The figure in the thumbnail is Shri yantra we can see this in many Hindu's temples. I know this figure because I'm from India.The Shri Yantra, Sri Yantra, or Shri Chakra is a form of mystical diagram used in the Shri Vidya school of Hinduism. It consists of nine interlocking triangles that surround a central point known as a bindu. These triangles represent the cosmos and the human body.
Hi, For fun: 2 "and so on and so forth", 1 "and then so on and so forth", 1 "daa, daa, daa...", 1 "let's go ahead and do that", 2 "let's go ahead and clean up this board", 1 "let's go ahead and clean this up", & "I'll may be go and clean this up", 1 "I'll go ahead and", 6 "great", 1 "Ok, great", 1 "what I want to do", 3 "the next thing that I want to do", 1 "the next thing that I want to notice", and the final "that's a good place to stop".
I was wondering the same thing. If (a,b,c) is a solution of (x,y,z) then no doubt all permutations of (a,b,c) are solutions. But there is no way to say a=b=c directly. We can apply symmetry to do stuff like wlog a>=b>=c or relations like that and permute the solutions. But a=b=c is unique solution - that's false for any generalized set of symmetric equations. I wonder if its because the symmetric equations are linear and hence he assumes it has only one unique solution - something on those lines.
@@thephysicistcuber175 Yes I agree with you, I was just thinking what Michael may have had in mind when he directly concluded about unique solution given equations are linear. Obviously not all set of linear equations have unique solution
22:39 I thought it would be a nice idea but posting my exercises is literally taking opportunities away for Michael to solve those in his videos, which is in fact counterproductive and harms his channel’s growth. So here’s the last exercise for now. Using the numbers from 1000 to 9999, what is the biggest sum between the sum of numbers containing even digits only (2020, 4286, 8008...) OR the sum of numbers containing odd digits only (1993, 5771, 7319...)?
I got that the sum of the odds is 3,402,500 and the sum of the evens is 2,272,200. It's a little complicated to show how I got it using only text, but if someone asks me I can send my solution too.
Odd is bigger. For every even digit number between 1000 and 9999 you can obtain a bigger odd digit number by adding 1111. Conclusion is trivial. EDIT: I forgot to say that the odd digit number you obtain will still be between 1000 and 9999. This isn't hard to prove.
15:00 - You showed an inequality concerning areas of triangles, not parabolic sectors. Besides that, what does this derivation have to do with symmetry?
The notation I learned was congruent: ≅ (equal sign with tilde above), similar: ~ (tilde). For congruence modulo a number, I use the 3 parallel lines. Curious what the etymology of that is. I guess equal sign with tilde makes sense for triangles because congruent triangles have equal sides (=) and equal angles (~).
Cool, but logically flawed. It sort of assumes beforehand that there is only one solution. Compare x^2+y+z=x+y^2+z=x+y+z^2=1, which has two solutions with x=y=z but also a number of others, like (1,0,0).
21:35 I think where the assumption isn't quite right. There are 6 permutations of x,y,z so there are some permutations not represented by one of the 3 equations.
Strictly taken, he just introduced a new symbol phi and defined it to be (sqrt(5)-1)/2. He is allowed to do that (although I have to admit that his chosen name is unlucky). But on the other hand: If I have to name the length of the edges of a pentagon, no one would cry if I name them a, b, c, d and e. Despite the fact that normally e=2,718...
@@Zarunias there are symbols you should not use unless it is for their math definition.(i do not think anyone would use nowadays Pi as a symbol for anything else) And especially not use with a "value" that is "relatively close" to their math definition (sorry but i do not know the exact term in english). This might also confuse students. And 'e' does not appear (a lot if any) in geometry problems.
I don't see why symmetry helps with the last one. To me, it only says that if (x,y,z) is a solution, then (y,x,z) is a solution, too, and so on. But why would they need to be the same solutions? This only makes sense for me here because it's a linear system of equations and the associated 3x3 matrix has rank 3, so there is indeed a unique solution, which has to satisfy x=y=z because of the symmetry.
