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.

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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
Difficulty rating: 1560
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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