# Maximize the Area of a Quadrilateral given Three Sides

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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Consider quadrilateral with angles 60,60,120,120 degrees. It's area bigger than square's. – Nurdin Takenov Nov 26 '11 at 4:48
In fact, the third side is known to be less than 60. by virtue of the triangle equality. – Henning Makholm Nov 26 '11 at 4:48
Calculus or no calculus? – André Nicolas Nov 26 '11 at 5:35
$$A=\frac{|\tan \theta|}{4}\cdot |a^{2}+c^{2}-b^{2}-d^{2}|$$ , where $\theta$ is intersection angle of the diagonals... – pedja Nov 26 '11 at 6:01

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:

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.

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Step 3 is a deus ex machina. – zyx Nov 26 '11 at 7:29
Well. perhaps. But it's quite common when you're avoiding use of calculus - you had to invent some tricks. – Nurdin Takenov Nov 26 '11 at 7:35
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. – zyx Nov 26 '11 at 20:43

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.

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