The easiest solution not involving brute-forcing (like directly using x=y=z=k) is to use the RMS-AM inequality. For x, y, z: sqrt((x^2 + y^2 + z^2)/3) >= (x + y + z)/3; with equality only if x = y = z. Calculating using given equations, LHS = RHS = 1/sqrt(3), which shows that x = y = z = 1/sqrt(3) is the only solution.
Sum of squares... Titu's lemma here I come! Solution: Using Titu's lemma we have: x^2+y^2+z^2 >= ((x+y+z)^2)/3 = 1 With equality iff x=y=z meaning that x=y=z=sqrt(3)/3 And so we are done!
Here, the plane touches the sphere at a single point as the perpendicular distance of the plane with the centre of the sphere (0,0,0) is exactly equal to the radius of the sphere (1 units). Due to symmetry this point should be in the form of (a,a,a) where a is unknown, which can be obtained as 1 by sqrt(3) by substituting in the equation of plane or sphere.
@@xwtek3505 changing plane equation means the intersection could be a circle. Then the solution will be infinite. Else no solution could also be present. But for plane to intersect sphere only at 1 place, then distance of plane from origin should be 1. If your equation satisfies that condition, then the point of intersection will change, thus changing the solution
You could have mentioned the beautiful geometric interpretation of this result at the end of the video. It will give you more views ;). The equations represent a plane and a sphere respectively and the single real solution indicates that the plane is tangent to the sphere.
I used a different way to find this solution. Expanding (x + y + z)^2, and replacing by what we know from the instructions, we get that xy + xz + yz = 1. Thus, setting xyz = k, we can use Vieta's theorem to get that u^3 - sqrt(3)*u^2 + u - k = 0 has x, y, z as solutions. However, by computing the derivative of f(u) = u^3 - sqrt(3)*u^2 + u - k, we see that it has no extremum. Therefore, we know that the only possibility is to have all three solutions being equal (x = y = z since it is thus stritcly increasing), which means that f(u) = (u - x)^3 u^3 - 3u^2*x + 3ux^2 - x^3 = u^3 - sqrt(3)u^2 + u - k. Two polynomials are equal, if and only if their coefficients are equal, and those 4 equations leads us to get (without any contradiction) that x = sqrt(3)/3 and k = 1/sqrt(27). Since we had that x = y = z, we have that x = y = z = sqrt(3)/3. QED :D
I started like this and got xy + yz + zx = 1. Then using the identity (x - y)^2 + (y - z)^2 + (z - x)^2 = 2(x^2 + y^2 + z^2) - 2(xy + yz + zx), I got (x - y)^2 + (y - z)^2 + (z - x)^2 = 2 . 1 - 2 . 1 = 0, so that (x - y)^2 + (y - z)^2 + (z - x)^2 = 0. From there immediately one gets x=y=z and then easily one gets the solution x=y=z=sqrt(3)/3 from the equation x + y + z =1
I haven't gone through all the 75 or so comments, so I don't know whether anyone has given the solution that I'm about to give. It's given that x+y+z=√3 -- (i) and x²+y²+z²=1 -- (ii) Squaring (i), we get: x²+y²+z²+2xy+2yz+2zx = 3 -- (iii) Multiplying (ii) throughout by 3, we get: 3x²+3y²+3z²=3 -- (iv) Subtracting (iii) from (iv), we get: (3x²+3y²+3z²)-(x²+y²+z²+2xy+2yz+2zx)=3-3=0 i.e, 2x²+2y²+2z²-2xy-2yz-2zx=0 The LHS can be written as (x-y)²+(y-z)²+(z-x)²=0 -- (v) If x, y, and z are all real, each component of LHS must be real. In that case, because each component is also a perfect square, it must be the case that each component is >=0. But the RHS=0, and thus each component of LHS must be 0. Thus (x-y)²=(y-z)²=(z-x)²=0 i.e., (x-y)=(y-z)=(z-x)=0 In other words, x=y=z Using this in (i), we get: 3x=√3 ==> x = (√3)/3=1/√3 Thus x=y=z=1/√3
Very nice problem, and interesting solution . I solved it in a shorter way I think. First, I squared the first equation: (x+y+z)^2=x^2+y^2+z^2+2(xy+xz+yz) 1+2(xy+xz+yz)=3 2(xy+xz+yz)=2 xy+xz+yz=1 Then, I created a new equation and made some math: x^2+y^2+z^2=xy+xz+yz x^2+y^2+z^2-xy-xz-yz=0 2*(x^2)+2*(y^2)+2*(z^2)-2xy-2xz-2yz=0 (x^2-2xy+y^2)+(x^2-2xz+z^2)+(y^2-2yz+z^2)=0 (x-y)^2+(x-z)^2+(y-z)^2=0 Each of the squares must be equal to zero, because that is the only way sum of square would be zero. Thus, we get x=y=z=sqrt(3)/3. The final step is simply a substitution in the first equation.
