Prove that if the orthocenter divides all heights of a triangle in the same proportion, the triangle is equilateral.

In triangle $ABC$, the medians and bisectors corresponding to sides $BC$, $CA$, $AB$ are $m_a$, $m_b$, $m_c$ and $w_a$, $w_b$, $w_c$ respectively. $P=w_a \cap m_b$, $Q=w_b \cap m_c$, $R=w_c \cap m_a$. Denote the areas of triangle $ABC$ and $PQR$ by $F_1$ and $F_2$ respectively. Find the least positive constant $m$ such that $\frac{F_1}{F_2}<m$ holds for any $\triangle{ABC}$.

Let $T$ be a triangle with side lengths $26$, $51$, and $73$. Let $S$ be the set of points inside $T$ which do not lie within a distance of $5$ of any side of $T$. Find the area of $S$.

A triangle ABC with $\angle A = 60^o$ is given. Points $M$ and $N$ on $AB$ and $AC$ respectively are such that the circumcenter of $ABC$ bisects segment $MN$. Find the ratio $AN:MB$. by E.Bakaev

A chess tournament is held with the participation of boys and girls. The girls are twice as many as boys. Each player plays against each other player exactly once. By the end of the tournament, there were no draws and the ratio of girl winnings to boy winnings was $\frac{7}{9}$. How many players took part at the tournament?

Let $n$ be an integer greater than two, and let $A_1,A_2, \cdots , A_{2n}$ be pairwise distinct subsets of $\{1, 2, ,n\}$. Determine the maximum value of \[\sum_{i=1}^{2n} \dfrac{|A_i \cap A_{i+1}|}{|A_i| \cdot |A_{i+1}|}\] Where $A_{2n+1}=A_1$ and $|X|$ denote the number of elements in $X.$

Many television screens are rectangles that are measured by the length of their diagonals. The ratio of the horizontal length to the height in a standard television screen is $ 4 : 3$. The horizontal length of a “$ 27$-inch” television screen is closest, in inches, to which of the following? [asy]import math; unitsize(7mm); defaultpen(linewidth(.8pt)+fontsize(8pt)); draw((0,0)--(4,0)--(4,3)--(0,3)--(0,0)--(4,3)); fill((0,0)--(4,0)--(4,3)--cycle,mediumgray); label(rotate(aTan(3.0/4.0))*"Diagonal",(2,1.5),NW); label(rotate(90)*"Height",(4,1.5),E); label("Length",(2,0),S);[/asy]$ \textbf{(A)}\ 20 \qquad \textbf{(B)}\ 20.5 \qquad \textbf{(C)}\ 21 \qquad \textbf{(D)}\ 21.5 \qquad \textbf{(E)}\ 22$

Given two triangles and such that the lengths of the sides of the first triangle are the lengths of the medians of the second triangle. Determine the ratio of the areas of these triangles.

Call a scalene triangle K [i]disguisable[/i] if there exists a triangle K′ similar to K with two shorter sides precisely as long as the two longer sides of K, respectively. Call a disguisable triangle [i]integral[/i] if the lengths of all its sides are integers. (a) Find the side lengths of the integral disguisable triangle with the smallest possible perimeter. (b) Let K be an arbitrary integral disguisable triangle for which no smaller integral disguisable triangle similar to it exists. Prove that at least two side lengths of K are perfect squares.

Let $ABCD$ be a parallelogram with $\angle BAC = 45^\circ,$ and $AC > BD .$ Let $w_1$ and $w_2$ be two circles with diameters $AC$ and $DC,$ respectively. The circle $w_1$ intersects $AB$ at $E$ and the circle $w_2$ intersects $AC$ at $O$ and $C$, and $AD$ at $F.$ Find the ratio of areas of triangles $AOE$ and $COF$ if $AO = a,$ and $FO = b .$

If the ratio of $2x-y$ to $x+y$ is $\frac{2}{3}$, what is the ratio of $x$ to $y$? $\text{(A)} \ \frac{1}{5} \qquad \text{(B)} \ \frac{4}{5} \qquad \text{(C)} \ 1 \qquad \text{(D)} \ \frac{6}{5} \qquad \text{(E)} \ \frac{5}{4}$

