Maths Olympiad Prep

Track / Stage 8 / 35 of 180 #1735 of 1964

Problem 1735

IMO Shortlist mid-range; USAMO P2/P5
Geometry Difficulty 8.1 Prove it

Consider a convex pentagon circumscribed about a circle. We name the lines that connect vertices of the pentagon with the opposite points of tangency with the circle gergonnians.
(a) Prove that if four gergonnians are conncurrent, the all five of them are concurrent.
(b) Prove that if there is a triple of gergonnians that are concurrent, then there is another triple of gergonnians that are concurrent.

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.

Official solution

Let's break down the problem and solution step-by-step, leveraging Brianchon's Theorem as suggested.

### Part (a)
1. Define the vertices and touch points:
Let the vertices of the pentagon be A,B,C,D,E A, B, C, D, E and the points where the incircle touches the sides be A′,B′,C′,D′,E′ A', B', C', D', E' respectively.

2. State Brianchon's Theorem:
Brianchon's Theorem states that for a hexagon circumscribed about a conic section, the three diagonals connecting opposite vertices are concurrent.

3. Apply Brianchon's Theorem to different hexagons:
- Consider the hexagon AEDA′ED′ AEDA'ED' . By Brianchon's Theorem, the diagonals AA′ AA' , DD′ DD' , and EE1 EE_1 are concurrent.
- Consider the hexagon ABCA′BC′ ABCA'BC' . By Brianchon's Theorem, the diagonals AA′ AA' , CC′ CC' , and BB1 BB_1 are concurrent.

4. Relate the concurrency conditions:
From the above, we have:
AA′,DD′,EE1 concurrentandAA′,CC′,BB1 concurrent AA', DD', EE_1 \text{ concurrent} \quad \text{and} \quad AA', CC', BB_1 \text{ concurrent}
This implies:
AA′,CC′,DD′ concurrent  ⟺  AA′,BB1,EE1 concurrent AA', CC', DD' \text{ concurrent} \iff AA', BB_1, EE_1 \text{ concurrent}

5. Apply Brianchon's Theorem to another hexagon:
- Consider the hexagon A1BE1A′BE A_1BE_1A'BE . By Brianchon's Theorem, the diagonals A′A1 A'A_1 , BB1 BB_1 , and EE1 EE_1 are concurrent.
- This implies:
AA′,BB1,EE1 concurrent  ⟺  A,A′,A1 collinear AA', BB_1, EE_1 \text{ concurrent} \iff A, A', A_1 \text{ collinear}

6. Final application of Brianchon's Theorem:
- Consider the hexagon ABE′A1B′E ABE'A_1B'E . By Brianchon's Theorem, the diagonals AA1 AA_1 , BB′ BB' , and EE′ EE' are concurrent.
- This implies:
A,A′,A1 collinear  ⟺  AA′,BB′,EE′ concurrent A, A', A_1 \text{ collinear} \iff AA', BB', EE' \text{ concurrent}

7. Conclusion:
Stringing together these results, we have shown that:
AA′,BB′,EE′ concurrent  ⟺  AA′,CC′,DD′ concurrent AA', BB', EE' \text{ concurrent} \iff AA', CC', DD' \text{ concurrent}
Therefore, if four gergonnians are concurrent, then all five of them are concurrent.

### Part (b)
1. Given condition:
Assume there is a triple of gergonnians that are concurrent.

2. Identify the concurrent triples:
Without loss of generality, assume AA′,BB′,CC′ AA', BB', CC' are concurrent.

3. Apply Brianchon's Theorem:
- Consider the hexagon AEDA′ED′ AEDA'ED' . By Brianchon's Theorem, the diagonals AA′ AA' , DD′ DD' , and EE1 EE_1 are concurrent.
- Consider the hexagon ABCA′BC′ ABCA'BC' . By Brianchon's Theorem, the diagonals AA′ AA' , CC′ CC' , and BB1 BB_1 are concurrent.

4. Relate the concurrency conditions:
From the above, we have:
AA′,DD′,EE1 concurrentandAA′,CC′,BB1 concurrent AA', DD', EE_1 \text{ concurrent} \quad \text{and} \quad AA', CC', BB_1 \text{ concurrent}
This implies:
AA′,CC′,DD′ concurrent  ⟺  AA′,BB1,EE1 concurrent AA', CC', DD' \text{ concurrent} \iff AA', BB_1, EE_1 \text{ concurrent}

5. Conclusion:
Since AA′,BB′,CC′ AA', BB', CC' are concurrent, by the above steps, we can find another triple of gergonnians that are concurrent, such as AA′,DD′,EE′ AA', DD', EE' .

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