In triangle $ABC$, the angles $\alpha=\angle A$ and $\beta=\angle B$ are acute. The isosceles triangle $ACD$ and $BCD$ with the bases $AC$ and $BC$ and $\angle ADC=\beta$, $\angle BEC=\alpha$ are constructed in the exterior of the triangle $ABC$. Let $O$ be the circumcenter of $\triangle ABC$. Prove that $DO+EO$ equals the perimeter of triangle $ABC$ if and only if $\angle ACB$ is right.

Regular hexagon $ABCDEF$ has side length $2$. Let $G$ be the midpoint of $\overline{AB}$, and let $H$ be the midpoint of $\overline{DE}$. What is the perimeter of $GCHF$? $ \textbf{(A)}\ 4\sqrt3 \qquad \textbf{(B)}\ 8 \qquad \textbf{(C)}\ 4\sqrt5 \qquad \textbf{(D)}\ 4\sqrt7 \qquad \textbf{(E)}\ 12$

Given an arbitrary triangle $ ABC$, denote by $ P,Q,R$ the intersections of the incircle with sides $ BC, CA, AB$ respectively. Let the area of triangle $ ABC$ be $ T$, and its perimeter $ L$. Prove that the inequality \[\left(\frac {AB}{PQ}\right)^3 \plus{}\left(\frac {BC}{QR}\right)^3 \plus{}\left(\frac {CA}{RP}\right)^3 \geq \frac {2}{\sqrt {3}} \cdot \frac {L^2}{T}\] holds.

One container of paint is exactly enough to cover the inside of an old rectangle which is three times as long as it is wide. If we make a new rectangle by shortening the old rectangle by $18$ feet and widening it by $8$ feet as shown below, one container of paint is also exactly enough to cover the inside of the new rectangle. Find the length in feet of the perimeter of the new rectangle. [asy] size(250); defaultpen(linewidth(0.8)); draw((-2,0)--(-2,5)--(13,5)--(13,0)--cycle^^(16,-1)--(16,6)--(27,6)--(27,-1)--cycle^^(9,5)--(9,0)^^(16,4)--(27,4)); path rect1=(13,5)--(13,0)--(9,0)--(9,5)--cycle,rect2=(16,6)--(16,4)--(27,4)--(27,6)--cycle; fill(rect1,lightgray); fill(rect2,lightgray); draw(rect1^^rect2); [/asy]

In a triangle $ABC$, the excircle $\omega_a$ opposite $A$ touches $AB$ at $P$ and $AC$ at $Q$, while the excircle $\omega_b$ opposite $B$ touches $BA$ at $M$ and $BC$ at $N$. Let $K$ be the projection of $C$ onto $MN$ and let $L$ be the projection of $C$ onto $PQ$. Show that the quadrilateral $MKLP$ is cyclic. ([i]Bulgaria[/i])

Let $ABCDEF$ be a regular hexagon with side 1. Point $X, Y$ are on sides $CD$ and $DE$ respectively, such that the perimeter of $DXY$ is $2$. Determine $\angle XAY$.

Consider a regular $2^k$-gon with center $O$ and label its sides clockwise by $l_1,l_2,...,l_{2^k}$. Reflect $O$ with respect to $l_1$, then reflect the resulting point with respect to $l_2$ and do this process until the last side. Prove that the distance between the final point and $O$ is less than the perimeter of the $2^k$-gon. [i]Proposed by Hesam Rajabzade[/i]

Two points $A$ and $B$ are selected independently and uniformly at random along the perimeter of a unit square with vertices at $(0,0)$, $(1,0)$, $(0,1)$, and $(1,1)$. The probability that the $y$-coordinate of $A$ is strictly greater than the $y$-coordinate of $B$ can be expressed as $\tfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Compute $100m+n$. [i]Proposed by Rajiv Movva[/i]

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: $\textbf{(A)}\ 8 \qquad \textbf{(B)}\ 10\qquad \textbf{(C)}\ 12 \qquad \textbf{(D)}\ 14 \qquad \textbf{(E)}\ 16$

Let two ants stand on the perimeter of a regular $2019$-gon of unit side length. One of them stands on a vertex and the other one is on the midpoint of the opposite side. They start walking along the perimeter at the same speed counterclockwise. The locus of their midpoints traces out a figure $P$ in the plane with $N$ corners. Let the area enclosed by the convex hull of $P$ be $\tfrac{A}{B}\tfrac{\sin^m\left(\tfrac{\pi}{4038}\right)}{\tan\left(\tfrac{\pi}{2019}\right)}$, where $A$ and $B$ are coprime positive integers, and $m$ is the smallest possible positive integer such that this formula holds. Find $A+B+m+N$. [i]Note:[/i] The [i]convex hull[/i] of a figure $P$ is the convex polygon of smallest area which contains $P$.

The hypotenuse of a right triangle is $ 10$ inches and the radius of the inscribed circle is $ 1$ inch. The perimeter of the triangle in inches is: $ \textbf{(A)}\ 15 \qquad \textbf{(B)}\ 22 \qquad \textbf{(C)}\ 24 \qquad \textbf{(D)}\ 26 \qquad \textbf{(E)}\ 30$

Given two circles $C_1,C_2$ with common center, the radius of $C_2$ is twice the radius of $C_1$. Quadrilateral $A_1A_2A_3A_4$ is inscribed in $C_1$. The extension of $A_4A_1$ meets $C_2$ at $B_1$; the extension of $A_1A_2$ meets $C_2$ at $B_2$; the extension of $A_2A_3$ meets $C_2$ at $B_3$; the extension of $A_3A_4$ meets $C_2$ at $B_4$. Prove that $P(B_1B_2B_3B_4)\ge 2P(A_1A_2A_3A_4)$, and in what case the equality holds? ($P(X)$ denotes the perimeter of quadrilateral $X$)

In triangle $ABC$, let points $M$ and $N$ be the midpoints of sides $AB$ and $BC$ respectively. It is known that the perimeter of the triangle $MBN$ is $12$ cm, and the perimeter of the quadrilateral $AMNC$ is $20$ cm. Find the length of the segment $MN$.

