Maths Olympiad Prep

Track / Stage 7 / 222 of 300 #1622 of 1964

Problem 1622

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

Point FF lies on the circumscribed circle around ΔABC\Delta ABC, PP and QQ are projections of point FF on ABAB and ACAC respectively. Prove that, if MM and NN are the middle points of BCBC and PQPQ respectively, then MNMN is perpendicular to FNFN.

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 Projections:
- Let FF be a point on the circumscribed circle of ΔABC\Delta ABC.
- Let PP and QQ be the projections of FF onto ABAB and ACAC, respectively.
- Let MM be the midpoint of BCBC.
- Let NN be the midpoint of PQPQ.

2. Simson Line:
- By the Simson line theorem, the points PP, QQ, and the foot of the perpendicular from FF to BCBC (let's call this point KK) are collinear.

3. Cyclic Quadrilateral:
- Since PP and QQ are projections of FF onto ABAB and ACAC, respectively, the quadrilateral PAQFPAQF is cyclic.
- This implies that ∠FPQ=∠FAQ\angle FPQ = \angle FAQ and ∠PFQ=∠PAQ\angle PFQ = \angle PAQ.

4. Angle Relationships:
- Since FF lies on the circumcircle of ΔABC\Delta ABC, ∠FBC=∠FAC\angle FBC = \angle FAC and ∠BFC=∠BAC\angle BFC = \angle BAC.
- Therefore, ∠FPQ=∠FBC\angle FPQ = \angle FBC and ∠PFQ=∠BFC\angle PFQ = \angle BFC.

5. Similarity of Triangles:
- From the above angle relationships, we have △PFQ∼△BFC\triangle PFQ \sim \triangle BFC by AA similarity criterion.

6. Medians:
- FNFN is the median of △PFQ\triangle PFQ.
- FMFM is the median of △BFC\triangle BFC.

7. **Cyclic Quadrilateral FNMKFNMK:**
- Since FNFN and FMFM are medians, and ∠FNK=∠FMK\angle FNK = \angle FMK, the quadrilateral FNMKFNMK is cyclic.
- This implies that ∠FNM=∠FKM=90∘\angle FNM = \angle FKM = 90^\circ.

8. Conclusion:
- Since ∠FNM=90∘\angle FNM = 90^\circ, MNMN is perpendicular to FNFN.

MN is perpendicular to FN \boxed{\text{MN is perpendicular to FN}}

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