Found problems: 1342
Rectangle $ABCD$ has sides $\overline {AB}$ of length 4 and $\overline {CB}$ of length 3. Divide $\overline {AB}$ into 168 congruent segments with points $A=P_0, P_1, \ldots, P_{168}=B$, and divide $\overline {CB}$ into 168 congruent segments with points $C=Q_0, Q_1, \ldots, Q_{168}=B$. For $1 \le k \le 167$, draw the segments $\overline {P_kQ_k}$. Repeat this construction on the sides $\overline {AD}$ and $\overline {CD}$, and then draw the diagonal $\overline {AC}$. Find the sum of the lengths of the 335 parallel segments drawn.
A $(3n + 1) \times (3n + 1)$ table $(n \in \mathbb{N})$ is given. Prove that deleting any one of its squares yields a shape cuttable into pieces of the following form and its rotations: ''L" shape formed by cutting one square from a $2 \times 2$ squares.
A rectangle of size $ m \times n$ has been filled completely by trominoes (a tromino is an L-shape consisting of 3 unit squares).
There are four ways to place a tromino
1st way: let the "corner" of the L be on top left
2nd way: let the "corner" of the L be on top right
3rd way: let the "corner" of the L be on bottom left
4th way: let the "corner" of the L be on bottom right
Prove that the difference between the number of trominoes placed in the 1st and the 4th way is divisible by $ 3$.
Prove that if the length and breadth of a rectangle are both odd integers, then there does not exist a point $P$ inside the rectangle such that each of the distances from $P$ to the 4 corners of the rectangle is an integer.
All the vertices of a convex pentagon are on lattice points. Prove that the area of the pentagon is at least $\frac{5}{2}$.
[i]Bogdan Enescu[/i]
Consider a right circular cylinder with radius $ r$ of the base, hight $ h$. Find the volume of the solid by revolving the cylinder about a diameter of the base.
In the rectangle $ABCD, BC = 5, EC = 1/3 CD$ and $F$ is the point where $AE$ and $BD$ are cut. The triangle $DFE$ has area $12$ and the triangle $ABF$ has area $27$. Find the area of the quadrilateral $BCEF$ .
[img]https://1.bp.blogspot.com/-4w6e729AF9o/XNY9hqHaBaI/AAAAAAAAKL0/eCaNnWmgc7Yj9uV4z29JAvTcWCe21NIMgCK4BGAYYCw/s400/may%2B2011%2Bl1.png[/img]
$ABCD$ is a cyclic quadrilateral with $AC \perp BD$; $AC$ meets $BD$ at $E$. Prove that \[ EA^2 + EB^2 + EC^2 + ED^2 = 4 R^2 \]
where $R$ is the radius of the circumscribing circle.
Can we find $N$ such that all $m\times n$ rectangles with $m,n>N$ can be tiled with $4\times6$ and $5\times7$ rectangles?
Let $ABCD$ be a cyclic convex quadrilateral and let $r_a,r_b,r_c,r_d$ be the radii of the circles inscribed in the triangles $BCD, ACD, ABD, ABC$, respectively. Prove that $r_a+r_c=r_b+r_d$.
Prove that a square of side 2.1 units can be completely covered by seven squares of side 1 unit.
Extra: Try to prove that 7 is the minimal amount.
Find the volume of the solid generated by a rotation of the region enclosed by the curve $y=x^3-x$ and the line $y=x$ about the line $y=x$ as the axis of rotation.
There are two paper figures: an equilateral triangle and a rectangle. The height of rectangle is equal to the height of the triangle and the base of the rectangle is equal to the base of the triangle. Divide the triangle into three parts and the rectangle into two, using straight cuts, so that with the five pieces can be assembled, without gaps or overlays, a equilateral triangle. To assemble the figure, each part can be rotated and / or turned around.
We are given a triangle $ABC$ and three rectangles $R_1,R_2,R_3$ with sides parallel to two fixed perpendicular directions and such that their union covers the sides $AB,BC$, and $CA$; i.e., each point on the perimeter of $ABC$ is contained in or on at least one of the rectangles. Prove that all points inside the triangle are also covered by the union of $R_1,R_2,R_3.$
a) Suppose that $ RBR'B'$ is a convex quadrilateral such that vertices $ R$ and $ R'$ have red color and vertices $ B$ and $ B'$ have blue color. We put $ k$ arbitrary points of colors blue and red in the quadrilateral such that no four of these $ k\plus{}4$ point (except probably $ RBR'B'$) lie one a circle. Prove that exactly one of the following cases occur?
1. There is a path from $ R$ to $ R'$ such that distance of every point on this path from one of red points is less than its distance from all blue points.
2. There is a path from $ B$ to $ B'$ such that distance of every point on this path from one of blue points is less than its distance from all red points.
We call these two paths the blue path and the red path respectively.
Let $ n$ be a natural number. Two people play the following game. At each step one player puts a point in quadrilateral satisfying the above conditions. First player only puts red point and second player only puts blue points. Game finishes when every player has put $ n$ points on the plane. First player's goal is to make a red path from $ R$ to $ R'$ and the second player's goal is to make a blue path from $ B$ to $ B'$.
b) Prove that if $ RBR'B'$ is rectangle then for each $ n$ the second player wins.
c) Try to specify the winner for other quadrilaterals.
