A Generalization Simson's line,carnot theorem, Collings-Carn

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A Generalization Simson's line,carnot theorem, Collings-Carn

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#1 h Nov 1, 2013, 5:45 AM • 3 Y

Y by mathmoGJ, Adventure10, Mango247

Dear Mathlinkers!

Let a conic through three point $A,B,C$ , and any point D on the plane. $DA,DB,DC$ respectively intersect the conic at $A_1,B_1,C_1$ . E is a point on the conic. $EA_1$ intersect $BC$ at $A_2$ , $EB_1$ intersect $AC$ at $B_2$ , $EC_1$ intersect $AB$ at $C_2$ . Then $A_2,B_2,C_2,D$ are colinear.

Converse: Let six points $A,B,C,A_1,B_1,C_1$ are on a conic. $AA_1,BB_1,CC_1$ concurrent at $D$ . Construct a line through $D$ intersect $BC,CA,AB$ at $A_2,B_2,C_2$ . Then $A_1A_2,B_1B_2,C_1C_2$ are concurrent on the conic.

Best regard

Dao Thanh Oai

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A Generalization Simson's line,carnot theorem, Collings-Carnot theorem, Adam-Bliss-Ibid problem.pdf (194kb)

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#2 h Nov 3, 2013, 4:25 PM • 2 Y

Y by Adventure10, Mango247

By Pascal theorem for the hexagon $ABCC_1EA_1,$ the intersections $C_2 \equiv AB \cap C_1E,$ $A_2 \equiv BC \cap EA_1$ and $D \equiv CC_1 \cap A_1A$ are collinear. By Pascal theorem for the hexagon $BCAA_1EB_1,$ the intersections $A_2 \equiv BC \cap A_1E,$ $B_2 \equiv CA \cap EB_1$ and $D \equiv AA_1 \cap B_1B$ are collinear $\Longrightarrow$ $A_2,B_2,C_2$ and $D$ are collinear.

The converse is proved in the same way, in fact, it was posted before at The point of concurrency lies on the circumcircle.

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#3 h Nov 7, 2013, 3:17 AM • 2 Y

Y by Adventure10, Mango247

Very cool thing here but I am unsure why D at infinity yields Simson lines...

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9775 posts

jayme

#4 h Jun 11, 2016, 1:38 PM • 2 Y

Y by Adventure10, Mango247

Dear Mathlinkers,
this result for a circle is well-known...

http://jl.ayme.pagesperso-orange.fr/Docs/La%20P-transversale%20de%20Q.pdf

Sincerely
Jean-Louis

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#5 h Jul 12, 2016, 4:42 PM • 2 Y

Y by Adventure10, Mango247

Dear Dr. Jayme,

Please give your proof the following result:

Let $ABC$ be a triangle, let $(S)$ be a circumconic of $ABC$ , let $P$ be a point on the plane. Let the lines $AP, BP, CP$ meet the conic again at $A', B', C'$ . Let $D$ be a point on the polar of point $P$ with respect to $(S)$ or $D$ lies on the conic $(S)$ . Let $A_0= DA' \cap BC$ , $B_0= DB' \cap AC$ ; $C_0= DC' \cap AB$ . Then $A_0, B_0, C_0$ are collinear. Further more four points $A_0, B_0, C_0, P$ are collinear if only if $D$ lie on the conic.

In the case D lies on the conic $(S)$ . Let $AP$ meets $BC$ at $P_a$ . Let the line through $D$ and parallel to $AP$ , this line meets $BC$ at $D_a$ . Let $D'_a$ on the ray $DD_a$ such that $\frac{\overline{DD_a}}{\overline{DD'_a}}=\frac{\overline{A'P_a}}{\overline{A'P}}$ . Define $D'_b, D'_c$ cyclically. Then seven points $A_0, B_0, C_0, D'_a, D'_b, D'_c$ and $P$ are collinear.

(Posed by Dao Thanh Oai)

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This post has been edited 3 times. Last edited by daothanhoai, Jul 12, 2016, 4:45 PM

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