It's a fun mystery game and I find it sad that some people don't see math in that perspective (tbh I can direct my lament at the way math is taught [heh], but that's a separate comment)
Weird?! I've been a fan of math puzzles since high-school (and a math enthusiast in general) and still am about 6 years later now, and not planning to change that, actually that's one of the reasons many of my friends respect me, and if the people around you think you're weird for that then I believe you shouldn't even sit with them, I may be an engineer and a math addict, doesn'tmean I'm creepy, still that doesn't mean I'll specifically bring up those talks in a conversation with just anyone cause most wouldn't be interested (I'd totally show it to others proudly if I'm solving a puzzle though even if they're not interested in math they might think it cool)
@Deathkyun me too, I find it sad that people don't see math as it actually is in general, they just see it as a nerdy subject only meant for geniuses and dummies will just never be able to understand or do it ever, they don't realize that it's a subject made for everyone and that it's merely an amazing arrangement of logical rules to reach awesome conclusions (that's not fully on grade teachers though you can't teach the logic and ideas of math rules for kids who barely understand how to apply it, but it should be taught at middle and high-school though)
It's also interesting that we don't need to calculate a and b separately. So, they can be anything as long as they're in that relationship and the area is still 108.
An alternate method is to use scale factors instead of using 3 variables. If we call the blue diameter d (we can solve for this as you did in the video but it's unnecessary), and then call the large red diameter dx, where x is the factor needed to scale d up to the width of the rectangle, scaling d up by x will scale the area up by x^2, so the large red semicircle has area 36x^2. The small red diameter is then dx-d or d(x-1) so x-1 is the scale factor this time, so area is scaled by (x-1)^2, so the small semicircle has area 36(x^2 -2x+1). Adding these gives the total red area as 36(2x^2 -2x+1), we just need to find x. I use the same right triangle you did, but this time I switch over to radii, so the left side is 2r, the large red radius is rx, and the small red redius is rx-r, subtracting them gives the bottom edge of the triangle as simply r, and adding them gives the hypotenuse of 2rx-r. We can reduce this triangle by a factor of r, giving us side lengths 1, 2, and 2x-1. Using pythagoras we get 5=4x^2 -4x+1, a little rearranging we get 4x^2 =4x+4, divide everything by 4 and we get x^2 =x+1. We might recognize that this is the equation for the golden ratio, but we don't actually need to finish solving for x, we can just use the fact that x^2 = x+1 so substitute into the area we want: 36[2(x+1)-2x+1] becomes 36[2x+2-2x+1] the x's actually cancel out, we just get 36(3)=108 for the area.
@@BlackEnigma908 low iq is brutal. no shame in it though nothing wrong with being perfectly average or even below average. nothing wrong with janitors we need them
Once you found 3c^2 = a^2 + b^2 you should have looked back at the picture and realized that the 3 semicircles totaled 3(pi/2 * c^2) which equaled (a^2 + b^2) * pi/2 which equaled 108.
We did our a & b backwards, so when you solved the known radius as (b - a) red alerts kept going off in my head, and it took me a second to reorient. Same answer though! And seeing that triangle was such a simpler method than solving for the sums of the sines and differences of the cosines. On the bright side, doing it that way brought out the golden ratio and its conjugate in my intermediate steps, so I'll keep that as a consolation prize. Thanks for the video!