Prove that every rectangle circumscribed by a square is itself a square.
(A rectangle is circumscribed by a square if there is exactly one corner point of the square on each side of the rectangle.)
We have a rectangle with it sides being a mirror.A light Ray enters from one of the corners of the rectangle and after being reflected several times enters to the opposite corner it started.Prove that at some time the light Ray passed the center of rectangle(Intersection of diagonals.)
Circles centered at $ A$ and $ B$ each have radius $ 2$, as shown. Point $ O$ is the midpoint of $ \overline{AB}$, and $ OA \equal{} 2\sqrt {2}$. Segments $ OC$ and $ OD$ are tangent to the circles centered at $ A$ and $ B$, respectively, and $ EF$ is a common tangent. What is the area of the shaded region $ ECODF$?
[asy]unitsize(6mm);
defaultpen(fontsize(10pt)+linewidth(.8pt));
dotfactor=3;
pair O=(0,0);
pair A=(-2*sqrt(2),0);
pair B=(2*sqrt(2),0);
pair G=shift(0,2)*A;
pair F=shift(0,2)*B;
pair C=shift(A)*scale(2)*dir(45);
pair D=shift(B)*scale(2)*dir(135);
pair X=A+2*dir(-60);
pair Y=B+2*dir(240);
path P=C--O--D--Arc(B,2,135,90)--G--Arc(A,2,90,45)--cycle;
fill(P,gray);
draw(Circle(A,2));
draw(Circle(B,2));
dot(A);
label("$A$",A,W);
dot(B);
label("$B$",B,E);
dot(C);
label("$C$",C,W);
dot(D);
label("$D$",D,E);
dot(G);
label("$E$",G,N);
dot(F);
label("$F$",F,N);
dot(O);
label("$O$",O,S);
draw(G--F);
draw(C--O--D);
draw(A--B);
draw(A--X);
draw(B--Y);
label("$2$",midpoint(A--X),SW);
label("$2$",midpoint(B--Y),SE);[/asy]$ \textbf{(A)}\ \frac {8\sqrt {2}}{3}\qquad \textbf{(B)}\ 8\sqrt {2} \minus{} 4 \minus{} \pi \qquad \textbf{(C)}\ 4\sqrt {2}$
$ \textbf{(D)}\ 4\sqrt {2} \plus{} \frac {\pi}{8}\qquad \textbf{(E)}\ 8\sqrt {2} \minus{} 2 \minus{} \frac {\pi}{2}$
Three rectangles, each of area $6$ square inches, are placed inside a $4$ in. by $4$ in. square. Prove that, no matter how the three rectangles are shaped and arranged, (for example, like in the picture below), one can find two of them which have a common area of at least $2/3$ square inches.
The pentagon below has three right angles. Find its area.
[asy]
size(150);
defaultpen(linewidth(1));
draw((4,10)--(0,10)--origin--(10,0)--(10,2)--cycle);
label("4",(2,10),N);
label("10",(0,5),W);
label("10",(5,0),S);
label("2",(10,1),E);
label("10",(7,6),NE);
[/asy]
Each point of a plane is colored in one of three colors: red, black and blue. Prove that there exists a rectangle in this plane whose vertices all have the same color.
Given $2009 \times 4018$ rectangular board. Frame is a rectangle $n \times n$ or $n \times(n + 2)$ for $ ( n \geq 3 )$ without all cells which don’t have any common points with boundary of rectangle. Rectangles $1\times1,1\times 2,1\times 3$ and $ 2\times 4$ are also frames. Two players by turn paint all cells of some frame that has no painted cells yet. Player that can't make such move loses. Who has a winning strategy?
Let $ABCD$ be a rectangle of center $O$ in the plane $\alpha$, and let $V \notin\alpha$ be a point in space such that $V O \perp \alpha$. Let $A' \in (V A)$, $B'\in (V B)$, $C'\in (V C)$, $D'\in (V D)$ be four points, and let $M$ and $N$ be the midpoints of the segments $A'C'$ and $B'D'$. .Prove that $MN \parallel \alpha$ if and only if $V , A', B', C', D'$ all lie on a sphere.
An interior point in a rectangle is connected by line segments to the midpoints of its four sides. Thus four domains (polygons) with the areas $a, b, c$ and $d$ appear (see the figure). Prove that $a + c = b + d$.
[img]https://1.bp.blogspot.com/-BipDNHELjJI/XzcCa68P3HI/AAAAAAAAMXY/H2Iqya9VItMLXrRqsdyxHLTXCAZ02nEtgCLcBGAsYHQ/s0/2002%2BMohr%2Bp1.png[/img]
Let $ABC$ a triangle and $P$ a point such that the triangles $PAB, PBC, PCA$ have the same area and the same perimeter. Prove that if:
a) $P$ is in the interior of the triangle $ABC$ then $ABC$ is equilateral.
b) $P$ is in the exterior of the triangle $ABC$ then $ABC$ is right angled triangle.