We call a tetrahedron a "trirectangular " if it has a vertex (we call this is called a "right-angled" vertex) in which the planes of the three sides of the tetrahedron intersect at right angles. Prove the "three-dimensional Pythagorean theorem": The square of the area of the opposite face of the "right-angled" vertex of the ""trirectangular " tetrahedron is equal to the sum of the squares of the areas of three other sides of the tetrahedron .

Let $A'$, $B'$ and $C'$ be points on the sides $BC$, $CA$ and $AB$, respectively, of an equilateral triangle $ABC$. If $AC' = 2C'B$, $BA' = 2A'C$ and $CB' = 2B'A$, prove that the lines $AA'$, $BB'$ and $CC'$ enclose a triangle whose area is $1/7$ that of $ABC$.

$ABCD$ is a rectangle. $I$ is the midpoint of $CD$. $BI$ meets $AC$ at $M$. Show that the line $DM$ passes through the midpoint of $BC$. $E$ is a point outside the rectangle such that $AE = BE$ and $\angle AEB = 90^o$. If $BE = BC = x$, show that $EM$ bisects $\angle AMB$. Find the area of $AEBM$ in terms of $x$.

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 $EFG$.

In triangle $ABC$, let $P$ and $R$ be the feet of the perpendiculars from $A$ onto the external and internal bisectors of $\angle ABC$, respectively; and let $Q$ and $S$ be the feet of the perpendiculars from $A$ onto the internal and external bisectors of $\angle ACB$, respectively. If $PQ = 7, QR = 6$ and $RS = 8$, what is the area of triangle $ABC$?

Let $ABC$ be a triangle with $AB = 26$, $BC = 51$, and $CA = 73$, and let $O$ be an arbitrary point in the interior of $\vartriangle ABC$. Lines $\ell_1$, $\ell_2$, and $\ell_3$ pass through $O$ and are parallel to $\overline{AB}$, $\overline{BC}$, and $\overline{CA}$, respectively. The intersections of $\ell_1$, $\ell_2$, and $\ell_3$ and the sides of $\vartriangle ABC$ form a hexagon whose area is $A$. Compute the minimum value of $A$.

A triangle $ABC$ is given. For every positive numbers $p,q,r$, let $A',B',C'$ be the points such that $\overrightarrow{BA'} = p\overrightarrow{AB}, \overrightarrow{CB'}=q\overrightarrow{BC} $, and $\overrightarrow{AC'}=r\overrightarrow{CA}$. Define $f(p,q,r)$ as the ratio of the area of $\vartriangle A'B'C'$ to that of $\vartriangle ABC$. Prove that for all positive numbers $x,y,z$ and every positive integer $n$, $\sum_{k=0}^{n-1}f(x+k,y+k,z+k) = n^3f\left(\frac{x}{n},\frac{y}{n},\frac{z}{n}\right)$.

Find the area of the figure on the coordinate plane bounded by the straight lines $x = 0$, $x = 2$ and the graphs of the functions $y =\sqrt{x^3+ 1}$ and $y = - \sqrt[3]{x^2+ 2x}$.

Seven polygons of area $1$ lie in the interior of a square with side length $2$. Show that there are two of these polygons whose intersection has an area of at least $1\slash 7.$

Prove that all quadrilaterals $ABCD$ where $\angle B = \angle D = 90^o$, $|AB| = |BC|$ and $|AD| + |DC| = 1$, have the same area. [img]https://1.bp.blogspot.com/-55lHuAKYEtI/XzRzDdRGDPI/AAAAAAAAMUk/n8lYt3fzFaAB410PQI4nMEz7cSSrfHEgQCLcBGAsYHQ/s0/2016%2Bmohr%2Bp3.png[/img]

Let $ABC$ be an acute-angled triangle, $C'$ and $A'$ be arbitrary points on the sides $AB$ and $BC$ respectively, and $B'$ be the midpoint of the side $AC$. (a) Prove that the area of triangle $A'B'C'$ is at most half the area of triangle $ABC$. (b) Prove that the area of triangle $A'B'C'$ is equal to one fourth of the area of triangle $ABC$ if and only if at least one of the points $A'$, $C'$ is the midpoint of the corresponding side. (E Cherepanov)

Durer drew a regular triangle and then poked at an interior point. He made perpendiculars from it sides and connected it to the vertices. In this way, $6$ small triangles were created, of which (moving clockwise) all the second one is painted gray, as shown in figure. Show that the sum of the gray areas is just half the area of the triangle. [img]https://cdn.artofproblemsolving.com/attachments/e/7/a84ad28b3cd45bd0ce455cee2446222fd3eac2.png[/img]

The goal of this problem is to show that the maximum area of a polygon with a fixed number of sides and a fixed perimeter is achieved by a regular polygon. (a) Prove that the polygon with maximum area must be convex. (Hint: If any angle is concave, show that the polygon’s area can be increased.) (b) Prove that if two adjacent sides have different lengths, the area of the polygon can be increased without changing the perimeter. (c) Prove that the polygon with maximum area is equilateral, that is, has all the same side lengths. It is true that when given all four side lengths in order of a quadrilateral, the maximum area is achieved in the unique configuration in which the quadrilateral is cyclic, that is, it can be inscribed in a circle. (d) Prove that in an equilateral polygon, if any two adjacent angles are different then the area of the polygon can be increased without changing the perimeter. (e) Prove that the polygon of maximum area must be equiangular, or have all angles equal. (f ) Prove that the polygon of maximum area is a regular polygon. PS. You had better use hide for answers.

