Found problems: 663
Consider the pyramid $ VABCD, $ where $ V $ is the top and $ ABCD $ is a rectangular base. If $ \angle BVD = \angle AVC, $ then prove that the triangles $ VAC $ and $ VBD $ share the same perimeter and area.
Prove that a line that divides a triangle into two polygons of equal area and equal perimeter passes through the center of the circle inscribed in the triangle. Prove an analogous property for a polygon that has an inscribed circle.
City of Mar del Plata is a square shaped $WSEN$ land with $2(n + 1)$ streets that divides it into $n \times n$ blocks, where $n$ is an even number (the leading streets form the perimeter of the square). Each block has a dimension of $100 \times 100$ meters. All streets in Mar del Plata are one-way. The streets which are parallel and adjacent to each other are directed in opposite direction. Street $WS$ is driven in the direction from $W$ to $S$ and the street $WN$ travels from $W$ to $N$. A street cleaning car starts from point $W$. The driver wants to go to the point $E$ and in doing so, he must cross as much as possible roads. What is the length of the longest route he can go, if any $100$-meter stretch cannot be crossed more than once? (The figure shows a plan of the city for $n=6$ and one of the possible - but not the longest - routes of the street cleaning car. See http://goo.gl/maps/JAzD too.)
[img]http://s14.postimg.org/avfg7ygb5/CPS_2012_P5.jpg[/img]
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?
In a unit square one hundred segments are drawn from the centre to the sides dividing the square into one hundred parts (triangles and possibly quadruilaterals). If all parts have equal perimetr $p$, show that $\dfrac{14}{10} < p < \dfrac{15}{10}$.
Let $ABCD$ be a rectangle with $AB = 30$ and $BC = 28$. Point $P$ and $Q$ lie on $\overline{BC}$ and $\overline{CD}$ respectively so that all sides of $\triangle{ABP}, \triangle{PCQ},$ and $\triangle{QDA}$ have integer lengths. What is the perimeter of $\triangle{APQ}$?
(A) 84 (B) 86 (C) 88 (D)90 (E)92
Prove that of all the quadrilaterals circuscribed around a given circle, the square has the smallest perimeter.
The lengths of the sides of a triangle are 6, 8 and 10 units. Prove that there is exactly one straight line which simultaneously bisects the area and perimeter of the triangle.
A rectangle with perimeter $ 176$ is divided into five congruent rectangles as shown in the diagram. What is the perimeter of one of the five congruent rectangles?
[asy]defaultpen(linewidth(.8pt));
draw(origin--(0,3)--(4,3)--(4,0)--cycle);
draw((0,1)--(4,1));
draw((2,0)--midpoint((0,1)--(4,1)));
real r = 4/3;
draw((r,3)--foot((r,3),(0,1),(4,1)));
draw((2r,3)--foot((2r,3),(0,1),(4,1)));[/asy]$ \textbf{(A)}\ 35.2\qquad \textbf{(B)}\ 76\qquad \textbf{(C)}\ 80\qquad \textbf{(D)}\ 84\qquad \textbf{(E)}\ 86$
Let $n\ge3$ be an integer. A circle is divided into $2n$ arcs by $2n$ points. Each arc has one of three possible lengths, and no two adjacent arcs have the same lengths. The $2n$ points are colored alternately red and blue. Prove that the $n$-gon with red vertices and the $n$-gon with blue vertices have the same perimeter and the same area.
Let $a$, $b$, and $c$ be pairwise distinct positive integers, which are side lengths of a triangle. There is a line which cuts both the area and the perimeter of the triangle into two equal parts. This line cuts the longest side of the triangle into two parts with ratio $2:1$. Determine $a$, $b$, and $c$ for which the product abc is minimal.
An $ABC$ triangle divide by a $d$ line such that, new two pieces' areas and perimeters are equal. Prove that $ABC$'s incenter lies $d$
Acute triangle $A_1A_2A_3$ is inscribed in a circle of radius $2$. Prove that one can choose points $B_1, B_2, B_3$ on the arcs $A_1A_2, A_2A_3, A_3A_1$ respectively, such that the numerical value of the area of the hexagon $A_1B_1A_2B_2A_3B_3$ is equal to the numerical value of the perimeter of the triangle $A_1A_2A_3.$
A triangle with perimeter $2p$ is inscribed in a circle of radius $R$ and also circumscribed on a circle of radius $r$. Prove that $p < 2(R+r)$.
A circle of radius $\rho$ is tangent to the sides $AB$ and $AC$ of the triangle $ABC$, and its center $K$ is at a distance $p$ from $BC$.
