Question

The two squares shown share the same center $O$ and have sides of length $1.$ The length of $\overline{AB}$ is $\frac{43}{99}$ and the area of octagon $ABCDEFGH$ is $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$

Answer

Solution:

The whole configuration is unchanged by rotating $90^\circ$ about $O,$ which cycles the octagon side $AB$ to $CD,$ $EF,$ $GH,$ and it is also unchanged by the reflection that swaps the two squares, which carries those sides to $BC,$ $DE,$ $FG,$ $HA.$ So all eight sides of the octagon have the same length, $\frac{43}{99}.$

Segments from $O$ to the eight vertices cut the octagon into $8$ triangles. Each has base $\frac{43}{99}$ lying on a side of one of the unit squares, so its height from $O$ is the distance from the center to that side, namely $\frac{1}{2}.$ The area is $8 \cdot \frac{1}{2} \cdot \frac{43}{99} \cdot \frac{1}{2} = \frac{86}{99}.$

Since $\gcd(86, 99) = 1,$ the answer is $86 + 99 = 185.$

1999 AIME Problem 4

Solution:

Problem 4 in Other Years