$(FRA 3)$ A polygon (not necessarily convex) with vertices in the lattice points of a rectangular grid is given. The area of the polygon is $S.$ If $I$ is the number of lattice points that are strictly in the interior of the polygon and B the number of lattice points on the border of the polygon, find the number $T = 2S- B -2I + 2.$

Two points are located $10$ units apart, and a circle is drawn with radius $ r$ centered at one of the points. A tangent line to the circle is drawn from the other point. What value of $ r$ maximizes the area of the triangle formed by the two points and the point of tangency?

Triangle $ABC$ has $AB = 3$, $BC = 4$, and $CA = 5$. Points $D$, $E$, $F$, $G$, $H$, and $I$ are the reflections of $A$ over $B$, $B$ over $A$, $B$ over $C$, $C$ over $B$, $C$ over $A$, and $A$ over $C$, respectively. Find the area of hexagon $EFIDGH$.

This figure is cut out from a sheet of paper. Folding the sides upwards along the dashed lines, one gets a (non-equilateral) pyramid with a square base. Calculate the area of the base. [img]https://1.bp.blogspot.com/-lPpfHqfMMRY/XzcBIiF-n2I/AAAAAAAAMW8/nPs_mLe5C8srcxNz45Wg-_SqHlRAsAmigCLcBGAsYHQ/s0/2005%2BMohr%2Bp1.png[/img]

The area of a convex polygon in the plane is equally shared by the four standard quadrants, and all non-zero lattice points lie outside the polygon. Show that the area of the polygon is less than $4$.

Given a triangle with $a,b,c$ sides and with the area $1$ ($a \ge b \ge c$). Prove that $b^2 \ge 2$.

Two squares are arranged as shown in the picture. Prove that the areas of shaded quadrilaterals are equal. [img]https://3.bp.blogspot.com/-W50DOuizFvY/XT6wh3-L6sI/AAAAAAAAKaw/pIW2RKmttrwPAbrKK3bpahJz7hfIZwM8QCK4BGAYYCw/s400/Oral%2BSharygin%2B2016%2B10.11%2Bp3.png[/img]

In a triangle $ABC$, the bisector of the largest angle $\angle A$ meets $BC$ at point $D$. Let $E$ and $F$ be the feet of perpendiculars from $D$ to $AC$ and $AB$, respectively. Let $R$ denote the ratio between the areas of triangles $DEB$ and $DFC$. (a) Prove that, for every real number $r > 0$, one can construct a triangle ABC for which $R$ is equal to $r$. (b) Prove that if $R$ is irrational, then at least one side length of $\vartriangle ABC$ is irrational. (c) Give an example of a triangle $ABC$ with exactly two sides of irrational length, but with rational $R$.

In the square, the midpoints of the two sides were marked and the segments shown in the figure on the left were drawn. Which of the shaded quadrilaterals has the largest area? [img]https://cdn.artofproblemsolving.com/attachments/d/f/2be7bcda3fa04943687de9e043bd8baf40c98c.png[/img]

In an acute-angled triangle $ABC$ the interior bisector of angle $A$ meets $BC$ at $L$ and meets the circumcircle of $ABC$ again at $N$. From $L$ perpendiculars are drawn to $AB$ and $AC$, with feet $K$ and $M$ respectively. Prove that the quadrilateral $AKNM$ and the triangle $ABC$ have equal areas.

$ABCDE$ is a convex pentagon such that the triangles $ABC, BCD, CDE, DEA$ and $EAB$ have equal areas. Show that $(1/4)$ area $(ABCDE) <$ area $(ABC) < (1/3)$ area $(ABCDE)$.

The circle touches the square and goes through its two vertices as shown in the figure. Find the area of the square. (Distance in the picture is measured horizontally from the midpoint of the side of the square.) [img]https://cdn.artofproblemsolving.com/attachments/7/5/ab4b5f3f4fb4b70013e6226ce5189f3dc2e5be.png[/img]

Let be the points $ K,L,M $ on the sides $ BC,CA,AB, $ respectively, of a triangle $ ABC. $ Show that at least one of the areas of the triangles $ MAL,KBM,LCK $ doesn't surpass a fourth of the area of $ ABC. $

