Found problems: 239
Let $ABC$ be an equilateral triangle and let $P$ be a point in its interior. Let the lines $AP$, $BP$, $CP$ meet the sides $BC$, $CA$, $AB$ at the points $A_1$, $B_1$, $C_1$, respectively. Prove that
$A_1B_1 \cdot B_1C_1 \cdot C_1A_1 \ge A_1B \cdot B_1C \cdot C_1A$.
$D, \: E , \: F$ are points on the sides $AB, \: BC, \: CA,$ respectively, of a triangle $ABC$ such that $AD=AF, \: BD=BE,$ and $DE=DF.$ Let $I$ be the incenter of the triangle $ABC,$ and let $K$ be the point of intersection of the line $BI$ and the tangent line through $A$ to the circumcircle of the triangle $ABI.$ Show that $AK=EK$ if $AK=AD.$
Consider the cyclic quadrilateral with side lengths $1$, $4$, $8$, $7$ in that order. What is its circumdiameter? Let the answer be of the form $a\sqrt b+c$, for $b$ squarefree. Find $a+b+c$.
Let $D$ and $E$ be points on $[BC]$ and $[AC]$ of acute $\triangle ABC$, respectively. $AD$ and $BE$ meet at $F$. If $|AF|=|CD|=2|BF|=2|CE|$, and $Area(\triangle ABF) = Area(\triangle DEC)$, then $Area(\triangle AFC)/Area(\triangle BFC) = ?$
$ \textbf{(A)}\ 4 \qquad \textbf{(B)}\ 2\sqrt2 \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ \sqrt2 \qquad \textbf{(E)}\ 1$
$ABC$ is a triangle. $M$ is the midpoint of $BC$. $\angle MAB = \angle C$, and $\angle MAC = 15^{\circ}$. Show that $\angle AMC$ is obtuse. If $O$ is the circumcenter of $ADC$, show that $AOD$ is equilateral.
Two circles with radius $2$ and radius $4$ have a common center at P. Points $A, B,$ and $C$ on the larger circle are the vertices of an equilateral triangle. Point $D$ is the intersection of the smaller circle and the line segment $PB$. Find the square of the area of triangle $ADC$.
In circle $\theta_1$ with radius $1$, circles $\phi_1, \phi_2, \dots, \phi_8$, with equal radii, are drawn such that for $1 \le i \le 8$, $\phi_i$ is tangent to $\omega_1$, $\phi_{i-1}$, and $\phi_{i+1}$, where $\phi_0 = \phi_8$ and $\phi_1 = \phi_9$. There exists a circle $\omega_2$ such that $\omega_1 \neq \omega_2$ and $\omega_2$ is tangent to $\phi_i$ for $1 \le i \le 8$. The radius of $\omega_2$ can be expressed in the form $a - b\sqrt{c} -d\sqrt{e - \sqrt{f}} + g \sqrt{h - j \sqrt{k}}$ such that $a, b, \dots, k$ are positive integers and the numbers $e, f, k, \gcd(h, j)$ are squarefree. What is $a+b+c+d+e+f+g+h+j+k$.
[i]Proposed by Eugene Chen
[/i]
Let $ABCD$ be a cyclic quadrilateral, and $E$ be the intersection of its diagonals. If $m(\widehat{ADB}) = 22.5^\circ$, $|BD|=6$, and $|AD|\cdot|CE|=|DC|\cdot|AE|$, find the area of the quadrilateral $ABCD$.
Given a triangle $ABC$ and external points $X, Y$ , and $Z$ such that $\angle BAZ = \angle CAY , \angle CBX = \angle ABZ$, and $\angle ACY = \angle BCX$, prove that $AX,BY$ , and $CZ$ are concurrent.
Find the maximum value of \[ \sin{2\alpha} + \sin{2\beta} + \sin{2\gamma} \] where $\alpha,\beta$ and $\gamma$ are positive and $\alpha + \beta + \gamma = 180^{\circ}$.
In a triangle $ABC$, let $D$ be the point on the segment $BC$ such that $AB+BD=AC+CD$. Suppose that the points $B$, $C$ and the centroids of triangles $ABD$ and $ACD$ lie on a circle. Prove that $AB=AC$.
An equilateral triangle with side length $6$ has a square of side length $6$ attached to each of its edges as shown. The distance between the two farthest vertices of this figure (marked $A$ and $B$ in the figure) can be written as $m + \sqrt{n}$ where $m$ and $n$ are positive integers. Find $m + n$.
[asy]
draw((0,0)--(1,0)--(1/2,sqrt(3)/2)--cycle);
draw((1,0)--(1+sqrt(3)/2,1/2)--(1/2+sqrt(3)/2,1/2+sqrt(3)/2)--(1/2,sqrt(3)/2));
draw((0,0)--(-sqrt(3)/2,1/2)--(-sqrt(3)/2+1/2,1/2+sqrt(3)/2)--(1/2,sqrt(3)/2));
dot((-sqrt(3)/2+1/2,1/2+sqrt(3)/2));
label("A", (-sqrt(3)/2+1/2,1/2+sqrt(3)/2), N);
draw((1,0)--(1,-1)--(0,-1)--(0,0));
dot((1,-1));
label("B", (1,-1), SE);
[/asy]
Let $a$, $b$, $c$ be the three sides of a triangle, and let $\alpha$, $\beta$, $\gamma$, be the angles opposite them. If $a^2+b^2=1989c^2$, find \[ \frac{\cot \gamma}{\cot \alpha+\cot \beta}. \]
In rhombus $NIMO$, $MN = 150\sqrt{3}$ and $\measuredangle MON = 60^{\circ}$. Denote by $S$ the locus of points $P$ in the interior of $NIMO$ such that $\angle MPO \cong \angle NPO$. Find the greatest integer not exceeding the perimeter of $S$.
