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.

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

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.

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("<math>A</math>",A,SW);
label("<math>B</math>",B,SE);
label("<math>C</math>",C,W);
label("<math>D</math>",D,S);
label("<math>E</math>",E,NNE);
label("<math>T_a</math>",Ta,N);
label("<math>T_b</math>",Tb,WNW);
label("<math>D_a</math>",Da,S);
label("<math>D_b</math>",Db,S);
label("<math>C_a</math>",Ca,WSW);
label("<math>C_b</math>",Cb,ENE);
 (Error making remote request. Unknown error_msg)

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, $CC_b - DD_b = BC - DB,$ so that $CE = \frac{CC_a + CC_b - D_aD - DD_b}{2} = \frac{AC + BC - (AD+DB)}{2} = \frac{AC + BC - AB}{2} .$ 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 $E$ . $\blacksquare$

Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

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1991 USAMO (Problems • Resources)
Preceded by Problem 4	Followed by Last Question
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1991 USAMO Problems/Problem 5

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