1999 AIME Problem 12

Attempt Problem 12 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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12.

The inscribed circle of triangle ABCABC is tangent to AB\overline{AB} at P,P, and its radius is 21.21. Given that AP=23AP = 23 and PB=27,PB = 27, find the perimeter of the triangle.

Answer: 345
Concepts:incircle, incenter, and inradiusHeron’s Formulatriangle area
Difficulty rating: 2390
Solution:

Tangent segments from a point are equal, so the tangent lengths from A,A, B,B, CC are 23,23, 27,27, and some z.z. Then the sides are 50,50, 23+z,23 + z, 27+z,27 + z, the semiperimeter is s=50+z,s = 50 + z, and Heron's formula gives area (50+z)z2327.\sqrt{(50 + z) \cdot z \cdot 23 \cdot 27}.

The area also equals rs=21(50+z).rs = 21(50 + z). Squaring 21(50+z)=621z(50+z)21(50 + z) = \sqrt{621 z (50 + z)} and dividing by 50+z50 + z gives 441(50+z)=621z,441(50 + z) = 621 z, so 180z=22050180 z = 22050 and z=2452.z = \frac{245}{2}.

The perimeter is 2s=2(50+2452)=345.2s = 2\left(50 + \frac{245}{2}\right) = 345.

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