Re: A Putnam Geometry Problem
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then we want to maximize cos t (cos t + sin t). i believe that you can write this as a single trigonometric function, but it's simple enough to differentiate and get cos 2t - sin 2t, which implies that 2t = pi/4 or t = pi/8. now all that remains is to check that this is optimal.
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(cos t)^2 + sin(t)cost(t)
= (1+cos(2t))/2 + (sin(2t)+sin(0))/2)
= 1/2 + 1/2 cos2t + 1/2 sin2t
= 1/2 + Sqrt[1/2] cos(2t-Pi/4)
Then obviously this is maximized when 2t-Pi/4 = 0.
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