1965 AHSME Problems/Problem 39
Contents
[hide]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 1
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
:
.
Solution 2 (Overkill?)
This is Descartes' formula for four mutually tangent circles with radii
Let denote
. This is the curvature.
Note: A negative radius means that the other circles are internally tangent to it.
Then,
First, scale the problem up by two, so the radii are 1",2",3", and d"
Next, let be
Using the formula, we get:
(You'll see why I didn't simplify )
Using the quadratic formula:
So the diameter , which is about
See Also
1965 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 38 |
Followed by Problem 40 | |
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