My Theorem?

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My Theorem?

I recently discovered this theorem. I don't know if it has ever been seen before. [Unfortunately, it was discovered long before I discovered it. See addendum, below]

three maximized quadrilaterals Given: The lengths of the four sides of a quadrilateral. Prove: If you adjust the shape of each of the quadrilaterals with these four sides, to maximize the area of each, that all such maximized quadrilaterals with those four sides have the same area. Notice that there are three different convex quadrilaterals which meet these conditions. All other maximized convex quadrilaterals with those four sides are rotations and reflections of those. One way to orient the three quadrilaterals is shown above left.

first two maximized quadrilaterals Proof: Let's take a look at the first two quadrilaterals (on the right). Call the angle formed by a and b X in both quadrilaterals. And draw the diagonals x as shown, in both quadrilaterals. Assume that both quadrilaterals are not necessarily maximized yet, and draw them so that both angle X's are equal. Then both diagonals x are equal, and the areas are equal because the two quadrilaterals are made up of congruent triangles. Any adjustment in the two angle X's which will decrease (or increase) the area of one quadrilateral will decrease (or increase) the area of the other quadrilateral. So the maximized quadrilaterals will have the same angle X and the same diagonal x and the same area.

first and third maximized quadrilaterals Now let's look at the first and third quadrilaterals (on the left). Call the angle formed by a and d Y in both quadrilaterals. And draw the diagonals y as shown, in both quadrilaterals. Then by the same argument as above, the maximized quadrilaterals have the same area.

And so, all three maximized quadrilaterals have the same area. QED.

Discovering a theorem:

I was considering a geometric puzzle: What is the maximum area of a quadrilateral with sides 1, 2, 3, 4? I used the geometry program Cinderella (which I used to draw the above diagrams, along with Paint Shop Pro) to experiment with such a rectangle. I found the maximum area to be 4.90. I drew the other three quadrilaterals with the same length sides in different orders, and found the maximum areas of these to be 4.90. This probably wasn't a coincidence, they were probably exactly the same. And the same probably applied to any four lengths, the maximized areas should all be the same. But how could that be proven?

I worked on finding an equation for the maximized area of a general quadrilateral a, b, c, d. I decided that I could use one angle X (as in the second diagram above), then deduce an equation for the area. Then I could take the derivative of this equation with respect to X, and set the derivative to zero (the area would be at a maximum then). There are several published equations for areas of general quadrilaterals. But some involve too much information, like A=sqrt(4p^2q^2-(b^2+d^2-a^2-c^2)^2)/4, where p and q are the diagonals and p^2 means p squared. Here is another equation: A=pq sin(theta)/2, where theta is the angle between the diagonals. That is simple, but the lengths of the diagonals vary in my puzzle. This one looked promising: A=(b^2+d^2-a^2-c^2) tan(theta)/4. Obviously, theta will never be 90 degrees, for a general quadrilateral, otherwise the area will be infinite. So, it is hard to predict the maximum value of theta. So, I decided to derive my own equation for the area, given a, b, c, d, and angle X.

Then I noticed that two of the quadrilaterals that I had drawn by hand (the second diagram above), had the same length diagonal, whenever the angle X was the same. And it was obvious that they had the same area, being composed of congruent triangles. And it followed that the angle X would be the same, when both quadrilaterals had maximum area. But I didn't see how the third quadrilateral related to these two, as it did not have the angle X or the diagonal x. Well, this was a symmetrical situation, so there must be some angle Y and some diagonal y, which this third quadrilateral had in common with one of the other quadrilateral. I drew the third diagram above. And I had my proof.

Note: Q.E.D. is short for quod erat demonstrandum, which is Latin for "Which was to be demonstrated." It just means that the proof is done.

Addendum:

I received a couple of emails, informing me that the above theorem was already known. I do not know who discovered it. The above maximized quadrilaterals are cyclic quadrilaterals. In other words, they can be inscribed in a circle (with all four vertices on the circle). It is known that a cyclic quadrilateral is the maximum possible quadrilateral with those given side lengths. And, the area of such a quadrilateral is A=sqrt((s-a)(s-b)(s-c)(s-d)), where s is the semiperimeter: s=(a+b+c+d)/2. See Hero's Formula for a similar formula for a triangle. From that formula, it is obvious that it doesn't matter what order the four lengths are arranged, the area is the same.

This theorem may have been known in India over 1000 years ago. And I hear that it was rediscovered in the west, in the 17th century.

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