@Adam Romanov Well, your system is symmetric equation-wise, but the entire system is not. If you flip x, y and z around in the first equation, you don't get the second one or something, so it's not the same as in the video. But x^2 = x, y^2 = y is a symmetric system like in the video, right? One solution is (0,1), so they indeed don't have to have the same components. I guess invariance can mean f(X) = X or f(x) = x for all x in X. Oh, and I guess I should have written (y,z,x) instead of (y,x,z) in my first comment.
I was hoping this would help me solve the following but I am still stuck If x,y,z are positive reals with x+y+z=1 prove xy(x+y)^2 + yz(y+z)^2 + zx(z+x)^2 GTE 4xyz
There's a really nice systematic approach to symmetric homogeneous inequalities here: stanleyrabinowitz.com/download/inequalities.pdf. The inequality is already symmetric, and you can make it homogeneous by using the constraint to change the 4xyz term to 4xyz*1 = 4xyz*(x+y+z), so that all terms are degree 4. Then through a straightforward computation, you get (using notation from paper) [3, 1, 0] + [2, 2, 0] >= 2*[2,1,1]. The left hand side majorizes the right hand side, so the original inequality (with the equality constraint) is true.
or a more manual way that takes a bit more cleverness (personally I like when I can get a computer to solve it) can be found here math.stackexchange.com/questions/1657914/if-x-y-z0-and-xyz-1-then-prove-that-xyxy2yzyz2zxzx2-geq After dividing by xyz, it's a direct application of Titu's Lemma (which follows from the Cauchy-Schwarz inequality).
The last example is broken. This is just not true, consider the same problem, but in the Z[9] ring, then one solution of the x+2*y+3*z=30, 2*x+3*y+z=30, 3*x+y+2*z=30 system is x=0, y=6, z=3, and for that obviously x=y=z is not true.
in fact, his previous assessment isn't true either; the solutions only work under CYCLIC permutations of (x, y, z) - not just any permutation. consider this system: x + y + 2z + 2t = 12 x + 2y + 2z + t = 12 2x + 2y + z + t = 12 2x + y + z + 2t = 12 obviously (1, 3, 1, 3) is a solution, as is (3, 1, 3, 1), however (1, 1, 3, 3) is NOT a solution.
Quy Khoi Le so technically he probably meant combinations instead of permutations, which explains why your system doesn’t work. If you want to apply symmetry such that x=y=z=t, you also need the equations x+2y+z+2t=12 and 2x+y+2z+t=12, which yields the solution x=y=z=t=2. In your example, x,y,z,t are not in identical environments, so the assumption that they’re all equal is invalid.
One could say the ABBA sequence goes On and On and On.
I do, I do, I do.
The figure in the thumbnail is Shri yantra we can see this in many Hindu's temples.
I know this figure because I'm from India.The Shri Yantra, Sri Yantra, or Shri Chakra is a form of mystical diagram used in the Shri Vidya school of Hinduism. It consists of nine interlocking triangles that surround a central point known as a bindu. These triangles represent the cosmos and the human body.
a+b>=2sqrt(ab) 3:54
That's valid for positive integers a and b.
@@MaxMathGames non-negative reals*
@@twistedsector oh yes 👍👍👍
All the questions discussed by you were simple and easy, but you gave us a new perspective of looking at questions.
Impressive 👍👍👍❤️❤️❤️👌👌👌
Is the lectures on problem solving using PAUL ZEITZ as primary material? Feeling great and excited to learn!
Yeeeessssss, I've been needing a video on this for ages thank you so much!!
Hi,
For fun:
2 "and so on and so forth",
1 "and then so on and so forth",
1 "daa, daa, daa...",
1 "let's go ahead and do that",
2 "let's go ahead and clean up this board",
1 "let's go ahead and clean this up",
& "I'll may be go and clean this up",
1 "I'll go ahead and",
6 "great",
1 "Ok, great",
1 "what I want to do",
3 "the next thing that I want to do",
1 "the next thing that I want to notice",
and the final "that's a good place to stop".
He's a amazingly scintillating teacher
22:03 you assumed uniqueness there... Is there a nice way to prove uniqueness?
I was wondering the same thing. If (a,b,c) is a solution of (x,y,z) then no doubt all permutations of (a,b,c) are solutions.
But there is no way to say a=b=c directly.
We can apply symmetry to do stuff like wlog a>=b>=c or relations like that and permute the solutions. But a=b=c is unique solution - that's false for any generalized set of symmetric equations.