[i]Proposed by Evan Chen[/i]
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$
Rectangle $ABCD$ has $AB=6$ and $BC=3$. Point $M$ is chosen on side $AB$ so that $\angle AMD = \angle CMD$. What is the degree measure of $\angle AMD$?
$ \textbf{(A)}\ 15 \qquad
\textbf{(B)}\ 30 \qquad
\textbf{(C)}\ 45 \qquad
\textbf{(D)}\ 60 \qquad
\textbf{(E)}\ 75 $
Quadrilateral $ABCD$ is both cyclic and circumscribed. Its incircle touches its sides $AB$ and $CD$ at points $X$ and $Y$, respectively. The perpendiculars to $AB$ and $CD$ drawn at $A$ and $D$, respectively, meet at point $U$; those drawn at $X$ and $Y$ meet at point $V$, and finally, those drawn at $B$ and $C$ meet at point $W$. Prove that points $U$, $V$ and $W$ are collinear.
[i]Proposed by A. Golovanov[/i]
Let $X$ be a point inside convex quadrilateral $ABCD$ with $\angle AXB+\angle CXD=180^{\circ}$. If $AX=14$, $BX=11$, $CX=5$, $DX=10$, and $AB=CD$, find the sum of the areas of $\triangle AXB$ and $\triangle CXD$.
[i]Proposed by Michael Kural[/i]
In triangle $ABC$ let $D$ be the foot of the altitude from $A$. Suppose that $AD = 4$, $BD = 3$, $CD = 2$, and $AB$ is extended past $B$ to a point $E$ such that $BE = 5$. Determine the value of $CE^2$.
[i]Ray Li.[/i]
[hide="Clarifications"][list=1][*]Triangle $ABC$ is acute.[/list][/hide]
The lengths of the sides of a triangle are consescutive integers, and the largest angle is twice the smallest angle. The cosine of the smallest angle is
$\textbf {(A) } \frac 34 \qquad \textbf {(B) } \frac{7}{10} \qquad \textbf {(C) } \frac 23 \qquad \textbf {(D) } \frac{9}{14} \qquad \textbf {(E) } \text{none of these}$
Two rays with common endpoint $O$ forms a $30^\circ$ angle. Point $A$ lies on one ray, point $B$ on the other ray, and $AB = 1$. The maximum possible length of $OB$ is
$\textbf{(A)}\ 1 \qquad
\textbf{(B)}\ \dfrac{1+\sqrt{3}}{\sqrt{2}} \qquad
\textbf{(C)}\ \sqrt{3} \qquad
\textbf{(D)}\ 2 \qquad
\textbf{(E)}\ \dfrac{4}{\sqrt{3}}$
Points $A, B, C$ and $D$ are on a circle of diameter $1$, and $X$ is on diameter $\overline{AD}$. If $BX=CX$ and $3 \angle BAC=\angle BXC=36^{\circ}$, then $AX=$
[asy]
draw(Circle((0,0),10));
draw((-10,0)--(8,6)--(2,0)--(8,-6)--cycle);
draw((-10,0)--(10,0));
dot((-10,0));
dot((2,0));
dot((10,0));
dot((8,6));
dot((8,-6));
label("A", (-10,0), W);
label("B", (8,6), NE);
label("C", (8,-6), SE);
label("D", (10,0), E);
label("X", (2,0), NW);
[/asy]
$ \textbf{(A)}\ \cos 6^{\circ}\cos 12^{\circ} \sec 18^{\circ} \qquad\textbf{(B)}\ \cos 6^{\circ}\sin 12^{\circ} \csc 18^{\circ} \qquad\textbf{(C)}\ \cos 6^{\circ}\sin 12^{\circ} \sec 18^{\circ} \\ \qquad\textbf{(D)}\ \sin 6^{\circ}\sin 12^{\circ} \csc 18^{\circ} \qquad\textbf{(E)}\ \sin 6^{\circ} \sin 12^{\circ} \sec 18^{\circ} $
$[BD]$ is a median of $\triangle ABC$. $m(\widehat{ABD})=90^\circ$, $|AB|=2$, and $|AC|=6$. $|BC|=?$
$ \textbf{(A)}\ 3
\qquad\textbf{(B)}\ 3\sqrt2
\qquad\textbf{(C)}\ 5
\qquad\textbf{(D)}\ 4\sqrt2
\qquad\textbf{(E)}\ 2\sqrt6
$
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.
$ABCD$ is a parallelogram in which angle $DAB$ is acute. Points $A, P, B, D$ lie on one circle in exactly this order. Lines $AP$ and $CD$ intersect in $Q$. Point $O$ is the circumcenter of the triangle $CPQ$. Prove that if $D \neq O$ then the lines $AD$ and $DO$ are perpendicular.