Found problems: 259
In this exercise we only consider convex quadrilaterals.
(a) For such a quadrilateral $ABCD$, determine the set of points $P$ contained within that quadrilateral for which $PA$ and $PC$ divide the quadrilateral into two pieces of equal areas.
(b) Prove that there is a point $P$ inside such a quadrilateral, such that the triangles $PAB$ and $PCD$ have equal areas, as well as the triangles $PBC$ and $PAD$.
(c) Find out which quadrilaterals $ABCD$ contains a point $P$, so that the triangles $PAB$, $PBC$, $PCD$ and $PDA$ have equal areas.
A semicircle of radius $r$ is divided into $n + 1$ equal parts and any point $k$ of the division with the ends of the semicircle forms a triangle $A_k$. Calculate the limit, as $n$ tends to infinity, of the arithmetic mean of the areas of the triangles.
In the plane we are given $3 \cdot n$ points ($n>$1) no three collinear, and the distance between any two of them is $\leq 1$. Prove that we can construct $n$ pairwise disjoint triangles such that: The vertex set of these triangles are exactly the given 3n points and the sum of the area of these triangles $< 1/2$.
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]
The circle $x^2 + y^2 = 25$ intersects the abscissa in points $A$ and $B$. Let $P$ be a point that lies on the line $x = 11$, $C$ is the intersection point of this line with the $Ox$ axis, and the point $Q$ is the intersection point of $AP$ with the given circle. It turned out that the area of the triangle $AQB$ is four times smaller than the area of the triangle $APC$. Find the coordinates of $Q$.
Suppose that $a, b$ are natural numbers so that all the roots of $x^2 + ax - b$ and $x^2 - ax + b$ are integers. Show that exists a right triangle with integer sides, with $a$ the length of the hypotenuse and $b$ the area .
Given a pentagon of area $1993$ and $995$ points inside the pentagon, let $S$ be the set containing the vertices of the pentagon and the $995$ points. Show that we can find three points of $S$ which form a triangle of area $\le 1$.
Let $\triangle PQR$ be a triangle with $\angle P = 75^\circ$ and $\angle Q = 60^\circ$. A regular hexagon $ABCDEF$ with side length 1 is drawn inside $\triangle PQR$ so that side $\overline{AB}$ lies on $\overline{PQ}$, side $\overline{CD}$ lies on $\overline{QR}$, and one of the remaining vertices lies on $\overline{RP}$. There are positive integers $a$, $b$, $c$, and $d$ such that the area of $\triangle PQR$ can be expressed in the form $\tfrac{a+b\sqrt c}d$, where $a$ and $d$ are relatively prime and $c$ is not divisible by the square of any prime. Find $a+b+c+d$.
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 $PQR$ .