AoPS - Problem

Source: USAMO 1991

Tags: geometry, geometry solved

MithsApprentice

27.10.2005 11:15

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.

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.