2022/23 AHSMC Part 1 Problem 2

Problem 2 of 16

Suppose $r, s > 0$ . A disk of radius $r$ cm is cut into nine pieces and a second disk of radius $(r+s)$ cm is cut into sixteen pieces. If each of the twenty-five pieces has equal area then $r/s$ is

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Solution

We have $\dfrac{\pi r^2}{9} = \dfrac{\pi(r+s)^2}{16}$ , thus $16r^2 = 9(r+s)^2$ giving $4r = \pm 3(r+s)$ . Since $r, s > 0$ , it must be that $4r = 3(r+s)$ implying that $r = 3s$ . Thus $r/s = 3$ . The answer is (D).

University of Alberta

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