I saw that c=a-b and was surprised when you squared (a-b) rather than squaring c. I wondered if I was wrong to think that could be substituted or maybe that approach was more complicated. I was actually surprised at ~4:38 when you noticed the equality and chose the c^2 method. Wow, maybe there's hope for this guy who hasn't been into algebra for several decades. 😎
I'm too lazy to do the first step, which is to find the radius of the blue circles in terms of pi. Hence, the radius of the blue circles is k. Let the radius of the big circle be y and the radius of the small circle be x. 2x+2k=2y, so y-x=k The centers of the small circle have a vertical displacement of 2k and a horizontal displacement of something. This something is the absolute value of y-x, which is k. Distance between the two: rt(k² + (2k)²) = krt5 x + y = krt5 2x = krt5-k x = ½(krt5-k) y = x+k = ½(krt5+k) ½(pir² + pir²) = ½pi(x² + y²) = ½pi(½)²((krt5-k)² + (krt5+k)²) = ⅛pi(k²)((rt5-1)²+(rt5+1)²) = ⅛pi(k²)(5-2rt5+1+5+2rt5+1) = ⅛pik²(12) = ½(3)pik² ½pik²=36. k² = 72/pi = ½(3)pi(72/pi) = 36(3) = 108
Bro, I started laughing and couldn't stop smiling after he made the triangle and well, thank god me and ex broke up because it would ruin her to see someone else make me smile at my phone like that
So the only problem with this is that we must assume that the line created by connecting both radiuses of the red circles is straight @2:37. Otherwise this does not work.
For two circles that touch eachother at a single point this will always be true.. right? The addition of the two radii will always be the shortest path between center 1 and 2
I don't think you showed it, but I found it beautiful how the rectangle can be sliced into 3 equal pieces with one blue semicircle fitting perfectly into each since the blue and pink areas are equal.
I dont get why you didn't replace the b-a in (b-a)^2 for c right then and there (at 3:53). Youve already proven b-a = c, you can do the substitution there.
BOOM! Proud I got the right answer. Took half a page of calculations. I did approximately the same thing, tho for some reason I needed to use the Quadratic equation along the way.
@OmModani-g3p yep, I caught that too (see above comment) also, to be clear I love all of his videos... Didn't mean it as a criticism, but a genuine question
Eu sou um brasileiro, e o seu video está maravilhoso, apesar de eu não compreender muito bem a língua inglesa, a sua linguagem matemática é tão boa que eu consegui entender tudo, sem legendas, sem nada. Não acho que você esperava esse tipo de elogio, mas você realmente fala matematicamente.
It's not obviously clear to me why the line from the center of the small circle to the center of the large circle should pass through the point that small circle and large circle touch each other.
At the point where two circles touch, they are tangent to each other, and a tangent line and radius are always perpendicular to each other. If both circles share a single tangent line and their radii meet at a single point, but from opposite directions, then i must be a straight line because both are at 90° to the tangent and 90+90=180° which is a straight line.
I think I missed something... how do we know that there's a straight line connecting the midpoints of the diameters of both semi-circles that passes through the point where they touch?
It is a basic circle theorem. If two circles are tangent to each other (internally or externally), then the centre of each circle and the point of tangency are colinear. If you think about how the chord bisector theorem works, you should be able to prove this easily yourself.
By the same reasoning that we can assume that the semicircles are semicircles, and that we use to assume that the 2 red semicircles are perfectly tangent to one another, and that we use to assume that the 36's in the diagram represent the areas of the blue semicircles. None of these things are explicitly given to us. But we assume all of them based simply on the fact that this is a problem that is intended to be solvable, which would be impossible without these obviously intended assumptions.
How do we know the semicircles are perfectly halved, with the center on the flat edge rather than slightly to one side? Just convention? I know right triangles often have a small square drawn in the corner.
A^2+{72/pi}a-72/pi=0 So a=-{72/pi}+-{72/PI+4*72/pi}✓2 => (+-{5}-1){18/pi} A must be positive so exclude - So area is pi/2*(2*({5}-1){18/pi})^2+72/pi+2*({5}-1){18/pi}{72/pi}) pi/2*(2*({5}-1)^2*18/pi+72/pi+2*({5}-1)36/pi) ({5}-1)^2*18+36+({5}-1)36 18*(({5}-1)^2+2+2({5}-1)) 18*((5-2{5}+1)+2+2({5}-1)) 18*6=36*3 90-18*{5}
They are, the radius is perpendicular to the tangent line. The tangent lines for two circles at the point where they touch is the same line. So, the radii are perpendicular to the same line on opposite sides, i.e. a straight line
How do we know for sure that 2a + 2c = 2b other than visually? Is it possible for 2a and 2c not to meet exactly at a point on the top edge of the rectangle?
