Found problems: 1342
A rectangle has area $A \text{ cm}^2$ and perimeter $P \text{ cm}$, where $A$ and $P$ are positive integers. Which of the following numbers cannot equal $A+P$?
$ \textbf{(A) }100\qquad\textbf{(B) }102\qquad\textbf{(C) }104\qquad\textbf{(D) }106\qquad\textbf{(E) }108 $
A square with side length $8$ is cut in half, creating two congruent rectangles. What are the dimensions of one of these rectangles?
$ \textbf{(A)}\ 2\text{ by }4
\qquad\textbf{(B)}\ 2\text{ by }6
\qquad\textbf{(C)}\ 2\text{ by }8
\qquad\textbf{(D)}\ 4\text{ by }4
\qquad\textbf{(E)}\ 4\text{ by }8
$
If $ABCD$ is a rectangle and $P$ is a point inside it such that $AP=33, BP=16, DP=63$.
Find $CP$.
The width of rectangle $ABCD$ is twice its height, and the height of rectangle $EFCG$ is twice its width. The point $E$ lies on the diagonal $BD$. Which fraction of the area of the big rectangle is that of the small one?
[img]https://1.bp.blogspot.com/-aeqefhbBh5E/XzcBjhgg7sI/AAAAAAAAMXM/B0qSgWDBuqc3ysd-mOitP1LarOtBdJJ3gCLcBGAsYHQ/s0/2004%2BMohr%2Bp1.png[/img]
Given a rectangle $ABCD$ with $AB = a$ and $BC = b$. Point $O$ is the intersection of the two diagonals. Extend the side of the $BA$ so that $AE = AO$, also extend the diagonal of $BD$ so that $BZ = BO.$ Assume that triangle $EZC$ is equilateral. Prove that
(i) $b = a\sqrt3$
(ii) $EO$ is perpendicular to $ZD$
Alice and Bob play the following game. It begins with a set of $1000$ $1\times 2$ rectangles. A [i]move[/i] consists of choosing two rectangles (a rectangle may consist of one or several $1\times 2$ rectangles combined together) that share a common side length and combining those two rectangles into one rectangle along those sides sharing that common length. The first player who cannot make a move loses. Alice moves first. Describe a winning strategy for Bob.
Two equal rectangles of area $10$ are arranged as follows. Find the area of the gray rectangle.
[img]https://cdn.artofproblemsolving.com/attachments/7/1/112b07530a2ef42e5b2cf83a2cb9fb11dfc9e6.png[/img]
You have a paper pentagon, $ABCDE$, such that $AB = BC = 3$ cm, $CD = DE= 5$ cm, $EA = 4$ cm, $\angle ABC = 100^o$ ,$ \angle CDE = 80^o$. You have to divide the pentagon into four triangles, by three straight cuts, so that with the four triangles assemble a rectangle, without gaps or overlays. (The triangles can be rotated and / or turned around.)
If $ABCD$ is a cyclic quadrilateral, then prove that the incenters of the triangles $ABC$, $BCD$, $CDA$, $DAB$ are the vertices of a rectangle.
A triangle $KLM$ is given in the plane together with a point $A$ lying on the half-line opposite to $KL$. Construct a rectangle $ABCD$ whose vertices $B, C$ and $D$ lie on the lines $KM, KL$ and $LM$, respectively. (We allow the rectangle to be a square.)
A polygon can be dissected into $100$ rectangles, but it cannot be dissected into $99$ rectangles. Prove that this polygon cannot be dissected into $100$ triangles.
Let $ABC$ be a triangle and $D$ the foot of the altitude from $A$. Let $E$ and $F$ lie on a line passing through $D$ such that $AE$ is perpendicular to $BE$, $AF$ is perpendicular to $CF$, and $E$ and $F$ are different from $D$. Let $M$ and $N$ be the midpoints of the segments $BC$ and $EF$, respectively. Prove that $AN$ is perpendicular to $NM$.
Two circles with different radii intersect in two points $A$ and $B$. Let the common tangents of the two circles be $MN$ and $ST$ such that $M,S$ lie on the first circle, and $N,T$ on the second. Prove that the orthocenters of the triangles $AMN$, $AST$, $BMN$ and $BST$ are the four vertices of a rectangle.
A convex, closed figure lies inside a given circle. The figure is seen from every point of the circumference at a right angle (that is, the two rays drawn from the point and supporting the convex figure are perpendicular). Prove that the center of the circle is a center of symmetry of the figure.
Given $120$ unit squares arbitrarily situated in the $20\times 25$ rectangle. Prove that you can place a circle with the unit diameter without intersecting any of the squares.
The hexagon has five $90^o$ angles and one $270^o$ angle (see picture). Use a straight-line ruler to divide it into two equal-sized polygons.
[img]https://cdn.artofproblemsolving.com/attachments/d/8/cdd4df68644bb8e04adbe1b265039b82a5382b.png[/img]
In a $2^n \times 2^n$ square with $n$ positive integer is covered with at least two non-overlapping rectangle pieces with integer dimensions and a power of two as surface. Prove that two rectangles of the covering have the same dimensions (Two rectangles have the same dimensions as they have the same width and the same height, wherein they, not allowed to be rotated.)
