Difference between revisions of "1961 AHSME Problems/Problem 38"
Rockmanex3 (talk | contribs) (Solution to Problem 38) |
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− | == Problem | + | == Problem == |
<math>\triangle ABC</math> is inscribed in a semicircle of radius <math>r</math> so that its base <math>AB</math> coincides with diameter <math>AB</math>. | <math>\triangle ABC</math> is inscribed in a semicircle of radius <math>r</math> so that its base <math>AB</math> coincides with diameter <math>AB</math>. | ||
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\textbf{(C)}\ s^2 \ge 8r^2 \qquad\ | \textbf{(C)}\ s^2 \ge 8r^2 \qquad\ | ||
\textbf{(D)}\ s^2\le4r^2 \qquad | \textbf{(D)}\ s^2\le4r^2 \qquad | ||
− | \textbf{(E)}\ s^2=4r^2 </math> | + | \textbf{(E)}\ s^2=4r^2 </math> |
==Solution== | ==Solution== |
Revision as of 23:27, 1 June 2018
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 |
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