Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 6.9 National olympiad Prove it

An isosceles triangle $A B C$ with $A C = B C$ is given. Point $D$ is chosen on the side $A C$ . The circle $S_{1}$ of radius $R$ with the center $O_{1}$ touches the segment $A D$ and the extensions of $B A$ and $B D$ over the points $A$ and $D$ , respectively. The circle $S_{2}$ of radius $2 R$ with the center $O_{2}$ touches the segment $D C$ and the extensions of $B D$ and $B C$ over the points $D$ and $C$ , respectively. Let the tangent to the circumcircle of the triangle $B O_{1} O_{2}$ at the point $O_{2}$ intersect the line $B A$ at point $F$ . Prove that $O_{1} F = O_{1} O_{2}$ .

Solution

In the triangle $ABC$ we have $\angle A = \angle B$ . It is evident that $\angle O_{1} B O_{2} = \angle B / 2$ . Let $\ell$ be the straight line passing through $O_{2}$ parallel to $AC$ . By the problem condition $\ell$ touches $S_{1}$ (say, at a point $N$ ). Let also $K$ be the tangency point of $S_{1}$ and $BA$ . Then the clockwise rotation about the point $O_{1}$ through the angle $N O_{1} K$ transposes $\ell$ to $BA$ and thus transposes the point $O_{2}$ to some point $O \in BA$ . Hence $O_{1} O = O_{1} O_{2}$ and $\angle O O_{1} O_{2} = \angle N O_{1} K = 180^{\circ} - \angle A = 180^{\circ} - \angle B$ , so $\angle O_{1} O_{2} O = \angle B / 2 = \angle O_{1} B O_{2}$ . The latter does mean that the line $O_{2} O$ is the tangent to the circumcircle of $\triangle B O_{1} O_{2}$ . Hence $F = O$ , and $O_{1} F = O_{1} O_{2}$ , as was to be proved.

Fig. 1

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