2008 iTest Problems/Problem 98

Problem

Convex quadrilateral $ABCD$ has side-lengths $AB=7$ , $BC=9$ , $CD=15$ , and there exists a circle, lying inside the quadrilateral and having center $I$ , that is tangent to all four sides of the quadrilateral. Points $M$ and $N$ are on the midpoints of $AC$ and $BD$ respectively. It can be proven that point $I$ always lies on segment $MN$ . Supposing further that $I$ is the midpoint of $MN$ , the area of quadrilateral $ABCD$ may be expressed as $p\sqrt q$ , where $p$ and $q$ are positive integers and $q$ is not divisible by the square of any prime. Compute $p\cdot q$ .

Solution (credit to official solution)

Let $P$ be the midpoint of $AD$ and $Q$ be the midpoint of $BC$ . Draw lines from $M$ and $N$ to $P$ and $Q$ , as seen in the diagram. By SAS Similarity, we find that $\triangle CBA \sim \triangle CQM$ and $\triangle DBA \sim \triangle DNP$ , so $\tfrac12 AB = QM = NP$ . Similarly, $\tfrac12 CD = QN = MP$ .

Because opposite sides of $QMPN$ have equal length, $QMPN$ is a parallelogram. Thus, the diagonals bisect each other, and $QI = IP$ .

Let $E$ be on $AB$ where $AB \perp EI$ , and let $F$ be on $CD$ where $IF \perp CD$ . Also, let $T$ be on $BC$ where $IT \perp BC$ , and let $S$ be on $AD$ where $IS \perp AD$ . By HL Congruency, $\triangle ITQ \cong \triangle ISP$ , so $PS = QT$ . Now we have information we can use to determine side lengths to compute the inradius so we can determine the area of $ABCD$ .

Let $a = AS = AE$ . That means $EB = BT = 7-a$ , $TC = CF = a+2$ , and $FD = SD = 13-a$ . That means $AD = 13$ . Additionally, since $AP = 6.5$ and $BQ = 4.5$ , $QT = 2.5-a$ and $PS = a-6.5$ . Since $QT = PS$ , we must have $a = 4.5$ .

We know that $\angle AIS + \angle DIS + \angle BIT + \angle CIT = 180^\circ$ , so $\begin{align*} \tan(\angle AIS + \angle DIS) &= -\tan(\angle BIT + \angle CIT) \\ \frac{\tan \angle AIS + \tan \angle DIS}{1 - \tan \angle AIS \tan \angle DIS} &= \frac{\tan \angle BIT + \tan \angle CIT}{1 - \tan \angle BIT \tan \angle CIT} \end{align*}$ Let $r$ be the radius of the incircle. That means $\begin{align*} \frac{\frac{4.5}{r} + \frac{8.5}{r}}{1 - \frac{4.5 \cdot 8.5}{r^2}} &= -\frac{\frac{2.5}{r} + \frac{6.5}{r}}{1 - \frac{2.5 \cdot 6.5}{r^2}} \\ \frac{13r}{r^2-\frac{153}{4}} &= \frac{-9r}{r^2 - \frac{65}{4}} \\ 13r^3 - \frac{13 \cdot 65}{4}r &= -9r^3 + \frac{153 \cdot 9}{4} \\ 22r^2 &= \frac{13 \cdot 65 + 153 \cdot 9}{4} \\ r^2 &= \frac{101}{4} \\ r &= \frac{\sqrt{101}}{2} \end{align*}$ Therefore, the area of the quadrilateral is $\tfrac12 \cdot (7 + 9 + 13 + 15) \cdot \tfrac{\sqrt{101}}{2} = 11\sqrt{101}$ , so $p \cdot q = \boxed{1111}$ .

2008 iTest (Problems)
Preceded by: Problem 97	Followed by: Problem 99
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2008 iTest Problems/Problem 98

Problem

Solution (credit to official solution)

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