The biggest triangle

Here is a general triangle, circumscribed in a circle of radius $r$ . Additional markings have been made to help us solve the problem.

Note that we can choose any one side as a base, because the circle is symmetrical from all angles of viewing.

Also note: $0 \leq h_1, h_2 \leq r$ .

Let us find the width, $w$ , and the height $h$ , of this triangle, so that we can find a formula for the area, $A$ .

Height is simple, $h = h_1 + h_2.$

To find width, we need to use the Pythagorean theorem,

$r^2 = h_1^2 + (w/2)^2 \\ \Rightarrow w/2 = \sqrt{r^2 - h_1^2} \\ \Rightarrow w = 2\sqrt{r^2 - h_1^2}.$

Thus we see that

$\displaystyle A = \frac{(h_1 + h_2) \cdot 2\sqrt{r^2 - h_1^2}}{2} = (h_1 + h_2)\sqrt{r^2 - h_1^2}.$

Note that $h_2$ is not affected by $h_1$ , so we are free to set it to its maximum value, $r$ .

$A = (h_1 + r)\sqrt{r^2 - h_1^2}.$

So now we have a formula describing $A$ . Now we will use some calculus to find its maximum value.

Since $A^2$ will have the same maximum as $A$ , and is a little easier to differentiate, we will use it to find our critical points (where $\frac{d(A^2)}{dh_1} = 0$ ).

Squaring, $A^2 = (h_1 + r)^2(r^2 - h_1^2).$

Taking the derivative,

$\frac{d(A^2)}{dh_1} = 2(h_1 + r)(r^2 - h_1^2) - 2h_1(h_1 + r)^2$

Factoring out $2(h_1 + r)$ ,

$= 2(h_1 + r) [(r^2 - h_1^2) - h_1(h_1 + r)] \\ = 2(h_1 + r) [r^2 - h_1^2 - h_1^2 - h_1r] \\ = 2(h_1 + r) [r^2 - h_1r - 2h_1^2]$

Factoring again,

$= 2(h_1 + r)(r - 2h_1)(r + h_1) \\ = 2(r + h_1)^2(r - 2h_1).$

Thus we see that $\frac{d(A^2)}{dh_1} = 0$ when $h_1 = r/2$ (the other values of $h_1$ are not between $0$ and $r$ ).

To find the absolute maximum value of $A$ , therefore, we must check $h_1 = r/2$ , as well as the bounds of $h_1 = 0$ and $h_1 = 1$ .

$A|_{h_1=0} = (0 + r)\sqrt{r^2 - 0^2} = r \cdot r = r^2.$

$A|_{h_1=r/2} = (r/2 + r) \sqrt{r^2 - (r/2)^2} \\ = (3r/2)r\sqrt{1 - 1/4} \\ = (3r^2/2)\sqrt{3/4} \\ = (3r^2/4)\sqrt{3} \\ = (3\sqrt{3}/4)r^2.$

$A|_{h_1=r} = (r + r)\sqrt{r^2 - r^2} = 0.$

Since, $(3\sqrt{3}/4)r^2 > r^2 > 0$ , we see that $A$ reaches an absolute maximum of $(3\sqrt{3}/4)r^2$ at $h_1 = r/2$ .

Plugging in $r = 7$ , we see that the maximum area is $(3\sqrt{3}/4)(7)^2 \approx 63.653$ .

QED.