Yes, actually. If solved a different way, we actually find that the scale factor between the blue semicircles and the large red semicircle is the golden ratio, φ so b=cφ . And similarly we would find that the scale factor for the small red semicircle is the φ-1, so a=c(φ-1). With c being sqrt(72/π), we can approximate all of these values: c is approx 4.787, b is approx 7.746, and a is approx 2.959
You're missing something important in between: Showing that the yellow triangle actually exists - you should mention that the radii at the tangent line between the two circles meet in a 180° angle.
This is another Catriona Agg puzzle. It is my goal to solve all of her puzzles!
Hey Andy I have tried to contact you several times through almost all the social media platforms, however you never responded
That subscribe button looks important, let's put a box around it.
How exciting
And move it over here.
You're not getting a heart lil bro
@caseydriscoll5331 😭 This is funnier
@@ABCD-v6v1jbro liked his own comment
People think I'm weird for watching these math videos, but they're so entertaining that I can't stop watching
you are one lucky mf, imagine you would love something like w33d or league of legends, very useless. Math will help you in life
What people? Personally I don't think you're weird.
It's a fun mystery game and I find it sad that some people don't see math in that perspective (tbh I can direct my lament at the way math is taught [heh], but that's a separate comment)
Weird?! I've been a fan of math puzzles since high-school (and a math enthusiast in general) and still am about 6 years later now, and not planning to change that, actually that's one of the reasons many of my friends respect me, and if the people around you think you're weird for that then I believe you shouldn't even sit with them, I may be an engineer and a math addict, doesn'tmean I'm creepy, still that doesn't mean I'll specifically bring up those talks in a conversation with just anyone cause most wouldn't be interested (I'd totally show it to others proudly if I'm solving a puzzle though even if they're not interested in math they might think it cool)
@Deathkyun me too, I find it sad that people don't see math as it actually is in general, they just see it as a nerdy subject only meant for geniuses and dummies will just never be able to understand or do it ever, they don't realize that it's a subject made for everyone and that it's merely an amazing arrangement of logical rules to reach awesome conclusions (that's not fully on grade teachers though you can't teach the logic and ideas of math rules for kids who barely understand how to apply it, but it should be taught at middle and high-school though)
the answer equals the areas of the 3 blue semi circles!!!! HOW EXCITING!!
It's also interesting that we don't need to calculate a and b separately. So, they can be anything as long as they're in that relationship and the area is still 108.
oh my god, you just SPOILED THE ENTIRE VIDEO FOR ME 😢
That’s super interesting, is it circumferencestantial or as good as a rule for always solving this problem with that method?😅
@@aleksapupovac i don't think it can be generalized. Maybe that's for Andy to research for another video
@@kaitek666 i don't think it can be generalized. it looks limited
An alternate method is to use scale factors instead of using 3 variables. If we call the blue diameter d (we can solve for this as you did in the video but it's unnecessary), and then call the large red diameter dx, where x is the factor needed to scale d up to the width of the rectangle, scaling d up by x will scale the area up by x^2, so the large red semicircle has area 36x^2. The small red diameter is then dx-d or d(x-1) so x-1 is the scale factor this time, so area is scaled by (x-1)^2, so the small semicircle has area 36(x^2 -2x+1). Adding these gives the total red area as 36(2x^2 -2x+1), we just need to find x.
