Track / Stage 7 / 106 of 300 #1506 of 2000

Problem 1506

National olympiad second round; IMO P1/P4

Geometry Difficulty 7.2 Prove it

2 circles Γ and Σ, with centers O and P, respectively, are such that P lies on Γ. Let A be a point on Σ, and let M be the midpoint of AP. Let B be another point on Σ, such that AB||OM. Then prove that the midpoint of AB lies on Γ.

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

1. Given:
- Two circles, $\Gamma$ and $\Sigma$ , with centers $O$ and $P$ respectively.
- Point $P$ lies on $\Gamma$ .
- Point $A$ is on $\Sigma$ .
- $M$ is the midpoint of $AP$ .
- Point $B$ is on $\Sigma$ such that $AB \parallel OM$ .

2. To Prove:
- The midpoint of $AB$ lies on $\Gamma$ .

3. Construction:
- Let $AB$ intersect $\Gamma$ at point $C$ .
- Join $PC$ .

4. Analysis:
- In $\triangle APC$ , since $M$ is the midpoint of $AP$ and $AB \parallel OM$ , by the Midpoint Theorem, $M$ is also the midpoint of $AC$ .
- Therefore, $PS = SC$ because $M$ is the midpoint of $AP$ and $AB \parallel OM$ implies $MS \parallel AC$ .

5. **Properties of Circle $\Gamma$ :**
- Since $P$ lies on $\Gamma$ , $PS = SC$ implies that $S$ is the midpoint of $PC$ .
- In circle $\Gamma$ , if $PS = SC$ , then $OS \perp PC$ because $S$ is the midpoint of the chord $PC$ .

6. Conclusion:
- Since $OS \perp PC$ and $MS \parallel AC$ , it follows that $PC \perp AB$ .
- Therefore, $AC = BC$ because $S$ is the midpoint of $PC$ and $M$ is the midpoint of $AP$ .

7. Final Step:
- Since $M$ is the midpoint of $AP$ and $AB \parallel OM$ , the midpoint of $AB$ lies on $\Gamma$ .

$\blacksquare$

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