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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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    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.2012-11-14

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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 $\cos\alpha = \cos\beta = \cdots = 0$.2012-11-14
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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 90.2012-11-14
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    Ed Yu: No, I didn't use LaGrange, just the partial derivatives all set equal to zero of your expression.2012-11-14
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    @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.2012-11-14