There might be a simple way. x + y + z = \sqrt{3} is a plane, which pass (0, \sqrt{3}, 0), (\sqrt{3}, 0, 0) and (0, 0, \sqrt{3}) x ^ 2 + y ^ 2 + z ^ 3 = 1 is a sphere, whose origin is in (0, 0, 0) and the radius is 1. Therefore if a plane cut a sphere, there can be three possibilities: 1) a circle 2) a tangent piont 3) they don't touch. In this case it is trivial to guess and verify it is the scenario 2.
I interpreted x+y+z as the dot product of (1,1,1) and (x,y,z). Given that (x,y,z) must be a unit vector and that the length of (1,1,1) is sqrt(3), (x,y,z) must be a (positive) multiple of (1,1,1) for the dot product to be sqrt(3)
At first blush, I see that you're intersecting a plane with a 2-sphere, so there is going to be a circle of solutions. With a little more playing you can also obtain the orientation of the sphere you're dealing with. The equation you obtain after getting rid of z is: (x-\sqrt{3}/2)^2+(y-\sqrt{3}/2)^2-xy=2 write u=x-\sqrt{3}/2\\y-\sqrt{3}/2, and then you have the LHS as: u^T*A*u=2, for some A Then what you do is diagonalise A using standard linear algebra techniques and you're left with your circle of solutions. It takes work, but it's easy work.
An intersection between a sphere and a plane must be: a circle (infinite real solutions), a point or nothing (infinite complex solutions). If you are looking for a unique solution, it must be a point. By simetry, x, y and z must be equal. The rest is trivial
We can use (x+y+z) ^2 = x^2+y^2+z^2+2xy+2yz+2zx, from that we get xy+yz+zx=1. Now as we know that x^2+y^2+z^2-(xy+yz+zx) = 1/2[(x-y) ^2+(y-z) ^2+(z-x) ^2]. As LHS is zero, this means RHS must be zero and that can happen only when x=y=z. Hence x+y+z=√3 will become 3x=√3 hence, x=1/√3. Hence x=y=z=1/√3 as x=y=z.
From the second equation, (x,y,z) is a point on the unit sphere centered at the origin in 3D. And notice (1/sqrt(3), 1/sqrt(3), 1/sqrt(3)) is also on this unit sphere. Now, the first equation can be interpreted as the cosine of the angle between (x,y,z) and (1/sqrt(3), 1/sqrt(3), 1/sqrt(3)) is equal to 1. But cosine value equals 1 only when the angle is 0 modulo 2pi. Therefore, (x,y,z) can only be (1/sqrt(3), 1/sqrt(3), 1/sqrt(3)).
Hey, anyone have thoughts on this? Assign any real positive values to a and b. We find a' and b' using the iterative function (x'/y') = (x+yn)/(x+y). Upon iteration, a/b tends to sqrt(n). Ex. say x = 7, y = 4, n = 3. One iteration gets us 19/11 which is closer to sqrt(3). Another gets us 26/15, and another 71/41. Applying this with n = 5 yields an interesting comparison to the Fibonacci series, of course.
