Find the number of all integer-sided [i]isosceles obtuse-angled[/i] triangles with perimeter $ 2008$. [16 points out of 100 for the 6 problems]

Triangle $ABC$ is isosceles, with $AB=AC$ and altitude $AM=11.$ Suppose that there is a point $D$ on $\overline{AM}$ with $AD=10$ and $\angle BDC=3\angle BAC.$ Then the perimeter of $\triangle ABC$ may be written in the form $a+\sqrt{b},$ where $a$ and $b$ are integers. Find $a+b.$ [asy] import graph; size(7cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-1.55,xmax=7.95,ymin=-4.41,ymax=5.3; draw((1,3)--(0,0)); draw((0,0)--(2,0)); draw((2,0)--(1,3)); draw((1,3)--(1,0)); draw((1,0.7)--(0,0)); draw((1,0.7)--(2,0)); label("$11$",(0.75,1.63),SE*lsf); dot((1,3),ds); label("$A$",(0.96,3.14),NE*lsf); dot((0,0),ds); label("$B$",(-0.15,-0.18),NE*lsf); dot((2,0),ds); label("$C$",(2.06,-0.18),NE*lsf); dot((1,0),ds); label("$M$",(0.97,-0.27),NE*lsf); dot((1,0.7),ds); label("$D$",(1.05,0.77),NE*lsf); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy]

Given an $ a \times b$ rectangle with $ a > b > 0,$ determine the minimum side of a square that covers the rectangle. (A square covers the rectangle if each point in the rectangle lies inside the square.)

Consider these two geoboard quadrilaterals. Which of the following statements is true? [asy] for (int a = 0; a < 5; ++a) { for (int b = 0; b < 5; ++b) { dot((a,b)); } } draw((0,3)--(0,4)--(1,3)--(1,2)--cycle); draw((2,1)--(4,2)--(3,1)--(3,0)--cycle); label("I",(0.4,3),E); label("II",(2.9,1),W); [/asy] $\text{(A)}\ \text{The area of quadrilateral I is more than the area of quadrilateral II.}$ $\text{(B)}\ \text{The area of quadrilateral I is less than the area of quadrilateral II.}$ $\text{(C)}\ \text{The quadrilaterals have the same area and the same perimeter.}$ $\text{(D)}\ \text{The quadrilaterals have the same area, but the perimeter of I is more than the perimeter of II.}$ $\text{(E)}\ \text{The quadrilaterals have the same area, but the perimeter of I is less than the perimeter of II.}$

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$

Let $s=\frac{AB+BC+AC}{2}$ be half-perimeter of triangle $ABC$. Let $L$ and $N$be a point's on ray's $AB$ and $CB$, for which $AL=CN=s$. Let $K$ is point, symmetric of point $B$ by circumcenter of $ABC$. Prove, that perpendicular from $K$ to $NL$ passes through incenter of $ABC$. Solution for problem [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=125&t=365714&p=2011659#p2011659]here[/url]

A square and an equilateral triangle have equal perimeters. The area of the triangle is $ 9 \sqrt{3}$ square inches. Expressed in inches the diagonal of the square is: $ \textbf{(A)}\ \frac{9}{2} \qquad \textbf{(B)}\ 2 \sqrt{5} \qquad \textbf{(C)}\ 4 \sqrt{2} \qquad \textbf{(D)}\ \frac{9 \sqrt{2}}{2} \qquad \textbf{(E)}\ \text{none of these}$

In an acute triangle $ ABC$, let $ D$ be a point on $ \left[AC\right]$ and $ E$ be a point on $ \left[AB\right]$ such that $ \angle ADB\equal{}\angle AEC\equal{}90{}^\circ$. If perimeter of triangle $ AED$ is 9, circumradius of $ AED$ is $ \frac{9}{5}$ and perimeter of triangle $ ABC$ is 15, then $ \left|BC\right|$ is $\textbf{(A)}\ 5 \qquad\textbf{(B)}\ \frac{24}{5} \qquad\textbf{(C)}\ 6 \qquad\textbf{(D)}\ 8 \qquad\textbf{(E)}\ \frac{48}{5}$

(B.Frenkin, A.Zaslavsky) A convex quadrilateral was drawn on the blackboard. Boris marked the centers of four excircles each touching one side of the quadrilateral and the extensions of two adjacent sides. After this, Alexey erased the quadrilateral. Can Boris define its perimeter?

