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: 698

Determine whether does exists a triangle with area $2004$ with his sides positive integers.
We are given convex quadrilateral $ABCD$. Each of its sides is divided into $N$ line segments of equal length. The points of division of side $AB$ are connected with the points of division of side $CD$ by straight lines (which we call the first set of straight lines), and the points of division of side BC are connected with the points of division of side $DA$ by straight lines (which we call the second set of straight lines) as shown in the diagram, which illustrates the case $N = 4$. This forms $N^2$ smaller quadrilaterals. From these we choose $N$ quadrilaterals in such a way that any two are at least divided by one line from the first set and one line from the second set. Prove that the sum of the areas of these chosen quadrilaterals is equal to the area of $ABCD$ divided by $N$. (A Andjans, Riga) [img]http://4.bp.blogspot.com/-8Qqk4r68nhU/XVco29-HzzI/AAAAAAAAKgo/UY8mXxg7tD0OrS6bEnoAw7Vuf31BuOE8wCK4BGAYYCw/s1600/TOT%2B1980%2BSpring%2BJ4.png[/img]
Let $a$ be a real number. 1) Prove that there is a triangle with side lengths $\sqrt{a^2-a + 1}$, $\sqrt{a^2+a + 1}$, and $\sqrt{4a^2 + 3}$. 2) Prove that the area of this triangle does not depend on $a$.
A point $O$ is chosen inside the square $ABCD$. The square $A'B'C'D'$ is the image of the square $ABCD$ under the homothety with center at point $O$ and coefficient $k> 1$ (points $A', B', C', D' $ are images of points $A, B, C, D$ respectively). Prove that the sum of the areas of the quadrilaterals $A'ABB'$ and $C'CDD'$ is equal to the sum of the areas quadrilaterals $B'BCC'$ and $D'DAA'$.
Two perpendiculars are drawn from the midpoints of each side of the acute-angle triangle to two other sides. Those six segments make hexagon. Prove that the hexagon area is a half of the triangle area.
The area of the triangle $ABC$ shown in the figure is $1$ unit. Points $D$ and $E$ lie on sides $AC$ and $BC$ respectively, and also are its ''one third'' points closer to $C$. Let $F$ be that $AE$ and $G$ are the midpoints of segment $BD$. What is the area of the marked quadrilateral $ABGF$? [img]https://cdn.artofproblemsolving.com/attachments/4/e/305673f429c86bbc58a8d40272dd6c9a8f0ab2.png[/img]
Mari and Juri ordered a round pizza. Juri cut the pizza into four pieces by two straight cuts, none of which passed through the centre point of the pizza. Mari can choose two pieces not aside of these four, and Juri gets the rest two pieces. Prove that if Mari chooses the piece that covers the centre point of the pizza, she will get more pizza than Juri.
From a paper quadrilateral like the one in the figure, you have to cut out a new quadrilateral whose area is equal to half the area of the original quadrilateral.You can only bend one or more times and cut by some of the lines of the folds. Describe the folds and cuts and justify that the area is half. [img]https://2.bp.blogspot.com/-btvafZuTvlk/XNY8nba0BmI/AAAAAAAAKLo/nm4c21A1hAIK3PKleEwt6F9cd6zv4XffwCK4BGAYYCw/s400/may%2B2012%2Bl1.png[/img]
For any pair $(x,y)$ of real numbers, a sequence $(a_{n}(x,y))$ is defined as follows: $$a_{0}(x,y)=x, \;\;\;\; a_{n+1}(x,y) =\frac{a_{n}(x,y)^{2} +y^2 }{2} \;\, \text{for}\, n\geq 0$$ Find the area of the region $\{(x,y)\in \mathbb{R}^{2} \, |\, (a_{n}(x,y)) \,\, \text{converges} \}$.
Let $P$ be a polygon that is convex and symmetric to some point $O$. Prove that for some parallelogram $R$ satisfying $P\subset R$ we have \[\frac{|R|}{|P|}\leq \sqrt 2\] where $|R|$ and $|P|$ denote the area of the sets $R$ and $P$, respectively. [i]Proposed by Witold Szczechla, Poland[/i]
Inside the parallelogram $ABCD$ is a point $X$. Make a line that passes through point $X$ and divides the parallelogram into two parts whose areas differ from each other the most.
Let $XYZ$ be a right triangle of area $1$ m$^2$ . Consider the triangle $X'Y'Z'$ such that $X'$ is the symmetric of X wrt side $YZ$, $Y'$ is the symmetric of $Y$ wrt side $XZ$ and $Z' $ is the symmetric of $Z$ wrt side $XY$. Calculate the area of the triangle $X'Y'Z'$.
