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
grobber
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 $UV=DU+DV=\frac{DA+DC-AC}2+\frac{DB+DC-BC}2$, 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.
darij grinberg
27.10.2005 14:37
Posted before at http://www.mathlinks.ro/Forum/viewtopic.php?t=28329 . darij
HamstPan38825
14.09.2022 02:27
I claim that $E$ moves along a circle centered at $A$. Let $\overline{AB}$ be tangent to the incircle of $ABD$ at $H$, and $\overline{BC}$ be tangent to the incircle of $ACD$ at $I$. Then $$AE = AH-EF=AH-DI= \frac 12(AB+AD-BD-CD-AD+AC) = \frac 12(AB+AC-BC)$$is constant, as needed.