# 1991 USAMO Problems/Problem 5

## 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] size(220); defaultpen(1); pair A=(0,0), B=(220,0), C=(18.7723,118.523); pair D=(72.6,0); pair Ia=incenter(A,D,C), Ib=incenter(B,D,C); pair Ta=(24.9758,52.5775),Tb=(86.6196,67.4129); pair E=IntersectionPoint((Ta--Tb),(C--D)); path Oa=circle(Ia,inradius(A,D,C)); path Ob=circle(Ib,inradius(B,D,C)); pair Da=IP(Oa,A--B), Db=IP(Ob,A--B); draw(D--C--A--B--C); draw(Ta--Tb); draw(Oa); draw(Ob); dot(A,linewidth(4)); dot(B,linewidth(4)); dot(C,linewidth(4)); dot(D,linewidth(4)); dot(E,linewidth(4)); dot(Ta,linewidth(4)); dot(Tb,linewidth(4)); label("$$A$$",A,SW); label("$$B$$",B,SE); label("$$C$$",C,W); label("$$D$$",D,S); label("$$E$$",E,NNE); [/asy]$

## Solution 1

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] size(220); defaultpen(1); pair A=(0,0), B=(220,0), C=(18.7723,118.523); pair D=(72.6,0); pair Ia=incenter(A,D,C), Ib=incenter(B,D,C); pair Ta=(24.9758,52.5775),Tb=(86.6196,67.4129); pair E=IntersectionPoint((Ta--Tb),(C--D)); path Oa=circle(Ia,inradius(A,D,C)); path Ob=circle(Ib,inradius(B,D,C)); pair Da=IP(Oa,A--B), Db=IP(Ob,A--B); pair Ca=IP(Oa,C--D), Cb=IP(Ob,C--D); draw(D--C--A--B--C); draw(Ta--Tb); draw(Oa); draw(Ob); dot(A,linewidth(4)); dot(B,linewidth(4)); dot(C,linewidth(4)); dot(D,linewidth(4)); dot(E,linewidth(4)); dot(Ta,linewidth(4)); dot(Tb,linewidth(4)); dot(Ca,linewidth(4)); dot(Cb,linewidth(4)); dot(Da,linewidth(4)); dot(Db,linewidth(4)); label("$$A$$",A,SW); label("$$B$$",B,SE); label("$$C$$",C,W); label("$$D$$",D,S); label("$$E$$",E,NNE); label("$$T_a$$",Ta,N); label("$$T_b$$",Tb,WNW); label("$$D_a$$",Da,S); label("$$D_b$$",Db,S); label("$$C_a$$",Ca,WSW); label("$$C_b$$",Cb,ENE); [/asy]$

We note that $$CE = CC_a - EC_a = CC_b - EC_b = \frac{CC_a + CC_b - (EC_a + EC_b)}{2} .$$ 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, $$CC_b - DD_b = BC - DB,$$ 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 $E$. $\blacksquare$

## Solution 2

Define the same points as in the first solution. First extend $T_aT_b$ to intersect $AB$ at a point $P$; without loss of generality let $A$ lie in between $B$ and $P$. Then the incircle of $\triangle ACD$ is also the incircle of $\triangle PED$, while the incircle of $\triangle BCD$ is the $P$-excircle of $\triangle PED$. It follows that $EC_a = C_bD$; denote this equality by $(*)$.

Now remark that $$CC_a + CC_b = \frac{AC+AD-CD}2 + \frac{BC+DC-BD}2 = \frac{AC+BC-AB}2 + CD.$$ Hence $$CC_a + CC_b - CD = CC_a - C_bD \stackrel{(*)}= CC_a - C_aE = CE$$ is a constant equal to $r := \tfrac{AC+BC-AB}2$, and so $C$ lies on the circle with center $C$ and radius $r$.

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 