If x² = y² = z² = ⅓, then they add to 1. Then x + y + z = 3*1/sqrt(3) = sqrt(3), so x = y = z = 1/sqrt(3) is one solution. Negative numbers won't work. Now maximize x + y + z with the constraint that x² + y² + z² - 1 = 0, using Lagrange multipliers. L = x + y + z - λ(x² + y² + z² - 1). 0 = ∂L/∂x = 1 - 2λx, and identical for y and z. That means that x = y = z maximizes their sum, and anything else gives a lower sum. So z = y = z = 1/sqrt(3) is the only solution.
Solving through, and substituting x=a+bi Then answers are permutations of (a+bi,0.5(sqrt(3)-a-sqrt(3)b-i(1-sqrt(3)a+b)),0.5(sqrt(3)-a+sqrt(3)b+i(1-sqrt(3)a-b))) Interesting values: (sqrt(3),i,-i) (2sqrt(3)+i,-sqrt(3)+2i,-3i)
second equation shows the solution is a sphere centered at (0,0,0). but you do not even need to use that, since, by symmetry, first equation shows that x=y=z=sqrt(3)/3
Nice problem Syber 🤩 though the solution was a bit tedious 😅. I hope you won't mind me saying like that 😁. I solved it in a geometrical/graphical approach. In a 3D space, x+y+z=√3 represents a plane passing through (0,0,√3),(0,√3,0),(√3,0,0) and x²+y²+z²=1 represents a sphere centred at origin having a radius of 1 unit. The intersection points of two curves represent the common solution to the equations corresponding to those curves respectively. I imagined that sphere and the plane and found it to be symmetric about the line x=y=z. When a sphere and a plane intersects each other it either gives a circle or a point (i.e. they just touch each other at that point) and all the solution(s) should lie on this circle (or a point). Since, the whole thing here is symmetric about the line x=y=z the centre of the circle should be on the line x=y=z and obviously on the plane x+y+z=√3. So, the centre is (1/√3,1/√3,1/√3). Now, this point also lies on the sphere. So this point is the only solution. Therefore, x=y=z=1/√3 .
It is solvable geometrically. Polar distance of plane is 1, therefore this plane is tangent for sphere, then there is unique solution, where x=y=z because of symmetry
Hi there ! thanks for this system. We can use a tricky way. We can easily notice that 1/sqrt(3) is a solution of this system. So we can then calculate (x-1/sqrt(3))² + (y-1/sqrt(3))² + (z-1/sqrt(3))². Using the 2 equations of the system we can easily find that this sum is equal to 0 ... as each member of the sum is a positive number then all the squares are equal to 0 ...
Actually when it's proven that there's only one solution, then automatically x = y = z because of symmetry. And there are various ways to do so - by geometry (distance between the sphere's centre and the plane) or by algebra (AM-GM).
@@phamnguyenductin , when I solved as the distance form the plane to the sphere's center is 1. We have only one solution. But before say that x=y=z by simetry we have to prove that we have only one solution.
I tried using Vieta’s formula to solve this. I got the sum and xy+xz+yz, but I couldn’t find a way to get xyz (except for noticing that 3xyz = sum of cubes). This was a cool solution!
