Found problems: 67
Let $A$ be a vertex of a regular star-shaped pentagon, the angle at $A$ being less than $180^o$ and the broken line $AA_1BB_1CC_1DD_1EE_1$ being its contour. Lines $AB$ and $DE$ meet at $F$. Prove that polygon $ABB_1CC_1DED_1$ has the same area as the quadrilateral $AD_1EF$.
Note: A regular star pentagon is a figure formed along the diagonals of a regular pentagon.
If a square is intersected by another square equal to it but rotated by $45^o$ around its centre, each side is divided into three parts in a certain ratio $a : b : a$ (which one can compute). Make the following construction for an arbitrary convex quadrilateral: divide each of its sides into three parts in this same ratio $a : b : a$, and draw a line through the two division points neighbouring each vertex. Prove that the new quadrilateral bounded by the four drawn lines has the same area as the original one.
(A. Savin, Moscow)
A terrain ( $ABCD$ ) has a rectangular trapezoidal shape. The angle in $A$ measures $90^o$. $AB$ measures $30$ m, $AD$ measures $20$ m and $DC$ measures 45 m. This land must be divided into two areas of the same area, drawing a parallel to the $AD$ side . At what distance from $D$ do we have to draw the parallel?
[img]https://1.bp.blogspot.com/-DnyNY3x4XKE/XNYvRUrLVTI/AAAAAAAAKLE/gohd7_S9OeIi-CVUVw-iM63uXE5u-WmGwCK4BGAYYCw/s400/image002.gif[/img]
Inside the square $ABCD$ there is the square $A'B' C'D'$ so that the segments $AA', BB', CC'$ and $DD'$ do not intersect each other neither the sides of the smaller square (the sides of the larger and the smaller square do not need to be parallel). Prove that the sum of areas of the quadrangles $AA'B' B$ and $CC'D'D$ is equal to the sum of areas of the quadrangles $BB'C'C$ and $DD'A'A$.
The sides $AB, BC, CD$ and $DA$ of the convex quadrilateral $ABCD$ have midpoints $E, F, G$ and $H$. Prove that the triangles $EFB, FGC, GHD$ and $HEA$ can be put together into a parallelogram equal to $EFGH$.
Given triangle $ABC$. Let $A_1B_1$, $A_2B_2$,$ ...$, $A_{2008}B_{2008}$ be $2008$ lines parallel to $AB$ which divide triangle $ABC$ into $2009$ equal areas. Calculate the value of $$ \left\lfloor \frac{A_1B_1}{2A_2B_2} + \frac{A_1B_1}{2A_3B_3} + ... + \frac{A_1B_1}{2A_{2008}B_{2008}} \right\rfloor$$
Consider a point $M$ inside triangle $ABC$ such that triangles $ABM, BCM$ and $CAM$ have equal areas. Prove that $M$ is the intersection point of the medians of triangle $ABC$.
Points $M$ and $N$ are given on sides $AD$ and $BC$ of rhombus $ABCD$, respectively. Line $MC$ intersects line $BD$ in point $T$, line $MN$ intersects line $BD$ in point $U$, line $CU$ intersects line $AB$ in point $Q$ and line $QT$ intersects line $CD$ in $P$. Prove that triangles $QCP$ and $MCN$ have equal area
Let $A_1A_2A_3$ be a triangle and $M$ an interior point. The straight lines $MA_1, MA_2, MA_3$ intersect the opposite sides at the points $B_1, B_2, B_3$ respectively (see Fig.). Show that if the areas of triangles $A_2B_1M, A_3B_2M$ and $A_1B_3M$ are equal, then $M$ coincides with the centroid of triangle $A_1A_2A_3$.
[img]https://cdn.artofproblemsolving.com/attachments/1/7/b29bdbb1f2b103be1f3cb2650b3bfff352024a.png[/img]
Point $M$ is the midpoint of the side $AB$ of an acute triangle $ABC$. Circle with center $M$ passing through point $ C$, intersects lines $AC ,BC$ for the second time at points $P,Q$ respectively. Point $R$ lies on segment $AB$ such that the triangles $APR$ and $BQR$ have equal areas. Prove that lines $PQ$ and $CR$ are perpendicular.
$ABCD$ is a convex quadrilateral. The midpoints of the diagonals and the midpoints of $AB$ and $CD$ form another convex quadrilateral $Q$. The midpoints of the diagonals and the midpoints of $BC$ and $CA$ form a third convex quadrilateral $Q'$. The areas of $Q$ and $Q'$ are equal. Show that either $AC$ or $BD$ divides $ABCD$ into two parts of equal area.
