AnExtremelyLongName

Thanks! Not gonna edit it because people can learn from my mistake.

Call the center of the circle \(L\)

\(SL=r\)

\(KL=r\)

\(OL=45-r\)

By pythagorean theorem:

(Hypotenuse of Triangle SOK)^2 - (Longer side of triangle SOK)^2= (Hypotenuse of triangle KOL)^2 - (Shorter side of triangle KOL)^2

\(52^2-45^2=r^2-(45-r)^2\)

Now it's your turn to solve.

Alien 2 and 3 sit on a side. While Alien 1 and 4 sit on the other side. So yes, these playful aliens can do it!

Since the square root goes on for infinity:

\(\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x...}}}}=\sqrt{x+\sqrt{x+\sqrt{x...}}}\)

Let us set \(\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x...}}}}=y\)

This means: \(y=9\)

This also means: \(\sqrt{x+y}=9\)

Substitute \(y\): \(\sqrt{x+9}=9\)

Square both sides: \(x+9=81\)

\(\boxed{x=72}\)

How did you bold your question?

The first video is what I think Melody wants you to see. As to Mathway, it is blocked on my browser :/

Example:

How many integers less than or equal to 20 are divisible by 2? 5? or 10?

20/2 = 10

2, 4, 6, 8, 10, 12, 14, 16, 18, 20

20/5 = 4

5, 10, 15, 20

20/10 = 2

10, 20

10 + (4/2)_{because every 2 numbers there is a even number} - 2 = 10

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Now apply the same to 10,000

Oh, I thought both could not be bisected, so my imagination went a little wild.

I am thinking of a non-regular octagon, but I still think it is not possible. If it has to be regular, then the only shape is a circle.

You can always use shoelace formula. I'm a little lazy today but I will provide you a resource:

https://www.youtube.com/watch?v=Ow24sLQXokU for shoelace formula