i got a better solution: (x+y+z)²=3 then after some replacements and simplifications, we get xy+xz+yz = 1, then x²+y²+z²=xy+xz+yz, then we multiply both sides by 2, and we have 2x²+2y²+2z²=2xy+2xz+2yz, then we substract the RHS from both sides, and we get 2x²+2y²+2z²-2xy-2xz-2yz=0, then x²+y²-2xy+x²+z²-2xz+y²+z²-2yz=0, which means (x-y)²+(x-z)²+(y-z)²=0, and we know that the square roots are always positive because x, y and z are real number, then since the sum of the square roots is zero, then x-y = 0 and x-z = 0 and y-z = 0, which give us x=y=z, and by replacing in the original equation, we get x=y=z=sqrt(3)/2
Good morning! I really liked so much this problem. x^2+y^2+z^2= R^2 and Ax+By+Cx=D. If d= |D|/SQR(A^2+B^2+C^2) < R we have a circumference as solution and if d = R we have just a a point. Otherwise we have no real solution. In the case of a circumference, the center O=(xo,yo,zo)= D /SQR(A^2+^B^2+C^2)(A,B,C) and the ratio r= SQR(R^2-d^2). Then we can change for Wt=(w1,w2,w3)=(x - xo, y- yo, z-zo) and for W we have the center of the circumference at (0,0,0) Taking (A,B,0) and rotating the axis X to align with vector (A,B,0), we have to multiply W by M, and rotating axis Z to align with (A,B,C) multipying by N. | cos(u) sen(u) 0 | | cos (v) 0 -sen(v) | M= | -sen(u) cos (u) 0 | N= | 0 1 0 | | 0 0 1 | | sen(v) 0 cos(v) | cos(u) = A/SQR(A^2+B^2) and Xt=(x-xo, y-yo,z-zo)cos(v) = C/SQR (A^2+B^2+C^2) and P= M*N so W= P*X. Is so easy to show that M and N are orthogonal, so P is also orthogonal and than P^-1= Pt as the axis W3 is orthogonal to the plane and we forced the plane pass to (0,0,0) for W. We have that w3=0 and w1^2+w2^2=r^2. So X= (x,y,z,)t = (xo,yo,zo) + Pt* (w1, w2, 0)t where w1^2+w2^2 = r^2 and we can find w1 and w2 and then return to (x,y,z). if d= R we have only Xt= D /SQR(A^2+^B^2+C^2)(A,B,C). E.g., x^2+y^2+z^2 = 25 and x+y+z= 4SQR(3) O=4sqr(3)/3 (1,1,1) and r= 3; we have | SQR(6)/6 -SQR(2)/2 SQR(3)/3 | | w1 | X= 4sqr(3)/3 (1,1,1)t + | SQR(6)/6 SQR(2)/2 SQR(3)/3 | * | w2 | | -SQR(6)/3 SQR(2)/2 SQR(3)/3 | | 0 | Where w1^2 + w2^2 = 9 E.g. , for w1=3 and w2=0; x= 4 SQR(3)/3 + SQR(6)/2 ; y= 4 SQR(3)/3 + SQR(6)/2; z= 4 SQR(3)/3 - SQR(6) w1=2 and w2=SQR(5) ; x= SQR(6)/3 - SQR(10)/3 + 4SQR(3)/3 ; y = SQR(6)/3 + SQR(10)/3 + 4SQR(3)/3; z= -2SQR(6)/3 + 4SQR(3)/3. Believe it, I tested and it works.
RMS(root mean square) of x,y,z = sqrt[(x²+y²+z²)/3] = sqrt(1/3) = 1/√3 AM of x,y,z = (x+y+z)/3 = √3/3 = 1/√3 So, RMS=AM But, RMS>=AM with equality occurring if all elements are equal => x=y=z But, x+y+z=√3 => x=y=z= *√3/3*
Best regards. From Brazil. I solved by Linear Álgebra. We have a surface of a ball with center (0,0,0) and radius 1. And a plane orthogonal with the vector (1,1,1) that have the point (0;0;3^0;5). So we have a circumference for solution if the distance d from the center of the ball were less than the radius, one point if d=R and no solution if d>R. For this case d=R=1. As if(a,b,c) was a solution, any permutation is also a solution. And as we have only one solution a=b=c=3^(1/2)/3. My batery is finishing, and I am not home. Tomorrow, I guess, I will post a solution for x^2+y^2+z^2= R^2 and Ax+By+Cz=D and |D|/(A^2+B^2+C^2)^0;5
There is a faster solution. x+y+z=sqrt(3), then x^2+y^2+z^2+2(xy+yz+xz)=3. Therefore, xy+yz+xz=1=x^2+y^2+z^2. As a result, 2(x^2+y^2+z^2-xy-yz-xz)=0. Then (x-y)^2+(y-z)^2+(z-x)^2=0, so x=y=z.