[i](a)[/i] Prove that $a(p - \rho) = 2s(r - \rho)$, where $r$ is the inradius and $2s$ the perimeter of $ABC$.
[i](b)[/i] Prove that if the circle intersect $BC$ at $D$ and $E$, then
\[DE=\frac{4\sqrt{rr_1(\rho-r)(r_1-\rho)}}{r_1-r}\]
where $r_1$ is the exradius corresponding to the vertex $A.$
A triangle $ ABC$ has integer sides, $ \angle A\equal{}2 \angle B$ and $ \angle C>90^{\circ}$. Find the minimum possible perimeter of this triangle.
Let $\alpha , \beta ,\gamma$ be the angles of a triangle $ABC$ of perimeter $ 2p $ and $R$ is the radius of its circumscribed circle.
$(a)$ Prove that
\[\cot^{2}\alpha +\cot^{2}\beta+\cot^{2}\gamma\geq 3\left(9\cdot \frac{R^{2}}{p^{2}} - 1\right).\]
$(b)$ When do we have equality?
Let $ABC$ be a triangle with $\angle A=\angle C=30^{\circ}.$ Points $D,E,F$ are chosen on the sides $AB,BC,CA$ respectively so that $\angle BFD=\angle BFE=60^{\circ}.$ Let $p$ and $p_1$ be the perimeters of the triangles $ABC$ and $DEF$, respectively. Prove that $p\le 2p_1.$
Given a convex polygon with $n$ sides and perimeter $S$, which has an incircle $\omega$ with radius $R$. A regular polygon with $n$ sides, whose vertices lie on $\omega$, has a perimeter $s$. Determine whether the following inequality holds:
\[ S \ge \frac{2sRn}{\sqrt{4n^2R^2-s^2}}. \]
The lengths of the diagonals of a rhombus are, in inches, two consecutive integers. The area of the rhombus is $210$ sq. in. Find its perimeter, in inches.
Triangle $ABC$ is equilateral with $AB=1$. Points $E$ and $G$ are on $\overline{AC}$ and points $D$ and $F$ are on $\overline{AB}$ such that both $\overline{DE}$ and $\overline{FG}$ are parallel to $\overline{BC}$. Furthermore, triangle $ADE$ and trapezoids $DFGE$ and $FBCG$ all have the same perimeter. What is $DE+FG$?
[asy]
size(180);
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps);
real s=1/2,m=5/6,l=1;
pair A=origin,B=(l,0),C=rotate(60)*l,D=(s,0),E=rotate(60)*s,F=m,G=rotate(60)*m;
draw(A--B--C--cycle^^D--E^^F--G);
dot(A^^B^^C^^D^^E^^F^^G);
label("$A$",A,SW);
label("$B$",B,SE);
label("$C$",C,N);
label("$D$",D,S);
label("$E$",E,NW);
label("$F$",F,S);
label("$G$",G,NW);
[/asy]
$\textbf{(A) }1\qquad
\textbf{(B) }\dfrac{3}{2}\qquad
\textbf{(C) }\dfrac{21}{13}\qquad
\textbf{(D) }\dfrac{13}{8}\qquad
\textbf{(E) }\dfrac{5}{3}\qquad$
A circle with radius $R$ is divided into twelve equal parts. The twelve dividing points are connected with the centre of the circle, producing twelve rays. Starting from one of the dividing points a segment is drawn perpendicular to the next ray in the clockwise sense; from the foot of this perpendicular another perpendicular segment is drawn to the next ray, and the process is continued [i]ad infinitum[/i]. What is the limit of the sum of these segments (in terms of $R$)?
[img]https://cdn.artofproblemsolving.com/attachments/2/6/83705b54ecc817b7d913468cd8467d7b8d9f8f.png[/img]
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 $\angle XAY$.
Nondegenerate $ \triangle ABC$ has integer side lengths, $ BD$ is an angle bisector, $ AD \equal{} 3$, and $ DC \equal{} 8$. What is the smallest possible value of the perimeter?
$ \textbf{(A)}\ 30 \qquad
\textbf{(B)}\ 33 \qquad
\textbf{(C)}\ 35 \qquad
\textbf{(D)}\ 36 \qquad
\textbf{(E)}\ 37$
Let $P$, $Q$, and $R$ be the points where the incircle of a triangle $ABC$ touches the sides $AB$, $BC$, and $CA$, respectively.
Prove the inequality $\frac{BC} {PQ} + \frac{CA} {QR} + \frac{AB} {RP} \geq 6$.