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# s50 - with hypotenuse of length 10 and leg BC of length 5...

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Solution to Problem 50 Congratulations to this week’s winners Nathan Pauli, Ray Kremer, Felice Kelly Correct solutions were also received from Bradley alumnus Brian Laughlin (’81), as well as from Daniel Statman, Steven Young, William Webb, Philippe Fondanaiche, Robert McQuaid, Javier Echavarri. Several incorrect submissions were received. [IMAGE] See the figure on the right. There are two ways to attack the problem, either via integration or by plane geometry. No matter how you proceed, what you want to find is four times the area of the blue region. This solution follows that of Felice Kelly and is geometric. The coordinates of some points are given. Note that A is at (5,5) so the length of OA is 2. Triangle OBC is a right triangle

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Unformatted text preview: with hypotenuse of length 10 and leg BC of length 5, so the length of OC is 5Ö3, and the angle BOC is p/6. This implies that the angle AOB is p/12, since angle AOC is clearly p/4. The area of a circular segment of radius 10 and central angle p/6 is 25p /3. From this it’s easy to compute that triangle AOB has base length 10 and height 5Ö2 sin(p/12) = 5(Ö3 - 1)/2, giving an area of (25Ö3 - 1) /2. Put this all together and you get an answer of 4(25p/3 - 25(Ö3 - 1)) @ 31.5146 square ft. William Webb also computed the area of regions where the professor is in the range of two or of three students. (Notice that the wet lecturer is never in the range of only one shooter.) Anyone care to compute these?...
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s50 - with hypotenuse of length 10 and leg BC of length 5...

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