x+y+z=√3 x^2+y^2+z^2=1 At first glance We understand x=y=z=√3/3 But we have to prove that x,y,z are real numbers.we take z=>y=>x, z=x+b, y=x+a above equation z must have positive by order numbers x,y,z We have a, b are non negative real numbers. We get x^2+y^2+z^2=xy+yz+zx x^2-xy+y^2-yz+z^2-zx=0 x(x-y)+y(y-z)+z(z-x)=0 x(-a)+(x+a)(a-b)+(x+b)b=0 -ax+ax-bx+a^2-ab+bx+b^2=0 a^2-ab+b^2=0 (a-b)^2=-ab As a and b are real numbers, square of a-b is nonnegative,-ab also nonnegative.Then ab is less than or equal to zero , buy here a,b both are non negative real numbers,only option ab=0. (a-b)^2=0 a-b=0, a=b=0 x=y=z=√3/3
I actually have a faster solution. I find xy+yz+xz thru (x+y+z)² - x²+y²+z², which is 1. therefore x²+y²+z²-xy-xz-yz= (x-y)²+(y-z)²+(x-z)²/2=1-1=0. So we have x=y=z. Substituting these back into the equation we have x=y=z= rt(3)/ 3. Help me kindly check if this is correct. Thank you!
I did it a different way First I squared first equ and expanded and got xy+yz+xz=1 We see that x^2+y^2+z^2=xy+yz+xz=1. 2x^2+2y^2+2z^2-2xy-2yz-2xz=0 (x-y)^2+(y-z)^2+(z-x)^2=0 So x=y=z Sub it in first equ we get x=y=z=√3/3
@@SyberMath Yup, Cauchy-Schwartz offers a quick solution to this (though this was a very fun video, especially the negative inside a radical at the end)
Let's make it simpler: We have the following system: x + y + z = √3 (1) x^2 + y^2 + z^2 = 1 (2) ⟹ (x+y+z)^2 - (x^2+y^2+z^2 ) = 2(xy+yz+zx) = 2 ⟹ xy + yz + zx = 1 (3) ⟹ (x^2+y^2+z^2 ) - (xy+yz+zx) = [(x-y)^2+(y-z)^2+(z-x)^2] / 2 = 0 which implies : (x-y)^2 + (y-z)^2 + (z-x)^2 = 0 The last equation is satisfied (for real x, y, z) if and only if: x = y = z = √3/3 = 1/√3 which is the single real solution to the given system of equations.
easier way: (x + y + z)^2 = x^2 + y^2 + z^2 + 2(xy + yz + xz) = 3 => xy + yz + xz = 1 but x^2 + y^2 + z^2 always > or = xy + yz + xz, in this case is equal => x = y = z = 1/sqrt(3)
I wonder if it can be proven that for a+b+c+... = sqrt(n) and a²+b²+c²+... = 1, where n is the amount of variables, the only solution is to for all variables be equal to sqrt(n)/n
Yes it can, using the Cauchy-Schwartz inequality, which says that the dot product of two vectors must be less than or equal to the product of their magnitudes. Equality hold if and only if one of the two vectors is a scalar multiple of the other, ie their corresponding terms are in the same ratio. Consider the vectors (a,b,c,...) and (1,1,1,...) of equal dimensions, and let a^2 + b^2 + c^2 +... = 1. The dot product of the two is a+b+c+..., and their magnitudes are, respectively, 1 and sqrt(n) (where n is the number of dimensions/quantities). Thus by C-S, a+b+c+...
Much easier solution: What if they're all the same value? Then they will be root 3 over 3, from the first equation. Try it in the second equation. Oh! It works. Done.
I understood the the discriminant setting equal to zero approach, but you always had the option to take minus sign common from the parentheses in the radical to cancel out the minus sign outside, leaving you with flipped terms in parentheses which could be further evaluated after the square and radical cancelled, so what is wrong with this approach !
