The key thing to observe here is that there are "arbitrary" parameters in this problem. Specifically, the white circle could be made larger or smaller, which would move the location of the 20-long line. And yet we're led to believe there is a well-defined value for the red area anyway. So that area must be the same regardless of the variant details. In cases like this the thing to do is take an extreme case - it's usually easy to solve. If the white circle is reduced to radius zero, then the radius of the large quarter circle becomes 20. The entire quarter circle will be red. That area calculation is easy - it's just pi*(20^2)/4 = 100*pi. That must be the right answer regardless of the radius of the white circle. Q.E.D. So, with a simple bit of reasoning it becomes possible to solve the problem in your head in a few seconds. This would have been much more tedious if we'd been given some other piece of information that restricted the problem to a particular case. We could still have solved it, but we'd have needed to work through all the geometry and trig and so on - we wouldn't have been able to conclude right off that bat that the result was independent of the details.
The problem with this type of reasoning in my view, is that you would need to show that the area is invariant with respect to the radius of the inscribed circle, and showing that is not always easy. If this were a Physics problem, we could invoke some physical law depending on the context to argue that the area is an invariant, but the problem under consideration is a purely mathematical problem. Saying that "the problem didn't give the radius of the inscribed circle, so it must be irrelevant" is not sufficient, because if the author wanted, he could have labelled the radius "r" just to make things look complicated.
@@User-jr7vf I agree - if I were to come onto a problem like this, I'd never be able to just jump to the idea that the result was an invariant. As you note - I'd have to prove that before relying on it. In a situation like this, though, if we assume the problem being presented to us actually has a well-defined answer, then that answer MUST be an invariant. So the only reason I could make that assumption was because of the context of the problem. Of course there would always be the possibility that they'd erred in presenting the problem to us, and failed to give us an important (and necessary) piece of information. So - you are right; I made a leap of faith so to speak - faith that they had given us a solvable problem and given it to us correctly. In a real world situation where this problem just "came up in research" that wouldn't have been possible, and we'd have had to do the hard work. I think your point is totally valid.
@@KipIngram I think you are very smart for thinking of a short route to solve the problem, this shows that you have a "geometry oriented" mind, i.e. you are able to manipulate the figure in your mind and figure out the result based on that. No worries, I'm not in a position to dismiss your method, I'm just a hobbyist when it comes to math :)
@@User-jr7vf I am too, really. I've had a decent amount of math. I studied engineering, but in graduate school I knew I wanted to "self-study" physics later, so I tried to load up on extra math. But I'd never really consider myself an expert of any sort. I feel kind of like my answer on this one was a "trick, since it relied on the nature of the question situation. I suppose taking advantage of favorable circumstances is a decent skill to have. And I THINK I could have worked this out the hard way - it just wouldn't have been as much fun. 🙂
Damn, that is cool. At thirst glance it doesn't look solvable. Because you know, you can't find r and R. But when you think about it, it turns out you don't need these radiuses to get the answer.
Area if the red shaded region=1/4πR^2-πr^2=π(1/4R^2-r^2) (2r)^2+20^2=R^2 4r^2+400=R^2 R^2-4r^2=400 1/4R^2-r^2=100 Red shaded region=100π square units.=314 square units.❤❤❤
Replace 20 by some fixed constant a. If you wanted to get R_(larger Radius) and r_(smaller radius) as a function of any arbitrary positive real number e let R = 2 * r + e implies r(e) = (a^2 - e^2)/(4 * e) and R(e) = (a^2 + e^2)/(2 * e) and this implies that the given area A = a^2 * pi/4. In this way we have all possible circles for any positive real number e so that the given area A = a^2 * pi/4.
I considered the bottom right corner of the figure to be (0, 0) and used x^2 + y^2 = r^2. The x coordinate is -2r and the y coordinate must be 20. Plug these in and you find that (R - 2r) = 400. The rest is the same as the video! :)
Don't be scared, be just excited! It's totally normal to feel a mix of emotions when diving into something new like geometry...we've all been there! 😊 You've got this!
@@ThePhantomoftheMath Thanks, a minute ago I tried to find R, or r, and ended up finding the area in a completely new way! Helps boost that excitedness because of the possibilities to solve.
@@darkphoenixzyzz oops. Should have said radius of white, not black. If you decrease the radius of the red quarter circle to 20 it will ‘squeeze’ the white circle down to nothing.
Great observation on the "Big R" being 20. Once again a black circle gets caught up in some red pi and makes racial accusations against the sine of a white variable from a different equation completely. BTW... The circle is black with a white circumference. I just can't stop thinking about American Pi.
Inutile de faire un calcul compliqué ! Puisqu'une seule donnée est disponible. On peut faire varier cette donnée, par exemple le rayon du petit cercle jusqu'à 0 . Et alors R= 20. Donc quelque que soit r on obtient A= pi(20^2)/4=100pi env. 314 !!!!!! Il m'a fallu 2 secondes...
The key thing to observe here is that there are "arbitrary" parameters in this problem. Specifically, the white circle could be made larger or smaller, which would move the location of the 20-long line. And yet we're led to believe there is a well-defined value for the red area anyway. So that area must be the same regardless of the variant details.
In cases like this the thing to do is take an extreme case - it's usually easy to solve. If the white circle is reduced to radius zero, then the radius of the large quarter circle becomes 20. The entire quarter circle will be red. That area calculation is easy - it's just pi*(20^2)/4 = 100*pi. That must be the right answer regardless of the radius of the white circle. Q.E.D.