I wonder if its because the symmetric equations are linear and hence he assumes it has only one unique solution - something on those lines.
@@caesar_cipher You can't assume uniqueness from linearity. An easy way to see this is to see what happens if the coefficients are all equal to 1.
@@thephysicistcuber175 Yes I agree with you, I was just thinking what Michael may have had in mind when he directly concluded about unique solution given equations are linear. Obviously not all set of linear equations have unique solution
@@hai-mel6815 No, he didn't. Where do you think he proved it?
@@thephysicistcuber175 Now that I think about it, yes, you're right. Sorry for that
22:39
I thought it would be a nice idea but posting my exercises is literally taking opportunities away for Michael to solve those in his videos, which is in fact counterproductive and harms his channel’s growth. So here’s the last exercise for now.
Using the numbers from 1000 to 9999, what is the biggest sum between the sum of numbers containing even digits only (2020, 4286, 8008...) OR the sum of numbers containing odd digits only (1993, 5771, 7319...)?
Hmm... i thought I had it, but that mine has all the
The odd sum is bigger, used python
I got that the sum of the odds is 3,402,500 and the sum of the evens is 2,272,200. It's a little complicated to show how I got it using only text, but if someone asks me I can send my solution too.
Odd is bigger. For every even digit number between 1000 and 9999 you can obtain a bigger odd digit number by adding 1111. Conclusion is trivial.
EDIT: I forgot to say that the odd digit number you obtain will still be between 1000 and 9999. This isn't hard to prove.
@@thephysicistcuber175 Yeah that’s the idea.
2000 + 2002 + ... + 8886 + 8888 (all even digit numbers) < 3111 + 3113 + ... + 9997 + 9999 (all even digits numbers + 1111) < 1111 + 1113 + .... + 1999 + 3111 + 3113 + ... + 9997 + 9999 (all odd digits numbers)
15:00 - You showed an inequality concerning areas of triangles, not parabolic sectors. Besides that, what does this derivation have to do with symmetry?
Isn't the symbol for congruence 3 parallel lines, like an equal with an extra line, while the symbol you used is approximately equal to?
The notation I learned was congruent: ≅ (equal sign with tilde above), similar: ~ (tilde). For congruence modulo a number, I use the 3 parallel lines. Curious what the etymology of that is. I guess equal sign with tilde makes sense for triangles because congruent triangles have equal sides (=) and equal angles (~).
13:11 Those inequalities only work for A ≥ 0. We can use symmetry to handle A < 0 though.
Nice presentation. The next to last problem (roots of quadratic) appeared at the web site AlmostSure.
19:18 I think 𝝋 = (1+√5)/2. Here, you have 1/𝝋, isn't it ? ... which doesn't change anything to the final answer 😀.
Yeah, to have 4 solns
4:40 a+b
Looks like he got the inequality reversed XD
10:00 the problem is a simplification of one of the famous "Coffin problems"
Please make a video about scale symmetry, if you get the chance.
Thank you for the video!!
Could someone explain the last problem how x=y=z
The way symmetry is used to solve the last set of equations is so cool
Cool, but logically flawed. It sort of assumes beforehand that there is only one solution. Compare x^2+y+z=x+y^2+z=x+y+z^2=1, which has two solutions with x=y=z but also a number of others, like (1,0,0).
vindex7 indeed.
21:35 I think where the assumption isn't quite right. There are 6 permutations of x,y,z so there are some permutations not represented by one of the 3 equations.
19:23 that is not Phi
Strictly taken, he just introduced a new symbol phi and defined it to be (sqrt(5)-1)/2. He is allowed to do that (although I have to admit that his chosen name is unlucky). But on the other hand: If I have to name the length of the edges of a pentagon, no one would cry if I name them a, b, c, d and e. Despite the fact that normally e=2,718...
@@Zarunias there are symbols you should not use unless it is for their math definition.(i do not think anyone would use nowadays Pi as a symbol for anything else) And especially not use with a "value" that is "relatively close" to their math definition (sorry but i do not know the exact term in english).
This might also confuse students.
And 'e' does not appear (a lot if any) in geometry problems.
In 20:05 you erased the ± but I don't think you should have... Anyway, that was a great video!
