Found problems: 698
A straight line passing through the center of the circumscribed circle and the intersection point of the heights of the non-equilateral triangle $ABC$ divides its perimeter and area in the same ratio.Find this ratio.
Each of the two opposite sides of a convex quadrilateral is divided into seven equal parts, and corresponding division points are connected by a segment, thus dividing the quadrilateral into seven smaller quadrilaterals. Prove that the area of at least one of the small quadrilaterals equals $1\slash 7$ slash of the area of the large quadrilateral.
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]
Let $ABC$ be a triangle and $AD$ be the bisector of the triangle ($D \in (BC)$) Assume that $AB =14$ cm,
$AC = 35$ cm and $AD = 12$ cm; which of the following is the area of triangle $ABC$ in cm$^2$?
[b]A.[/b] $\frac{1176}{5}$ [b]B.[/b] $\frac{1167}{5}$ [b]C.[/b] $234$ [b]D.[/b] $\frac{1176}{7}$ [b]E.[/b] $236$
Triangle $ABC$ of area $1$ is given. Point $A'$ lies on the extension of side $BC$ beyond point $C$ with $BC = CA'$. Point $B'$ lies on extension of side $CA$ beyond $A$ and $CA = AB'$. $C'$ lies on extension of $AB$ beyond $B$ with $AB = BC'$. Find the area of triangle $A'B'C'$.
Let $D$ be the midpoint of the side $BC$ of a triangle $ABC$ and let $G$ be the point of the segment $AD$ such that $AG = 2GD$. Let $E$ and $F$ be points on the sides $AB$ and $AC$, respectively, such that$ G$ lies on the segment $EF$. Let $M$ and $N$ be points of the segments $AE$ and $AF$, respectively, such that $ME = EB$ and $NF = FC$.
a) Prove that the area of the quadrilateral $BMNC$ is equal to four times the area of the triangle $DEF$.
b) Prove that the quadrilaterals $MNFE$ and $AMDN$ have the same area.
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 $PQR$.
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 $4$.
The inscribed circle of $\triangle ABC$ touches $BC, AC$ and $AB$ at $D,E$ and $F$ respectively. Denote the perpendicular foots from $F, E$ to $BC$ by $K, L$ respectively. Let the second intersection of these perpendiculars with the incircle be $M, N$ respectively. Show that $\frac{{{S}_{\triangle BMD}}}{{{S}_{\triangle CND}}}=\frac{DK}{DL}$
by Mahdi Etesami Fard
The star shown is symmetric with respect to each of the six diagonals shown. All segments connecting the points $A_1, A_2, . . . , A_6$ with the centre of the star have the length $1$, and all the angles at $B_1, B_2, . . . , B_6$ indicated in the figure are right angles. Calculate the area of the star.
[img]https://1.bp.blogspot.com/-Rso2aWGUq_k/XzcAm4BkAvI/AAAAAAAAMW0/277afcqTfCgZOHshf_6ce2XpinWWR4SZACLcBGAsYHQ/s0/2006%2BMohr%2Bp1.png[/img]
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 .
What is the largest area of a regular hexagon that can be drawn inside the equilateral triangle of side $3$?
A. $3\sqrt7$ B. $\frac{3 \sqrt3}{2}$ C. $2\sqrt5$ D. $\frac{3\sqrt3}{8}$ E. $3\sqrt5$
Prove that in every convex hexagon of area $S$ one can draw a diagonal that cuts off a triangle of area not exceeding $\frac{1}{6}S.$
A rectangular sheet of paper (white on one side and gray on the other) was folded three times, as shown in the figure:
Rectangle $1$, which was white after the first fold, has $20$ cm more perimeter than rectangle $2$, which was white after the second fold, and this in turn has $16$ cm more perimeter than rectangle $3$, which was white after the third fold. Determine the area of the sheet.
