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We have three sides of a quadrilateral given, each of side length 20.The third side length is known to be less than length 100. Determine the maximum area of such a quadrilateral.

I would guess the answer is when it is a square, but I have no proof. How would we do this?

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    $A=\frac{|\tan \theta|}{4}\cdot |a^{2}+c^{2}-b^{2}-d^{2}|$ , where $\theta$ is intersection angle of the diagonals...2011-11-26

2 Answers 2

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OK, here is the plan how to solve this problem:

1) The quadrilateral with maximum area exists. It's not hard to show, you're looking for a maximum of a continuous function on a compact set.

2) This quadrilateral is convex - again it's not hard to show; here is the hint:

concave quadrilateral

3) Let ABCD be the quadrilateral with the maximal area. If we denote known sides as $AB$, $BC$, $CD$, so $|AB|=|BC|=|CD|=20$, then $\angle ABD=\angle ACD=\frac{\pi}{2}$:

4) Now you will be able to find $AD$ and angles.

quadrilateral diagram

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    How does one invent the trick in step 3? It is a reduction of the original problem to a more difficult problem (to which the answer is just stated as a fact). For example, in the case where the sides of the quadrilateral are $a,b$ and $c$, the optimum is the solution of a cubic equation, so it is not clear why it should also have a characterization in terms of angles. The reason for $a=b=c$ in the question is that there is a simpler form of the answer and one could use it as a step toward guessing the more general principles.2011-11-26
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This is a polygonal case of Dido's Problem, and has the same solution using the isoperimetric principle.

For an elementary solution, not assuming knowledge of the isoperimetric problem (or its polygonal analogue), one can argue that if the three sides are AB, BC, and CD, then:

  • ABCD is convex

  • BC is parallel to AD

  • BC and AD have the same perpendicular bisector

so that the only free parameter is the angle CBA, which can be chosen to maximize the area.