For $i=1,2,$ let $T_i$ be a triangle with side length $a_i,b_i,c_i,$ and area $A_i.$ Suppose that $a_1\le a_2, b_1\le b_2, c_1\le c_2,$ and that $T_2$ is an acute triangle. Does it follow that $A_1\le A_2$?

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$

Prove that a triangle with angles $\alpha, \beta, \gamma$, circumradius $R$, and area $A$ satisfies \[\tan \frac{ \alpha}{2}+\tan \frac{ \beta}{2}+\tan \frac{ \gamma}{2} \leq \frac{9R^2}{4A}.\] [hide="Remark."]Remark. Can we determine [i]all[/i] of equality cases ?[/hide]

A trapezoid has side lengths $3, 5, 7,$ and $11$. The sum of all the possible areas of the trapezoid can be written in the form of $r_1 \sqrt{n_1} + r_2 \sqrt{n_2} + r_3$, where $r_1, r_2,$ and $r_3$ are rational numbers and $n_1$ and $n_2$ are positive integers not divisible by the square of a prime. What is the greatest integer less than or equal to \[r_1 + r_2 + r_3 + n_1 + n_2?\] $ \textbf{(A)}\ 57\qquad\textbf{(B)}\ 59\qquad\textbf{(C)}\ 61\qquad\textbf{(D)}\ 63\qquad\textbf{(E)}\ 65 $

Prove that for every triangle the following inequality holds: \[\frac{ab+bc+ca}{4S} \geq \cot \frac{\pi}{6}.\] where $a, b, c$ are lengths of the sides and $S$ is the area of the triangle.

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 $a,b,c$).

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})}$

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

Point $P$ is in the interior of $\triangle ABC$. The side lengths of $ABC$ are $AB = 7$, $BC = 8$, $CA = 9$. The three foots of perpendicular lines from $P$ to sides $BC$, $CA$, $AB$ are $D$, $E$, $F$ respectively. Suppose the minimal value of $\frac{BC}{PD} + \frac{CA}{PE} + \frac{AB}{PF}$ can be written as $\frac{a}{b}\sqrt{c}$, where $\gcd(a,b) = 1$ and $c$ is square free, calculate $abc$. [asy] size(120); pathpen = linewidth(0.7); pointfontpen = fontsize(10); // pointpen = black; pair B=(0,0), C=(8,0), A=IP(CR(B,7),CR(C,9)), P = (2,1.6), D=foot(P,B,C), E=foot(P,A,C), F=foot(P,A,B); D(A--B--C--cycle); D(P--D); D(P--E); D(P--F); D(MP("A",A,N)); D(MP("B",B)); D(MP("C",C)); D(MP("D",D)); D(MP("E",E,NE)); D(MP("F",F,NW)); D(MP("P",P,SE)); [/asy]

Triangle $ABC$ has $AB=27$, $AC=26$, and $BC=25$. Let $I$ denote the intersection of the internal angle bisectors of $\triangle ABC$. What is $BI$? $ \textbf{(A)}\ 15\qquad\textbf{(B)}\ 5+\sqrt{26}+3\sqrt{3}\qquad\textbf{(C)}\ 3\sqrt{26}\qquad\textbf{(D)}\ \frac{2}{3}\sqrt{546}\qquad\textbf{(E)}\ 9\sqrt{3} $

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.

Let $ABC$ be a triangle such that $AB=2$, $CA=3$, and $BC=4$. A semicircle with its diameter on $BC$ is tangent to $AB$ and $AC$. Compute the area of the semicircle.

Given a sequence $(a_n)$, with $a_1 = 4$ and $a_{n+1} = a_n^2-2 (\forall n \in\mathbb{N})$, prove that there is a triangle with side lengths $a_{n-1}, a_n, a_{n+1},$ and that its area is equal to an integer.

Let $P$ be an arbitrary point inside $\triangle ABC$ with sides $3,7,8$. What is the probability that the distance of $P$ to at least one vertices of the triangle is less than $1$? $ \textbf{(A)}\ \frac{\pi}{36}\sqrt 2 \qquad\textbf{(B)}\ \frac{\pi}{36}\sqrt 3 \qquad\textbf{(C)}\ \frac{\pi}{36} \qquad\textbf{(D)}\ \frac12 \qquad\textbf{(E)}\ \frac 34 $

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 $T = (a,b,c)$ 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 $\triangle$.

A triangle has side lengths $a$, $b$, $c$, and the altitudes have lengths $h_a$, $h_b$, $h_c$. Prove that \[ \left( \frac{a}{h_a} \right)^2 + \left( \frac{b}{h_b} \right)^2 + \left( \frac{c}{h_c} \right)^2 \ge 4. \]

(1) Let $a,b,c$ be positive real numbers satisfying $(a^2+b^2+c^2)^2>2(a^4+b^4+c^4)$. Prove that $a,b,c$ can be the lengths of three sides of a triangle respectively. (2) Let $a_1,a_2,\dots ,a_n$ be $n$ ($n>3$) positive real numbers satisfying $(a_1^2+a_2^2+\dots +a_n^2)^2>(n-1)(a_1^4+ a_2^4+\dots +a_n^4)$. Prove that any three of $a_1,a_2,\dots ,a_n$ can be the lengths of three sides of a triangle respectively.

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.

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$

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 $T$ 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 $S$.

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 $1/2$

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]