Maximizing the area of a cyclic trapezoid whose long base is the circumdiameter. Non-trigonometric solution?

Question

A half circle with a radius of R encompasses an isosceles trapezoid such that the large base of the trapezoid is the diameter of the circle encompassing it.

In terms of R, what is the length of the smaller base of all the possible trapezoids as described, whose area is maximal?

After some attempts at the problem, I managed to solve it using unit circle trigonometry, but I am curious if there are purely geometric solutions for this (which is what I was trying to find when I first attempted the problem).

Here is my solution:

Let $x$ be $\measuredangle AOD$
Let $h$ be the height of the trapezoid
Assume $0 < x < 90^\circ$

$$h = AO\sin x = R\sin x$$ $$AB = 2AO\cos x = 2R\cos x$$

Trapezoid area formula: $$\frac{AB + DC}{2} \cdot h $$ $$\downarrow$$ $$S_{(x)} = \frac{2R\cos x + 2R}{2} \cdot R\sin x$$

From here, we find our function's derivative, get the $x$ for which there is a maxima, and plug it into our definition of AB to get it in terms of R, which would be AB = R.

Is there an alternative?

Answer 1

You don't need any trigonometry here.

Ellegant approach: Try to think outside of the box! Ask yourself a more general question:

Of all quadrilaterals $ABCD$ inscribed in a semicricle with diameter AB, which one has the maximum area?

We'll prove that for optimal quadrilateral:

$$BC=CD=DA=R\tag{1}$$

Suppose that (1) is not true. In other words, suppose that the optimal quadrilateral $ABCD$ looks like in the picture above and $AD\neq CD$. Find point $D'$ on arc AC such that $AD'=CD'$.

Note that triangle $\triangle ACD'$ has bigger area than $\triangle ACD$ because $h_{D'}>h_{D}$. So obviously:

$$P_{ABCD}=P_{ABC}+P_{ACD}<P_{ABC}+P_{ACD'}<P_{ABCD'}$$

So quadrialteral $ABCD$ has smaller area compared to $ABCD'$ and cannot be optimal. In other words, if (1) is not true, we can always find a quadrilateral with a bigger area. Consequentially, the quadrialteral with the biggest area must have sides $BC$, $CD$ and $DA$ of equal lengths which is possible only if (1) is true.

And such quadrialteral is also a trapezoid. Any other quadrilateral, being it a trapezoid or not must have a smaller area.

No so ellegant approach, but still without trigonometry:

Denote the height of the trapezod with $h$ and the length of the smaller base with $b$:

You have to maximize the following expression:

$$A=\frac{(2R+b)h}2$$

...or:

$$B=A^2=\left(R+\frac b2\right)^2h^2\tag{1}$$

...with the following constraint:

$$h^2=R^2-\left(\frac b2\right)^2\tag{2}$$

Replace (2) into (1) and you get:

$$B=\left(R+\frac b2\right)^2\left(R^2-\left(\frac b2\right)^2\right)$$

$$B=\left(R+\frac b2\right)^3\left(R-\frac b2\right)$$

For simplicity, introduce expression $c=R+\frac b2$:

$$B(c)=c^3(2R-c)$$

$$B'(c)=3c^2(2R-c)-c^3=0\implies c=\frac{3R}2\implies b=R$$