A rectangle has sides of length $a$ and $36$. A hinge is installed at each vertex of the rectangle and at the midpoint of each side of length $36$. The sides of length $a$ can be pressed toward each other keeping those two sides parallel so the rectangle becomes a convex hexagon as shown. When the figure is a hexagon with the sides of length $a$ parallel and separated by a distance of $24,$ the hexagon has the same area as the original rectangle. Find $a^2$. [asy] pair A,B,C,D,E,F,R,S,T,X,Y,Z; dotfactor = 2; unitsize(.1cm); A = (0,0); B = (0,18); C = (0,36); // don't look here D = (12*2.236, 36); E = (12*2.236, 18); F = (12*2.236, 0); draw(A--B--C--D--E--F--cycle); dot(" ",A,NW); dot(" ",B,NW); dot(" ",C,NW); dot(" ",D,NW); dot(" ",E,NW); dot(" ",F,NW); //don't look here R = (12*2.236 +22,0); S = (12*2.236 + 22 - 13.4164,12); T = (12*2.236 + 22,24); X = (12*4.472+ 22,24); Y = (12*4.472+ 22 + 13.4164,12); Z = (12*4.472+ 22,0); draw(R--S--T--X--Y--Z--cycle); dot(" ",R,NW); dot(" ",S,NW); dot(" ",T,NW); dot(" ",X,NW); dot(" ",Y,NW); dot(" ",Z,NW); // sqrt180 = 13.4164 // sqrt5 = 2.236 [/asy]

A circle is inscribed in a triangle $ABC$ with sides $a,b,c$. Tangents to the circle parallel to the sides of the triangle are contructe. Each of these tangents cuts off a triagnle from $\triangle ABC$. In each of these triangles, a circle is inscribed. Find the sum of the areas of all four inscribed circles (in terms of $a,b,c$).

Let a, b, c be the sidelengths and S the area of a triangle ABC. Denote $x=a+\frac{b}{2}$, $y=b+\frac{c}{2}$ and $z=c+\frac{a}{2}$. Prove that there exists a triangle with sidelengths x, y, z, and the area of this triangle is $\geq\frac94 S$.

In $\triangle ABC$, $AC = BC$, and point $D$ is on $\overline{BC}$ so that $CD = 3 \cdot BD$. Let $E$ be the midpoint of $\overline{AD}$. Given that $CE = \sqrt{7}$ and $BE = 3$, the area of $\triangle ABC$ can be expressed in the form $m\sqrt{n}$, where $m$ and $n$ are positive integers and $n$ is not divisible by the square of any prime. Find $m+n$.

Denote by $S(a,b,c)$ the area of a triangle whose lengthes of three sides are $a,b,c$ Prove that for any positive real numbers $a_{1},b_{1},c_{1}$ and $a_{2},b_{2},c_{2}$ which can serve as the lengthes of three sides of two triangles respectively ,we have $ \sqrt{S(a_{1},b_{1},c_{1})}+\sqrt{S(a_{2},b_{2},c_{2})}\le\sqrt{S(a_{1}+a_{2},b_{1}+b_{2},c_{1}+c_{2})}$

The sides of $\triangle ABC$ measure 11,20, and 21 units. We fold it along $PQ,QR,RP$ where $P,Q,R$ are the midpoints of its sides until $A,B,C$ coincide. What is the volume of the resulting tetrahedron?

Let $a,b,c$ be positive real numbers such that $abc(a+b+c)=3.$ Prove that we have \[(a+b)(b+c)(c+a)\geq 8.\] Also determine the case of equality.

A circle passes through the vertices of a triangle with side-lengths of $7\tfrac{1}{2},10,12\tfrac{1}{2}$. The radius of the circle is: $\textbf{(A)}\ \dfrac{15}{4} \qquad \textbf{(B)}\ 5 \qquad \textbf{(C)}\ \dfrac{25}{4} \qquad \textbf{(D)}\ \dfrac{35}{4} \qquad \textbf{(E)}\ \dfrac{15\sqrt2}{2} $

An equilateral triangle and a regular hexagon have equal perimeters. If the area of the triangle is 4, what is the area of the hexagon? $\textbf{(A)}\hspace{.05in}4 \qquad \textbf{(B)}\hspace{.05in}5 \qquad \textbf{(C)}\hspace{.05in}6 \qquad \textbf{(D)}\hspace{.05in}4\sqrt3 \qquad \textbf{(E)}\hspace{.05in}6\sqrt3 $

Point $P$ is inside $\triangle ABC$. Line segments $APD$, $BPE$, and $CPF$ are drawn with $D$ on $BC$, $E$ on $AC$, and $F$ on $AB$ (see the figure at right). Given that $AP=6$, $BP=9$, $PD=6$, $PE=3$, and $CF=20$, find the area of $\triangle ABC$. [asy] size(200); pair A=origin, B=(7,0), C=(3.2,15), D=midpoint(B--C), F=(3,0), P=intersectionpoint(C--F, A--D), ex=B+40*dir(B--P), E=intersectionpoint(B--ex, A--C); draw(A--B--C--A--D^^C--F^^B--E); pair point=P; label("$A$", A, dir(point--A)); label("$B$", B, dir(point--B)); label("$C$", C, dir(point--C)); label("$D$", D, dir(point--D)); label("$E$", E, dir(point--E)); label("$F$", F, dir(point--F)); label("$P$", P, dir(0));[/asy]