I use the same right triangle you did, but this time I switch over to radii, so the left side is 2r, the large red radius is rx, and the small red redius is rx-r, subtracting them gives the bottom edge of the triangle as simply r, and adding them gives the hypotenuse of 2rx-r. We can reduce this triangle by a factor of r, giving us side lengths 1, 2, and 2x-1. Using pythagoras we get 5=4x^2 -4x+1, a little rearranging we get 4x^2 =4x+4, divide everything by 4 and we get x^2 =x+1. We might recognize that this is the equation for the golden ratio, but we don't actually need to finish solving for x, we can just use the fact that x^2 = x+1 so substitute into the area we want: 36[2(x+1)-2x+1] becomes 36[2x+2-2x+1] the x's actually cancel out, we just get 36(3)=108 for the area.
I'm so disappointed in myself that i can never solve any of these questions 😔
FR
it's just low iq. nothing wrong with being perfectly average or even below average . the world needs janitors too after all
Same 😢
Don't be. You'll learn! Gotta to believe in yourself.
@@BlackEnigma908 low iq is brutal. no shame in it though nothing wrong with being perfectly average or even below average. nothing wrong with janitors we need them
At first I was like, what possessed you to divide by 2…??? then it clicked! 😂
Once you found 3c^2 = a^2 + b^2 you should have looked back at the picture and realized that the 3 semicircles totaled 3(pi/2 * c^2) which equaled (a^2 + b^2) * pi/2 which equaled 108.
I just noticed the red area is equal to the blue area. How exciting!
I have always been terrible at math but this really makes me want to get better. This just seems so much more FUN
We did our a & b backwards, so when you solved the known radius as (b - a) red alerts kept going off in my head, and it took me a second to reorient. Same answer though! And seeing that triangle was such a simpler method than solving for the sums of the sines and differences of the cosines. On the bright side, doing it that way brought out the golden ratio and its conjugate in my intermediate steps, so I'll keep that as a consolation prize. Thanks for the video!
I have a faster solution: 36 + 36 + 36 = 108 ;-)
Nintendo is going to sue you bro
I'm dying bro
Wow, the way to the answer was so satisfying
How exciting!
That was breathtakingly stunning 😮😊
that was sooo satisfying
I saw that c=a-b and was surprised when you squared (a-b) rather than squaring c. I wondered if I was wrong to think that could be substituted or maybe that approach was more complicated. I was actually surprised at ~4:38 when you noticed the equality and chose the c^2 method. Wow, maybe there's hope for this guy who hasn't been into algebra for several decades. 😎
I'm too lazy to do the first step, which is to find the radius of the blue circles in terms of pi. Hence, the radius of the blue circles is k.
Let the radius of the big circle be y and the radius of the small circle be x.
2x+2k=2y, so y-x=k
The centers of the small circle have a vertical displacement of 2k and a horizontal displacement of something.
This something is the absolute value of y-x, which is k.
Distance between the two: rt(k² + (2k)²) = krt5
x + y = krt5
2x = krt5-k
x = ½(krt5-k)
y = x+k
= ½(krt5+k)
½(pir² + pir²)
= ½pi(x² + y²)
= ½pi(½)²((krt5-k)² + (krt5+k)²)
= ⅛pi(k²)((rt5-1)²+(rt5+1)²)
= ⅛pi(k²)(5-2rt5+1+5+2rt5+1)
= ⅛pik²(12)
= ½(3)pik²
½pik²=36. k² = 72/pi
= ½(3)pi(72/pi)
= 36(3) = 108
a+c=b , b-a=c
πc^2/2=36, c^2=36(2/π)
(2c)^2+(b-a)^2=(b+a)^2 , 4c^2= (b+a)^2 - (b-a)^2= 4ab , c^2=ab
Answer= (a^2+b^2)π/2 =π/2[ (b-a)^2+2ab] = π/2[3c^2] =(36)3
Bro, I started laughing and couldn't stop smiling after he made the triangle and well, thank god me and ex broke up because it would ruin her to see someone else make me smile at my phone like that
That really snuck up on me! BAM there it is! How exciting.
4:33 The Best part 😅
Oop!
