Finding maximal area of triangle.

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Homework Statement



Consider the points P(1,0) and Q(0,1) on the unit circle. For which point R(cos(theta),sin(theta)) on the unit circle, is the area of the triangle PQR maximal?

Homework Equations



Hint - It might be easier to use vectors to calculate the area of the triangle.

The Attempt at a Solution



I'm really not sure how to approach this question. It's part of an integration question set. A wild guess would assume cos(pi/4) and sin(pi/4) would be the co-ordinates that would give a maximal area but I'm not sure if that's right or how to prove it.

Thanks for any help.
 
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KaliBanda said:

Homework Statement



Consider the points P(1,0) and Q(0,1) on the unit circle. For which point R(cos(theta),sin(theta)) on the unit circle, is the area of the triangle PQR maximal?

Homework Equations



Hint - It might be easier to use vectors to calculate the area of the triangle.

The Attempt at a Solution



I'm really not sure how to approach this question. It's part of an integration question set. A wild guess would assume cos(pi/4) and sin(pi/4) would be the co-ordinates that would give a maximal area but I'm not sure if that's right or how to prove it.

Thanks for any help.

Guesses don't count, even if they are wild. Make vectors out of two sides of the triangle and remember that the area is ##\frac 1 2|\vec A \times \vec B|##. Use a little calculus on that. You can make the third component of the vectors 0 to use the cross product.

[Edit] I should have mentioned the determinant form for the area of a triangle given its vertices in case you haven't had 3D vectors. If the vertices are ##(a,b),(c,d),(e,f)## the area is also the absolute value of the determinant$$
A = \frac 1 2\left |\begin{array}{ccc}
1 & a & b\\

1 & c & d\\

1 & e & f
\end{array}\right|$$
 
Last edited:
LCKurtz said:
Guesses don't count, even if they are wild. Make vectors out of two sides of the triangle and remember that the area is ##\frac 1 2|\vec A \times \vec B|##. Use a little calculus on that. You can make the third component of the vectors 0 to use the cross product.

We haven't learned the cross product as it isn't part of our couse, but I'll give it a go.

Thanks for the reply!
 
KaliBanda said:
We haven't learned the cross product as it isn't part of our couse, but I'll give it a go.

Thanks for the reply!

I edited my post with another 2D suggestion.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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