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

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?

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 $$0.366:1$$.

• sorry not to be precise. What I'm asking here is the arc length :D, so your answer has helped me to find the length segment anyway by using the angle i.e you've confirmed it's $1:3$ then the angle must be $\pi/6$ and the formula I've mentioned can be used finally. Commented Nov 30, 2021 at 1:05

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.

• The question is in regard to the arc lengths. Commented Nov 30, 2021 at 0:45
• Ah I see what you mean. The question was quite ambiguous regarding that and I added an edit now. Commented Nov 30, 2021 at 23:17