In a triangle, the edges are extended past both vertices by the length of the edge opposite to the respective vertex. Show that the area of the resulting hexagon is at least $13$ times the area of the original triangle.

Given some square carpets with the total area $4$. Prove that they can fully cover the unit square.

In the figure below, parallelograms $ABCD$ and $BFEC$ have areas $1234$ cm$^2$ and $2804$ cm$^2$, respectively. Points $M$ and $N$ are chosen on sides $AD$ and $FE$, respectively, so that segment $MN$ passes through $B$. Find the area of $\vartriangle MNC$. [img]https://cdn.artofproblemsolving.com/attachments/b/6/8b57b632191bdb3a27ab7c59e2376dab23950b.png[/img]

From a square of side length $1$, four identical triangles are removed, one at each corner, leaving a regular octagon. What is the area of the octagon?

Two squares are arranged as shown. Prove that the area of the black triangle equal to the sum of the gray areas. [img]https://2.bp.blogspot.com/-byhWqNr1ras/XTq-NWusg2I/AAAAAAAAKZA/1sxEZ751v_Evx1ij7K_CGiuZYqCjhm-mQCK4BGAYYCw/s400/Oral%2BSharygin%2B2017%2B8.9%2Bp5.png[/img]

Let $ABC$ be an equilateral triangle. $N$ is a point on the side $AC$ such that $\vec{AC} = 7\vec{AN}$, $M$ is a point on the side $AB$ such that $MN$ is parallel to $BC$ and $P$ is a point on the side $BC$ such that $MP$ is parallel to $AC$. Find the ratio of areas $\frac{ (MNP)}{(ABC)}$

In a triangle $ABC$ with $AC = b \ne BC = a$, points $E,F$ are taken on the sides $AC,BC$ respectively such that $AE = BF =\frac{ab}{a+b}$. Let $M$ and $N$ be the midpoints of $AB$ and $EF$ respectively, and $P$ be the intersection point of the segment $EF$ with the bisector of $\angle ACB$. Find the ratio of the area of $CPMN$ to that of $ABC$.

In a convex quadrilateral $ ABCD $ with area $ S $, each side was divided into 3 equal parts and segments were drawn connecting the appropriate points of division of the opposite sides in such a way that the quadrilateral was divided into 9 quadrilaterals. Prove that the sum of the areas of the following three quadrilaterals resulting from the division: the one containing the vertex $ A $, the middle one and the one containing the vertex $ C $ is equal to $ \frac{S}{3} $.

Triangle $XYZ$ is inside square $KLMN$ shown below so that its vertices each lie on three different sides of the square. It is known that: $\bullet$ The area of square $KLMN$ is $1$. $\bullet$ The vertices of the triangle divide three sides of the square up into these ratios: $KX : XL = 3 : 2$ $KY : YN = 4 : 1$ $NZ : ZM = 2 : 3$ What is the area of the triangle $XYZ$? (Note that the sketch is not drawn to scale). [img]https://cdn.artofproblemsolving.com/attachments/8/0/38e76709373ba02346515f9949ce4507ed4f8f.png[/img]

In triangle $ABC$, angle $C$ is right and the two catheti are both length $1$. For one given the choice of the point $P$ on the cathetus $BC$, the point $Q$ on the hypotenuse and the point $R$ are plotted on the second cathetus so that $PQ$ is parallel to $AC$ and $QR$ is parallel to $BC$. Thereby the triangle is divided into three parts. Determine the locations of point $P$ for which the rectangular part has a larger area than each of the other two parts.

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]

In a convex quadrilateral $ABCD$, $E$ is the midpoint of $CD$, $F$ is midpoint of $AD$, $K$ is the intersection point of $AC$ with $BE$. Prove that the area of triangle $BKF$ is half the area of triangle $ABC$.

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]

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]

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.

