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Problem: What is the area of the largest equilateral triangle that can be inscribed in a rectangle with sides $10$ and $11$?

(The problem comes from an old high school math contest. I believe it's either the AIME or ARML, but I have lost the source.)

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The answer is $l=\frac{10}{\cos x}=\frac{11}{\cos \frac{\pi}{6}-x}$. (Solve it $x=\tan^{-1}(\frac{11}{5}-\sqrt{3})$ http://www.wolframalpha.com/input/?i=arctan+%28%5Cfrac%7B11%7D%7B5%7D-%5Csqrt%7B3%7D%29, l=http://www.wolframalpha.com/input/?i=10%2Fcos%28arctan+%28%5Cfrac%7B11%7D%7B5%7D-%5Csqrt%7B3%7D%29%29)

We have to find $\max l$ with $l\cos x\le 10 $ and $l\cos \frac{\pi}{6}-x\le 11$ for $0\le x\le \frac{\pi}{6}$. We shall consider $\min_x\max\{ \frac{\cos x}{10}, \frac{\cos\frac{\pi}{6}-x}{11}\}$. Since $f(x)=\frac{\cos x}{10}$ decreases and $g(x)=\frac{\cos\frac{\pi}{6}-x}{11}$ increases, and $f(0)>g(0), f(\frac{\pi}{6})<g(\frac{\pi}{6})$, the $\min\max $ shall be obtained at the intersection point $f(x)=g(x)$.

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