Olympiad Maths Prep

Track / Stage 7 / 135 of 300 #1535 of 2000

Problem 1535

National olympiad second round; IMO P1/P4
Geometry Difficulty 7.3 Prove it

A,B,C,DA, B, C, D are consecutive vertices of a regular 77-gon. ALAL and AMAM are tangents to the circle center CC radius CBCB. NN is the intersection point of ACAC and BDBD. Show that L,M,NL, M, N are collinear.

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. Identify the given elements and their properties:
- A,B,C,DA, B, C, D are consecutive vertices of a regular 77-gon.
- ALAL and AMAM are tangents to the circle centered at CC with radius CBCB.
- NN is the intersection point of ACAC and BDBD.

2. Establish the cyclic nature of the polygons:
- Since A,B,C,DA, B, C, D are vertices of a regular 77-gon, they lie on a common circle (the circumcircle of the 77-gon).
- Points L,B,M,DL, B, M, D lie on a circle with center CC and radius CBCB because LL and MM are points where tangents from AA touch the circle centered at CC.

3. **Prove that ALCMALCM is cyclic:**
- Since ALAL and AMAM are tangents to the circle centered at CC, we have CLALCL \perp AL and CMAMCM \perp AM.
- This implies that CLA=CMA=90\angle CLA = \angle CMA = 90^\circ.
- Therefore, quadrilateral ALCMALCM is cyclic because opposite angles sum to 180180^\circ.

4. Apply the Radical Axis Theorem:
- The Radical Axis Theorem states that for three circles, the radical axes of each pair of circles are concurrent.
- Consider the three circles:
- The circumcircle of ABCDABCD.
- The circle with center CC and radius CBCB (containing L,B,M,DL, B, M, D).
- The circle containing ALCMALCM.
- The radical axis of the circumcircle of ABCDABCD and the circle containing L,B,M,DL, B, M, D is line BDBD.
- The radical axis of the circumcircle of ABCDABCD and the circle containing ALCMALCM is line ACAC.
- The radical axis of the circle containing L,B,M,DL, B, M, D and the circle containing ALCMALCM is line LMLM.

5. **Show concurrency of ACAC, BDBD, and LMLM:**
- By the Radical Axis Theorem, the lines ACAC, BDBD, and LMLM are concurrent.
- Since NN is the intersection point of ACAC and BDBD, it must also lie on LMLM.

6. **Conclude that L,M,NL, M, N are collinear:**
- Since NN lies on LMLM, points L,M,andNL, M, and N are collinear.

None \boxed{\text{None}}

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