Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.3 AIME, harder Prove it United States

Problem:

Let rectangle $A B C D$ have lengths $A B = 20$ and $B C = 12$ . Extend ray $B C$ to $Z$ such that $C Z = 18$ . Let $E$ be the point in the interior of $A B C D$ such that the perpendicular distance from $E$ to $\overline{A B}$ is $6$ and the perpendicular distance from $E$ to $\overline{A D}$ is $6$ . Let line $E Z$ intersect $A B$ at $X$ and $C D$ at $Y$ . Find the area of quadrilateral $A X Y D$ .

Solution

Solution:

Answer: $72$

Draw the line parallel to $\overline{A D}$ through $E$ , intersecting $\overline{A B}$ at $F$ and $\overline{C D}$ at $G$ . It is clear that $X F E$ and $Y G E$ are congruent, so the area of $A X Y D$ is equal to that of $A F G D$ . But $A F G D$ is simply a $12$ by $6$ rectangle, so the answer must be $72$ .

(Note: It is also possible to directly compute the values of $A X$ and $D Y$ , then use the formula for the area of a trapezoid.)

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