In one of the adjoining figures a square of side $2$ is dissected into four pieces so that $E$ and $F$ are the midpoints of opposite sides and $AG$ is perpendicular to $BF$. These four pieces can then be reassembled into a rectangle as shown in the second figure. The ratio of height to base, $XY$ / $YZ$, in this rectangle is [asy] size(180); defaultpen(linewidth(0.7)+fontsize(10)); pair A=(0,1), B=(0,-1), C=(2,-1), D=(2,1), E=(1,-1), F=(1,1), G=(.8,.6); pair X=(4,sqrt(5)), Y=(4,-sqrt(5)), Z=(4+2/sqrt(5),-sqrt(5)), W=(4+2/sqrt(5),sqrt(5)), T=(4,0), U=(4+2/sqrt(5),-4/sqrt(5)), V=(4+2/sqrt(5),1/sqrt(5)); draw(A--B--C--D--A^^B--F^^E--D^^A--G^^rightanglemark(A,G,F)); draw(X--Y--Z--W--X^^T--V--X^^Y--U); label("A", A, NW); label("B", B, SW); label("C", C, SE); label("D", D, NE); label("E", E, S); label("F", F, N); label("G", G, E); label("X", X, NW); label("Y", Y, SW); label("Z", Z, SE); label("W", W, NE); [/asy] $ \textbf{(A)}\ 4\qquad\textbf{(B)}\ 1+2\sqrt{3}\qquad\textbf{(C)}\ 2\sqrt{5}\qquad\textbf{(D)}\ \frac{8+4\sqrt{3}}{3}\qquad\textbf{(E)}\ 5 $

Let $ABC$ be a triangle with area $1$ (cm$^2$). Points $D,E$ and $F$ lie on the sides $AB, BC$ and CA, respectively. Prove that $min\{$area of $\vartriangle ADF,$ area of $\vartriangle BED,$ area of $\vartriangle CEF\} \le \frac14$ (cm$^2$).

In a triangle $ABC$ the area is $18$, the length $AB$ is $5$, and the medians from $A$ and $B$ are orthogonal. Find the lengths of the sides $BC,AC$.

A trapezoid has side lengths as indicated in the figure (the sides with length $11$ and $36$ are parallel). Calculate the area of the trapezoid.[img]https://1.bp.blogspot.com/-5PKrqDG37X4/XzcJtCyUv8I/AAAAAAAAMY0/tB0FObJUJdcTlAJc4n6YNEaVIDfQ91-eQCLcBGAsYHQ/s0/1995%2BMohr%2Bp1.png[/img]

Let $ABCD$ be a parallelogram, and let $P$ be a point on the side $AB$. Let the line through $P$ parallel to $BC$ intersect the diagonal $AC$ at point $Q$. Prove that $$|DAQ|^2 = |PAQ| \times |BCD| ,$$ where $|XY Z|$ denotes the area of triangle $XY Z$.

In a convex quadrilateral (in the plane) with the area of $32 \text{ cm}^{2}$ the sum of two opposite sides and a diagonal is $16 \text{ cm}$. Determine all the possible values that the other diagonal can have.

The triangle $ABC$ is right in $A$ and $R$ is the midpoint of the hypotenuse $BC$ . On the major leg $AB$ the point $P$ is marked such that $CP = BP$ and on the segment $BP$ the point $Q$ is marked such that the triangle $PQR$ is equilateral. If the area of triangle $ABC$ is $27$, calculate the area of triangle $PQR$ .

Points $ X $ and $ Y $ are the midpoints of the sides $ AB $ and $ AC $ of the triangle $ ABC $, $ I $ is the center of its inscribed circle, $ K $ is the point of tangency of the inscribed circles with side $ BC $. The external angle bisector at the vertex $ B $ intersects the line $ XY $ at the point $ P $, and the external angle bisector at the vertex of $ C $ intersects $ XY $ at $ Q $. Prove that the area of the quadrilateral $ PKQI $ is equal to half the area of the triangle $ ABC $.

From the midpoints of the sides of an acute-angled triangle, perpendiculars are drawn to the adjacent sides. The resulting six straight lines bound the hexagon. Prove that its area is half the area of the original triangle.

Suppose that there is a point $P$ inside a convex quadrilateral $ABCD$ such that the triangles $PAB$, $PBC$, $PCD$, $PDA$ have equal areas. Prove that one of the diagonals bisects the area of $ABCD$.

