1991 USAMO Problems/Problem 5
Problem
Let be an arbitrary point on side of a given triangle and let be the interior point where intersects the external common tangent to the incircles of triangles and . As assumes all positions between and , prove that the point traces the arc of a circle.
Solution
Let the incircle of and the incircle of touch line at points , respectively; let these circles touch at , , respectively; and let them touch their common external tangent containing at , 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 On the other hand, since and are tangents from the same point to a common circle, , and similarly , so On the other hand, the segments and evidently have the same length, and , so . Thus If we let be the semiperimeter of triangle , then , and , so Similarly, so that Thus lies on the arc of the circle with center and radius intercepted by segments and . If we choose an arbitrary point on this arc and let be the intersection of lines and , then becomes point in the diagram, so every point on this arc is in the locus of .
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
Resources
1991 USAMO (Problems • Resources) | ||
Preceded by Problem 4 |
Followed by Last Question | |
1 • 2 • 3 • 4 • 5 | ||
All USAMO Problems and Solutions |