# Optimization - Find area of triangle in a circle

1. Jul 30, 2008

### lizzie

a chord AB of a circle subtends an angle that is not equal to 60 degees at a point C on the circumference. ABC has maximum area. then find A & B in terms of the angle.

2. Jul 30, 2008

### HallsofIvy

Staff Emeritus
Re: circle

??? I don't understand the problem. "A chord AB of a circle subtends an angle that is not equal to 60 degrees" is okay but what does "at a point C on the circumference" mean?

3. Jul 31, 2008

### chaoseverlasting

Re: circle

I think it means that on a given circle, there's an arbitrary chord AB and there's a point C on the circumference which when joined with points A and B give rise to a triangle ABC, and you have to maximize the area of the triangle. The triangle cant be equilateral.

4. Jul 31, 2008

### chaoseverlasting

Re: circle

Can you try to find the area of the triangle in terms of the radius of the circle and the angle subtended? If you can do so, then its a simple problem in maximization.

For example, say the radius of the circle is r and the angle subtended at the point C is $$\theta$$ and the center of the circle is at the point $$O$$. Also, the area of the triangle depends on the base and the height of the triangle, for the height to be maximum, the triangle MUST be isosceles for a given base, can you see why?

5. Jul 31, 2008

### Defennder

Re: circle

The relative position of A and B are not given and neither is C. It seems all I can understand from this question is
"Find the relative position of A,B given that the area enclosed by the triangle in the circle is maximised and angle ACB is not 60 degrees".

This looks like it could be solved by Lagrange multipliers, except I can't quite come up with the constraints. I mean all we know is that the angle is not 60 degrees.

6. Aug 1, 2008

### lizzie

Re: circle

I feel in an isosceles triangle the base will be less but height will be more so how do we know that the area is maximum.

7. Aug 1, 2008

### Defennder

Re: circle

lizzie, it appears that you probably omitted some conditions for this triangle ABC in a circle. Symmetry and mathematical intuition demands that the triangle with the max area in a circle must exhibit symmetry in some way, so that leaves us with isosceles and equilateral triangles (yeah yeah I know that's not a rigorous proof but still...). And I can find the following on the largest triangle inscribed in a circle:
http://demonstrations.wolfram.com/LargestTriangleInscribedInACircle/

which clearly indicates that it has to be equilateral. So this question is a bit like asking "What's the smallest real number larger than 2?".

8. Aug 2, 2008

### chaoseverlasting

Re: circle

Well, the triangle you get will not have the maximum area because an equilateral triangle will have the maximum area for a given circle and your constraints omit that possibility.

So what WILL happen is that when you get the function which defines the area of a triangle, you will have to take the case that comes closest to it. The problem so far is that you have to start defining the area.

Since the circle is constant (i.e. the radius is constant), this is the case so far:

You have a chord AB and a point C which completes the triangle ABC. Angle ACB is $$\theta$$.

Now, say we join the center (point O) with the points A and B, and we get an angle AOB. You can easily prove that this angle AOB is $$2\theta$$.

You also know that AO and OB are radii of the circle and their length is r.

Can you use this information to find the length of the chord AB and the height of the triangle? Also, because the triangle is isosceles, the perpendicular from C to AB (the height) will pass through the center O. You now have everything you need to define the length of the base (AB) and the height of the triangle using simple trigonometry. What do you think these will be?

9. Sep 3, 2008

### lizzie

Re: circle

Thanks chaoseverlasting i got the answer.