The angle bisector of $A$ in triangle $ABC$ intersects $BC$ in the point $D$. The point $E$ lies on the side $AC$, and the lines $AD$ and $BE$ intersect in the point $F$. Furthermore, $\frac{|AF|}{|F D|}= 3$ and $\frac{|BF|}{|F E|}=\frac{5}{3}$. Prove that $|AB| = |AC|$. [img]https://1.bp.blogspot.com/-evofDCeJWPY/XzT9dmxXzVI/AAAAAAAAMVY/ZN87X3Cg8iMiULwvMhgFrXbdd_f1f-JWwCLcBGAsYHQ/s0/2013%2BMohr%2Bp5.png[/img]

Given distinct straight lines $ OA$ and $ OB$. From a point in $ OA$ a perpendicular is drawn to $ OB$; from the foot of this perpendicular a line is drawn perpendicular to $ OA$. From the foot of this second perpendicular a line is drawn perpendicular to $ OB$; and so on indefinitely. The lengths of the first and second perpendiculars are $ a$ and $ b$, respectively. Then the sum of the lengths of the perpendiculars approaches a limit as the number of perpendiculars grows beyond all bounds. This limit is: $ \textbf{(A)}\ \frac {b}{a \minus{} b} \qquad \textbf{(B)}\ \frac {a}{a \minus{} b} \qquad \textbf{(C)}\ \frac {ab}{a \minus{} b} \qquad \textbf{(D)}\ \frac {b^2}{a \minus{} b} \qquad \textbf{(E)}\ \frac {a^2}{a \minus{} b}$

A lake contains $250$ trout, along with a variety of other fish. When a marine biologist catches and releases a sample of $180$ fish from the lake, $30$ are identified as trout. Assume that the ratio of trout to the total number of fish is the same in both the sample and the lake. How many fish are there in the lake? $\textbf{(A)}~1250\qquad \textbf{(B)}~1500\qquad \textbf{(C)}~1750\qquad \textbf{(D)}~1800\qquad \textbf{(E)}~2000$

Let A be be a point outside the circle C, and AB and AC be the two tangents from A to this circle C. Let L be an arbitrary tangent to C that cuts AB and AC in P and Q. A line through P parallel to AC cuts BC in R. Prove that while L varies, QR passes through a fixed point. :)

In the acute angle triangle $ABC$ the point $O$ is the center of the circumscribed circle and the lines $OA,OB,OC$ intersect sides $BC,CA,AB$ respectively in points $M,N,P$ such that $\angle NMP=90^o$. [b](a)[/b] Find the ratios $\frac{\angle AMN}{\angle NMC}$,$\frac{\angle AMP}{\angle PMB}$. [b](b)[/b] If any of the angles of the triangle $ABC$ is $60^o$, find the two other angles.

For geometrical sequence $(a_n)$, the first term $a_1=1536$, common ratio $q=-\frac{1}{2}$. Let $\pi_n=\prod_{i=1}^n a_i$, so the lagerest one in $(\pi_n)$ is $\text{(A)} \pi_9\qquad\text{(B)} \pi_{11}\qquad\text{(C)} \pi_{12}\qquad\text{(D)} \pi_{13}$

An angle is drawn on a set of equally spaced parallel lines as shown. The ratio of the area of shaded region $\mathcal{C}$ to the area of shaded region $\mathcal{B}$ is $11/5$. Find the ratio of shaded region $\mathcal{D}$ to the area of shaded region $\mathcal{A}$. [asy] defaultpen(linewidth(0.7)+fontsize(10)); for(int i=0; i<4; i=i+1) { fill((2*i,0)--(2*i+1,0)--(2*i+1,6)--(2*i,6)--cycle, mediumgray); } pair A=(1/3,4), B=A+7.5*dir(-17), C=A+7*dir(10); draw(B--A--C); fill((7.3,0)--(7.8,0)--(7.8,6)--(7.3,6)--cycle, white); clip(B--A--C--cycle); for(int i=0; i<9; i=i+1) { draw((i,1)--(i,6)); } label("$\mathcal{A}$", A+0.2*dir(-17), S); label("$\mathcal{B}$", A+2.3*dir(-17), S); label("$\mathcal{C}$", A+4.4*dir(-17), S); label("$\mathcal{D}$", A+6.5*dir(-17), S);[/asy]

