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
:
.
See Also
1965 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 38 |
Followed by Problem 40 | |
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