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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)

2008 iTest Problem 98 Part 1.png

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 $QI = IP$.

2008 iTest Problem 98 Part 2.png

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 $p \cdot q = \boxed{1111}$.

See Also

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