1959 AMC 12 Problem 43

Attempt Problem 43 of the 1959 AMC 12 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 1959 AMC 12 solutions, or check the answer key.

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

43.

The sides of a triangle are 25,25, 39,39, and 40.40. The diameter of the circumscribed circle is:
1333\dfrac{133}{3}
1253\dfrac{125}{3}
4242
4141
4040
Answer: B
Concepts:Heron’s Formulacircumcircle, circumcenter, and circumradiustriangle area
Small hint:
Use Heron’s formula with semiperimeter 5252
Big hint:
After finding the area K,K, use abc=4KRabc=4KR and double the circumradius
Solution:
The semiperimeter is 52,52, so Heron’s formula gives K=52⋅27⋅13⋅12=468. K=\sqrt{52\cdot27\cdot13\cdot12}=468. If RR is the circumradius, then R=25⋅39⋅404⋅468=1256. R=\frac{25\cdot39\cdot40}{4\cdot468}=\frac{125}{6}. Hence the diameter is 2R=1253.2R=\frac{125}{3}.
Thus, the correct answer is B.
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