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Geometry Difficulty 4.5 AIME Prove it Japan

$AB$ is a segment on a plane with length $7$ , and $P$ is a point such that the distance between $P$ and line $AB$ is $3$ . Find the smallest possible value of $AP \times BP$ .

Solution

Take $\angle APB = \theta$ and let $S$ be the area of $APB$ . Then $\frac{1}{2} \times AP \times BP \times \sin \theta = S = \frac{3 \times 7}{2} = \frac{21}{2}$ . Since $\sin \theta$ is positive, $AP \times BP$ takes minimum value when $\sin \theta$ takes maximum. Since $\frac{7}{2} > 3$ , we can take $P$ on the circle with diameter $AB$ . Then $\sin \theta$ takes maximum value $1$ and $AP \times BP$ takes minimum value $21$ .

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