Question

Let $ABCDE$ be a convex pentagon with $\overline{AB} \parallel \overline{CE},$ $\overline{BC} \parallel \overline{AD},$ $\overline{AC} \parallel \overline{DE},$ $\angle ABC = 120^\circ,$ $AB = 3,$ $BC = 5,$ and $DE = 15.$ Given that the ratio between the area of triangle $ABC$ and the area of triangle $EBD$ is $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers, find $m + n.$

Answer

Solution:

By the law of cosines, $AC^2 = 3^2 + 5^2 - 2 \cdot 3 \cdot 5 \cos 120^\circ = 49,$ so $AC = 7.$ Let $F$ be the intersection of $\overline{AD}$ and $\overline{CE}.$ Since $AF \parallel BC$ and $CF \parallel AB,$ quadrilateral $ABCF$ is a parallelogram, so $F$ lies at the same distance $h$ from line $AC$ as $B,$ on the opposite side, where $[ABC] = \frac{1}{2} \cdot 7h.$

Since $\overline{AC} \parallel \overline{DE},$ triangles $FAC$ and $FDE$ are similar with ratio $AC : DE = 7 : 15,$ so the distance from $F$ to line $DE$ is $\frac{15h}{7},$ with $DE$ on the far side of $F$ from $AC.$ The distance from $B$ to line $DE$ is therefore $h + h + \frac{15h}{7} = \frac{29h}{7},$ giving $[EBD] = \frac{1}{2} \cdot 15 \cdot \frac{29h}{7} = \frac{435h}{14}.$

Thus $\frac{[ABC]}{[EBD]} = \frac{7h/2}{435h/14} = \frac{49}{435},$ which is in lowest terms since $435 = 3 \cdot 5 \cdot 29.$ The answer is $m + n = 49 + 435 = 484.$

2004 AIME II Problem 13

Solution:

Problem 13 in Other Years