Let D be an arbitrary point on side AB of a given triangle ABC, and let E be the interior point where CD intersects the external common tangent to the incircles of triangles ACD and BCD. As D assumes all positions between A and B, prove that the point E traces the arc of a circle.
Problem
Source: USAMO 1991
Tags: geometry, geometry solved
27.10.2005 13:48
Let U,V be the points where the incircles of ADC,BDC touch AB, respectively. Suppose we manage to show that UV=ED. Then CE=CD−ED=CD−UV, while UV=DU+DV=DA+DC−AC2+DB+DC−BC2, which proves that CE=CD−UV is constant, so E moves on a circle centered at C. This means that all that's left to prove is that UV=ED. Let T,W be the points where the second external common tangent touches the incircles of ADC,BDC respectively. We have ED=TE+UD=EW+DV, and also UV=TW=UD+DV=TE+EW, meaning that TE=DV and UD=EW, from which the conclusion follows.
27.10.2005 14:37
Posted before at http://www.mathlinks.ro/Forum/viewtopic.php?t=28329 . darij
14.09.2022 02:27
I claim that E moves along a circle centered at A. Let ¯AB be tangent to the incircle of ABD at H, and ¯BC be tangent to the incircle of ACD at I. Then AE=AH−EF=AH−DI=12(AB+AD−BD−CD−AD+AC)=12(AB+AC−BC)is constant, as needed.