The sides of a triangle are $ 30$, $ 70$, and $ 80$ units. If an altitude is dropped upon the side of length $ 80$, the larger segment cut off on this side is: $ \textbf{(A)}\ 62\qquad \textbf{(B)}\ 63\qquad \textbf{(C)}\ 64\qquad \textbf{(D)}\ 65\qquad \textbf{(E)}\ 66$

Let $ABC$ be an equilateral triangle, and $P$ be an arbitrary point within the triangle. Perpendiculars $PD,PE,PF$ are drawn to the three sides of the triangle. Show that, no matter where $P$ is chosen, \[ \frac{PD+PE+PF}{AB+BC+CA}=\frac{1}{2\sqrt{3}}. \]

In a triangle $ABC$, the incircle touches the sides $BC, CA, AB$ at $D, E, F$ respectively. If the radius if the incircle is $4$ units and if $BD, CE , AF$ are consecutive integers, find the sides of the triangle $ABC$.

The sides of a triangle are of lengths $13$, $14$, and $15$. The altitudes of the triangle meet at point $H$. If $AD$ is the altitude to the side length $14$, what is the ratio $HD:HA$? $\textbf{(A) } 3 : 11\qquad \textbf{(B) } 5 : 11\qquad \textbf{(C) } 1 : 2\qquad \textbf{(D) }2 : 3\qquad \textbf{(E) }25 : 33$

The lengths of the sides of a triangle are integers, and its area is also an integer. One side is $21$ and the perimeter is $48$. The shortest side is: $\textbf{(A)}\ 8 \qquad \textbf{(B)}\ 10\qquad \textbf{(C)}\ 12 \qquad \textbf{(D)}\ 14 \qquad \textbf{(E)}\ 16$

Find the perimeter of a triangle whose altitudes are $3,4,$ and $6$. $ \textbf{(A)}\ 12\sqrt\frac35 \qquad \textbf{(B)}\ 16\sqrt\frac35 \qquad \textbf{(C)}\ 20\sqrt\frac35 \qquad \textbf{(D)}\ 24\sqrt\frac35 \qquad \textbf{(E)}\ \text{None}$

Let line $ AC$ be perpendicular to line $ CE$. Connect $ A$ to $ D$, the midpoint of $ CE$, and connect $ E$ to $ B$, the midpoint of $ AC$. If $ AD$ and $ EB$ intersect in point $ F$, and $ \overline{BC} \equal{} \overline{CD} \equal{} 15$ inches, then the area of triangle $ DFE$, in square inches, is: $ \textbf{(A)}\ 50 \qquad \textbf{(B)}\ 50\sqrt {2} \qquad \textbf{(C)}\ 75 \qquad \textbf{(D)}\ \frac {15}{2}\sqrt {105} \qquad \textbf{(E)}\ 100$

Let $T = (a,b,c)$ be a triangle with sides $a,b$ and $c$ and area $\triangle$. Denote by $T' = (a',b',c')$ the triangle whose sides are the altitudes of $T$ (i.e., $a' = h_a, b' = h_b, c' = h_c$) and denote its area by $\triangle '$. Similarly, let $T'' = (a'',b'',c'')$ be the triangle formed from the altitudes of $T'$, and denote its area by $\triangle ''$. Given that $\triangle ' = 30$ and $\triangle '' = 20$, find $\triangle$.

On a $4 \times 4 \times 3$ rectangular parallelepiped, vertices $A$, $B$, and $C$ are adjacent to vertex $D$. The perpendicular distance from $D$ to the plane containing $A$, $B$, and $C$ is closest to $\text{(A)}\ 1.6 \qquad \text{(B)}\ 1.9 \qquad \text{(C)}\ 2.1 \qquad \text{(D)}\ 2.7 \qquad \text{(E)}\ 2.9$

Let $ a$, $ b$, $ c$ be the sides of a triangle, and $ S$ its area. Prove: \[ a^{2} \plus{} b^{2} \plus{} c^{2}\geq 4S \sqrt {3} \] In what case does equality hold?

A triangle has sides of length $\sqrt{13}$, $\sqrt{17}$, and $2 \sqrt{5}$. Compute the area of the triangle.

Find the radius of a circle inscribed in a triangle with side lengths $4$, $5$, and $6$

The side lengths of a triangle are the roots of a cubic polynomial with rational coefficients. Prove that the altitudes of this triangle are roots of a polynomial of sixth degree with rational coefficients.

In a circle of radius 42, two chords of length 78 intersect at a point whose distance from the center is 18. The two chords divide the interior of the circle into four regions. Two of these regions are bordered by segments of unequal lenghts, and the area of either of them can be expressed uniquley in the form $m\pi-n\sqrt{d},$ where $m, n,$ and $d$ are positive integers and $d$ is not divisible by the square of any prime number. Find $m+n+d.$

Let $V_n=\sqrt{F_n^2+F_{n+2}^2}$, where $F_n$ is the Fibonacci sequence ($F_1=F_2=1,F_{n+2}=F_{n+1}+F_{n}$) Show that $V_n,V_{n+1},V_{n+2}$ are the sides of a triangle with area $1/2$