2 Circles 1 Square

I was never good at maths. But this was a delight to watch.

I am 30+ years out of algebra, and just stumbled across this in my feed and I still watched the whole thing. That was a terrific explanation. Good teaching and very clear.

You lost me after “hey guys”, but I watched the whole thing.

Just out of 7 years of engineering education and watching these videos reminded me of why I chose to follow this path at the first place. Don't forget guys we are not machines enjoy yourselves

x=10 is actually a fairly reasonable result, it just refers to the square which is bounded by the upper arcs of the two circles (not by their lower arcs like the original case we are dealing with) ..

I love the logic in math. When I saw the thumbnail, I couldn’t fathom a way to calculate that. 3 min and 34 seconds later it totally makes sense and it’s not even difficult math. Just logic and reasoning. Right on man!

Excellent work! I think another way of looking at it is to determine that the half lengths of the square is x/2, and by using the line you drew for the triangle you would quickly find that y actually is 5 - x/2, thus making the equation being: (5 - x/2)squared + (5-x) squared = 5 squared.

Me at 2am for no reason

Im not sure how i feel about the title of this video

Wow. I love how this used such simple math in combination to solve a bigger problem

This is why i always loved maths!!! Its logic always felt almost magical! Like anything can be explained and solved with Mathematics!!

For a general solution, using: "r" for Radius; "L" for lenght of the square side; the triangle drawn at 00:26, Use sin^2+cos^2=1. Sin=(r-L/2)/r ; Cos=(r-L)/r. Solving for L, you get a 2nd degree equation in which the solutions are 2r and 2r/5. As you know the side L is smaller than R, the only valid answer is 2r/5. For radius=5, L becomes 2, thus the Area (L²) = 4.

A simpler solution (what I did) is to draw a perpendicular from the base to the point where the circles meet, so y = 5 - x/2. Then the Pythagoras and quadratic method is just the same but there’s only one variable

Choose a coordinate system such that its origin is at the center of the bottom edge of the red square, y and x axes point towards up and to the right respectively. Since the top right corner (x,y) is on the circle we can write (x-r)²+(y-r)²=r² and if that point (and its mirror image on the other circle) would be moved a bit on the circle, it would distort the red square into a rectangle. The only way it will remain square is when y=2x. Substitute y into the equation of the circle and solve for x: (x-r)²+(2x-r)²=r². x²-2rx+r²+4x²-4rx+r²=r². 5x²-6rx+r²=0. Solutions: x=r and x=r/5. Reject the first and the area is A=2xy=4x²=4r²/25=4

I always find the little tricks used to determine side lengths so fun. Good Job!

that was exciting! I no longer have the brain for hard math stuff like this but watching people solve problems like this is very fun thanks for bringing some joy to my night

You reminded me of my childhood. Those were some of the most tensed moment preparing for competitive exams.

I'm 44 years old, Maths was my subject in school and this came in my feed, I watch the whole thing and was genuinely rivetted. Andy you are a gift my friend

Even though I just watched you do it, I still dont know how you did it. I'm slowly starting to understand bits. You're significantly better than any teacher I ever had.

I graduated in electrical engineering Bsc in 2021 and like algebra. But I always felt bad when had to solve at geometry problems. Your videos make me feel the happyness of understading and solving problems again. Thank you