1961 AHSME Problems/Problem 38
Contents
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 1
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 .
Solution 2 (Mean Inequality Chain)
First . .
By quickly considering extreme cases (when coincides with or ) we can suspect that the maximum value of is achieved when is perpendicular to , where is the origin of the semicircle. We will prove this using the Mean Inequality Chain. Let . . By QM-AM, . Plug in , we have . Square both sides, we find that is correct.
~hastapasta
Solution 3 (Calculus Optimization)
This solution utilizes optimization techniques taught in Calculus 1 classes.
Let the equation of the semicircle be . Notice that in this semicircle, is a function of . .
Now let . Using the Pythagorean Theorem with , finding in terms of , differentiating and finding the absolute maximum, we can find that is correct.
P.S.: The process of bash is omitted here since anybody that actually learned optimization problems and paid attention to it (this is core curriculum in regular college math) should be able to bash it out.
~hastapasta
See Also
1961 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 37 |
Followed by Problem 39 | |
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