Difference between revisions of "1954 AHSME Problems/Problem 26"
| Line 1: | Line 1: | ||
==Problem 26== | ==Problem 26== | ||
https://artofproblemsolving.com/community/c4h256249s1_lines_and_circles | https://artofproblemsolving.com/community/c4h256249s1_lines_and_circles | ||
| − | The straight line <math> \overline{AB}</math> is divided at <math> C</math> so that <math> AC | + | The straight line <math> \overline{AB}</math> is divided at <math> C</math> so that <math> AC=3CB</math>. Circles are described on <math> \overline{AC}</math> and <math> \overline{CB}</math> as diameters and a common tangent meets <math> AB</math> produced at <math> D</math>. Then <math> BD</math> equals: |
<math> \textbf{(A)}\ \text{diameter of the smaller circle} \\ | <math> \textbf{(A)}\ \text{diameter of the smaller circle} \\ | ||
Revision as of 17:03, 22 April 2020
Problem 26
https://artofproblemsolving.com/community/c4h256249s1_lines_and_circles
The straight line 
is divided at 
so that 
. Circles are described on 
and 
as diameters and a common tangent meets 
produced at 
. Then 
equals:
Solution
See Also
| 1954 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 25 |
Followed by Problem 27 | |
| 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. 