The diagram below shows a rectangle with side lengths $4$ and $8$ and a square with side length $5$. Three vertices of the square lie on three different sides of the rectangle, as shown. What is the area of the region inside both the square and the rectangle?
[asy]
size(5cm);
filldraw((4,0)--(8,3)--(8-3/4,4)--(1,4)--cycle,mediumgray);
draw((0,0)--(8,0)--(8,4)--(0,4)--cycle,linewidth(1.1));
draw((1,0)--(1,4)--(4,0)--(8,3)--(5,7)--(1,4),linewidth(1.1));
label("$4$", (8,2), E);
label("$8$", (4,0), S);
label("$5$", (3,11/2), NW);
draw((1,.35)--(1.35,.35)--(1.35,0),linewidth(.4));
draw((5,7)--(5+21/100,7-28/100)--(5-7/100,7-49/100)--(5-28/100,7-21/100)--cycle,linewidth(.4));
[/asy]
$\textbf{(A) } 15\dfrac{1}{8} \qquad \textbf{(B) } 15\dfrac{3}{8} \qquad \textbf{(C) } 15\dfrac{1}{2} \qquad \textbf{(D) } 15\dfrac{5}{8} \qquad \textbf{(E) } 15\dfrac{7}{8}$
Let $n\geq 3$ be a positive integer. Inside a $n\times n$ array there are placed $n^2$ positive numbers with sum $n^3$. Prove that we can find a square $2\times 2$ of 4 elements of the array, having the sides parallel with the sides of the array, and for which the sum of the elements in the square is greater than $3n$.
[i]Radu Gologan[/i]
Given a rectangle $ABCD$ such that $AB = b > 2a = BC$, let $E$ be the midpoint of $AD$. On a line parallel to $AB$ through point $E$, a point $G$ is chosen such that the area of $GCE$ is
$$(GCE)= \frac12 \left(\frac{a^3}{b}+ab\right)$$
Point $H$ is the foot of the perpendicular from $E$ to $GD$ and a point $I$ is taken on the diagonal $AC$ such that the triangles $ACE$ and $AEI$ are similar. The lines $BH$ and $IE$ intersect at $K$ and the lines $CA$ and $EH$ intersect at $J$. Prove that $KJ \perp AB$.
Let $ABC$ be an acute-angled triangle and $D$ a point on the altitude through $C$. Let $E$, $F$, $G$ and $H$ be the midpoints of the segments $AD$, $BD$, $BC$ and $AC$. Show that $E$, $F$, $G$, and $H$ form a rectangle.
(G. Anegg, Innsbruck)
Find the number of rectangles that can be formed inside a fixed regular dodecagon ($12$-gon) where each side of the rectangle lies on either a side or a diagonal of the dodecagon. The diagram below shows three of those rectangles.
[asy]
unitsize(40);
real r = pi/6;
pair A1 = (cos(r),sin(r));
pair A2 = (cos(2r),sin(2r));
pair A3 = (cos(3r),sin(3r));
pair A4 = (cos(4r),sin(4r));
pair A5 = (cos(5r),sin(5r));
pair A6 = (cos(6r),sin(6r));
pair A7 = (cos(7r),sin(7r));
pair A8 = (cos(8r),sin(8r));
pair A9 = (cos(9r),sin(9r));
pair A10 = (cos(10r),sin(10r));
pair A11 = (cos(11r),sin(11r));
pair A12 = (cos(12r),sin(12r));
draw(A1--A2--A3--A4--A5--A6--A7--A8--A9--A10--A11--A12--cycle);
filldraw(A2--A1--A8--A7--cycle, mediumgray, linewidth(1.2));
draw(A4--A11);
draw(0.365*A3--0.365*A12, linewidth(1.2));
dot(A1);
dot(A2);
dot(A3);
dot(A4);
dot(A5);
dot(A6);
dot(A7);
dot(A8);
dot(A9);
dot(A10);
dot(A11);
dot(A12);
[/asy]
Cut $2003$ disjoint rectangles from an acute-angled triangle $ABC$, such that any of them has a parallel side to $AB$ and the sum of their areas is maximal.
[asy]
draw((-2,1)--(2,1)--(2,-1)--(-2,-1)--cycle);
draw((0,0)--(0,-1)--(-2,-1)--(-2,0)--cycle);
label("$F$",(0,0),E);
label("$A$",(-2,1),W);
label("$B$",(2,1),E);
label("$C$", (2,-1),E);
label("$D$",(-2,-1),WSW);
label("$E$",(-2,0),W);
label("$G$",(0,-1),S);
//Credit to TheMaskedMagician for the diagram
[/asy]
If rectangle $ABCD$ has area $72$ square meters and $E$ and $G$ are the midpoints of sides $AD$ and $CD$, respectively, then the area of rectangle $DEFG$ in square meters is
$\textbf{(A) }8\qquad\textbf{(B) }9\qquad\textbf{(C) }12\qquad\textbf{(D) }18\qquad\textbf{(E) }24$
Given a rectangle $ABCD$, side $AB$ is longer than side $BC$. Find all the points $P$ of the side line $AB$ from which the sides $AD$ and $DC$ are seen from the point $P$ at an equal angle (i.e. $\angle APD = \angle DPC$)