Difference between revisions of "1959 AHSME Problems/Problem 40"
Tecilis459 (talk | contribs) (Add problem statement & Unify answer) |
m (see also box) |
||
| Line 9: | Line 9: | ||
Since ADG is similar to ACB, we know that AG is 10. | Since ADG is similar to ACB, we know that AG is 10. | ||
| − | Thus, AF = 10 + 5 = 15, which is answer <math>\ | + | Thus, AF = 10 + 5 = 15, which is answer <math>\fbox{\textbf{(C)}}</math>. |
| + | |||
| + | == See also == | ||
| + | {{AHSME 50p box|year=1959|num-b=39|num-a=41}} | ||
| + | {{MAA Notice}} | ||
Revision as of 15:53, 21 July 2024
Problem
On the same side of a straight line three circles are drawn as follows: a circle with a radius of 
inches is tangent to the line, the other two circles are equal, and each is tangent to the line and to the other two circles. The radius of the equal circles is: 
Solution
Make a line DG which is parallel to FC. We know that GF = BF = 5, since BFE is similar to BGD.
Since ADG is similar to ACB, we know that AG is 10.
Thus, AF = 10 + 5 = 15, which is answer 
.
See also
| 1959 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 39 |
Followed by Problem 41 | |
| 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 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
