HMMT 二月 2021 · 团队赛 · 第 7 题

HMMT February 2021 — Team Round — Problem 7

[70] In triangle ABC , let M be the midpoint of BC and D be a point on segment AM . Distinct points Y − → − − → and Z are chosen on rays CA and BA , respectively, such that ∠ DY C = ∠ DCB and ∠ DBC = ∠ DZB . Prove that the circumcircle of 4 DY Z is tangent to the circumcircle of 4 DBC .

解析

[70] In triangle ABC , let M be the midpoint of BC and D be a point on segment AM . Distinct points Y − → − − → and Z are chosen on rays CA and BA , respectively, such that ∠ DY C = ∠ DCB and ∠ DBC = ∠ DZB . Prove that the circumcircle of 4 DY Z is tangent to the circumcircle of 4 DBC . Proposed by: Joseph Heerens Solution 1: We first note that the circumcircles of DBZ and Y DC are tangent to BC from our angle criteria. By power of a point, we obtain that M lies on the radical axis of the two circles and clearly D does as well. Therefore, we find that A lies on the radical axis so AY · AC = AB · AZ implying that BY ZC is a cyclic quadrilateral. ′ ′ Next, by Reim’s Theorem on ( BY ZC ) and ( DY Z ), we get that ( DY Z ) intersects AB, AC at B , C ′ ′ ′ ′ where BC, B C are parallel. Then a negative homothety maps B to B and C to C , so ( DBC ) gets ′ ′ mapped to ( DB C ), and we have tangent circles. ′ ′ ′ ′ Solution 2: Let ( DY Z ) intersect AB and AC at B and C , respectively. We see that ] Y C B = ′ ′ ′ ] Y ZB = ] Y ZB = ] Y CB. Thus, BC ‖ B C . This means that there exists a negative homothety ′ ′ ′ ′ taking B to B and C to C which will map ( DBC ) to ( DB C ) which is also ( DY Z ).