1959 AHSME Problems/Problem 41
Problem
On the same side of a straight line three circles are drawn as follows: a circle with a radius of 
inches is tangent to the line, the other two circles are equal, and each is tangent to the line and to the other two circles. The radius of the equal circles is: 
Solution
![[asy] import geometry; point A = (0,4); point B = (-16,16); point C = (16,16); point D = (-16,0); point E = (16,0); point F = (0,16); // The line line l = line((-20,0),(20,0)); draw(l, Arrows); // Circles draw(circle(A,4)); dot(A); label("A",A,(0,-3)); draw(circle(B,16)); dot(B); label("B",B,W); draw(circle(C,16)); dot(C); label("C",C,(1,0)); //Tangency points dot(D); label("D",D,S); dot(E); label("E",E,S); dot(F); label("F",F,NE); // Triangle AFB, Segment BD draw(triangle(A,F,B)); draw(B--D); // Right angle labels markscalefactor=0.3; draw(rightanglemark(F,B,D)); draw(rightanglemark(B,D,E)); draw(rightanglemark(B,F,A)); [/asy]](http://latex.artofproblemsolving.com/b/a/4/ba4b4fcd6b43412b4e142d0bd51e9d5d17adff4a.png)
Let the radius of the two large circles be 
. Also, let the center of the small circle be 
and the centers of the two large circles be 
and 
, as in the diagram. Further, let the large circles intersect at 
, let the circle centered at be tangent to the line at 
, and let the circle centered at be tangent to the line at 
. Because 
and 
, 
. Because the two large circles are tangent at , is on the segment connecting their centers, 
. Thus, is a distance from 
. Because the small circle (with radius ) is tangent to the line, 
. Also, because the small circle and the circle centered at are tangent, 
. Because 
, by the Pythagorean Theorem, we have the following equation:
\begin{align*}
BF^2+AF^2 &= AB^2 \\
r^2+(r-4)^2 &= (r+4)^2 \\
r^2+r^2-8r+16 &= r^2+8r+16 \\
r^2-16r &= 0 \\
r(r-16) &= 0
\end{align*}
Because 
, we are left with 
. Thus, the radius of the two large circles is 
.
See also
| 1959 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 40 |
Followed by Problem 42 | |
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