1999 AIME Problem 4

Attempt Problem 4 of the 1999 AIME below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1999 AIME solutions, or check the answer key.

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4.

The two squares shown share the same center OO and have sides of length 1.1. The length of AB\overline{AB} is 4399\frac{43}{99} and the area of octagon ABCDEFGHABCDEFGH is mn,\frac{m}{n}, where mm and nn are relatively prime positive integers. Find m+n.m + n.

Answer: 185
Concepts:square (geometry)area decompositionsymmetry
Difficulty rating: 2350
Solution:

The whole configuration is unchanged by rotating 9090^\circ about O,O, which cycles the octagon side ABAB to CD,CD, EF,EF, GH,GH, and it is also unchanged by the reflection that swaps the two squares, which carries those sides to BC,BC, DE,DE, FG,FG, HA.HA. So all eight sides of the octagon have the same length, 4399.\frac{43}{99}.

Segments from OO to the eight vertices cut the octagon into 88 triangles. Each has base 4399\frac{43}{99} lying on a side of one of the unit squares, so its height from OO is the distance from the center to that side, namely 12.\frac{1}{2}. The area is 812439912=8699.8 \cdot \frac{1}{2} \cdot \frac{43}{99} \cdot \frac{1}{2} = \frac{86}{99}.

Since gcd(86,99)=1,\gcd(86, 99) = 1, the answer is 86+99=185.86 + 99 = 185.

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