2005 AMC 8 Problem 23

Attempt Problem 23 of the 2005 AMC 8 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 2005 AMC 8 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

23.

Isosceles right triangle ABCABC encloses a semicircle of area 2π.2\pi. The circle has its center OO on hypotenuse AB‾\overline{AB} and is tangent to sides AC‾\overline{AC} and BC‾.\overline{BC}. What is the area of triangle ABC?ABC?
66
88
3π3 \pi
1010
4π4 \pi
Answer: B
Concepts:circle areasymmetry
Small hint:
Reflect the triangle and semicircle across the hypotenuse.
Big hint:
This makes a circle inscribed in a square.
Solution:
Reflect the triangle and semicircle across hypotenuse AB‾\overline{AB}. This forms a full circle inscribed in a square.
The semicircle has area 2π2\pi, so the full circle has area 4π4\pi and radius 22. The square side length is therefore 44.
The square area is 1616, so the original triangle has half that area, 88.
Thus, B is the correct answer.
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