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
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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