Finding the area of a square inside a quarter of a circle

Question

Here's the problem:

This problem could be easy, were I to know if the small pink square divided the arc length of a quarter circle into 3 pieces (identical).

What I'm trying to say is, if my guess is correct, the ratio of the length of $\dfrac{\alpha\beta}{BD}$ is $\frac13$.

But, is it correct? I want to assure myself if this hypothesis is correct. How do you prove it? This is the important key to find the small square. Because, if that's so, I can use this formula (below) to find the side of the small square:

$$\text{side} = 2r\sin\left(\frac{t}{2}\right)$$

Where $r$ is the radius of the circle and $t$ is the angle of the sector circle excribed the small square.

In conclusion, my point is just I'm asking whether it's true or not that the ratio $\dfrac{\alpha\beta}{BD}$ is $\frac13$.

Or perhaps you have another simpler way to find the small pink square?

Answer 1

The answer to your small question is yes, the arcs of the three divisions of the quarter-circles are equal

The easiest way to see this is that $\alpha$ is the same distance from $C$ as $D$ is, and the same distance from $D$ as $C$ is, because they lie on the same circles. So $\triangle \alpha CD$ is equilateral, as is $\triangle \beta BC$, and that leads to the trisection

So the ratio of the arcs $\alpha \beta:BD $ is $1:3$. But the ratio of the line segments $\alpha \beta:BD $ is not $1:3$; it is $\frac{3}{\sqrt{2}}(\sqrt{3}-1):3\sqrt{2}$ or $\frac12(\sqrt{3}-1):1$, about $0.366:1$.

Answer 2

No, $\dfrac{\alpha\beta}{BD}$ is not equal to $\frac13$.

This can be seen from the fact that $B, C$ and $\beta$ are all in the same distance from each other it can be seen that the angle $\phi = △\beta B C = 60°$ and consequently the angle $\theta = △\beta B D = 15°$.

Knowing that $\phi = 60°$ and from symmetry it can be seen that all three arc segments you mention are over an angle of 30° and will therefore have the same length.

If $\dfrac{\alpha\beta}{BD}$ would be equal to $\frac13$. the following had to hold:

$x = s$

$\tan(\theta ) = \frac{s/2}{s+s} = \frac{1}{4}$

where s is the side length of our small square.

This is not the case (since $\tan(15°) = 2 - \sqrt{3} $) and we therefore have a proof by contradiction.