A circle is inscribed in an equilateral triangle, and a square is inscribed in the circle. The ratio of the area of the triangle to the area of the square is: $\textbf{(A)}\ \sqrt{3}:1 \qquad \textbf{(B)}\ \sqrt{3}:\sqrt{2} \qquad \textbf{(C)}\ 3\sqrt{3}:2 \qquad \textbf{(D)}\ 3:\sqrt{2} \qquad \textbf{(E)}\ 3:2\sqrt{2}$

Let $P,Q,R$ be points on the sides $BC,CA,AB$ respectively of a triangle $ABC$. Suppose that $BQ$ and $CR$ meet at $A', AP$ and $CR$ meet at $B'$, and $AP$ and $BQ$ meet at $C'$, such that $AB' = B'C', BC' =C'A'$, and $CA'= A'B'$. Compute the ratio of the area of $\triangle PQR$ to the area of $\triangle ABC$.

Consider a triangle $ABC$, inscribed in $O$ and $\angle A < \angle B$. Some point $P$ outside the circle satisfies $$\angle A=\angle PBA =180^{\circ}- \angle PCB$$ Let $D$ be the intersection of line $PB$ and $O$(different from $B$), and $Q$ the intersection of the tangent line of $O$ passing through $A$ and line $CD$. Show that $CQ : AB=AQ^2:AD^2$.

Show that the number of integral-sided right triangles whose ratio of area to semi-perimeter is $p^{m}$, where $p$ is a prime and $m$ is an integer, is $m+1$ if $p=2$ and $2m+1$ if $p \neq 2$.

In acute-angled triangle $ABC$, a semicircle with radius $r_a$ is constructed with its base on $BC$ and tangent to the other two sides. $r_b$ and $r_c$ are defined similarly. $r$ is the radius of the incircle of $ABC$. Show that \[ \frac{2}{r} = \frac{1}{r_a} + \frac{1}{r_b} + \frac{1}{r_c}. \]

If $AB$ and $CD$ are perpendicular diameters of circle $Q$, $P$ in $\overline{AQ}$, and $\measuredangle QPC = 60^\circ$, then the length of $PQ$ divided by the length of $AQ$ is [asy] size(200); defaultpen(linewidth(0.7)+fontsize(10)); pair A=(-1,0), B=(1,0), C=(0,1), D=(0,-1), Q=origin, P=(-0.5,0); draw(P--C--D^^A--B^^Circle(Q,1)); label("$A$", A, W); label("$B$", B, E); label("$C$", C, N); label("$D$", D, S); label("$P$", P, S); label("$Q$", Q, SE); label("$60^\circ$", P+0.0.5*dir(30), dir(30));[/asy] $ \textbf{(A)} \ \frac{\sqrt{3}}{2} \qquad \textbf{(B)} \ \frac{\sqrt{3}}{3} \qquad \textbf{(C)} \ \frac{\sqrt{2}}{2} \qquad \textbf{(D)} \ \frac12 \qquad \textbf{(E)} \ \frac23 $

A triangle has a fixed base $AB$ that is $2$ inches long. The median from $A$ to side $BC$ is $ 1\frac{1}{2}$ inches long and can have any position emanating from $A$. The locus of the vertex $C$ of the triangle is: $\textbf{(A)}\ \text{A straight line }AB,1\dfrac{1}{2}\text{ inches from }A \qquad\\ \textbf{(B)}\ \text{A circle with }A\text{ as center and radius }2\text{ inches} \qquad\\ \textbf{(C)}\ \text{A circle with }A\text{ as center and radius }3\text{ inches} \qquad\\ \textbf{(D)}\ \text{A circle with radius }3\text{ inches and center }4\text{ inches from }B\text{ along } BA \qquad\\ \textbf{(E)}\ \text{An ellipse with }A\text{ as focus}$