Let $ABCD$ be a convex quadrilateral and $S$ an arbitrary point in its interior. Let also $E$ be the symmetric point of $S$ with respect to the midpoint $K$ of the side $AB$ and let $Z$ be the symmetric point of $S$ with respect to the midpoint $L$ of the side $CD$. Prove that $(AECZ) = (EBZD) = (ABCD)$.
Consider a trapezoid $ABCD$ with bases $AB$ and $CD$ satisfying $| AB | > | CD |$. Let $M$ be the midpoint of $AB$. Let the point $P$ lie inside $ABCD$ such that $| AD | = | PC |$ and $| BC | = | PD |$. Prove that if $| \angle CMD | = 90^o$, then the quadrilaterals $AMPD$ and $BMPC$ have the same area.
An arbitrary triangle $ABC$ is given and a point $P$ lies on the side $AB$. It is requested to draw through $P$ a line that divides the triangle into two figures of the same area.
A terrain ( $ABCD$ ) has a rectangular trapezoidal shape. The angle in $A$ measures $90^o$. $AB$ measures $30$ m, $AD$ measures $20$ m and $DC$ measures 45 m. This land must be divided into two areas of the same area, drawing a parallel to the $AD$ side . At what distance from $D$ do we have to draw the parallel?
[img]https://1.bp.blogspot.com/-DnyNY3x4XKE/XNYvRUrLVTI/AAAAAAAAKLE/gohd7_S9OeIi-CVUVw-iM63uXE5u-WmGwCK4BGAYYCw/s400/image002.gif[/img]
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.
Suppose that each of the diagonals $AD,BE,CF$ divides the hexagon $ABCDEF$ into two parts of the same area and perimeter. Does the hexagon necessarily have a center of symmetry?
The points $A_1,B_1,C_1$ belong to $[BC],[CA],[AB]$ sides of the $ABC$ triangle respectively. The $[AA_1], [BB_1], [CC_1]$ segments split the $ABC$ onto $4$ smaller triangles and $3$ quadrangles. It is known, that the smaller triangles have the same area. Prove that the quadrangles have equal areas. What is the quadrangle area, it the small triangle has the unit area?
A square is divided into $5$ rectangles in such a way that its $4$ vertices belong to $4$ of the rectangles , whose areas are equal , and the fifth rectangle has no points in common with the side of the square (see diagram) . Prove that the fifth rectangle is a square.
[img]https://3.bp.blogspot.com/-TQc1v_NODek/XWHHgmONboI/AAAAAAAAKi4/XES55OJS5jY9QpNmoURp4y80EkanNzmMwCK4BGAYYCw/s1600/TOT%2B1985%2BSpring%2BJ4.png[/img]
Given two non-parallel segments $AB$ and $CD$ on the plane, find the locus of points $P$ on the plane such that the area of triangle $ABP$ is equal to the area of triangle $CDP$.
The sides $AB, BC, CD$ and $DA$ of the convex quadrilateral $ABCD$ have midpoints $E, F, G$ and $H$. Prove that the triangles $EFB, FGC, GHD$ and $HEA$ can be put together into a parallelogram equal to $EFGH$.
The hexagon $ABCDEF$ is convex and opposite sides are parallel. Show that the triangles $ACE$ and $BDF$ have equal area
Points $C_1, A_1$ and $B_1$ are marked on the sides $AB, BC$ and $CA$ of a triangle $ABC$ so that the segments $AA_1, BB_1$, and $CC_1$ are concurrent (see the fig.). It is known that the area of the white part of the triangle $ABC$ is equal to the area of its black part. Prove that at least one of the segments $AA_1, BB_1, CC_1$ is a median of the triangle $ABC$.
[img]https://1.bp.blogspot.com/-nVVhqdRdf0s/X-WVmt_gyqI/AAAAAAAAM40/943sCRGyCPwT-vqIilTCtXOXHByRLIvPwCLcBGAsYHQ/s0/2014%2Bbelarus%2B11.6.png[/img]
Let $ABC$ be a triangle with $\angle BAC \ne 45^o$ and $\angle ABC \ne 135^o$. Let $P$ be the point on the line $AB$ with $\angle CPB = 45^o$. Let $O_1$ and $O_2$ be the centers of the circumcircles of the triangles $ACP$ and $BCP$ respectively. Show that the area of the square $CO_1P O_2$ is equal to the area of the triangle $ABC$.
Consider a triangle $ABC$ and a point $P$ in its interior. Lines $PA$, $PB$, $PC$ intersect $BC$, $CA$, $AB$ at $A', B', C'$ , respectively. Prove that $$\frac{BA'}{BC}+ \frac{CB'}{CA}+ \frac{AC'}{AB}= \frac32$$ if and only if at least two of the triangles $PAB$, $PBC$, $PCA$ have the same area.