Maximizing area of the triangle in a quarter circle

Question

The radius of the quarter circle is $6\sqrt 5$ and we assume that $OA= 5$ and $OC=10$. What is the maximum area of the blue triangle?

Interpreting the problem statement, I believe that points $A$ and $C$ are fixed and point $B$ can move on the arc. To solve this problem, I assumed that the coordinate of $O$ is $(0,0)$ and then assigned coordinates for each vertex of the triangle: $A(5,0), C(0,10), B(x,\sqrt{180-x^2})$ where $x \in [0, 6\sqrt5]$. Then I applied the formula for the area of the triangle given its vertices, and the problem is reduced to maximizing

$$A(x)= \left|25-(\frac52\sqrt{180-x^2}+5x)\right|\quad \text{for}\quad x \in [0, 6\sqrt5]$$

Which is easy to continue and I got $50$ as the answer.

I'm looking for other approaches to solve this problem. I'm particularly interested in geometric approaches.

Answer 1

The area of the triangle ABC is equal to its side AC multiplied by the height, divided by $2$. The height will be maximal if we draw a radius OB such that OB is perpendicular to AC. Let H be the intersection of AC and OB.

Then $$AC=\sqrt{OC^2+OA^2}=\sqrt{25+100}=5\sqrt5.$$

It is known that $$OH=\frac{OA\cdot OC}{AC}=\frac{5\cdot 10}{5\sqrt5}=2\sqrt5.$$

Then $$HB=OB-OH=6\sqrt5-2\sqrt5=4\sqrt5.$$

Then the area of the triangle ABC is $$\frac{BH\cdot AC}2=\frac{4\sqrt5\cdot 5\sqrt5}2=50.$$

Answer 2

Just extend the circle, call $A'$ and $C'$ the intersection of the line AC with the circle. The middle $B$ of the arc $A'C'$ is the solution.

Answer 3

Applying Heron's formula for the area of a triangle as a function of the value of the length of its sides and finding the value of x for which this value has a maximum, we obtain that:

$AC=5\sqrt{5};AB=\sqrt{85};CB=4\sqrt{10}$

being the coordinates of the point B : $x=12, y=6$.