1952 AMC 12 Problem 27

Attempt Problem 27 of the 1952 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 1952 AMC 12 solutions, or check the answer key.

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

27.

The ratio of the perimeter of an equilateral triangle having an altitude equal to the radius of a circle, to the perimeter of an equilateral triangle inscribed in the circle is:
1:21:2
1:31:3
1:31:\sqrt3
3:2\sqrt3:2
2:32:3
Answer: E
Concepts:equilateral trianglecircleratio and proportion
Small hint:
Express each equilateral triangle’s side length in terms of the same circle radius RR
Big hint:
An equilateral triangle of altitude RR has side 2R3\frac{2R}{\sqrt3}, while an inscribed one has side 3R\sqrt3R
Solution:
An equilateral triangle with side ss has altitude s32.\frac{s\sqrt3}{2}. Thus the first triangle has side 2R3\frac{2R}{\sqrt3} and perimeter 23R.2\sqrt3R. An equilateral triangle inscribed in a circle of radius RR has side 3R\sqrt3R and perimeter 33R.3\sqrt3R.
The perimeter ratio is therefore 2:3,2:3, so the correct answer is E.
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