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 $\omega=-\tfrac{1}{2}+\tfrac{1}{2}i\sqrt3.$ Let $S$ denote all points in the complex plane of the form $a+b\omega+c\omega^2,$ where $0\leq a \leq 1,0\leq b\leq 1,$ and $0\leq c\leq 1.$ What is the area of $S$? $\textbf{(A) } \frac{1}{2}\sqrt3 \qquad\textbf{(B) } \frac{3}{4}\sqrt3 \qquad\textbf{(C) } \frac{3}{2}\sqrt3\qquad\textbf{(D) } \frac{1}{2}\pi\sqrt3 \qquad\textbf{(E) } \pi$

Given a (not necessarily regular) tetrahedron, all of its sides are equal in area. Prove that the following points then coincide: a) the center of the inscribed sphere, i.e. all four side surfaces internally touching sphere, b) the center of the surrounding sphere, i.e. the sphere passing through the four vertixes.

Let $ABC$ be a triangle, and $M$ be the midpoint of $BC$, Let $N$ be the point on the segment $AM$ with $AN = 3NM$, and $P$ be the intersection point of the lines $BN$ and $AC$. What is the area in cm$^2$ of the triangle $ANP$ if the area of the triangle $ABC$ is $40$ cm$^2$?

Consider an arbitrary (optional convex) polygon. It's [i]chord [/i] is a segment whose ends lie on the boundary of the polygon, and itself belongs entirely to the polygon. Will there always be a chord of a polygon that divides it into two equal parts? Is it true that any polygon can be divided by some chord into parts, the area of each of which is not less than $\frac13$ the area of the polygon?

Prove that for every tetrahedron there exist two planes such that the projection areas on those planes ratio is not less than $\sqrt 2$.

A triangle is given. Its side a is of length $20$ cm, and its area is $125$ cm$^2$. It is also known that one of the angles lying on side a is twice as large as the other one. We cut the triangle into two parts at the median belonging to side a. Then we move the so-obtained two parts towards each other, such that the two segments of side a remain on the same line (i.e., the line initially occupied by side a). We move the two parts towards each other until we first reach a moment when the common part of the two segments is of length $4$ cm. What is the area of the so-obtained shape in cm$^2$? The so-obtained shape is the union of the two parts, which is a heptagon. [img]https://cdn.artofproblemsolving.com/attachments/3/0/3d45e2df6a0043dfa4fe5ccf64865da8879b42.png[/img]

$O$ is the intersection point of the diagonals of the trapezoid $ABCD$. A line passing through $C$ and a point symmetric to $B$ with respect to $O$, intersects the base $AD$ at the point $K$. Prove that $S_{AOK} = S_{AOB} + S_{DOK}$.

Let $K$ be a convex polygon in the plane and suppose that $K$ is positioned in the coordinate system in such a way that \[\text{area } (K \cap Q_i) =\frac 14 \text{area } K \ (i = 1, 2, 3, 4, ),\] where the $Q_i$ denote the quadrants of the plane. Prove that if $K$ contains no nonzero lattice point, then the area of $K$ is less than $4.$

A convex quadrilateral $ABCD$ is given. Let $O_a$ be the circumcenter of the triangle $DBC$, and define $O_b,O_c$ and $O_d$ similarly. The points $O_a, O_b, O_c, O_d$ are the vertices of a convex quadrilateral. Prove that its area is equal to half of the absolute value of the difference between the areas of $AO_bCO_d$ and $BO_cDO_a$. [i]Proposed by V. Dubrovsky[/i]

The polyhedron $PABCDQ$ has the form shown in the figure. It is known that $ABCD$ is parallelogram, the planes of the triangles of the $PAC$ and $PBD$ mutually perpendicular, and also mutually perpendicular are the planes of triangles $QAC$ and $QBC$. Each face of this polyhedron is painted black or white so that the faces that have a common edge are painted in different colors. Prove that the sum of the squares of the areas of the black faces is equal to the sum of the squares of the areas of the white faces. [img]https://1.bp.blogspot.com/-UM5PKEGGWqc/X1V2cXAFmwI/AAAAAAAAMdw/V-Qr94tZmqkj3_q-5mkSICGF1tMu-b_VwCLcBGAsYHQ/s0/2007.5%2Bchampions%2Btourn.png[/img]

An equilateral triangle of side$ 1$ is rotated around its center, yielding another equilareral triangle. Find the area of the intersection of these two triangles.