2007/08 AHSMC Part 1 Problem 13

Problem 13 of 16

The diameter $AC$ divides a circle into two semicircular arcs. $B$ is the midpoint of one these arcs, and $D$ is any point on the other arc. If the area of $ABCD$ is $16$ square centimetres, the distance, in centimetres, from $B$ to $AD$ is

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Let $E$ be the point on $AD$ such that $BE$ is perpendicular to $AD$ . Complete the rectangle $BEDF$ . Now $AB = BC$ , $\angle AEB = 90^\circ = \angle CFB$ and $\angle ABE = 90^\circ - \angle CBE = \angle CBF$ . Hence $ABE$ and $CBF$ are congruent triangles and they have equal area. It follows that $BEDF$ is a square, and its area is also $16$ . Hence $BE = 4$ . The answer is (C).

University of Alberta

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