Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.2 AIME, harder Prove it United States

Problem:

Circles $\omega_{1}$ , $\omega_{2}$ , and $\omega_{3}$ are centered at $M$ , $N$ , and $O$ , respectively. The points of tangency between $\omega_{2}$ and $\omega_{3}$ , $\omega_{3}$ and $\omega_{1}$ , and $\omega_{1}$ and $\omega_{2}$ are tangent at $A$ , $B$ , and $C$ , respectively. Line $MO$ intersects $\omega_{3}$ and $\omega_{1}$ again at $P$ and $Q$ respectively, and line $AP$ intersects $\omega_{2}$ again at $R$ . Given that $ABC$ is an equilateral triangle of side length $1$ , compute the area of $PQR$ .

Solution

Solution:

$\boxed{2 \sqrt{3}}$ . Note that $ONM$ is an equilateral triangle of side length $2$ , so $m \angle BPA = m \angle BOA / 2 = \pi / 6$ . Now $BPA$ is a $30$ - $60$ - $90$ triangle with short side length $1$ , so $AP = \sqrt{3}$ . Now $A$ and $B$ are the midpoints of segments $PR$ and $PQ$ , so

$[PQR] = \frac{PR}{PA} \cdot \frac{PQ}{PB} [PBA] = 2 \cdot 2 [PBA] = 2 \sqrt{3}$

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.