Track / Stage 7 / 206 of 300 #1606 of 2000

Problem 1606

National olympiad second round; IMO P1/P4

Geometry Difficulty 7.4 Prove it

The quadrilateral $ABCD$ is inscribed in a circle with center $O$ . The diagonals $AC$ and $BD$ do not pass through $O$ . If the circumcentre of triangle $AOC$ lies on the line $BD$ , prove that the circumcentre of triangle $BOD$ lies on the line $AC$ .

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. Define the circumcenters and reflection:
Let $P$ and $Q$ be the circumcenters of triangles $AOC$ and $BOD$ respectively. Let $R$ be the reflection of $O$ in the line $BD$ .

2. **Circumcenter $P$ on line $BD$ :**
Since $P$ is the circumcenter of $\triangle AOC$ and lies on the line $BD$ , we have $PR = PO$ . This implies that $R$ lies on the circumcircle of $\triangle AOC$ .

3. **Inversion about circle $(ABCD)$ :**
Consider the inversion about the circumcircle $(ABCD)$ . This inversion maps the point $R$ to the point $Q$ and the circumcircle $(AOC)$ to the line $AC$ .

4. **Conclusion about point $Q$ :**
Since $R$ lies on the circumcircle $(AOC)$ , under the inversion, $Q$ must lie on the line $AC$ .

Thus, we have shown that if the circumcenter of $\triangle AOC$ lies on the line $BD$ , then the circumcenter of $\triangle BOD$ lies on the line $AC$ .

$\blacksquare$

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