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: 1679

In the rectangle there is a broken line, the neighboring links of which are perpendicular and equal to the smaller side of the rectangle (see the figure). Find the ratio of the sides of the rectangle. [img]https://2.bp.blogspot.com/-QYj53KiPTJ8/XT_mVIw876I/AAAAAAAAKbE/gJ1roU4Bx-kfGVfJxYMAuLE0Ax0glRbegCK4BGAYYCw/s1600/oral%2Bmoscow%2B2016%2B8.9%2Bp2.png[/img]
Let $ABCDV$ be a regular quadrangular pyramid with $V$ as the apex. The plane $\lambda$ intersects the $VA$, $VB$, $VC$ and $VD$ at $M$, $N$, $P$, $Q$ respectively. Find $VQ : QD$, if $VM : MA = 2 : 1$, $VN : NB = 1 : 1$ and $VP : PC = 1 : 2$.
What is the ratio of the area of the shaded square to the area of the large square? (The figure is drawn to scale) [asy] draw((0,0)--(0,4)--(4,4)--(4,0)--cycle); draw((0,0)--(4,4)); draw((0,4)--(3,1)--(3,3)); draw((1,1)--(2,0)--(4,2)); fill((1,1)--(2,0)--(3,1)--(2,2)--cycle,black);[/asy] $ \text{(A)}\ \frac{1}{6}\qquad\text{(B)}\ \frac{1}{7}\qquad\text{(C)}\ \frac{1}{8}\qquad\text{(D)}\ \frac{1}{12}\qquad\text{(E)}\ \frac{1}{16} $
Three bags of jelly beans contain 26, 28, and 30 beans. The ratios of yellow beans to all beans in each of these bags are $50\%$, $25\%$, and $20\%$, respectively. All three bags of candy are dumped into one bowl. Which of the following is closest to the ratio of yellow jelly beans to all beans in the bowl? $\textbf{(A)}\ 31\% \qquad \textbf{(B)}\ 32\% \qquad \textbf{(C)}\ 33\% \qquad \textbf{(D)}\ 35\% \qquad \textbf{(E)}\ 95\%$
What is the maximum number of terms in a geometric progression with common ratio greater than 1 whose entries all come from the set of integers between 100 and 1000 inclusive?
Let $P$ be a convex $2006$-gon. The $1003$ diagonals connecting opposite vertices and the $1003$ lines connecting the midpoints of opposite sides are concurrent, that is, all $2006$ lines have a common point. Prove that the opposite sides of $P$ are parallel and congruent.
In a triangle $ABC$, $\angle C=45$. $AD$ is the altitude of the triangle. $X$ is on $AD$ such that $\angle XBC=90-\angle B$ ($X$ is in the triangle). $AD$ and $CX$ cut the circumcircle of $ABC$ in $M$ and $N$ respectively. if tangent to circumcircle of $ABC$ at $M$ cuts $AN$ at $P$, prove that $P$,$B$ and $O$ are collinear.(25 points) the exam time was 4 hours and 30 minutes.
In an acute-angled triangle $ABC$ with circumcenter $O$, points $P$ and $Q$ are taken on sides $AC$ and $BC$ respectively such that $\frac{AP}{PQ} = \frac{BC}{AB}$ and $\frac{BQ}{PQ} =\frac{AC}{AB}$ . Prove that the points $O, P,Q,C$ lie on a circle.
The extensions of the sides $AB$ and $CD$ of the trapezoid $ABCD$ intersect at point $E$. Denote by $H$ and $G$ the midpoints of $BD$ and $AC$. Find the ratio of the area $AEGH$ to the area $ABCD$.