Let $ ABCD$ a cuadrilateral inscribed in a circumference. Suppose that there is a semicircle with its center on $ AB$, that is tangent to the other three sides of the cuadrilateral. (i) Show that $ AB \equal{} AD \plus{} BC$. (ii) Calculate, in term of $ x \equal{} AB$ and $ y \equal{} CD$, the maximal area that can be reached for such quadrilateral.

Let $ABC$ be a triangle with $AB=13$, $BC=14$, and $CA=15$. Points $P$, $Q$, and $R$ are chosen on segments $BC$, $CA$, and $AB$, respectively, such that triangles $AQR$, $BPR$, $CPQ$ have the same perimeter, which is $\frac{4}{5}$ of the perimeter of $PQR$. What is the perimeter of $PQR$? [i]2021 CCA Math Bonanza Individual Round #2[/i]

A triangle is circumscribed about a circle of radius $r$ inches. If the perimeter of the triangle is $P$ inches and the area is $K$ square inches, then $\frac{P}{K}$ is: $ \text{(A)}\text{independent of the value of} \; r\qquad\text{(B)}\ \frac{\sqrt{2}}{r}\qquad\text{(C)}\ \frac{2}{\sqrt{r}}\qquad\text{(D)}\ \frac{2}{r}\qquad\text{(E)}\ \frac{r}{2} $

Triangular grid divides an equilateral triangle with sides of length $n$ into $n^2$ triangular cells as shown in figure for $n=12$. Some cells are infected. A cell that is not yet infected, ia infected when it shares adjacent sides with at least two already infected cells. Specify for $n=12$, the least number of infected cells at the start in which it is possible that over time they will infected all the cells of the original triangle. [asy] unitsize(0.25cm); path p=polygon(3); for(int m=0; m<=11;++m){ for(int n=0 ; n<= 11-m; ++n){ draw(shift((n+0.5*m)*sqrt(3),1.5*m)*p); } } [/asy]

The number of inches in the perimeter of an equilateral triangle equals the number of square inches in the area of its circumscribed circle. What is the radius, in inches, of the circle? $ \textbf{(A)}\ \frac{3\sqrt2}{\pi} \qquad \textbf{(B)}\ \frac{3\sqrt3}{\pi} \qquad \textbf{(C)}\ \sqrt3 \qquad \textbf{(D)}\ \frac{6}{\pi} \qquad \textbf{(E)}\ \sqrt3\pi$

Let $D$ be the point on the side $BC$ of the triangle $ABC$ such that $AD$ is the bisector of $\angle CAB$. Let $I$ be the incenter of$ ABC.$ [i](a)[/i] Construct the points $P$ and $Q$ on the sides $AB$ and $AC$, respectively, such that $PQ$ is parallel to $BC$ and the perimeter of the triangle $APQ$ is equal to $k \cdot BC$, where $k$ is a given rational number. [i](b) [/i]Let $R$ be the intersection point of $PQ$ and $AD$. For what value of $k$ does the equality $AR = RI$ hold? [i](c)[/i] In which case do the equalities $AR = RI = ID$ hold?

