Track / Stage 6 / 61 of 400 #1061 of 2000

Problem 1061

National olympiad, first round

Geometry Difficulty 6.0 Prove it

9. 7 (IMO 42 Pre) Let $O$ be a point inside an acute-angled $\triangle ABC$ , and let $OA_1$ be perpendicular to $BC$ , with the foot of the perpendicular being $A_1$ . Similarly, define $B_1$ on $CA$ and $C_1$ on $AB$ . Prove that $O$ is the circumcenter of $\triangle ABC$ if and only if the perimeter of $\triangle A_1B_1C_1$ is not less than the perimeter of any of $\triangle AB_1C_1$ , $\triangle BC_1A_1$ , and $\triangle CA_1B_1$ .

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

9.7 The necessity is easy to prove: If $O$ is the circumcenter of $\triangle ABC$ , then the perimeters of the four smaller triangles are all equal. The conclusion holds.

Conversely, assuming the perimeter of $\triangle ABC$ is not less than that of any of the other three triangles, we can use proof by contradiction, addressing each case individually. Construct the parallelogram $A B_{1} A_{2} C_{1}$ , and let $\angle CAB = \alpha, \angle CA_{1}B_{1} = \alpha_{1}, \angle BA_{1}C_{1} = \alpha_{2}, \angle CB_{1}A_{1} = \beta_{2}$ , $\angle BC_{1}A_{1} = \gamma_{1}$ , etc.

Assume $\gamma_{1} \geqslant \alpha, \beta_{2} \geqslant \alpha$ . If either of these inequalities is strict, then $A_{1}$ is inside $\triangle B_{1}C_{1}A_{2}$ , and not a vertex. Then the perimeter of $\triangle A_{1}B_{1}C_{1}$ , $CO, CO > BO, BO > AO$ , is a contradiction.

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