2008/09 AHSMC Part 1 Problem 13

Problem 13 of 16

On a vertical wall are drawn four geometric figures with equal area. The first is an equilateral triangle with a side along the floor. The second is an isosceles right triangle with one of the short sides along the floor. The third is a square with a side along the floor. The fourth is a circle tangent to the floor. The figure which reaches the highest point above the floor is

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Solution

The height of the circle is $2r$ where $r$ is its radius. From $\pi r^2 = 1$ , we have $2r = \dfrac{2}{\sqrt{\pi}}$ . Since $2 < \pi < 4$ , this is greater than $\dfrac{2}{\sqrt{4}} = 1$ , the height of the square, and less than $\dfrac{2}{\sqrt{2}} = \sqrt{2}$ , the height of the right isosceles triangle. The height of the equilateral triangle is $\sqrt{3}\,s$ where $2s$ is its side length. From $\sqrt{3}\,s^2 = 1$ , we have $\sqrt{3}\,s = \sqrt[4]{3}$ . Since $3 < 4$ , $\sqrt[4]{3} < \sqrt[4]{4} = \sqrt{2}$ . The answer is (B).

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