Maths Olympiad Prep

Track / Stage 7 / 218 of 300 #1618 of 1964

Problem 1618

National Olympiad second round; IMO P1/P4
Geometry Difficulty 7.4 Prove it

Given is a triangle ABCABC and MM is the midpoint of the minor arc BCBC. Let M1M_1 be the reflection of MM with respect to side BCBC. Prove that the nine-point circle bisects AM1AM_1.

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

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Official solution

1. Identify Key Points and Definitions:
- Let HH be the orthocenter of △ABC\triangle ABC.
- MM is the midpoint of the minor arc BCBC on the circumcircle (ABC)(ABC).
- M1M_1 is the reflection of MM with respect to side BCBC.

2. Symmetry and Reflection:
- Reflecting MM across BCBC to get M1M_1 implies that M1M_1 lies on the circumcircle (BHC)(BHC) because reflecting MM across BCBC maps the circumcircle (ABC)(ABC) to (BHC)(BHC).

3. Homothety Argument:
- Consider a homothety centered at AA with a coefficient of 12\frac{1}{2}. This homothety maps the circumcircle (BHC)(BHC) to the nine-point circle of △ABC\triangle ABC.
- Under this homothety, the point M1M_1 maps to the midpoint of AM1AM_1.

4. Nine-Point Circle:
- The nine-point circle of △ABC\triangle ABC passes through the midpoints of the sides of △ABC\triangle ABC, the feet of the altitudes, and the midpoints of the segments joining the orthocenter HH to the vertices of △ABC\triangle ABC.
- Since the homothety maps M1M_1 to the midpoint of AM1AM_1, and the nine-point circle is invariant under this homothety, the midpoint of AM1AM_1 must lie on the nine-point circle.

5. Conclusion:
- Therefore, the nine-point circle bisects AM1AM_1.

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.