An equilateral triangle is drawn with a side of length $ a$. A new equilateral triangle is formed by joining the midpoints of the sides of the first one. Then a third equilateral triangle is formed by joining the midpoints of the sides of the second; and so on forever. The limit of the sum of the perimeters of all the triangles thus drawn is: $ \textbf{(A)}\ \text{Infinite} \qquad\textbf{(B)}\ 5\frac {1}{4}a \qquad\textbf{(C)}\ 2a \qquad\textbf{(D)}\ 6a \qquad\textbf{(E)}\ 4\frac {1}{2}a$

We are given some similar triangles. Their areas are $1^2,3^2,5^2,\cdots,$ and $49^2$. If the smallest triangle has a perimeter of $4$, what is the sum of all the triangles' perimeters?

A large rectangle is tiled by some $1\times1$ tiles. In the center there is a small rectangle tiled by some white tiles. The small rectangle is surrounded by a red border which is five tiles wide. That red border is surrounded by a white border which is five tiles wide. Finally, the white border is surrounded by a red border which is five tiles wide. The resulting pattern is pictured below. In all, $2900$ red tiles are used to tile the large rectangle. Find the perimeter of the large rectangle. [asy] import graph; size(5cm); fill((-5,-5)--(0,-5)--(0,35)--(-5,35)--cycle^^(50,-5)--(55,-5)--(55,35)--(50,35)--cycle,red); fill((0,30)--(0,35)--(50,35)--(50,30)--cycle^^(0,-5)--(0,0)--(50,0)--(50,-5)--cycle,red); fill((-15,-15)--(-10,-15)--(-10,45)--(-15,45)--cycle^^(60,-15)--(65,-15)--(65,45)--(60,45)--cycle,red); fill((-10,40)--(-10,45)--(60,45)--(60,40)--cycle^^(-10,-15)--(-10,-10)--(60,-10)--(60,-15)--cycle,red); fill((-10,-10)--(-5,-10)--(-5,40)--(-10,40)--cycle^^(55,-10)--(60,-10)--(60,40)--(55,40)--cycle,white); fill((-5,35)--(-5,40)--(55,40)--(55,35)--cycle^^(-5,-10)--(-5,-5)--(55,-5)--(55,-10)--cycle,white); for(int i=0;i<16;++i){ draw((-i,-i)--(50+i,-i)--(50+i,30+i)--(-i,30+i)--cycle,linewidth(.5)); } [/asy]

We have four white equilateral triangles of $3$ cm on each side and join them by their sides to obtain a triangular base pyramid. At each edge of the pyramid we mark two red dots that divide it into three equal parts. Number the red dots, so that when you scroll them in the order they were numbered, result a path with the smallest possible perimeter. How much does that path measure?

Prove: If $x, y, z$ are the lengths of the angle bisectors of a triangle with perimeter 6, than we have: \[\frac{1}{x^2} + \frac{1}{y^2} + \frac{1}{z^2} \geq 1.\]

A large square is divided into a small square surrounded by four congruent rectangles as shown. The perimeter of each of the congruent rectangles is 14. What is the area of the large square? [asy]unitsize(3mm); defaultpen(linewidth(.8pt)); draw((0,0)--(7,0)--(7,7)--(0,7)--cycle); draw((1,0)--(1,6)); draw((7,1)--(1,1)); draw((6,7)--(6,1)); draw((0,6)--(6,6));[/asy]$ \textbf{(A)}\ \ 49 \qquad \textbf{(B)}\ \ 64 \qquad \textbf{(C)}\ \ 100 \qquad \textbf{(D)}\ \ 121 \qquad \textbf{(E)}\ \ 196$

The side lengths $a,b,c$ of a triangle $ABC$ are positive integers. Let:\\ \[T_{n}=(a+b+c)^{2n}-(a-b+c)^{2n}-(a+b-c)^{2n}+(a-b-c)^{2n}\] for any positive integer $n$. If $\frac{T_{2}}{2T_{1}}=2023$ and $a>b>c$ , determine all possible perimeters of the triangle $ABC$.

Let $ABC$ be an equilateral triangle and $n{}, n>1$ an integer. Let $S{}$ be the set of the $n-1$ lines parallel with $BC$ that cut $ABC$ in $n{}$ figures with equal areas and $S^{'}$ be the set of the $n-1$ lines parallel with $BC$ that cut $ABC$ in $n{}$ figures with equal perimeters. Show that $S{}$ and $S^{'}$ are disjunctive.

Let $M$ be the midpoint of side $BC$ of triangle $ABC$ ($AB>AC$), and let $AL$ be the bisector of the angle $A$. The line passing through $M$ perpendicular to $AL$ intersects the side $AB$ at the point $D$. Prove that $AD+MC$ is equal to half the perimeter of triangle $ABC$.

On sides $CB$ and $CD$ of square $ABCD$ are chosen points $M$ and $K$ so that the perimeter of triangle $CMK$ equals double the side of the square. Find angle $\angle MAK$.

Find all triangles with integer sidelenghts such that their perimeter and area are equal.

Prove that there exist infinitely many positive integers $ n$ such that $ p \equal{} nr,$ where $ p$ and $ r$ are respectively the semiperimeter and the inradius of a triangle with integer side lengths.

Points $A$ and $B$ are the endpoints of a diameter of a circle with center $C$. Points $D$ and $E$ lie on the same diameter so that $C$ bisects segment $\overline{DE}$. Let $F$ be a randomly chosen point within the circle. The probability that $\triangle DEF$ has a perimeter less than the length of the diameter of the circle is $\tfrac{17}{128}$. There are relatively prime positive integers m and n so that the ratio of $DE$ to $AB$ is $\tfrac{m}{n}.$ Find $m + n$.

In equilateral triangle $ABC$ with side length $2$, let the parabola with focus $A$ and directrix $BC$ intersect sides $AB$ and $AC$ at $A_1$ and $A_2$, respectively. Similarly, let the parabola with focus $B$ and directrix $CA$ intersect sides $BC$ and $BA$ at $B_1$ and $B_2$, respectively. Finally, let the parabola with focus $C$ and directrix $AB$ intersect sides $CA$ and $C_B$ at $C_1$ and $C_2$, respectively. Find the perimeter of the triangle formed by lines $A_1A_2$, $B_1B_2$, $C_1C_2$.

Consider a square of sidelength $ n$ and $ (n\plus{}1)^2$ interior points. Prove that we can choose $ 3$ of these points so that they determine a triangle (eventually degenerated) of area at most $ \frac12$.