Let $ABC$ be triangle with $H$ is the orthocenter and $I$ is incenter. Denote $A_{1}, A_{2}, B_{1}, B_{2}, C_{1}, C_{2}$ be the points on the rays $AB, AC, BC, CA, CB$, respectively such that $$AA_{1} = AA_{2} = BC, BB_{1} = BB_{2} = CA, CC_{1} = CC_{2} = AB.$$ Suppose that $B_{1}B_{2}$ cuts $C_{1}C_{2}$ at $A'$, $C_{1}C_{2}$ cuts $A_{1}A_{2}$ at $B'$ and $A_{1}A_{2}$ cuts $B_{1}B_{2}$ at $C'$. a) Prove that area of triangle $A'B'C'$ is smaller than or equal to the area of triangle $ABC$. b) Let $J$ be circumcenter of triangle $A'B'C'$. $AJ$ cuts $BC$ at $R$, $BJ$ cuts $CA$ at $S$ and $CJ$ cuts $AB$ at $T$. Suppose that $(AST), (BTR), (CRS)$ intersect at $K$. Prove that if triangle $ABC$ is not isosceles then $HIJK$ is a parallelogram.

Two medians of a triangle with unequal sides are $ 3$ inches and $ 6$ inches. Its area is $ 3 \sqrt{15}$ square inches. The length of the third median in inches, is: $ \textbf{(A)}\ 4 \qquad \textbf{(B)}\ 3 \sqrt{3} \qquad \textbf{(C)}\ 3 \sqrt{6} \qquad \textbf{(D)}\ 6 \sqrt{3} \qquad \textbf{(E)}\ 6 \sqrt{6}$

In triangle $ ABC$, $ AB \equal{} 13$, $ BC \equal{} 14$, and $ AC \equal{} 15$. Let $ D$ denote the midpoint of $ \overline{BC}$ and let $ E$ denote the intersection of $ \overline{BC}$ with the bisector of angle $ BAC$. Which of the following is closest to the area of the triangle $ ADE$? $ \textbf{(A)}\ 2 \qquad \textbf{(B)}\ 2.5 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 3.5 \qquad \textbf{(E)}\ 4$

Let $ABCD$ be a trapezium (trapezoid) with $AD$ parallel to $BC$ and $J$ be the intersection of the diagonals $AC$ and $BD$. Point $P$ a chosen on the side $BC$ such that the distance from $C$ to the line $AP$ is equal to the distance from $B$ to the line $DP$. [i]The following three questions 1, 2 and 3 are independent, so that a condition in one question does not apply in another question.[/i] 1.Suppose that $Area( \vartriangle AJB) =6$ and that $Area(\vartriangle BJC) = 9$. Determine $Area(\vartriangle APD)$. 2. Find all points $Q$ on the plane of the trapezium such that $Area(\vartriangle AQB) = Area(\vartriangle DQC)$. 3. Prove that $PJ$ is the angle bisector of $\angle APD$.

Let $ABC$ be an equilateral triangle of area $1998$ cm$^2$. Points $K, L, M$ divide the segments $[AB], [BC] ,[CA]$, respectively, in the ratio $3:4$ . Line $AL$ intersects the lines $CK$ and $BM$ respectively at the points $P$ and $Q$, and the line $BM$ intersects the line $CK$ at point $R$. Find the area of the triangle $PQR$.

Each of the sides $AB$ and $CD$ of a convex quadrilateral $ABCD$ is divided into three equal parts, $|AE| = |EF| = |F B|$ , $|DP| = |P Q| = |QC|$. The diagonals of $AEPD$ and $FBCQ$ intersect at $M$ and $N$, respectively. Prove that the sum of the areas of $\vartriangle AMD$ and $\vartriangle BNC$ is equal to the sum of the areas of $\vartriangle EPM$ and $\vartriangle FNQ$.

