Super Popular Problem

A lot of people sent me requests to go over this problem. I hope you love it!

[CantoMando]

Love watching these. Hated working as an engineer, but I always had a love for math and this takes me back to solving fun puzzles like these in school

[joshuafeichter]

My wife doesn’t understand why I love these videos

[alexandersvensson]

I have not done math for 3 years since my studies are in different fields, but this was truly exciting. Nice video!

[davidrosenthal]

Since the answer mustn’t depend on the exact values of x and r, there is a very fast solution: Assume r is very, very small. This trivially makes the blue rectangle a square with a side of length 5.

[zosanabria]

this guy solved the problem before i even knew what was happening

[MoholtSb]

It's a pity when i was at university i struggled a lot on passing math exams and i didn't appreciate the little things like this.
Now after a few years i appreciate it

[_umarro_]

you know a problem is even more exciting than usual when he hit us with the "isn't that cool?"

[srini]

I missed that the 5cm line was tangent to the semi-circle. Cool problem. I have enjoyed. Thanks once again!

[bradballinger]

Another great video.

This problem is a wonderful opportunity to use the Power of a Point Theorem. This says that if a line through point P intersects a circle and we measure the near and far distances from P to the circle along that line, the product of the two measurements is the same...no matter which line it was!

In this case, P is the bottom left corner of the rectangle, and the circle I care about is the one with radius r.

Measuring horizontally, we see that the "near distance" to the circle is x, while the "far distance" to the circle is x+2r. The product of these measurements is x(x+2r).

Measuring along the red segment, the "near distance" is 5cm, and so is the "far distance". Their product is 25cm^2.

By PoP, x(x+2r)=25cm^2.

[laincoubert]

also you can use circle properties. idk the name of the theorem, but it states that tangent^2 = secant * secant's external segment. this problem is a unique case when the secant (the rectangle's lower side) passes through the circle's center.

[lelouch]

Never could have thought the process would be this simple
I mean I looked at it and was like calculus for sure

[Reskamo]

Your enthusiasm when saying "isn't that cool? :)" is really lovely

[notryangosling]

I've tried this question before but I made the assumption that the radius of semicircle and quater circle are same.
It's amazing to see how that would not affect the answer at all, we still get the same result even if the radius was not same.

[VCOTABFONDD]

I’ve been looking for channels like this for ages because I *love* math! Just hearing one of the math terms just makes me smile!

[ClaudioBrogliato]

My way of solving: 5 is a leg of a 30 - 60 - 90 triangle which hypotenuse is 2 r and the other leg r. So r is 5 over sqrt 3. The rectangle has a side which is 3r long and the other which is as long as r. The area is 5 over sqrt 3 multiplied by 3 times 5 over sqrt 3 which is equal to 75 over 3 which is 25.

[jonathanmartin]

I literally screamed in excitement when you said that tangents of circles are perpendicular to the radius that touches at the same point. I completely forgot about that when I was trying to solve it on my own 😂

[kingsonamadi]

I tried it by considering the radius of the semicircle and the quater circle to be same and got the same answer.

However when you consider x and r, I was really curious to see where i went wrong. Amazing sum. Thanks for explaining.

[strangeboysam]

By actually applying Power of point Theorem you can end on the same result directly.

[yucalvin]

If the question didn't mention the line to be a tangent line, how am I supposed to prove that?

[nicholascalabria]

I loved math in high school. I thought I wanted to be a mathematician cause I loved stuff like this. Ended up in accounting and slowly got away from math like this. This video reignited that love. That's so cool. I love this. Thank you!