Thứ Sáu, 12 tháng 9, 2014

7-Maybe Thebault have ovelook

Let ABC be a triangle, any P on the plane. Construct a circle tangent CP, ABcircumcircle at C_A; Construct another circle tangent CP,AB circumcircle at C_B. Prove that C_AC_B through fixed point, when P moved


http://www.geogebratube.org/student/m68511

Solution by: Luis González:

Label (J_A),(J_B) the Thebault circles of the cevian CP. (J_A) touches \overline{AB} at Xand the circumcircle (O) at C_A while (J_B) touches \overline{AB} at Y and the circumcircle at C_B. From internal tangencies of (O),(J_A) and (O),(J_B), we deduce that C_AX, C_BY bisect \angle AC_AB, \angle AC_BB \Longrightarrow M \equiv XC_A \cap YC_B is midpoint of the arc AB of (O). 

By Sawayama's lemma, the incenter I of \triangle ABC verifies IX \perp IY \Longrightarrow pencilsIX,IY are in involution \Longrightarrow pencils MX \equiv MC_A, MY \equiv MC_B are in involution \Longrightarrow C_A \mapsto  C_B is an involutive homography on (O) \Longrightarrow all lines C_AC_B pass through the fixed pole of the involution. Making A \equiv Y and B \equiv X, we figure out that the fixed point is the exsimilicenter X_{56} of (I) \sim (O).

Solution 2 by Telv Cohl https://www.facebook.com/telv.cohl?fref=ts

Another proof of this problem

http://www.cut-the-knot.org/m/Geometry/ThebaultMiss.shtml

First, rewrite the problem as following:

Giving a triangle ABC and point D on AB,Construct two Thebault circles (O_1)and (O_2) and their points of tangency with the circumcircle, X and Y.Prove that line XY passes through a fixed point(when D changes).

S = intersection of O_1O_2 and AB
I = incenter of triangle ABC
O = circumcenter of triangle ABC



By Thebault theorem,we deduce that O_1, I, O_2 are collinear,so the external center of similitude of (I) and (O1) and the external center of similitude of (O_1) and (O_2) coincide with each other.Because X is the external center of similitude of (O_1) and (O),S is the external center of similitude of (I) and (O_1).By D'Alembert theorem (consider (I),(O) and (O_1)),we deduce that SX pass through the external center ofsimilitude of (I) and (O),namely X(56).By same reason we can deduce that S, Y, X(56) are collinear,so XY pass through X(56),which is a fixed point. Q.E.D

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