Track / Stage 6 / 113 of 400 #1113 of 2000

Problem 1113

National olympiad, first round

Geometry Difficulty 6.1 Prove it

391. Given two intersecting circles and a line passing through their points of intersection. Prove that the square of the tangent segment, drawn from any point on one circle to the other, is proportional to the distance from this point to the line.

This one wants a proof. Work it on paper, read the official solution, then mark yourself honestly — the ladder only means something if the record is true.

Official solution

391. Connect an arbitrary point $P$ of circle $O_{1}$ (from which a tangent $P T$ to circle $O_{2}$ is drawn) with the centers $O_{1}$ and $O_{2}$ (Fig. 47). Denoting the projection of segment $P O_{2}$ onto the line of centers by $K O_{2}$ , and the projection of radius $O_{2} A$ , drawn to the point of intersection of the circles, onto the line of centers by $L O_{2}$ , we have that the distance
$h$ from point $P$ to the line is the difference between segments $K O_{2}$ and $L O_{2}$ . The latter can be determined using the cosine theorem from triangles $O_{1} P O_{2}$ and $O_{1} A O_{2}$ . Considering that $P O_{2}^{2}=O_{2} T^{2}+P T^{2}$ , we get that $P T^{2}=2 O_{2} O_{1} \cdot h$

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.