Assign to each side $b$ of a convex polygon $P$ the maximum area of a triangle that has $b$ as a side and is contained in $P$. Show that the sum of the areas assigned to the sides of $P$ is at least twice the area of $P$.
Let $P,Q,R$ be points on the sides $BC,CA,AB$ respectively of a triangle $ABC$. Suppose that $BQ$ and $CR$ meet at $A', AP$ and $CR$ meet at $B'$, and $AP$ and $BQ$ meet at $C'$, such that $AB' = B'C', BC' =C'A'$, and $CA'= A'B'$. Compute the ratio of the area of $\triangle PQR$ to the area of $\triangle ABC$.
$PS$ is a line segment of length $4$ and $O$ is the midpoint of $PS$. A semicircular arc is drawn with $PS$ as diameter. Let $X$ be the midpoint of this arc. $Q$ and $R$ are points on the arc $PXS$ such that $QR$ is parallel to $PS$ and the semicircular arc drawn with $QR$ as diameter is tangent to $PS$. What is the area of the region $QXROQ$ bounded by the two semicircular arcs?
An arbitrary triangle $ABC$ is given and a point $P$ lies on the side $AB$. It is requested to draw through $P$ a line that divides the triangle into two figures of the same area.
Among all orthogonal projections of a regular tetrahedron to all possible planes, find the projection of the greatest area.
Let $ABC$ be a triangle and $P$ an exterior point in the plane of the triangle. Suppose the lines $AP$, $BP$, $CP$ meet the sides $BC$, $CA$, $AB$ (or extensions thereof) in $D$, $E$, $F$, respectively. Suppose further that the areas of triangles $PBD$, $PCE$, $PAF$ are all equal. Prove that each of these areas is equal to the area of triangle $ABC$ itself.
On a straight line $\ell$ there are four points, $A$, $B$, $C$ and $D$ in that order, such that $AB=BC=CD$. A point $E$ is chosen outside the straight line so that when drawing the segments $EB$ and $EC$, an equilateral triangle $EBC$ is formed . Segments $EA$ and $ED$ are drawn, and a point $F$ is chosen so that when drawing the segments $FA$ and $FE$, an equilateral triangle $FAE$ is formed outside the triangle $EAD$. Finally, the lines $EB$ and $FA$ are drawn , which intersect at the point $G$. If the area of triangle $EBD$ is $10$, calculate the area of triangle $EFG$.
Let $AB$ and $CD$ be two parallel chords in a circle with radius $5$ such that the centre $O$ lies between these chords. Suppose $AB = 6, CD = 8$. Suppose further that the area of the part of the circle lying between the chords $AB$ and $CD$ is $(m\pi + n) / k$, where $m, n, k$ are positive integers with gcd$(m, n, k) = 1$. What is the value of $m + n + k$ ?
Trapezium $ABCD$ and parallelogram $MBDK$ are located so that the sides of the parallelogram are parallel to the diagonals of the trapezoid (see fig.). Prove that the area of the gray part is equal to the sum of the areas of the black part. (Yu. Blinkov) [img]https://cdn.artofproblemsolving.com/attachments/b/9/cfff83b1b85aea16b603995d4f3d327256b60b.png[/img]
Given a quadrilateral $ABCD$ with $AB = BC =3$ cm, $CD = 4$ cm, $DA = 8$ cm and $\angle DAB + \angle ABC = 180^o$. Calculate the area of the quadrilateral.
Let $M,N$ be the midpoints of the sides $AD,BC$ respectively of the convex quadrilateral $ABCD$, $K=AN \cap BM$, $L=CM \cap DN$. Find the smallest possible $c\in R$ such that $S(MKNL)<c \cdot S(ABCD)$ for any convex quadrilateral $ABCD$. I. Voronovich
Let $ABCD$ be a cyclic quadrilateral. Let $E$ and $F$ be variable points on the sides $AB$ and $CD$, respectively, such that $AE:EB=CF:FD$. Let $P$ be the point on the segment $EF$ such that $PE:PF=AB:CD$. Prove that the ratio between the areas of triangles $APD$ and $BPC$ does not depend on the choice of $E$ and $F$.
Quadrilateral $ABCD$ satisfies $\angle ABC = \angle ACD = 90^{\circ}, AC = 20$, and $CD = 30$. Diagonals $\overline{AC}$ and $\overline{BD}$ intersect at point $E$, and $AE = 5$. What is the area of quadrilateral $ABCD$? $\textbf{(A) } 330 \qquad\textbf{(B) } 340 \qquad\textbf{(C) } 350 \qquad\textbf{(D) } 360 \qquad\textbf{(E) } 370$