I don't see why symmetry helps with the last one. To me, it only says that if (x,y,z) is a solution, then (y,x,z) is a solution, too, and so on. But why would they need to be the same solutions? This only makes sense for me here because it's a linear system of equations and the associated 3x3 matrix has rank 3, so there is indeed a unique solution, which has to satisfy x=y=z because of the symmetry.
@Adam Romanov Well, your system is symmetric equation-wise, but the entire system is not. If you flip x, y and z around in the first equation, you don't get the second one or something, so it's not the same as in the video. But x^2 = x, y^2 = y is a symmetric system like in the video, right? One solution is (0,1), so they indeed don't have to have the same components. I guess invariance can mean f(X) = X or f(x) = x for all x in X. Oh, and I guess I should have written (y,z,x) instead of (y,x,z) in my first comment.
@Adam Romanov Never mind, your system is symmetric, the permutation is just the identity.
Saying that any solution is invariant under permutations of x,y and z is stronger than what you say. It means that (x,y,z)=(z,x,y)=(y,z,x).
@@hai-mel6815 Yeah, that's the problem.
You could use the rearrangement inequality to show that it must be that x=y=z here. I agree that symmetry alone doesn't suffice.
You missed an opportunity to have a triangle named ODB
would you explain why that would be relevant?
Kacem Abd El Aziz He was a rapper in the 90s 😄
Nice
good shit
I was hoping this would help me solve the following but I am still stuck
If x,y,z are positive reals with x+y+z=1 prove
xy(x+y)^2 + yz(y+z)^2 + zx(z+x)^2 GTE 4xyz
There's a really nice systematic approach to symmetric homogeneous inequalities here: stanleyrabinowitz.com/download/inequalities.pdf. The inequality is already symmetric, and you can make it homogeneous by using the constraint to change the 4xyz term to 4xyz*1 = 4xyz*(x+y+z), so that all terms are degree 4. Then through a straightforward computation, you get (using notation from paper) [3, 1, 0] + [2, 2, 0] >= 2*[2,1,1]. The left hand side majorizes the right hand side, so the original inequality (with the equality constraint) is true.
or a more manual way that takes a bit more cleverness (personally I like when I can get a computer to solve it) can be found here
math.stackexchange.com/questions/1657914/if-x-y-z0-and-xyz-1-then-prove-that-xyxy2yzyz2zxzx2-geq After dividing by xyz, it's a direct application of Titu's Lemma (which follows from the Cauchy-Schwarz inequality).
@@MrDowntownjbrown Thank you very much. I've been stuck on that one for so long!
triangle AOC might have upset about 30% of your viewers
That warranted a backflip.
Hi
When u r early but don't know what to say
@@ahuman6546 yup
@@ahuman6546 i think i saw some roots of unity in the algebraic part. Thats all i can say
Watch pendyala education and entertainment channel
no more fkin ABBA...
Why did i see shri yantra symbol on thumbnail....
The last example is broken. This is just not true, consider the same problem, but in the Z[9] ring, then one solution of the x+2*y+3*z=30, 2*x+3*y+z=30, 3*x+y+2*z=30 system is x=0, y=6, z=3, and for that obviously x=y=z is not true.
in fact, his previous assessment isn't true either; the solutions only work under CYCLIC permutations of (x, y, z) - not just any permutation.
consider this system:
x + y + 2z + 2t = 12
x + 2y + 2z + t = 12
2x + 2y + z + t = 12
2x + y + z + 2t = 12
obviously (1, 3, 1, 3) is a solution, as is (3, 1, 3, 1), however (1, 1, 3, 3) is NOT a solution.
Your solution doesn’t work - try plugging it in and you’ll see that (x,y,z)=(0,6,3) is not a solution
@@SMTsquidge 0+2*6+3*3=21==30 mod 9 for the first equation, you can check the rest 2 also holds.
Robert Gerbicz what does modular arithmetic have to do with this problem?
Quy Khoi Le so technically he probably meant combinations instead of permutations, which explains why your system doesn’t work. If you want to apply symmetry such that x=y=z=t, you also need the equations x+2y+z+2t=12 and 2x+y+2z+t=12, which yields the solution x=y=z=t=2. In your example, x,y,z,t are not in identical environments, so the assumption that they’re all equal is invalid.