This website contains problems from math contests. Problems and corresponding tags were obtained from the Art of Problem Solving website.

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Found problems: 1342

Ali Barber, the carpet merchant, has a rectangular piece of carpet whose dimensions are unknown. Unfortunately, his tape measure is broken and he has no other measuring instruments. However, he finds that if he lays it flat on the floor of either of his storerooms, then each corner of the carpet touches a different wall of that room. He knows that the sides of the carpet are integral numbers of feet and that his two storerooms have the same (unknown) length, but widths of 38 feet and 50 feet respectively. What are the carpet dimensions?
Decide whether any rectangle that can be covered by 25 circles of radius 2 can also be covered by 100 circles of radius 1.
A flea jumps in a straight numbered line. It jumps first from point $0$ to point $1$. Afterwards, if its last jump was from $A$ to $B$, then the next jump is from $B$ to one of the points $B + (B - A) - 1$, $B + (B - A)$, $B + (B-A) + 1$. Prove that if the flea arrived twice at the point $n$, $n$ positive integer, then it performed at least $\lceil 2\sqrt n\rceil$ jumps.
Is it possible to divide a $6 \times 6$ chessboard into $18$ rectangles, each either $1 \times 2$ or $2 \times 1$, and to draw exactly one diagonal on each rectangle such that no two of these diagonals have a common endpoint? (A Shapovalov)
We call a rectangle of the size $1 \times 2$ a domino. Rectangle of the $2 \times 3$ removing two opposite (under center of rectangle) corners we call tetramino. These figures can be rotated. It requires to tile rectangle of size $2008 \times 2010$ by using dominoes and tetraminoes. What is the minimal number of dominoes should be used?
For a rectangle $R$ with integral side lengths, denote by $D(a, b)$ the number of ways of covering $R$ by congruent rectangles with integral side lengths formed by a family of cuts parallel to one side of $R$. Determine the perimeter $P$ of the rectangle $R$ for which $\frac{D(a,b)}{a+b}$ is maximal.
A rectangular table with 2001 rows and 2002 columns is partitioned into $1\times 2$ rectangles. It is known that any other partition of the table into $1\times 2$ rectangles contains a rectangle belonging to the original partition. Prove that the original partition contains two successive columns covered by 2001 horizontal rectangles. [i]Proposed by S. Volchenkov[/i]
[b]a)[/b] Show that if four distinct complex numbers have the same absolute value and their sum vanishes, then they represent a rectangle. [b]b)[/b] Let $ x,y,z,t $ be four real numbers, and $ k $ be an integer. Prove the following implication: $$ \sum_{j\in\{ x,y,z,t\}} \sin j = 0 = \sum_{j\in\{ x,y,z,t\}} \cos j\implies \sum_{j\in\{ x,y,z,t\}} \sin (1+2n)j. $$
