1991 USAMO Problems/Problem 5

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Problem

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.

[Asy_image]

Solution

Let the incircle of $ACD$ and the incircle of $BCD$ touch line $AB$ at points $D_a,D_b$ , respectively; let these circles touch $CD$ at $C_a$ , $C_b$ , respectively; and let them touch their common external tangent containing $E$ at $T_a,T_b$ , respectively, as shown in the diagram below.

[Asy_image]

We note that $CE = CC_a - EC_a = CC_b - EB_b = \frac{CC_a + CC_b - (EC_a + EC_b)}{2} .$ On the other hand, since $EC_a$ and $ET_a$ are tangents from the same point to a common circle, $EC_a = T_aE$ , and similarly $EC_b = ET_b$ , so EC_a + EC_b = T_aE + ET_b = T_a T_b . On the other hand, the segments $T_a T_b$ and $D_a D_b$ evidently have the same length, and $D_a D_b = D_aD + DD_b$ , so $EC_a + EC_b = D_aD + DD_b$ . Thus $CE = \frac{CC_a + CC_b - (EC_a + EC_b)}{2} = \frac{CC_a + CC_b - D_aD - DD_b}{2} .$ If we let $s_a$ be the semiperimeter of triangle $ACD$ , then $CC_a = s_a - AD$ , and $D_aD = s_a - AC$ , so CC_a - D_aD = (s_a - AD) - (s_a - AC) = AC - AD . Similarly, so that $\begin{align*}CE &= \frac{CC_a + CC_b - D_aD - DD_b}{2} = \frac{AC + BC - (AD+DB)}{2} \\&= \frac{AC + BC - AB}{2} .\end{align*}$ Thus $E$ lies on the arc of the circle with center $C$ and radius $(AB+BC-AB)/2$ intercepted by segments $CA$ and $CB$ . If we choose an arbitrary point $X$ on this arc and let $D$ be the intersection of lines $CX$ and $AB$ , then $X$ becomes point $E$ in the diagram, so every point on this arc is in the locus of $E$ . $\blacksquare$