Let $ ABC$ be a triangle, and let $ P$, $ Q$, $ R$ be three points in the interiors of the sides $ BC$, $ CA$, $ AB$ of this triangle. Prove that the area of at least one of the three triangles $ AQR$, $ BRP$, $ CPQ$ is less than or equal to one quarter of the area of triangle $ ABC$. [i]Alternative formulation:[/i] Let $ ABC$ be a triangle, and let $ P$, $ Q$, $ R$ be three points on the segments $ BC$, $ CA$, $ AB$, respectively. Prove that $ \min\left\{\left|AQR\right|,\left|BRP\right|,\left|CPQ\right|\right\}\leq\frac14\cdot\left|ABC\right|$, where the abbreviation $ \left|P_1P_2P_3\right|$ denotes the (non-directed) area of an arbitrary triangle $ P_1P_2P_3$.

Triangle $ABC$ is divided into six smaller triangles by lines that pass through the vertices and through a common point inside of the triangle. The areas of four of these triangles are indicated. Calculate the area of triangle $ABC$. [img]https://cdn.artofproblemsolving.com/attachments/9/2/2013de890e438f5bf88af446692b495917b1ff.png[/img]

In a quadrilateral $ABCD$ let $E$ be the intersection of the two diagonals, I the center of the parallelogram whose vertices are the midpoints of the four sides of the quadrilateral, and K the center of the parallelogram whose sides pass through the points. divide the four sides of the quadrilateral into three equal parts (see illustration ). [img]https://cdn.artofproblemsolving.com/attachments/1/c/8f2617103edd8361b8deebbee13c6180fa848b.png[/img] a) Prove that $\overrightarrow{EK} =\frac43 \overrightarrow{EI}$. b) Prove that $$\lambda_A \overrightarrow{KA} +\lambda_B \overrightarrow{KB} + \lambda_C \overrightarrow{KC} + \lambda_D \overrightarrow{KD} = \overrightarrow{0}$$ , where $$\lambda_A=1+\frac{S(ADB)}{S(ABCD)},\lambda_B=1+\frac{S(BCA)}{S(ABCD)},\lambda_C=1+\frac{S(CDB)}{S(ABCD)},\lambda_D=1+\frac{S(DAC)}{S(ABCD)}$$ , where $S$ is the area symbol.

On the side $AB$ of a triangle $ABC$ is chosen a point $P$. Let $Q$ be the midpoint of $BC$ and let $CP$ and $AQ$ intersect at $R$. If $AB + AP = CP$, prove that $CR = AB$.

Let $ABCD$ be a parallelogram, $E$ a point on the line $AB$, beyond $B, F$ a point on the line $AD$, beyond $D$, and $K$ the point of intersection of the lines $ED$ and $BF$. Prove that quadrilaterals $ABKD$ and $CEKF$ have the same area.

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 $XYZ$ 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 $X'Y'Z'$.

Let $ABC$ be a triangle with side lengths $AB = 2011$, $BC = 2012$, $AC = 2013$. Create squares $S_1 =ABB'A''$, $S_2 = ACC''A'$ , and $S_3 = CBB''C'$ using the sides $AB$, $AC$, $BC$ respectively, so that the side $B'A''$ is on the opposite side of $AB$ from $C$, and so forth. Let square $S_4$ have side length $A''A' $, square $S_5$ have side length $C''C'$, and square $S_6$ have side length $B''B'$. Let $A(S_i)$ be the area of square $S_i$ . Compute $\frac{A(S_4)+A(S_5)+A(S_6)}{A(S_1)+A(S_2)+A(S_3)}$?

The pairs $(a,b)$ such that $|a+bt+ t^2 |\leq 1$ for $0\leq t \leq 1$ fill a certain region in the plane. What is the area of this region?

Given triangle $A_0B_0C_0$. On the segment $A_0B_0$ points $A_1$, $A_2$, $...$, $A_n$, and on the segment $B_0C_0$ - points $C_1$, $C_2$, $...$, $Cn$ so that all segments $A_iC_{i+1}$ ($i = 0$, $1$, $...$,$n-1$) are parallel to each other and all segments $ C_iA_{i+1}$ ($i = 0$, $1$, $...$,$n-1$) are too. Segments $C_0A_1$, $A_1C_2$, $A_2C_1$ and $C_1A_0$ bound a certain parallelogram, segments $C_1A_2$, $A_2C_3$, $A_3C_2$ and $C_2A_1$ too, etc. Prove that the sum of the areas of all $n -1$ resulting parallelograms less than half the area of triangle $A_0B_0C_0$.