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 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 | ||
All AHSME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.