Geometry Puzzle: What’s the Radius?

It is lovely that you leave out not a single step; this is a great refresher for things I learned more than half-a-century back.

x = r/sin45 and x = d - r. Therefore d - r = r/sin45. Rearranging for r gives r = dsin45/(1+sin45). dsin45 is the side of the small square = 2. Thus r = 2/(1+sin45) = 1.17

With your help, I'm trying to keep my 85 y/o brain in shape. I think it works!😄

Susanne, thanks for helping me exercise my 84 y/o brain. Many of your maths principles take me back to my geometry class of 1958.

A simpler solution. You don’t need to solve for d or x. After you show the center of the circle and the new radius you draw at 1:20 are both on the diagonal, you know that new radius is at an angle of 45 degrees. So that new radius’ horizontal projection plus the radius from the circle’s center to the left side add to 2 units. r*sqrt(2)/2 + r = 2 Then solve for r to get r = 2/[1+sqrt(2)/2] Looks very different from your solution of r = sqrt(8)/[1+sqrt(2)] but they are the same. Since sqrt(8) = 2*sqrt(2) , just multiply my solution's numerator and denominator by sqrt(2).

I use your videos to pull myself out of occasional bouts of depression. You’re just what I needed today. Thanks!

I got the half-diagonal as √8 in the same manner, then... Construct another half-diagonal from the centre to the top-left corner. We now have converging tangents, so the left side of the square from the corner to the point of tangency is also length √8. So the entire side is r + √8 From there it's just algebra: r + √8 = 4 r = 4 − √8 r = 1.17 (rounded)

Math can be so beautiful when explained in such a way. So glad I found your channel.

Ladies & gents: The “Bob Ross” of math! 🎉 I’ve never loved math so much…and I LOVE algebra & calculus! ❤

I made a right triangle with hypotenuse from the center of circle to center of big square. The legs are both (2-r) and the hypotenuse is r. Solved 2(2-r)^2=r^2 and got 4-2srt(2).

Amazing. 40 years no dealing with these math problems but all coming back.

So enjoy your methodical 'stepping through the problem and calm descriptive narrative' process. It is a joy to watch, learn, and participate in the problem solving.

Loved it because it is not obvious. But draw a diagonal for the square which is sqr (32). Half of that is sqr (8) which is part of the diagonal and a tangent to the circle. Equal tangents determine that the upper part of the LHS of the square is also sqr(8). So the radius must be 4 (side of big square) minus sqr(8) which is about 1.17157.

I knew that there would be right triangles and Pythagoras, but I didn't see it all the way to the end. Thank you for this video.

You used a long method that encouraged everyone to show all possible shortcut methods. Appreciable job.

I just used 1 / (sin(45)) A circle that inscribes the remaining quarter of the big square would have r=1, but wouldn't touch the midpoint (2,2) We know the midpoint would touch at 45 degrees so we calculate how much we need to blow up the circle to make it toch by dividing 1 by sin(45) ≈1.17 In

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Hi, the short version is: 2=r+vertical component of r. The vertical component of are=0.7071 r, because 45 degre angel. 2=r+0.7071 r. 2=1.7071r; r=2/1.7071=1.1716

Diagonal of the small rectangle = 2√2 And distance from the center of the circle to the corner of the large rectangle = r√2 (According to Pythagoras) -> r = 2√2 - r√2 -> r = (2√2)/(1+√2) = 1,17

4√2 = r√2+r+2√2 --> r = 4 - 2√2