Alternative Geometric Proof ideas?

[Gib Z]

The question from my textbook was "Give a complete proof that the largest triangle that can be inscribed in a circle is an equilateral triangle ", largest meaning that with the greatest area.

I found a proof that finds the area of the general triangle in terms of the angles, and through quite a bit of calculus trickery showed that it reached a maximum when all angles were equal to 60 degrees, ie Equilateral triangle, but my teacher said that there was a different, geometric based proof that didn't require calculus at all, though he could not produce it. This question came from a chapter of the textbook that was past basic Differentiation, but not yet up to the derivatives of trig functions, so there probably should be a more elementary proof right? If anyone has any ideas, please share !

 

[tiny-tim]

Hi Gib Z!

You can easily prove that the area of a triangle in a circle of radius r is 4r²sinAsinBsinC, or r²(sin2A + sin2B + sin2C).

Now, if A+B is fixed, then sinAsinB = cos(A-B)/2 - cos(A+B)/2, or sin2A + sin2B = 2sin(A+B)cos(A_B), which are both obviously greatest for A = B.

Now fill in the gaps and make good!

 

[Gib Z]

Umm. I don't really understand the "If A+B is fixed" part. Basically, I got the same formula and then, since C= pi - (A+B), the formula could be expressed as 4r^2 sin A sin B sin (A+B).

Then I held B constant and differentiated, showing the maximum was when 2a+b = pi. Then I subbed that back into the formula, and maximized for b again. That took quite a while :(

 

[tiny-tim]

Umm. I don't really understand the "If A+B is fixed" part.

That means C is fixed, but you're free to change A and B … and the result is that A = B is best.

If a triangle is not equilateral, then two of the angles must be unequal (A and B, say).

So making them equal would be better.

So a non-equilateral triangle cannot be the largest.

 

[Gib Z]

Ahh I see! Thanks. And I think I ran into a similar expression using my slightly different method, perhaps I can just show its obviously a maximum rather than use calculus. Thank you!

 

[tiny-tim]

Just thought of a much simpler proof …

forget trigonometry …

suppose the largest triangle is not equilateral, then two sides must be unequal …

let the vertex where they meet be P, and the opposite side be QR …

Let S be where the bisector of QR meets the larger arc QR.

Then S is clearly further from QR than P is …

and area = QR x height …

so areaSQR > areaPQR.

So the largest triangle must be equilateral.

EDIT: interestingly, if you follow this procedure, of progressively making isoceles triangles, starting with angle A, the first two steps give angles 90º - A/2 and 45º + A/4, and after 2n steps 45º(1 + 1/4 + … + (1/4)ⁿ) + A/4ⁿ, of which the limit is 45º x 4/3 = 60º​

 

[Gib Z]

Awesome! That was the sort of thing I was looking for =D. Thanks heaps!