An angle is drawn on a set of equally spaced parallel lines as shown. The ratio of the area of shaded region $\mathcal{C}$ to the area of shaded region $\mathcal{B}$ is $11/5$. Find the ratio of shaded region $\mathcal{D}$ to the area of shaded region $\mathcal{A}$. [asy] defaultpen(linewidth(0.7)+fontsize(10)); for(int i=0; i<4; i=i+1) { fill((2*i,0)--(2*i+1,0)--(2*i+1,6)--(2*i,6)--cycle, mediumgray); } pair A=(1/3,4), B=A+7.5*dir(-17), C=A+7*dir(10); draw(B--A--C); fill((7.3,0)--(7.8,0)--(7.8,6)--(7.3,6)--cycle, white); clip(B--A--C--cycle); for(int i=0; i<9; i=i+1) { draw((i,1)--(i,6)); } label("$\mathcal{A}$", A+0.2*dir(-17), S); label("$\mathcal{B}$", A+2.3*dir(-17), S); label("$\mathcal{C}$", A+4.4*dir(-17), S); label("$\mathcal{D}$", A+6.5*dir(-17), S);[/asy]
Volume $ A$ equals one fourth of the sum of the volumes $ B$ and $ C$, while volume $ B$ equals one sixth of the sum of the volumes $ C$ and $ A$. Find the ratio of the volume $ C$ to the sum of the volumes $ A$ and $ B$.
Let $M$ be the midpoint of side $AB$ of the triangle $ABC$. Let$P$ be a point on $AB$ between $A$ and $M$, and let $MD$ be drawn parallel to $PC$ and intersecting $BC$ at $D$. If the ratio of the area of the triangle $BPD$ to that of triangle $ABC$ is denoted by $r$, then $\text{(A)}\ \tfrac{1}{2}<r<1\text{ depending upon the position of }P \qquad\\ \text{(B)}\ r=\tfrac{1}{2}\text{ independent of the position of }P\qquad\\ \text{(C)}\ \tfrac{1}{2}\le r<1\text{ depending upon the position of }P \qquad\\ \text{(D)}\ \tfrac{1}{3}<r<\tfrac{2}{3}\text{ depending upon the position of }P \qquad\\ \text{(E)}\ r=\tfrac{1}{3} \text{ independent of the position of }P$
In the figure, $\triangle ABC$ has $\angle A =45^{\circ}$ and $\angle B =30^{\circ}$. A line $DE$, with $D$ on $AB$ and $\angle ADE =60^{\circ}$, divides $\triangle ABC$ into two pieces of equal area. (Note: the figure may not be accurate; perhaps $E$ is on $CB$ instead of $AC$.) The ratio $\frac{AD}{AB}$ is [asy] size((220)); draw((0,0)--(20,0)--(7,6)--cycle); draw((6,6)--(10,-1)); label("A", (0,0), W); label("B", (20,0), E); label("C", (7,6), NE); label("D", (9.5,-1), W); label("E", (5.9, 6.1), SW); label("$45^{\circ}$", (2.5,.5)); label("$60^{\circ}$", (7.8,.5)); label("$30^{\circ}$", (16.5,.5)); [/asy] $ \textbf{(A)}\ \frac{1}{\sqrt{2}} \qquad\textbf{(B)}\ \frac{2}{2+\sqrt{2}} \qquad\textbf{(C)}\ \frac{1}{\sqrt{3}} \qquad\textbf{(D)}\ \frac{1}{\sqrt[3]{6}} \qquad\textbf{(E)}\ \frac{1}{\sqrt[4]{12}} $
(1) $D$ is an arbitary point in $\triangle{ABC}$. Prove that: \[ \frac{BC}{\min{AD,BD,CD}} \geq \{ \begin{array}{c} \displaystyle 2\sin{A}, \ \angle{A}< 90^o \\ \\ 2, \ \angle{A} \geq 90^o \end{array} \] (2)$E$ is an arbitary point in convex quadrilateral $ABCD$. Denote $k$ the ratio of the largest and least distances of any two points among $A$, $B$, $C$, $D$, $E$. Prove that $k \geq 2\sin{70^o}$. Can equality be achieved?
Show that the number of integral-sided right triangles whose ratio of area to semi-perimeter is $p^{m}$, where $p$ is a prime and $m$ is an integer, is $m+1$ if $p=2$ and $2m+1$ if $p \neq 2$.
