Mock AIME 3 Pre 2005 Problems/Problem 7

Problem

$ABCD$ is a cyclic quadrilateral that has an inscribed circle. The diagonals of $ABCD$ intersect at $P$ . If $AB = 1, CD = 4,$ and $BP : DP = 3 : 8,$ then the area of the inscribed circle of $ABCD$ can be expressed as $\frac{p\pi}{q}$ , where $p$ and $q$ are relatively prime positive integers. Determine $p + q$ .

Solution

Let $BP=3x$ and $PD=8x$ . Angle-chasing can be used to prove that $\triangle ABP \sim \triangle DCP$ . Therefore $\frac{AB}{DC}=\frac{AP}{DP}=\frac{BP}{PC}=\frac{1}{4}$ . This shows that $AP=2x$ and $CP=12x$ . More angle-chasing can be used to prove that $\triangle APD \sim \triangle BPC$ . This shows that $\frac{BC}{AD}=\frac{BP}{AP}=\frac{CP}{DP}=\frac{3}{2}$ . It is a well-known fact that if $ABCD$ is circumscriptable around a circle then $AB+CD=AD+BC$ . Therefore $BC+AD=5$ . We also know that $\frac{BC}{AD}=\frac{3}{2}$ , so we can solve (algebraically or by inspection) to get that $BC=3$ and $AD=2$ .

Brahmagupta's Formula states that the area of a cyclic quadrilateral is $\sqrt{(s-a)(s-b)(s-c)(s-d)}$ , where $s$ is the semiperimeter and $a$ , $b$ , $c$ , and $d$ are the side lengths of the quadrilateral. Therefore the area of $ABCD$ is $\sqrt{4\cdot 3\cdot 2\cdot 1}=\sqrt{24}$ . It is also a well-known fact that the area of a circumscriptable quadrilateral is $sr$ , where $r$ is the inradius. Therefore $5r=\sqrt{24}\Rightarrow r=\frac{\sqrt{24}}{5}$ . Therefore the area of the inscribed circle is $\frac{24\pi}{25}$ , and $p+q=\boxed{049}$ .

Page

Toolbox

Search

Mock AIME 3 Pre 2005 Problems/Problem 7

Problem

Solution

See also