Difference between revisions of "1961 AHSME Problems/Problem 38"
(→Solution) |
Rockmanex3 (talk | contribs) m (Undo revision 139965 by Quantomaticguy -- alternate solution not correct) (Tag: Undo) |
||
Line 31: | Line 31: | ||
Therefore, <math>s^2 \le 8r^2</math>, so the answer is <math>\boxed{\textbf{(A)}}</math>. | Therefore, <math>s^2 \le 8r^2</math>, so the answer is <math>\boxed{\textbf{(A)}}</math>. | ||
− | |||
− | |||
− | |||
==See Also== | ==See Also== |
Latest revision as of 15:44, 23 February 2021
Problem
is inscribed in a semicircle of radius so that its base coincides with diameter . Point does not coincide with either or . Let . Then, for all permissible positions of :
Solution
Since , . Since is inscribed and is the diameter, is a right triangle, and by the Pythagorean Theorem, . Thus, .
The area of is , so . That means . The area of can also be calculated by using base and the altitude from . The maximum possible value of the altitude is , so the maximum area of is .
Therefore, , so the answer is .
See Also
1961 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 37 |
Followed by Problem 39 | |
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.