On the sides $AB$, $BC$, $CD$, $DA$ of the square $ABCD$ points $P, Q, R, T$ are chosen such that $$\frac{AP}{PB}=\frac{BQ}{QC}=\frac{CR}{RD}=\frac{DT}{TA}=\frac12.$$ The segments $AR$, $BT$, $CP$, $DQ$ in the intersection form the quadrilateral $KLMN$ (see figure). [img]https://cdn.artofproblemsolving.com/attachments/f/c/587a2358734c300fe7082c520f90c91f872b49.png[/img] a) Prove that $KLMN$ is a square. b) Find the ratio of the areas of the squares $KLMN$ and $ABCD$. (Alexander Shkolny)
In a triangle $ ABC$ with $ \angle A\equal{}90^{\circ}$, $ D$ is the midpoint of $ BC$, $ F$ that of $ AB$, $ E$ that of $ AF$ and $ G$ that of $ FB$. Segment $ AD$ intersects $ CE,CF$ and $ CG$ in $ P,Q$ and $ R$, respectively. Determine the ratio: $ \frac{PQ}{QR}$.
Given triangle $ ABC$ with base $ AB$ fixed in length and position. As the vertex $ C$ moves on a straight line, the intersection point of the three medians moves on: $ \textbf{(A)}\ \text{a circle} \qquad \textbf{(B)}\ \text{a parabola} \qquad \textbf{(C)}\ \text{an ellipse} \qquad \textbf{(D)}\ \text{a straight line} \qquad \textbf{(E)}\ \text{a curve here not listed}$
In right $ \triangle ABC$ with hypotenuse $ \overline{AB}$, $ AC \equal{} 12$, $ BC \equal{} 35$, and $ \overline{CD}$ is the altitude to $ \overline{AB}$. Let $ \omega$ be the circle having $ \overline{CD}$ as a diameter. Let $ I$ be a point outside $ \triangle ABC$ such that $ \overline{AI}$ and $ \overline{BI}$ are both tangent to circle $ \omega$. The ratio of the perimeter of $ \triangle ABI$ to the length $ AB$ can be expressed in the form $ \displaystyle\frac{m}{n}$, where $ m$ and $ n$ are relatively prime positive integers. Find $ m\plus{}n$.
Let $ABC$ be a triangle with sides of length $a$, $b$ and $c$. Let the bisector of the angle $C$ cut $AB$ in $D$. Prove that the length of $CD$ is \[ \frac{2ab\cos \frac{C}{2}}{a+b}. \]
Give a geometric series $(a_n)$ with common ratio of $q$, let $b_1=a_1+a_2+a_3,b_2=a_4+a_5+a_6,\cdots,b_n=a_{3n}+a_{3n+1}+a_{3n+2}$, then sequence $(b_n)$ $\text{(A)}$ is an arithmetic sequence $\text{(B)}$ is a geometric series with common ratio of $q$ $\text{(C)}$ is a geometric series with common ratio of $q^3$ $\text{(D)}$ is neither an arithmetic sequence nor a geometric series
If $p$ is the perimeter of an equilateral triangle inscribed in a circle, the area of the circle is: $ \textbf{(A)}\ \frac{\pi p^2}{3} \qquad\textbf{(B)}\ \frac{\pi p^2}{9}\qquad\textbf{(C)}\ \frac{\pi p^2}{27}\qquad\textbf{(D)}\ \frac{\pi p^2}{81} \qquad\textbf{(E)}\ \frac{\pi p^2 \sqrt3}{27} $
The sum of two positive numbers is $5$ times their difference. What is the ratio of the larger number to the smaller? $\textbf{(A) }\dfrac54\qquad\textbf{(B) }\dfrac32\qquad\textbf{(C) }\dfrac95\qquad\textbf{(D) }2\qquad\textbf{(E) }\dfrac52$
Given an acute triangle $ABC$ with circumcenter $O$ and orthocenter $H$. Let $K$ be a point inside $ABC$ which is not $O$ nor $H$. Point $L$ and $M$ are located outside the triangle $ABC$ such that $AKCL$ and $AKBM$ are parallelogram. At last, let $BL$ and $CM$ intersects at $N$, and let $J$ be the midpoint of $HK$. Show that $KONJ$ is also a parallelogram. [i]Raja Oktovin, Pekanbaru[/i]
Points $D$ and $E$ lie on the lines $BC$ and $AC$ respectively so that $B$ is between $C$ and $D$, $C$ is between $A$ and $E$, $BC = BD$ and $\angle BAD = \angle CDE$. It is known that the ratio of the perimeters of the triangles $ABC$ and $ADE$ is $2$. Find the ratio of the areas of these triangles.