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Problem 685

AMC 10/12, early questions

Geometry Difficulty 3.8 Prove it Macedonian Junior Mathematical Olympiad · North Macedonia

A circle $k$ with center at $O$ and radius $r$ and a line $p$ which doesn't have a common point with $k$ are given. Let $E$ be the foot of the perpendicular from $O$ to $p$ . An arbitrary point $M$ different from $E$ is chosen on $p$ and the two tangents are drawn from $M$ to $k$ which touch the circle $k$ at points $A$ and $B$ . If $H$ is the intersection of $AB$ and $OE$ , prove that $\overline{OH} = \frac{r^2}{OE}$ .

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

Let $G$ be the intersection of $OM$ and $AB$ . Since $\triangle OGH \sim \triangle OEM$ we get $\overline{OG} = \overline{OE}$ hence $\overline{OE} \cdot \overline{OH} = \overline{OM} \cdot \overline{OG}$ . On the other hand, since $\triangle AOG \sim \triangle MOA$ , we have $\overline{OA} = \overline{OM}/\overline{OA}$ . Therefore $\overline{OM} \cdot \overline{OG} = \overline{OA}^2$ . We get $\overline{OH} = \overline{OA}^2 / \overline{OE}$ .

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