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Harvard-MIT Mathematics Tournament

Geometry Difficulty 4.7 AIME Prove it United States

Problem:

In trapezoid $ABCD$ , $AD$ is parallel to $BC$ . $\angle A = \angle D = 45^{\circ}$ , while $\angle B = \angle C = 135^{\circ}$ . If $AB = 6$ and the area of $ABCD$ is $30$ , find $BC$ .

Solution

Solution:

Draw altitudes from $B$ and $C$ to $AD$ and label the points of intersection $X$ and $Y$ , respectively. Then $ABX$ and $CDY$ are $45^{\circ}$ - $45^{\circ}$ - $90^{\circ}$ triangles with $BX = CY = 3\sqrt{2}$ . So, the area of $ABX$ and the area of $CDY$ are each $9$ , meaning that the area of rectangle $BCYX$ is $12$ . Since $BX = 3\sqrt{2}$ , $BC = 12/(3\sqrt{2}) = 2\sqrt{2}$ .

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