Oop!
So the only problem with this is that we must assume that the line created by connecting both radiuses of the red circles is straight @2:37. Otherwise this does not work.
For two circles that touch eachother at a single point this will always be true.. right? The addition of the two radii will always be the shortest path between center 1 and 2
there will always be a staight line
I don't think you showed it, but I found it beautiful how the rectangle can be sliced into 3 equal pieces with one blue semicircle fitting perfectly into each since the blue and pink areas are equal.
Just started watching, looks exciting.
Edit: it was brilliant
I dont get why you didn't replace the b-a in (b-a)^2 for c right then and there (at 3:53). Youve already proven b-a = c, you can do the substitution there.
BOOM! Proud I got the right answer. Took half a page of calculations. I did approximately the same thing, tho for some reason I needed to use the Quadratic equation along the way.
the way the solution was backwards was really cool
How exciting
I was thinking of using the property that all semicircles are, by definition, similar.
I like how you think of solutions outside the box like wow. I could have NOT thought of that AT ALL! Props to you!
You are a genius .You changed Maths to entertainment 🎉
Your problem solving skills are insane
Congratulations!
Since we already know c is (b-a) can't we simplify that sooner?
Oh nevermind... You caught it 😅
i thought the same too lol
Yes you can, he just didn’t realise it. Its still an amazing video though
@OmModani-g3p yep, I caught that too (see above comment) also, to be clear I love all of his videos... Didn't mean it as a criticism, but a genuine question
108🙏
Wait that’s so cool! What a fun problem! How exciting!!
I love how in fact we didn’t have to know the radius of a or b👍
i found this video to be very interesting :) thanks, Andy!
Eu sou um brasileiro, e o seu video está maravilhoso, apesar de eu não compreender muito bem a língua inglesa, a sua linguagem matemática é tão boa que eu consegui entender tudo, sem legendas, sem nada. Não acho que você esperava esse tipo de elogio, mas você realmente fala matematicamente.
It's not obviously clear to me why the line from the center of the small circle to the center of the large circle should pass through the point that small circle and large circle touch each other.
At the point where two circles touch, they are tangent to each other, and a tangent line and radius are always perpendicular to each other. If both circles share a single tangent line and their radii meet at a single point, but from opposite directions, then i must be a straight line because both are at 90° to the tangent and 90+90=180° which is a straight line.
that was fun
I feel like it took a turn in the middle, there
I think I missed something... how do we know that there's a straight line connecting the midpoints of the diameters of both semi-circles that passes through the point where they touch?
It is a basic circle theorem.
If two circles are tangent to each other (internally or externally), then the centre of each circle and the point of tangency are colinear.
If you think about how the chord bisector theorem works, you should be able to prove this easily yourself.
@@Grizzly01-vr4pn Huh. Did not realize that. Thank you.
Great puzzle. I noticed that the blue area is also 108.
I've been playing around with this, and also found that c / a = b / c = golden ratio
Best problem Ivh seen here
right off the bat we can tell 2b = 2a+36 which simplifies to b = a+18
and i am wrong immediately, would have worked if 36 was the diameter of the blue semicircles, but at least a+c = b
and he mentioned it later in the video, my comment is useless and inaccurate
This was amazing ❤
I feel like there's a much simpler and less roundabout way of solving this, but I'm also not very good at math so I'd be unable to figure it out.
Nice solution, could you show how to continue the 3 equations?
Top is not equal bottom!
Me for no reason at 2 am 😂
one of your top vids
What a great solution, what can I say? How …..
You didn't have to calculate c = b - a. It's clear from the picture.
I wonder if you could somehow get to equation a*a + b*b = 3c*c
If you stop at 5:19 then instead of plugging in the numerical value, you can arrive there in one step ;)
Good GRIEF! Brilliant!
But how can you assume it to be a rectangle?
