2013/14 AHSMC Part 1 Problem 5

Problem 5 of 16

$ABCD$ is a quadrilateral with $AB = 12$ and $CD = 18$ . Moreover, $AB$ is parallel to $CD$ and both $\angle ADC$ and $\angle BCD$ are less than $90^\circ$ . $P$ and $Q$ are points on side $CD$ such that $AD = AP$ and $BC = BQ$ . The length of $PQ$ is

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Solution

From $A$ and $B$ , drop perpendiculars onto $CD$ at the points $E$ and $F$ respectively. Since both $\angle ADC$ and $\angle BCD$ are less than $90^\circ$ , $E$ and $F$ do lie on the segment $CD$ . Note that $DE + FC = CD - EF = CD - AB = 6.$ Since $ED = EP$ and $FC = FQ$ , $EP + FQ = 6 < 12 = EF$ , $P$ is closer to $E$ than $Q$ and $Q$ is closer to $F$ than $P$ . Therefore, $PQ = EF - (EP + FQ) = 12 - 6 = 6$ . The answer is (A).

University of Alberta

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