A quick and easy geometry puzzle solved in two ways

This video is about four semicircles inscribed in a unit square
If you need to post a picture of your solution or idea:
#GeometryPuzzles

EXPLORE:

[georgesbv]

We apply Pitagora's theorem in a square triangle from the centers of 2 tangent circles.
(r)^2 + (1-r)^2 = (2r)^2
Only one positive solution: r = (sqrt(3)-1) / 2

[ashishpradhan]

Thankyou so much. You know the reason behind this, there is no need to mention. 😍😍😍

[charlesbromberick]

You´re right - it´s nice if a bit easy.

[kurtlichtenstein]

If you want difficult, how about the ratio of the area of the center section to the area of one of the smaller sections?

[Muzann]

Can we apply unitary method in geometry 🤔🤔

[tonyhaddad]

And i have a challenging problems i suggest you to solve
1)find all positive integers n such that every positive integer with n didgits, one of which is 7 and the others 1, is prime .

2) prove that for any convex quadrilateral of area 1, the sum of the sides and diagonals is not less than 4+sqrt(8)

Until now i have not solved this problems but i think you like the second problem becaus you like geomtric problems

[Shreyas_Jaiswal]

After seeing r and 2r, the special triangle immediately came to my mind. And the base becomes √3r.

[tonyhaddad]

This problem is like the proof of phytagrien theorm

[shatishankaryadav]

It could also possible 60-30-90 triangle

[tonyhaddad]

Helloo this seems a challenging problem !! I will try to solved it

[rishipatel]

I solved it in first 30 seconds that too orally

[Qermaq]

Interesting how root3 turns up with a square.

Won't work in an equilateral triangle like this, as its internal angle is acute.

I tried this with a unit hexagon and found r = (root13 - 1)/6. Yeah, root13. Ugh.

I generalized this by observing that this "r 2r (1-r)" triangle occurs in all the possibilities of an n-gon, it's just not a right triangle unless n=4. So I used the law of cos, setting 2r to c because I know the opposite angle is pi - (2pi/n). It neatens up to something like

2r^2 + 2r - 1 + (2r)(1-r)(cos(pi-(2pi/n))) = 0

So I sub in the cos, simplify, and we wind up with a simple quadratic. Don't want to try this with a pentagon though. I might hurt myself ;)

Edit: I did solve for the pentagon. Very nice solution, actually. Might consider it for another video....