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If $P$ is a point inside quadrilateral $ABCD$ with $P A = 2$, $P B = 3, P C = 5$ and $P D = 6$, find the maximum possible area of $ABCD$.

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What have you tried? –  Cameron Buie Nov 14 '12 at 1:39
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Hint: given a triangle with adjacent side lengths of M and N, you can maximize the area of the triangle by setting M and N as legs of a right triangle. –  Ben Nov 14 '12 at 2:04
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up vote 3 down vote accepted

Here is Ben's hint made more precise.

Consider the triangles $\triangle APB,\ \triangle BPC,\ \triangle CPD$ and $\triangle DPA$. Let them each subtend a central angle as shown in the figure below

                                    figure

The area of a triangle is given by $\frac{1}{2}ab\sin\theta$ where $a$ and $b$ are the two sides of the triangle containing angle $\theta$. Applying this to our quadrilateral yields $$\rm{Area} = 3\sin\alpha + \frac{15}{2}\sin\beta + 15\sin\gamma + 6\sin(2\pi - \alpha - \beta - \gamma)$$ It shouldn't be too hard to see what are the values of $\alpha,\ \beta$ and $\gamma$ such that the sines are maximized.

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@coffeemath Hmm, that's strange. Are you using lagrange multipliers? I guess it corresponds to $\cos\alpha = \cos\beta = \cdots = 0$. –  EuYu Nov 14 '12 at 2:23
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By what Ben said, each individual subtriangle's area ia maximized when its angle is 90 degrees. And making all the angles 90 is possible, so that at the max all the angles are 90. –  coffeemath Nov 14 '12 at 2:30
    
Ed Yu: No, I didn't use LaGrange, just the partial derivatives all set equal to zero of your expression. –  coffeemath Nov 14 '12 at 2:31
    
@coffeemath I don't think there's any need to even look at the partials. Right angles for each angle maximizes each sine at $1$. That is clearly the maximum possible. –  EuYu Nov 14 '12 at 2:41
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