Let $\mathbf T$ be a triangle with $[\mathbf T]=1.$ Show that there is a right triangle $\mathbf R$ such that $[\mathbf R]\le\sqrt3$ and $\mathbf T\subseteq\mathbf R.$ ($[-]$ denotes area of a triangle.)

The hexagon has five $90^o$ angles and one $270^o$ angle (see picture). Use a straight-line ruler to divide it into two equal-sized polygons. [img]https://cdn.artofproblemsolving.com/attachments/d/8/cdd4df68644bb8e04adbe1b265039b82a5382b.png[/img]

Let $ABCDEF$ be a convex hexagon such that $BCEF$ is a parallelogram and $ABF$ an equilateral triangle. Given that $BC = 1, AD = 3, CD+DE = 2$, compute the area of $ABCDEF$

In the square $ABCD$ the point $E$ is located on the side $[AB]$, and $F$ is the foot of the perpendicular from $B$ on the line $DE$. The point $L$ belongs to the line $DE$, such that $F$ is between $E$ and $L$, and $FL = BF$. $N$ and $P$ are symmetric of the points $A , F$ with respect to the lines $DE, BL$, respectively. Prove that: a) The quadrilateral $BFLP$ is square and the quadrilateral $ALND$ is rhombus. b) The area of the rhombus $ALND$ is equal to the difference between the areas of the squares $ABCD$ and $BFLP$.

Through an arbitrary point inside a triangle, lines parallel to the sides of the triangle are drawn, dividing the triangle into three triangles with areas $T_1,T_2,T_3$ and three parallelograms. If $T$ is the area of the original triangle, prove that $$T=(\sqrt{T_1}+\sqrt{T_2}+\sqrt{T_3})^2$$ .

Let $ABC$ be an acute-angled triangle with incenter $I$. Consider the point $A_1$ on $AI$ different from $A$, such that the midpoint of $AA_1$ lies on the circumscribed circle of $ABC$. Points $B_1$ and $C_1$ are defined similarly. (a) Prove that $S_{A_1B_1C_1}=(4R+r)p$, where $p$ is the semi-perimeter, $R$ is the circumradius and $r$ is the inradius of $ABC$. (b) Prove that $S_{A_1B_1C_1}\ge9S_{ABC}$.

Given is a triangle $ABC$ with side lengths $a, b, c$ ($a = \overline{BC}$, $b = \overline{CA}$, $c = \overline{AB}$) and area $F$. The side $AB$ is extended beyond $A$ by a and beyond $B$ by $b$. Correspondingly, $BC$ is extended beyond $B$ and $C$ by $b$ and $c$, respectively. Eventually $CA$ is extended beyond $C$ and $A$ by $c$ and $a$, respectively. Connecting the outer endpoints of the extensions , a hexagon if formed with area $G$. Prove that $\frac{G}{F}>13$.

Let $\vartriangle ABC$ be an equilateral triangle with side $1$. Four points are marked $P_1$, $P_2$, $P_3$, $P_4$ on side $AC$ and four points $Q_1$, $Q_2$, $Q_3$, $Q_4$ on side $AB$ (see figure) of such a way to generate $9$ triangles of equal area. Find the length of segment $AP_4$. [img]https://cdn.artofproblemsolving.com/attachments/5/f/29243932262cb963b376244f4c981b1afe87c6.png[/img] PS. Easier version of [url=https://artofproblemsolving.com/community/c6h3323141p30741525]2023 Chile NMO L2 P3[/url]

In the triangle $ABC$, the median $BD$ is drawn and through its midpoint and vertex $A$ the line $\ell$. Thus the triangle $ABC$ is divided into three triangles and one quadrilateral. Determine the areas of these figures if the area of ​​triangle $ABC$ is equal to $S$.

In the plane, given an angle $Axy$. a) Given a triangle $MNP$ of area $T$, describe how to construct a triangle of given area $T$ and altitude $h$. Using this, describe how to construct parallelogram A$BCD$ with two sides lying on $Ax$ and $Ay$, the area $T$ and the distance between the two opposite sides equal to d given. b) From an arbitrary point $I$ on the line $CD$, construct a line that intersects the lines $A$B, $BC$, $AD$ at $E$, $G$ and $F$ respectively so that the area of triangle $AEF$ is equal to the area of parallelogram $ABCD$. c) Apply the above two sentences: Given any point $O$ in the plane. From $O$, construct a line that intersects two rays $Ax$ and $Ay$ at $E$ and $F$ respectively so that the area of triangle $AEF$ is equal to the area of any given triangle.

The two sides $BC$ and $CD$ of an inscribed quadrangle $ABCD$ are of equal length. Prove that the area of this quadrangle is equal to $S =\frac12 \cdot AC^2 \cdot \sin \angle A$

Let $D$ be the midpoint of side $BC$ of triangle $ABC$ and $E$ the midpoint of median $AD$. Line $BE$ intersects side $CA$ at $F$. Prove that the area of quadrilateral $CDEF$ is $\frac{5}{12}$ the area of triangle $ABC$.