1965 AHSME Problems/Problem 39
Problem
A foreman noticed an inspector checking a 
"-hole with a 
"-plug and a 
"-plug and suggested that two more gauges
be inserted to be sure that the fit was snug. If the new gauges are alike, then the diameter, 
, of each, to the nearest hundredth of an inch, is:
Solution
![[asy] import geometry; point O = (0,0); point A = (-1/2,0); point B = (1,0); point C, D, E; // 1", 2", and 3" circles draw(circle(O,3/2)); dot(O); label("O", O, S); draw(circle(A,1)); dot(A); label("A", A, SW); draw(circle(B,1/2)); dot(B); label("B", B, SE); // Other two circles C=(15/14*3/5,15/14*4/5); D=(15/14*3/5,15/14*-4/5); dot(C); label("C",C,NW); draw(circle(C, 3/7)); dot(D); label("D",D,S); draw(circle(D, 3/7)); // Point E pair[] e=intersectionpoints(line(O,C), circle(O, 3/2)); E=e[1]; dot(E); label("E", E, NE); // Segments draw(A--B--C--A); draw(O--E); [/asy]](http://latex.artofproblemsolving.com/6/4/d/64d0e5f9bda8c00511478b958e86f17978067863.png)
Let the center of the " circle be 
, that of the " circle be 
, that of the " circle be 
, and those of the circles of unknown radius (let their radii have length 
) be 
and 
, as in the diagram. Also, in this problem, no two circles share a center, so let the circles be named by their corresponding centers (so, the " circle is circle , etc.). Extend 
past to intersect circle at point 
. Because circle has radius 
and circle has radius , 
. Likewise, because has radius 
, 
. Thus, 
. Furthermore, because the line connecting the centers of two tangent circles goes through their point of tangency, 
and 
. Because 
and 
, 
. With this information, we can now apply Stewart's Theorem to 
to solve for :
\begin{align*}
OB*OA*AB+OC^2*AB&=AC^2*OB+BC^2*OA \\
4[(1)(\frac{1}{2})(\frac{3}{2})+(\frac{3}{2}-r)^2(\frac{3}{2})]&=4[(r+1)^2(1)+(r+\frac{1}{2})^2(\frac{1}{2})] \\
3+6(\frac{9}{4}-3r+r^2)&=4(r^2+2r+1)+2(r^2+r+\frac{1}{4}) \\
3+\frac{27}{2}-18r+6r^2&=4r^2+8r+4+2r^2+2r+\frac{1}{2} \\
3+\frac{27-1}{2}-4+6r^2-6r^2&=18r+8r+2r \\
3+13-4&=28r \\
28r&=12 \\
r&=\frac{3}{7}
\end{align*}
Because the question asks for the diameter of the circle, we calculate 
.
See Also
| 1965 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 38 |
Followed by Problem 40 | |
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