Two triangles $ABC$ and $DEF$ have the same incircle. If a circle passes through $A,B,C,D,E$ prove that it also passes through $F$.
Problem
Source: India Postal Set 2 P 5 2016
Tags: geometry, incircle, 3D geometry, prism
09.03.2017 20:29
Bumpp...
09.03.2017 21:37
This is the rather well-known Poncelet's Theorem Let $\triangle A_1B_1C_1$ and $\triangle D_1E_1F_1$ be the contact triangles of $\triangle ABC$ and $\triangle DEF$, respectively in the common incircle $\omega$. Set $P_{XY}$ as the midpoint of $X_1Y_1$ where $X, Y \in \{A, B, C, D, E, F\}$. Lemma Points $A, B, C, D, E, F$ lie on a conic. (Proof) Note that $$C(DE, AB) \overset{C}{=} (D, E; CA \cap DE, CB \cap DE) \overset{I}{=} (P_{FD}, P_{FE}; P_{FB}, P_{FA}) \overset{F_1}{=} (D_1, E_1; B_1, A_1).$$It follows that $C(DE, AB)=F(DE, AB)$ and so the six points lie on a conic. $\square$ Finally, we remark that a non-degenerate conic is uniquely determined by five points on it, so $F$ lies on the circle passing through $A, B, C, D, E$ and we may conclude.
10.03.2017 11:16
anantmudgal09 wrote: This is the rather well-known Poncelet's Theorem Let $\triangle A_1B_1C_1$ and $\triangle D_1E_1F_1$ be the contact triangles of $\triangle ABC$ and $\triangle DEF$, respectively in the common incircle $\omega$. Set $P_{XY}$ as the midpoint of $X_1Y_1$ where $X, Y \in \{A, B, C, D, E, F\}$. Lemma Points $A, B, C, D, E, F$ lie on a conic. (Proof) Note that $$C(DE, AB) \overset{C}{=} (D, E; CA \cap DE, CB \cap DE) \overset{I}{=} (P_{FD}, P_{FE}; P_{FB}, P_{FA}) \overset{F_1}{=} (D_1, E_1; B_1, A_1).$$It follows that $C(DE, AB)=F(DE, AB)$ and so the six points lie on a conic. $\square$ Finally, we remark that a non-degenerate conic is uniquely determined by five points on it, so $F$ lies on the circle passing through $A, B, C, D, E$ and we may conclude. A special case of it I must say. My solution was to just quote Poncelet's theorem and then prove it and use it's very very special case as the problem.
10.03.2017 13:14
@above, they aren't that different. Use projective transformation to send the inconic to the incircle and proceed with the same proof as shown here
10.03.2017 13:23
Yeah precisely. My solution used the Poncelet's theorem for circles and used $n=3$.
23.03.2017 21:07
where can i get an article for it.