[asy] //rectangles above problem statement size(15cm); for(int i=0;i<5;++i){ draw((6*i-14,-1.2)--(6*i-14,1.2)--(6*i-10,1.2)--(6*i-10,-1.2)--cycle); } label("$A$", (-12,2.25)); label("$B$", (-6,2.25)); label("$C$", (0,2.25)); label("$D$", (6,2.25)); label("$E$", (12,2.25)); //top numbers label("$1$", (-12,1.25),dir(-90)); label("$0$", (-6,1.25),dir(-90)); label("$8$", (0,1.25),dir(-90)); label("$5$", (6,1.25),dir(-90)); label("$2$", (12,1.25),dir(-90)); //bottom numbers label("$9$", (-12,-1.25),dir(90)); label("$6$", (-6,-1.25),dir(90)); label("$2$", (0,-1.25),dir(90)); label("$8$", (6,-1.25),dir(90)); label("$0$", (12,-1.25),dir(90)); //left numbers label("$4$", (-14,0),dir(0)); label("$1$", (-8,0),dir(0)); label("$3$", (-2,0),dir(0)); label("$7$", (4,0),dir(0)); label("$9$", (10,0),dir(0)); //right numbers label("$6$", (-10,0),dir(180)); label("$3$", (-4,0),dir(180)); label("$5$", (2,0),dir(180)); label("$4$", (8,0),dir(180)); label("$7$", (14,0),dir(180)); [/asy] Each of the sides of the five congruent rectangles is labeled with an integer, as shown above. These five rectangles are placed, without rotating or reflecting, in positions $I$ through $V$ so that the labels on coincident sides are equal. [asy] //diagram below problem statement size(7cm); for(int i=-3;i<=1;i+=2){ for(int j=-1;j<=0;++j){ if(i==1 && j==-1) continue; draw((i,j)--(i+2,j)--(i+2,j-1)--(i,j-1)--cycle); }} label("$I$",(-2,-0.5)); label("$II$",(0,-0.5)); label("$III$",(2,-0.5)); label("$IV$",(-2,-1.5)); label("$V$",(0,-1.5)); [/asy] Which of the rectangles is in position $I$? $\textbf{(A)} \ A \qquad \textbf{(B)} \ B \qquad \textbf{(C)} \ C \qquad \textbf{(D)} \ D \qquad \textbf{(E)} \ E$
For a real number $ a$, let $ \lfloor a \rfloor$ denominate the greatest integer less than or equal to $ a$. Let $ \mathcal{R}$ denote the region in the coordinate plane consisting of points $ (x,y)$ such that \[\lfloor x \rfloor ^2 \plus{} \lfloor y \rfloor ^2 \equal{} 25.\] The region $ \mathcal{R}$ is completely contained in a disk of radius $ r$ (a disk is the union of a circle and its interior). The minimum value of $ r$ can be written as $ \tfrac {\sqrt {m}}{n}$, where $ m$ and $ n$ are integers and $ m$ is not divisible by the square of any prime. Find $ m \plus{} n$.
What is the perimeter of a rectangle of area $32$ inscribed in a circle of radius $4$?
Let $AB CD$ be a rectangle with $AB=1$. If $m ( \angle BDC) = 82^o30'$, compute the length of$ BD$ and the cosine of $82^o30'$.
A collection of cardboard circles, each with a diameter of at most $1$, lie on a $5\times 8$ table without overlapping or overhanging the edge of the table. A cardboard circle of diameter $2$ is added to the collection. Prove that this new collection of cardboard circles can be placed on a $7\times 7$ table without overlapping or overhanging the edge.
Prove that for any real $r>0$, one can cover the circumference of a $1\times r$ rectangle with non-intersecting disks of unit radius.
The diagonals $AC,BD$ of the quadrilateral $ABCD$ intersect at $E$. Let $M,N$ be the midpoints of $AB,CD$ respectively. Let the perpendicular bisectors of the segments $AB,CD$ meet at $F$. Suppose that $EF$ meets $BC,AD$ at $P,Q$ respectively. If $MF\cdot CD=NF\cdot AB$ and $DQ\cdot BP=AQ\cdot CP$, prove that $PQ\perp BC$.
Suppose $l(M, XYZ)$ is a Simson line of the triangle $XYZ$ that passes through $M$. Suppose $ABCDEF$ is a cyclic hexagon such that $l(A, BDF), l(B, ACE), l(D, ABF), l(E, ABC)$ intersect at a single point. Prove that $CDEF$ is a rectangle. [color=blue]Should the first sentence read: Suppose $l(M, XYZ)$ is a Simson line of the triangle $XYZ$ [u]with respect to[/u] $M$. ? Since it appears weird that a Simson line that passes a point is to be constructed. However, this is Unsolved after all, so I'm not sure.[/color]
