2015/16 AHSMC Part 1 Problem 12

Problem 12 of 16

In the diagram below, which is not drawn to scale, the circles are tangent at $A$ , the centre of the larger circle is at $O$ and the lines $AB$ and $CD$ are perpendicular.

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If $EB = 3$ and $FC = 2$ then the radius of the smaller circle is

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Solution

Let $R$ , $r$ denote the lengths of the large and respectively the small radius and $EB = a$ , $FC = b$ . First $a = 2R - 2r$ so that $R = r + a/2$ . If $O'$ denotes the centre of the smaller circle then $O'F^2 = O'O^2 + OF^2$ hence $r^2 = \frac{a^2}{4} + \Bigl(r + \frac{a}{2} - b\Bigr)^2.$ Solving for $r$ we get $r = \frac{(2b-a)^2 + a^2}{4(2b-a)}.$ Taking $a = 3$ , $b = 2$ we get $r = \dfrac{5}{2}$ . The answer is (C).

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