So, with a simple bit of reasoning it becomes possible to solve the problem in your head in a few seconds.
This would have been much more tedious if we'd been given some other piece of information that restricted the problem to a particular case. We could still have solved it, but we'd have needed to work through all the geometry and trig and so on - we wouldn't have been able to conclude right off that bat that the result was independent of the details.
The problem with this type of reasoning in my view, is that you would need to show that the area is invariant with respect to the radius of the inscribed circle, and showing that is not always easy. If this were a Physics problem, we could invoke some physical law depending on the context to argue that the area is an invariant, but the problem under consideration is a purely mathematical problem. Saying that "the problem didn't give the radius of the inscribed circle, so it must be irrelevant" is not sufficient, because if the author wanted, he could have labelled the radius "r" just to make things look complicated.
@@User-jr7vf I agree - if I were to come onto a problem like this, I'd never be able to just jump to the idea that the result was an invariant. As you note - I'd have to prove that before relying on it. In a situation like this, though, if we assume the problem being presented to us actually has a well-defined answer, then that answer MUST be an invariant. So the only reason I could make that assumption was because of the context of the problem. Of course there would always be the possibility that they'd erred in presenting the problem to us, and failed to give us an important (and necessary) piece of information. So - you are right; I made a leap of faith so to speak - faith that they had given us a solvable problem and given it to us correctly. In a real world situation where this problem just "came up in research" that wouldn't have been possible, and we'd have had to do the hard work.
I think your point is totally valid.
@@KipIngram I think you are very smart for thinking of a short route to solve the problem, this shows that you have a "geometry oriented" mind, i.e. you are able to manipulate the figure in your mind and figure out the result based on that. No worries, I'm not in a position to dismiss your method, I'm just a hobbyist when it comes to math :)
@@User-jr7vf I am too, really. I've had a decent amount of math. I studied engineering, but in graduate school I knew I wanted to "self-study" physics later, so I tried to load up on extra math. But I'd never really consider myself an expert of any sort. I feel kind of like my answer on this one was a "trick, since it relied on the nature of the question situation. I suppose taking advantage of favorable circumstances is a decent skill to have. And I THINK I could have worked this out the hard way - it just wouldn't have been as much fun. 🙂
Damn, that is cool. At thirst glance it doesn't look solvable. Because you know, you can't find r and R. But when you think about it, it turns out you don't need these radiuses to get the answer.
Yeah, it's a "trick of the eye" problem. But in the end, it's much easier than you initially thought.
Love this, thank you!
Area if the red shaded region=1/4πR^2-πr^2=π(1/4R^2-r^2)
(2r)^2+20^2=R^2
4r^2+400=R^2
R^2-4r^2=400
1/4R^2-r^2=100
Red shaded region=100π square units.=314 square units.❤❤❤
Epic dude!
Ty friend. Glad you liked it!
Replace 20 by some fixed constant a.
If you wanted to get R_(larger Radius) and r_(smaller radius) as a function of any arbitrary positive real number e let
R = 2 * r + e implies r(e) = (a^2 - e^2)/(4 * e) and R(e) = (a^2 + e^2)/(2 * e) and this implies that the given area A = a^2 * pi/4.
In this way we have all possible circles for any positive real number e so that the given area A = a^2 * pi/4.
Exactly wright!
Impressive
@@AslamSheikh-c2r Thank you!
I considered the bottom right corner of the figure to be (0, 0) and used x^2 + y^2 = r^2. The x coordinate is -2r and the y coordinate must be 20. Plug these in and you find that (R - 2r) = 400. The rest is the same as the video! :)
That's a method using analytic geometry, and it's a clever one. Good job!
I instantly knew that I have to drop the circle to be tangent to the horizontal R. 20 frustrating minutes later I decided to watch the video ...
❤❤❤i am math student in school i like this video❤. Are you in university
Hi! I'm glad you like the video! ❤ No, I've finished university... I'm actually a math teacher now 🙂
@@ThePhantomoftheMath ok 👌
(Pre alg) I am now very scared, and very excited for geometry.
Don't be scared, be just excited! It's totally normal to feel a mix of emotions when diving into something new like geometry...we've all been there! 😊 You've got this!
@@ThePhantomoftheMath Thanks, a minute ago I tried to find R, or r, and ended up finding the area in a completely new way! Helps boost that excitedness because of the possibilities to solve.
@@axolotlinanutshell7629 You just defined why I love math! 😄Good job man!
100pi. Two seconds in my head.
Assume the radius of red is 20. The radius of black becomes zero. Quarter circle of radius 20 is one hundred pi.
@@konradhunter1407 wo wo wo, can you explain what do you mean by radius of red or black ?
@@darkphoenixzyzz oops. Should have said radius of white, not black. If you decrease the radius of the red quarter circle to 20 it will ‘squeeze’ the white circle down to nothing.
@@konradhunter1407 oh I see it now, that's smart
Great observation on the "Big R" being 20. Once again a black circle gets caught up in some red pi and makes racial accusations against the sine of a white variable from a different equation completely. BTW... The circle is black with a white circumference. I just can't stop thinking about American Pi.
Without watching.... push the line to the R edge, making the quadrant radius 20 and the circle zero.... :)
Inutile de faire un calcul compliqué ! Puisqu'une seule donnée est disponible. On peut faire varier cette donnée, par exemple le rayon du petit cercle jusqu'à 0 . Et alors R= 20. Donc quelque que soit r on obtient A= pi(20^2)/4=100pi env. 314 !!!!!! Il m'a fallu 2 secondes...
100pi