Some days ago, a friend presented me a problem: If you have a circle of radius $r$, Which is the inscribed quadrilateral with the biggest area.

Currently I don’t have an answer for that question, but I decided to try something more easy. Which is the inscribed triangle with the biggest area?

Even if it seems more easy, I don’t have a simple solution. The intuition tells me that probably is the quadrilateral triangle, but in math intuition is not enough, it’s necessary a proof.

To attack the problem, I decided to take some suppositions. First, the circle will be of radius $$. Second the triangle will have three vertices, $A$, $B$ and $C$, that lies in the circle.Third $C$ will be the point $(-1,0)$ and last I will express $A$ and $B$ with polar coordinates, as follows:

$$A = (1; \alpha), \qquad B = (1; \beta), \quad -\pi < \beta < \alpha < \pi.$$

The first thought I had was: How I can get the area of the triangle? Well, this is an interesting question, what I will do is to know how to get the area of the parallelogram that defines two vectors. If I know this area, then I can just divide by two and get the area of the triangle that defines two vector. And how this can help me? Easy, if I know how to calculate the area of this triangle I have two options to calculate the area of the triangle $\triangle ABC$:

1. Use $C-A$ and $C-B$ as the vectors, since the triangle defined by these two vectors is congruent of the triangle $\triangle ABC$.
2. If $O$ is the origin $(0,0)$, then calculate the area of the triangles $\triangle AOB$, $\triangle BOC$ and $\triangle AOC$.

The area of the parallelogram defined by two vectors

Suppose that we have two vectors $\gdef\u{\bec u} \u$ and $\gdef\v{\bec v} \v$, together with the origin and the vector $\u + \v$ form a parallelogram. If $\theta$ is the angle defined by $\u$ and $\v$ a easy way to calculate the area $A_{\u\v}$ of the parallelogram defined by $\u$ and $\v$ is

$$A_{\u\v}= \norm{\u} \norm{\v} \sin\theta.$$

The proof of this property is simple, the formula for the area of a parallelogram it’s just base times height. We can use any vector as the base, and for the height, we can just use trigonometry and it will be the other vector times the sine of the angle between the vectors.

It’s possible to calculate $\sin\theta$ in terms of $\u$ and $\v$. To do this, first, it’s easies to calculate $A_{\u\v}^2$, because we can convert $\sin^2\theta = 1 - \cos^2\theta$. This way we got the next formula:

$$A_{\u\v} = \norm{\u}^2 \norm{\v^2} (1 - \cos^2\theta)$$

It might looks worse than the last formula, the change of sine to cosine makes the formula more easiest, just because if we use the cosine law we got the next formula

$$\cos\theta = \frac{\u\cdot\v}{\norm{\u}\norm{\v}}.$$

This way, if we join both formulas, we finally got a really easy way of calculate the area of the parallelogram defined by $\u$ and $\v$:

$$A_{\u\v}^2 = \norm{\u}^2 \norm{v}^2 - (u\cdot v)^2.$$

Some people may look this formula familiar. And that’s because is related to the Cauchy-Swartz inequality. Indeed this is a nice proof of the Cauchy-Swartz inequality, if we asume that we are in a Euclidean space (since we use terms like area or angle between two vectors).