I'm also curious
probably given by the question
Probably because the top red semi-circle starts after the top blue semi-circle stops. Therefore the height is not equal to the width and not square
@@SpcyToast no the point is we can't assume that the edges are right angle
By the same reasoning that we can assume that the semicircles are semicircles, and that we use to assume that the 2 red semicircles are perfectly tangent to one another, and that we use to assume that the 36's in the diagram represent the areas of the blue semicircles. None of these things are explicitly given to us. But we assume all of them based simply on the fact that this is a problem that is intended to be solvable, which would be impossible without these obviously intended assumptions.
Damm that looks hard😊
How do we know the semicircles are perfectly halved, with the center on the flat edge rather than slightly to one side? Just convention? I know right triangles often have a small square drawn in the corner.
Can we truly assume that the top is constrained to the diameter of 2a+2c? It looks that way, but is it explicitly defined?
R+a={72/pi}+a=b
B^2=72/pi*2a{72/pi}+a^2
Area is Pi/2*(a^2+b^2)=pi/2*(a^2+72/pi+2a{72/pi}+a^2)
Not sure how to solve a
Wait okay. There's a right triangle where (b+a)^2=(b-a)^2+4*72/pi
B2+2ab+a2=b2-2ab+a2+4*72/pi
4ab=4*72/pi
Ab=72/pi=({72/pi}+a)a
A^2+{72/pi}a-72/pi=0
So a=-{72/pi}+-{72/PI+4*72/pi}✓2 => (+-{5}-1){18/pi}
A must be positive so exclude -
So area is
pi/2*(2*({5}-1){18/pi})^2+72/pi+2*({5}-1){18/pi}{72/pi})
pi/2*(2*({5}-1)^2*18/pi+72/pi+2*({5}-1)36/pi)
({5}-1)^2*18+36+({5}-1)36
18*(({5}-1)^2+2+2({5}-1))
18*((5-2{5}+1)+2+2({5}-1))
18*6=36*3
90-18*{5}
I evaluated it and got 3*36 but I think my comment was filtered?
Love this
Is there a way to know that A+B is a line? Two radii that touch aren't necessarily straight or am I wrong?
They are, the radius is perpendicular to the tangent line. The tangent lines for two circles at the point where they touch is the same line. So, the radii are perpendicular to the same line on opposite sides, i.e. a straight line
Thank you
2:27 is this triangle method the only way to solve this? Creating a right triangle wasn't immediately obvious to me, or at all lol
how exiting 🗣🔥
Plz sir arrange playlist according to grade..
Aahhhhh.....clev-er!
or 3 blue area = 2 red area
3x36 = 108
is this coincidence ?
108
why does it look like hes stuck in the backrooms?
Cute problem
How do we know for sure that 2a + 2c = 2b other than visually? Is it possible for 2a and 2c not to meet exactly at a point on the top edge of the rectangle?
Is that the Nintendo Switch?
New switch 2 leak just dropped
How do you know they’re all half circles?
It is a given of the question.
Does anyone who solved it know what A and B would be in this exercise?
Yes, actually. If solved a different way, we actually find that the scale factor between the blue semicircles and the large red semicircle is the golden ratio, φ so b=cφ . And similarly we would find that the scale factor for the small red semicircle is the φ-1, so a=c(φ-1).
With c being sqrt(72/π), we can approximate all of these values: c is approx 4.787, b is approx 7.746, and a is approx 2.959
did you just leak Switch 2
So I'm the only one who sees a Nintendo switch? Really?
..Ok just making sure
Cool!👍👍
You're missing something important in between: Showing that the yellow triangle actually exists - you should mention that the radii at the tangent line between the two circles meet in a 180° angle.
Hello, new sub here, please use black background for eye comfort, thank you. :D
but was it exciting?
How does she come up with these?
PSP issue
Wow
Please don't spend long time simplifying equations.
Spend time on theory if you wanna spend more time explaining.
69th like. How exciting!
first
First
How exciting
How exciting