In the figure, $ABCD$ is a square of side length 1. The rectangles $JKHG$ and $EBCF$ are congruent. What is $BE$? [asy] unitsize(150); pair A,B,C,D,E,F,G,H,J,K; A=(1,0); B=(0,0); C=(0,1); D=(1,1); draw(A--B--C--D--A); E=(2-sqrt(3),0); F=(2-sqrt(3),1); draw(E--F); G=(1,sqrt(3)/2); H=(2.5-sqrt(3),1); K=(2-sqrt(3),1-sqrt(3)/2); J=(0.5,0); draw(G--H--K--J--G); label("$A$",A,SE); label("$B$",B,SW); label("$C$",C,NW); label("$D$",D,NE); label("$E$",E,S); label("$F$",F,N); label("$G$",G,E); label("$H$",H,N); label("$K$",K,W); label("$J$",J,S); [/asy] $ \textbf{(A) }\dfrac{1}{2}(\sqrt{6}-2)\qquad\textbf{(B) }\dfrac{1}{4}\qquad\textbf{(C) }2-\sqrt{3}\qquad\textbf{(D) }\dfrac{\sqrt{3}}{6}\qquad\textbf{(E) }1-\dfrac{\sqrt{2}}{2} $
One container of paint is exactly enough to cover the inside of an old rectangle which is three times as long as it is wide. If we make a new rectangle by shortening the old rectangle by $18$ feet and widening it by $8$ feet as shown below, one container of paint is also exactly enough to cover the inside of the new rectangle. Find the length in feet of the perimeter of the new rectangle. [asy] size(250); defaultpen(linewidth(0.8)); draw((-2,0)--(-2,5)--(13,5)--(13,0)--cycle^^(16,-1)--(16,6)--(27,6)--(27,-1)--cycle^^(9,5)--(9,0)^^(16,4)--(27,4)); path rect1=(13,5)--(13,0)--(9,0)--(9,5)--cycle,rect2=(16,6)--(16,4)--(27,4)--(27,6)--cycle; fill(rect1,lightgray); fill(rect2,lightgray); draw(rect1^^rect2); [/asy]
In parallelogram $ ABCD$, point $ M$ is on $ \overline{AB}$ so that $ \frac{AM}{AB} \equal{} \frac{17}{1000}$ and point $ N$ is on $ \overline{AD}$ so that $ \frac{AN}{AD} \equal{} \frac{17}{2009}$. Let $ P$ be the point of intersection of $ \overline{AC}$ and $ \overline{MN}$. Find $ \frac{AC}{AP}$.
A rectangle, $HOMF$, has sides $HO=11$ and $OM=5$. A triangle $\Delta ABC$ has $H$ as orthocentre, $O$ as circumcentre, $M$ be the midpoint of $BC$, $F$ is the feet of altitude from $A$. What is the length of $BC$ ? [asy] unitsize(0.3 cm); pair F, H, M, O; F = (0,0); H = (0,5); O = (11,5); M = (11,0); draw(H--O--M--F--cycle); label("$F$", F, SW); label("$H$", H, NW); label("$M$", M, SE); label("$O$", O, NE); [/asy]
Let $ ABC$ be an acute-angled triangle such that there exist points $ D,E,F$ on side $ BC,CA,AB$, respectively such that the inradii of triangle $ AEF,BDF,CDE$ are all equal to $ r_0$. If the inradii of triangle $ DEF$ and $ ABC$ are $ r$ and $ R$, respectively, prove that \[ r\plus{}r_0\equal{}R.\] [i]Soewono, Bandung[/i]
The equiangular convex hexagon $ ABCDEF$ has $ AB \equal{} 1$, $ BC \equal{} 4$, $ CD \equal{} 2$, and $ DE \equal{} 4$. The area of the hexagon is $ \textbf{(A)}\ \frac{15}{2}\sqrt{3}\qquad \textbf{(B)}\ 9\sqrt{3}\qquad \textbf{(C)}\ 16\qquad \textbf{(D)}\ \frac{39}{4}\sqrt{3}\qquad \textbf{(E)}\ \frac{43}{4}\sqrt{3}$
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] $\textbf{(A) } 15\dfrac{1}{8} \qquad \textbf{(B) } 15\dfrac{3}{8} \qquad \textbf{(C) } 15\dfrac{1}{2} \qquad \textbf{(D) } 15\dfrac{5}{8} \qquad \textbf{(E) } 15\dfrac{7}{8}$
Point $K$ is marked on the diagonal $AC$ in rectangle $ABCD$ so that $CK = BC$. On the side $BC$, point $M$ is marked so that $KM = CM$. Prove that $AK + BM = CM$.
Find the geometric locus of the center of a rectangle whose vertices lie on the perimeter of a given triangle.