Thứ Sáu, 12 tháng 9, 2014

23-Dao-6 points circle X(5569)

Theorem: Let ABC be a triangle. Circumcenter of six circle tangent three median at the centroid and through three vertex are on a circle

Solution by: Vslmat


 Let A', B', C' are the midpoint of AG, BG, CG, respectively and let the perpendicular bisectors of AG, BG, CG meet at K, L, H. Let A_{1}, A_{2}, B_{1}, B_{2}, C_{1}, C_{2} be the centers of the six mentioned circles.
Easy to see that \angle A_{1}C_{2}H + \angle C_{2}A_{1}H = \angle AGE + \angle EGC (1)
As E is the midpoint of AC, we have \frac{sin AGE}{sin EGC} = \frac{GC}{GA} = \frac{GC'}{GA'}=\frac{GC_{2}}{GA_{1}} (as \Delta GA'A_{1}\sim\ \Delta GC'C_{2})

=\frac{HA_{1}}{HC_{2}} = \frac{sin A_{1}C_{2}H}{sin C_{2}A_{1}H} (2)

(1) and (2) give \angle AGE = \ange A_{1}C_{2}H as well as \angle EGC = \angle C_{2}A_{1}H
But easy to see that \angle EGC = \angle KLH, hence A_{1}C_{2}LK is cyclic and as A_{2}C_{1}\parallel KL, quadrilateral A_{1}C_{2}C_{1}A_{2} is cyclic.
Similarly we have B_{1}A_{2}A_{1}B_{2} is cyclic and at the same time A_{2}A_{1}C_{2B_{1} is cyclic (see the diagram). Therefore, repeating the steps once more we get all the six points A_{1}, A_{2}, B_{1}, B_{2}, C_{1}, C_{2} lie on a circle

Please see X(5569) in Kimberling center
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