What is the perimeter of trapezoid $ ABCD$? [asy]defaultpen(linewidth(0.8));size(3inch, 1.5inch); pair a=(0,0), b=(18,24), c=(68,24), d=(75,0), f=(68,0), e=(18,0); draw(a--b--c--d--cycle); draw(b--e); draw(shift(0,2)*e--shift(2,2)*e--shift(2,0)*e); label("30", (9,12), W); label("50", (43,24), N); label("25", (71.5, 12), E); label("24", (18, 12), E); label("$A$", a, SW); label("$B$", b, N); label("$C$", c, N); label("$D$", d, SE); label("$E$", e, S);[/asy] $ \textbf{(A)}\ 180\qquad\textbf{(B)}\ 188\qquad\textbf{(C)}\ 196\qquad\textbf{(D)}\ 200\qquad\textbf{(E)}\ 204 $

Let $ABC$ be a triangle of perimeter $100$ and $I$ be the point of intersection of its bisectors. Let $M$ be the midpoint of side $BC$. The line parallel to $AB$ drawn by$ I$ cuts the median $AM$ at point $P$ so that $\frac{AP}{PM} =\frac73$. Find the length of side $AB$.

Let $ABC$ be an equilateral triangle. A point $T$ is chosen on $AC$ and on arcs $AB$ and $BC$ of the circumcircle of $ABC$, $M$ and $N$ are chosen respectively, so that $MT$ is parallel to $BC$ and $NT$ is parallel to $AB$. Segments $AN$ and $MT$ intersect at point $X$, while $CM$ and $NT$ intersect in point $Y$. Prove that the perimeters of the polygons $AXYC$ and $XMBNY$ are the same.

Let $I$ be the incenter of the triangle $ABC$. Let $A_1,A_2$ be two distinct points on the line $BC$, let $B_1,B_2$ be two distinct points on the line $CA$, and let $C_1,C_2$ be two distinct points on the line $BA$ such that $AI = A_1I = A_2I$ and $BI = B_1I = B_2I$ and $CI = C_1I = C_2I$. Prove that $A_1A_2+B_1B_2+C_1C_2 = p$ where $p$ denotes the perimeter of $ABC.$ Pierre.

A square of side $1$ is covered with $m^2$ rectangles. Show that there is a rectangle with perimeter at least $\frac{4}{m}$.

Triangle $ABC$ has perimeter $1$. Its three altitudes form the side lengths of a triangle. Find the set of all possible values of $\min(AB,BC,CA)$.

a) Prove that the line dividing the triangle onto two polygons with equal perimeters and equal areas passes through the centre of the inscribed circle. b) Prove the same statement for the arbitrary tangential polygon. c) Prove that all the lines halving its perimeter and area simultaneously, intersect in one point.

Let $ d$ be the sum of the lengths of all the diagonals of a plane convex polygon with $ n$ vertices (where $ n>3$). Let $ p$ be its perimeter. Prove that: \[ n\minus{}3<{2d\over p}<\Bigl[{n\over2}\Bigr]\cdot\Bigl[{n\plus{}1\over 2}\Bigr]\minus{}2,\] where $ [x]$ denotes the greatest integer not exceeding $ x$.

Let $L$ be a positive constant. For a point $P(t,\ 0)$ on the positive part of the $x$ axis on the coordinate plane, denote $Q(u(t),\ v(t))$ the point at which the point reach starting from $P$ proceeds by　distance $L$ in counter-clockwise on the perimeter of a circle passing the point $P$ with center $O$. (1) Find $u(t),\ v(t)$. (2) For real number $a$ with $0<a<1$, find $f(a)=\int_a^1 \sqrt{\{u'(t)\}^2+\{v'(t)\}^2}\ dt$. (3) Find $\lim_{a\rightarrow +0} \frac{f(a)}{\ln a}$. [i]2011 Tokyo University entrance exam/Science, Problem 3[/i]

The ratio of the perimeter of an equilateral triangle having an altitude equal to the radius of a circle, to the perimeter of an equilateral triangle inscribed in the circle is: $ \textbf{(A)}\ 1: 2 \qquad\textbf{(B)}\ 1: 3 \qquad\textbf{(C)}\ 1: \sqrt {3} \qquad\textbf{(D)}\ \sqrt {3}: 2 \qquad\textbf{(E)}\ 2: 3$