Three non-overlapping regular plane polygons, at least two of which are congruent, all have sides of length $ 1$. The polygons meet at a point $ A$ in such a way that the sum of the three interior angles at $ A$ is $ 360^\circ$. Thus the three polygons form a new polygon with $ A$ as an interior point. What is the largest possible perimeter that this polygon can have? $ \textbf{(A)}\ 12\qquad \textbf{(B)}\ 14\qquad \textbf{(C)}\ 18\qquad \textbf{(D)}\ 21\qquad \textbf{(E)}\ 24$

In $\triangle ABC$, $AB=AC=28$ and $BC=20$. Points $D,E,$ and $F$ are on sides $\overline{AB}$, $\overline{BC}$, and $\overline{AC}$, respectively, such that $\overline{DE}$ and $\overline{EF}$ are parallel to $\overline{AC}$ and $\overline{AB}$, respectively. What is the perimeter of parallelogram $ADEF$? [asy] size(180); pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); real r=5/7; pair A=(10,sqrt(28^2-100)),B=origin,C=(20,0),D=(A.x*r,A.y*r); pair bottom=(C.x+(D.x-A.x),C.y+(D.y-A.y)); pair E=extension(D,bottom,B,C); pair top=(E.x+D.x,E.y+D.y); pair F=extension(E,top,A,C); draw(A--B--C--cycle^^D--E--F); dot(A^^B^^C^^D^^E^^F); label("$A$",A,NW); label("$B$",B,SW); label("$C$",C,SE); label("$D$",D,W); label("$E$",E,S); label("$F$",F,dir(0)); [/asy] $\textbf{(A) }48\qquad \textbf{(B) }52\qquad \textbf{(C) }56\qquad \textbf{(D) }60\qquad \textbf{(E) }72\qquad$

Let $H$ be the orthocenter of an acute-angled triangle $ABC$. Show that the triangles $ABH,BCH$ and $CAH$ have the same perimeter if and only if the triangle $ABC$ is equilateral.

Let $P$ be an interior point of triangle $ABC$. The lines $AP$, $BP$ and $CP$ divide each of the three sides into two segments. If the so-obtained six segments all have distinct integer lengths, what is the minimum possible perimeter of $ABC$?

In convex quadrilateral $ABCD$, $\angle A \cong \angle C$, $AB = CD = 180$, and $AD \neq BC$. The perimeter of $ABCD$ is 640. Find $\lfloor 1000 \cos A \rfloor$. (The notation $\lfloor x \rfloor$ means the greatest integer that is less than or equal to $x$.)

A square-shaped field is divided into $n$ rectangular farms whose sides are parallel to the sides of the field. What is the greatest value of $n$, if the sum of the perimeters of the farms is equal to $100$ times of the perimeter of the field? $ \textbf{(A)}\ 10000 \qquad\textbf{(B)}\ 20000 \qquad\textbf{(C)}\ 50000 \qquad\textbf{(D)}\ 100000 \qquad\textbf{(E)}\ 200000 $

What is the perimeter of a rectangle of area $32$ inscribed in a circle of radius $4$?

A right triangle has perimeter $ 32$ and area $ 20$. What is the length of its hypotenuse? $ \textbf{(A)}\ \frac{57}{4} \qquad \textbf{(B)}\ \frac{59}{4} \qquad \textbf{(C)}\ \frac{61}{4} \qquad \textbf{(D)}\ \frac{63}{4} \qquad \textbf{(E)}\ \frac{65}{4}$

In trapezoid $ABCD$ with $AB || CD$, $\omega_1$ and $\omega_2$ are two circles with diameters $AD$ and $BC$, respectively. Let $X$ and $Y$ be two arbitrary points on $\omega_1$ and $\omega_2$, respectively. Show that the length of segment $XY$ is not more than half the perimeter of $ABCD$. [i]Proposed by Mahdi Etesami Fard[/i]

Prove that for any real $r>0$, one can cover the circumference of a $1\times r$ rectangle with non-intersecting disks of unit radius.

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]

Let $ABCD$ be a convex quadrilateral, and $M$, $N$, and $P$ be the midpoints of diagonals $AC$ and $BD$, and side $AD$, respectively. Also, suppose that $\angle{ABC} + \angle{DCB} = 90$ and that $AB = 6$, $CD = 8$. Calculate the perimeter of triangle $MNP$.

In rectangle $ABCD$, $AD=1$, $P$ is on $AB$ and $DB$ and $DP$ trisect $\angle ADC$. What is the perimeter $\triangle BDP$ [list=1] [*] $3+\frac{\sqrt{3}}{3}$ [*] $2+\frac{4\sqrt{3}}{3}$ [*] $2+2\sqrt{2}$ [*] $\frac{3+3\sqrt{5}}{2}$ [/list]

Let $P$ be a polygon (not necessarily convex) with $n$ vertices, such that all its sides and diagonals are less or equal with 1 in length. Prove that the area of the polygon is less than $\dfrac {\sqrt 3} 2$.

Let $ABC$ be a non isosceles triangle inscribed in a circle $(O)$ and $BE, CF$ are two angle bisectors intersect at $I$ with $E$ belongs to segment $AC$ and $F$ belongs to segment $AB$. Suppose that $BE, CF$ intersect $(O)$ at $M,N$ respectively. The line $d_1$ passes through $M$ and perpendicular to $BM$ intersects $(O)$ at the second point $P,$ the line $d_2$ passes through $N$ and perpendicular to $CN$ intersect $(O)$ at the second point $Q$. Denote $H, K$ are two midpoints of $MP$ and $NQ$ respectively. 1. Prove that triangles $IEF$ and $OKH$ are similar. 2. Suppose that S is the intersection of two lines $d_1$ and $d_2$. Prove that $SO$ is perpendicular to $EF$.

Show that in a non-obtuse triangle the perimeter of the triangle is always greater than two times the diameter of the circumcircle.