB$, respectively, such that \[\angle{AFE}=\angle{BFD}\mbox{,}\quad\angle{BDF}=\angle{CDE}\quad\mbox{and}\quad\angle{CED}=\angle{AEF}\mbox{.}\] Prove that $D$, $E$ and $F$ are the feet of the perpendiculars through $A$, $B$ and $C$ on $BC$, $CA$ and $AB$, respectively.
[i](Swiss Mathematical Olympiad 2011, Final round, problem 2)[/i]
In triangle $ABC$, $\angle ABC = 120^{\circ}$, $AB = 3$ and $BC = 4$. If perpendiculars constructed to $\overline{AB}$ at $A$ and to $\overline{BC}$ at $C$ meet at $D$, then $CD = $
$ \textbf{(A)}\ 3\qquad\textbf{(B)}\ \frac{8}{\sqrt{3}}\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ \frac{11}{2}\qquad\textbf{(E)}\ \frac{10}{\sqrt{3}} $
Triangle $ABC$ is isosceles, with $AB=AC$ and altitude $AM=11.$ Suppose that there is a point $D$ on $\overline{AM}$ with $AD=10$ and $\angle BDC=3\angle BAC.$ Then the perimeter of $\triangle ABC$ may be written in the form $a+\sqrt{b},$ where $a$ and $b$ are integers. Find $a+b.$
[asy] import graph; size(7cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-1.55,xmax=7.95,ymin=-4.41,ymax=5.3; draw((1,3)--(0,0)); draw((0,0)--(2,0)); draw((2,0)--(1,3)); draw((1,3)--(1,0)); draw((1,0.7)--(0,0)); draw((1,0.7)--(2,0)); label("$11$",(0.75,1.63),SE*lsf); dot((1,3),ds); label("$A$",(0.96,3.14),NE*lsf); dot((0,0),ds); label("$B$",(-0.15,-0.18),NE*lsf); dot((2,0),ds); label("$C$",(2.06,-0.18),NE*lsf); dot((1,0),ds); label("$M$",(0.97,-0.27),NE*lsf); dot((1,0.7),ds); label("$D$",(1.05,0.77),NE*lsf); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy]
Equilateral triangle $ABC$ has been creased and folded so that vertex $A$ now rests at $A'$ on $\overline{BC}$ as shown. If $BA' = 1$ and $A'C = 2$ then the length of crease $\overline{PQ}$ is
[asy]
size(170);
defaultpen(linewidth(0.7)+fontsize(10));
pair B=origin, A=(1.5,3*sqrt(3)/2), C=(3,0), D=(1,0), P=B+1.6*dir(B--A), Q=C+1.2*dir(C--A);
draw(B--P--D--B^^P--Q--D--C--Q);
draw(Q--A--P, linetype("4 4"));
label("$A$", A, N);
label("$B$", B, W);
label("$C$", C, E);
label("$A'$", D, S);
label("$P$", P, W);
label("$Q$", Q, E);
[/asy]
$ \textbf{(A)}\ \frac{8}{5}\qquad\textbf{(B)}\ \frac{7}{20}\sqrt{21}\qquad\textbf{(C)}\ \frac{1+\sqrt{5}}{2}\qquad\textbf{(D)}\ \frac{13}{8}\qquad\textbf{(E)}\ \sqrt{3} $
The base of an isosceles triangle is $ 6$ inches and one of the equal sides is $ 12$ inches. The radius of the circle through the vertices of the triangle is:
$ \textbf{(A)}\ \frac{7\sqrt{15}}{5} \qquad\textbf{(B)}\ 4\sqrt{3} \qquad\textbf{(C)}\ 3\sqrt{5} \qquad\textbf{(D)}\ 6\sqrt{3} \qquad\textbf{(E)}\ \text{none of these}$
Triangle $ABC$ is a right triangle with $AC=7,$ $BC=24,$ and right angle at $C.$ Point $M$ is the midpoint of $AB,$ and $D$ is on the same side of line $AB$ as $C$ so that $AD=BD=15.$ Given that the area of triangle $CDM$ may be expressed as $\frac{m\sqrt{n}}{p},$ where $m,$ $n,$ and $p$ are positive integers, $m$ and $p$ are relatively prime, and $n$ is not divisible by the square of any prime, find $m+n+p.$