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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 o$f$ M and N, you can ma$x$imize 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 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