A rectangular field is half as wide as it is long and is completely enclosed by $ x$ yards of fencing. The area in terms of $ x$ is: $ \textbf{(A)}\ \frac {x^2}{2} \qquad\textbf{(B)}\ 2x^2 \qquad\textbf{(C)}\ \frac {2x^2}{9} \qquad\textbf{(D)}\ \frac {x^2}{18} \qquad\textbf{(E)}\ \frac {x^2}{72}$

On a cyclic quadrilateral $ABCD$, there is a point $P$ on side $AD$ such that the triangle $CDP$ and the quadrilateral $ABCP$ have equal perimeters and equal areas. Prove that two sides of $ABCD$ have equal lengths.

The vertices of a given square are clockwise lettered $A,B,C,D$. On the side $AB$ is situated a point $E$ such that $AE = AB/3$. Starting from an arbitrarily chosen point $P_0$ on segment $AE$ and going clockwise around the perimeter of the square, a series of points $P_0, P_1, P_2, \ldots$ is marked on the perimeter such that $P_iP_{i+1} = AB/3$ for each $i$. It will be clear that when $P_0$ is chosen in $A$ or in $E$, then some $P_i$ will coincide with $P_0$. Does this possibly also happen if $P_0$ is chosen otherwise?

Inside a rectangle is inscribed a quadrilateral, which has a vertex on each side of the rectangle. Prove that the perimeter of the inscribed quadrilateral is not smaller than double the length of a diagonal of the rectangle. (V. V . Proizvolov , Moscow)

The perimeter of triangle $APM$ is $152,$ and the angle $PAM$ is a right angle. A circle of radius $19$ with center $O$ on $\overline{AP}$ is drawn so that it is tangent to $\overline{AM}$ and $\overline{PM}.$ Given that $OP=m/n,$ where $m$ and $n$ are relatively prime positive integers, find $m+n.$

Let $P(z) = z^8 + (4\sqrt{3} + 6) z^4 - (4\sqrt{3}+7)$. What is the minimum perimeter among all the 8-sided polygons in the complex plane whose vertices are precisely the zeros of $P(z)$? $ \textbf{(A)}\ 4\sqrt{3}+4 \qquad \textbf{(B)}\ 8\sqrt{2} \qquad \textbf{(C)}\ 3\sqrt{2}+3\sqrt{6} \qquad \textbf{(D)}\ 4\sqrt{2}+4\sqrt{3} \qquad $ $\textbf{(E)}\ 4\sqrt{3}+6 $

A circle and a square have the same perimeter. Then: $ \textbf{(A)}\ \text{their areas are equal} \qquad\textbf{(B)}\ \text{the area of the circle is the greater}$ $ \textbf{(C)}\ \text{the area of the square is the greater}$ $ \textbf{(D)}\ \text{the area of the circle is } \pi \text{ times the area of the square} \\ \qquad\textbf{(E)}\ \text{none of these}$

Prove the inequality \[(-a+b+c)^2(a-b+c)^2(a+b-c)^2 \geq (-a^2+b^2+c^2)(a^2-b^2+c^2)(a^2+b^2-c^2)\] for all real numbers $a, b, c.$

A square and a triangle have equal perimeters. The lengths of the three sides of the triangle are $ 6.2$ cm, $ 8.3$ cm, and $ 9.5$ cm. The area of the square is \[ \textbf{(A)}\ 24 \text{ cm}^2 \qquad \textbf{(B)}\ 36 \text{ cm}^2 \qquad \textbf{(C)}\ 48 \text{ cm}^2 \qquad \textbf{(D)}\ 64 \text{ cm}^2 \qquad \textbf{(E)}\ 144 \text{ cm}^2 \]

Suppose a square piece of paper is folded in half vertically. The folded paper is then cut in half along the dashed line. Three rectangles are formed-a large one and two small ones. What is the ratio of the perimeter of one of the small rectangles to the perimeter of the large rectangle? $\text{(A)}\ \frac{1}{2} \qquad \text{(B)}\ \frac{2}{3} \qquad \text{(C)}\ \frac{3}{4} \qquad \text{(D)}\ \frac{4}{5} \qquad \text{(E)}\ \frac{5}{6}$ [asy] draw((0,0)--(0,8)--(6,8)--(6,0)--cycle); draw((0,8)--(5,9)--(5,8)); draw((3,-1.5)--(3,10.3),dashed); draw((0,5.5)..(-.75,4.75)..(0,4)); draw((0,4)--(1.5,4),EndArrow); [/asy]

What is the minimum possible perimeter of a triangle two of whose sides are along the x- and y-axes and such that the third contains the point $(1,2)$?

Let $a_1,a_2,a_3,a_4$ be the sides of an arbitrary quadrilateral of perimeter $2s$. Prove that \[ \sum\limits^4_{i=1} \dfrac 1{a_i+s} \leq \dfrac 29\sum\limits_{1\leq i<j\leq 4} \dfrac 1{ \sqrt { (s-a_i)(s-a_j)}}. \] When does the equality hold?