Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 7.5 National olympiad, round 2 Find the answer

Let $ABC$ be a fixed acute triangle inscribed in a circle $\omega$ with center $O$ . A variable point $X$ is chosen on minor arc $AB$ of $\omega$ , and segments $CX$ and $AB$ meet at $D$ . Denote by $O_1$ and $O_2$ the circumcenters of triangles $ADX$ and $BDX$ , respectively. Determine all points $X$ for which the area of triangle $OO_1O_2$ is minimized.

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

Let $E$ be midpoint $AD.$ Let $F$ be midpoint $BD \implies$ $EF = ED + FD = \frac {AD}{2} + \frac {BD}{2} = \frac {AB}{2}.$ $E$ and $F$ are the bases of perpendiculars dropped from $O_1$ and $O_2,$ respectively.
Therefore $O_1O_2 \ge EF = \frac {AB}{2}.$
$CX \perp O_1O_2, AX \perp O_1O \implies \angle O O_1O_2 = \angle AXC$ $\angle AXC = \angle ABC (AXBC$ is cyclic) $\implies \angle O O_1O_2 = \angle ABC.$
Similarly $\angle BAC = \angle O O_2 O_1 \implies \triangle ABC \sim \triangle O_2 O_1O.$
The area of $\triangle OO_1O_2$ is minimized if $CX \perp AB$ because $\frac {[OO_1O_2]} {[ABC]} = \left(\frac {O_1 O_2} {AB}\right)^2 \ge \left(\frac {EF} {AB}\right)^2 = \frac {1}{4}.$ [email protected], vvsss

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