[img]https://cdn.artofproblemsolving.com/attachments/d/f/8e363b40654ad0d8e100eac38319ee3784a7a7.png[/img]
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 $\frac{A(S_4)+A(S_5)+A(S_6)}{A(S_1)+A(S_2)+A(S_3)}$?
Let $ABCDE$ be a convex pentagon inscribed in a circle $\Gamma$, such that $AB = BC = CD$. Let $F$ and $G$ be the intersections of $BE$ with $AC$ and of $CE$ with $BD$, respectively. Show that:
a) $[ABC] = [FBCG]$
b) $\frac{[EFG]}{[EAD]} = \frac{BC}{AD}$
[b]Note: [/b] $[X]$ denotes the area of polygon $X$.
Let $X, Y$ , and $Z$ be concentric circles with radii $1$, $13$, and $22$, respectively. Draw points $A, B$, and $C$ on $X$, $Y$ , and $Z$, respectively, such that the area of triangle $ABC$ is as large as possible. If the area of the triangle is $\Delta$, find $\Delta^2$.
From vertex $A$ in square $ABCD$ (of side length $1$ ) two lines are drawn , one intersecting side $BC$ and the other intersecting side $CD$. The angle between these lines is $\theta$. From vertices $B$ and $D$ we construct perpendiculars to each of these lines . Find the area of the quadrilateral whose vertices are the four feet of these perpendiculars.
Find the total area of the shaded area in the figure if all circles have an equal radius $R$ and the centers of the outer circles divide into six equal parts of the middle circle.
[img]http://3.bp.blogspot.com/-Ax0QJ38poYU/XovXkdaM-3I/AAAAAAAALvM/DAZGVV7TQjEnSf2y1mbnse8lL6YIg-BQgCK4BGAYYCw/s400/estonia%2B2000%2Bo.j.1.5.png[/img]
Let $ABCD$ be a convex quadrilateral with $BA = BC$ and $DA = DC$. Let $E$ and $F$ be the midpoints of $BC$ and $CD$ respectively, and let$ BF$ and $DE$ intersect at $G$. If the area of $CEGF$ is $50$, what is the area of $ABGD$?
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 $.
A rectangle is divided into $9$ small rectangles if by parallel lines to its sides, as shown in the figure.
[img]https://cdn.artofproblemsolving.com/attachments/e/d/1fd545862a3c7950249ec54a631c74e59fb9ed.png[/img]
The four numbers written indicate the areas of the four corresponding rectangles.
Prove that the total area of the rectangle is greater than or equal to $90$.
In a convex hexagon $ABCDEF, AB$ is parallel to $DE, BC$ is parallel to $EF$ and $CD$ is parallel to $FA$. Prove that the triangles $ACE$ and $BDF$ have the same area.
Daniel finds a rectangular index card and measures its diagonal to be 8 centimeters. Daniel then cuts out equal squares of side 1 cm at two opposite corners of the index card and measures the distance between the two closest vertices of these squares to be $4\sqrt{2}$ centimeters, as shown below. What is the area of the original index card?
[asy]
unitsize(0.6 cm);
pair A, B, C, D, E, F, G, H;
real x, y;
x = 9;
y = 5;
A = (0,y);
B = (x - 1,y);
C = (x - 1,y - 1);
D = (x,y - 1);
E = (x,0);
F = (1,0);
G = (1,1);
H = (0,1);
draw(A--B--C--D--E--F--G--H--cycle);
draw(interp(C,G,0.03)--interp(C,G,0.97), dashed, Arrows(6));
draw(interp(A,E,0.03)--interp(A,E,0.97), dashed, Arrows(6));
label("$1$", (B + C)/2, W);
label("$1$", (C + D)/2, S);
label("$8$", interp(A,E,0.3), NE);
label("$4 \sqrt{2}$", interp(G,C,0.2), SE);
[/asy]
$\textbf{(A) }14\qquad\textbf{(B) }10\sqrt{2}\qquad\textbf{(C) }16\qquad\textbf{(D) }12\sqrt{2}\qquad\textbf{(E) }18$
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 $CR = AB$.