Points $A, B,$ and $C$ are on a circle such that $\triangle ABC$ is an acute triangle. $X, Y ,$ and $Z$ are on the circle such that $AX$ is perpendicular to $BC$ at $D$, $BY$ is perpendicular to $AC$ at $E$, and $CZ$ is perpendicular to $AB$ at $F$. Find the value of \[ \frac{AX}{AD}+\frac{BY}{BE}+\frac{CZ}{CF}, \] and prove that this value is the same for all possible $A, B, C$ on the circle such that $\triangle ABC$ is acute. [asy] pathpen = linewidth(0.7); pair B = (0,0), C = (10,0), A = (2.5,8); path cir = circumcircle(A,B,C); pair D = foot(A,B,C), E = foot(B,A,C), F = foot(C,A,B), X = IP(D--2*D-A,cir), Y = IP(E--2*E-B,cir), Z = IP(F--2*F-C,cir); D(MP("A",A,N)--MP("B",B,SW)--MP("C",C,SE)--cycle); D(cir); D(A--MP("X",X)); D(B--MP("Y",Y,NE)); D(C--MP("Z",Z,NW)); D(rightanglemark(B,F,C,12)); D(rightanglemark(A,D,B,12)); D(rightanglemark(B,E,C,12));[/asy]

Let $ABC$ be a triangle. Let $B'$ be the symmetric of $B$ with respect to $C, C'$ the symmetry of $C$ with respect to $A$ and $A'$ the symmetry of $A$ with respect to $B$. a) Prove that the area of triangle $AC'A'$ is twice the area of triangle $ABC$. b) If we delete points $A, B, C$, how can they be reconstituted? Justify your reasoning.

Given a rectangle $ABCD$ such that $AB = b > 2a = BC$, let $E$ be the midpoint of $AD$. On a line parallel to $AB$ through point $E$, a point $G$ is chosen such that the area of $GCE$ is $$(GCE)= \frac12 \left(\frac{a^3}{b}+ab\right)$$ Point $H$ is the foot of the perpendicular from $E$ to $GD$ and a point $I$ is taken on the diagonal $AC$ such that the triangles $ACE$ and $AEI$ are similar. The lines $BH$ and $IE$ intersect at $K$ and the lines $CA$ and $EH$ intersect at $J$. Prove that $KJ \perp AB$.

The trapezoid $ABCD$ of bases $AB$ and $CD$, has $\angle A = 90^o, AB = 6, CD = 3$ and $AD = 4$. Let $E, G, H$ be the circumcenters of triangles $ABC, ACD, ABD$, respectively. Find the area of the triangle $EGH$.

A triangle has side lengths $a$, $b$, $c$, and the altitudes have lengths $h_a$, $h_b$, $h_c$. Prove that \[ \left( \frac{a}{h_a} \right)^2 + \left( \frac{b}{h_b} \right)^2 + \left( \frac{c}{h_c} \right)^2 \ge 4. \]

(a) Let $ABC$ be any triangle. Let $X, Y, Z$ be points on the sides $BC, CA, AB$ respectively. Suppose that $BX \leq XC, CY \leq YA, AZ \leq ZB$. Show that the area of the triangle $XYZ$ $\geq 1\slash 4$ times the area of $ABC.$ (b) Let $ABC$ be any triangle, and let $X, Y, Z$ be points on the sides $BC, CA, AB$ respectively. Using (a) or by any other method, show: One of the three corner triangles $AZY, BXZ, CYX$ has an area $\leq$ area of the triangle $XYZ.$

Side $AB$ of triangle $ABC$ was divided into $n$ equal parts (dividing points $B_0 = A, B_1, B_2, ..., B_n = B$), and side $AC$ of this triangle was divided into $(n + 1)$ equal parts (dividing points $C_0 = A, C_1, C_2, ..., C_{n+1} = C$). Colored are the triangles $C_iB_iC_{i+1}$ (where $i = 1,2, ..., n$). What part of the area of the triangle is painted over?

Given $7$ points inside or on a regular hexagon, show that three of them form a triangle with area $\le 1/6$ the area of the hexagon.

The Fibonacci sequence is given by equalities $$F_1=F_2=1, F_{k+2}=F_k+F_{k+1}, k\in N$$. a) Prove that for every $m \ge 0$, the area of ​​the triangle $A_1A_2A_3$ with vertices $A_1(F_{m+1},F_{m+2})$, $A_2 (F_{m+3},F_{m+4})$, $A_3 (F_{m+5},F_{m+6})$ is equal to $0.5$. b) Prove that for every $m \ge 0$ the quadrangle $A_1A_2A_4$ with vertices $A_1(F_{m+1},F_{m+2})$, $A_2 (F_{m+3},F_{m+4})$, $A_3 (F_{m+5},F_{m+6})$, $A_4 (F_{m+7},F_{m+8})$ is a trapezoid, whose area is equal to $2.5$. c) Prove that the area of ​​the polygon $A_1A_2...A_n$ , $n \ge3$ with vertices does not depend on the choice of numbers $m \ge 0$, and find this area.