@@SyberMath hey, you are quite expert maths, you might have heard of Photomath, right ?, Could you make videos on how to clear their test for teaching, will be highly appreciated, personally I cleared the knowledge part but failed the teaching skill section, don't know where did I go wrong, would really love any kind of advice or help, I am also enthusiastic when it comes to math......
@@anshumanagrawal346 That is the dumbest way to show that. Its a sphere and a plane. Show they're tangential at that point. There are so many other ways to show thats the only solultion
@@kouverbingham5997 The solution he shows is meant to be simple, using only some basic algebra, as most importantly, nothing "smart". His goal to is to present a solution everyone can understand
I think you could still use this method. To make things easier, I made the substitution x = u/sqrt(3), y = v/sqrt(3), z = w/sqrt(3). to get u+v+w = 3 and u^2+v^2+w^2 = 3. I get the polynomial P(x) = x^3 - 3x^2 + 3x + a_0. If you look at the graph and try various values of a_0 (or find where the derivative is 0), you can see that you get 3 real roots only when a_0 =-1, so P(x) = (x-1)^3, so u=v=w=1, meaning x=y=z=1/sqrt(3). If you don't restrict yourself to reals, you can get other answers by choosing different a_0. For example, if a_0 = 0, then { 0, (sqrt(3) +- i) / 2} satisfies the original equations.
@@SyberMath I squared the first equation and found out while expanding the whole thing that xy+yz+xz=1, which I then set equal to the second equation. I was able to conclude y=x, z=y, and x=z. Thus, by the transitive property x=y=z. So then I plugged in one of the letters to have it equal to sqrt(3)/3, making the only ordered triple solution (sqrt(3)/3, sqrt(3)/3, sqrt(3)/3).
Nice logic of solution for 3 variables by just two equations. Straight away if we replace x=y=z=k also we get values MURTHY RETIRED MATHS TEACHER SANGAREDDY 2O21O624TH
The easiest solution not involving brute-forcing (like directly using x=y=z=k) is to use the RMS-AM inequality. For x, y, z: sqrt((x^2 + y^2 + z^2)/3) >= (x + y + z)/3; with equality only if x = y = z. Calculating using given equations, LHS = RHS = 1/sqrt(3), which shows that x = y = z = 1/sqrt(3) is the only solution.
Exactly, when seeing a sum of squares Titu's lemma usually saves the day!
Pls see the solution given by Rajan Wadke
Sum of squares...
Titu's lemma here I come!
Solution:
Using Titu's lemma we have:
x^2+y^2+z^2 >= ((x+y+z)^2)/3 = 1
With equality iff x=y=z
meaning that x=y=z=sqrt(3)/3
And so we are done!
That’s way i like it 🎹🎹 😎
Really nice 👍
Here, the plane touches the sphere at a single point as the perpendicular distance of the plane with the centre of the sphere (0,0,0) is exactly equal to the radius of the sphere (1 units). Due to symmetry this point should be in the form of (a,a,a) where a is unknown, which can be obtained as 1 by sqrt(3) by substituting in the equation of plane or sphere.
great, but sahab ham logo ko thoda calculation wala method sain leke giya
Underrated comment. Saves 9 minutes of youtube🥰🥰
I had a similar thought process.
Using this logic, If I changed the plane equation to anything other than sqrt(3), won't it produce an invalid answer of (1, 1, 1) too?
@@xwtek3505 changing plane equation means the intersection could be a circle. Then the solution will be infinite. Else no solution could also be present. But for plane to intersect sphere only at 1 place, then distance of plane from origin should be 1. If your equation satisfies that condition, then the point of intersection will change, thus changing the solution
You could have mentioned the beautiful geometric interpretation of this result at the end of the video. It will give you more views ;). The equations represent a plane and a sphere respectively and the single real solution indicates that the plane is tangent to the sphere.
The spheare I understand but the plane?
@@gavasiarobinssson5108 x^2+y^2+z^2=1 defines a sphere with radius 1. x+y+z=root3 defines a plane angled in xyz space to cross each axis at +root3.
Nice!
I used a different way to find this solution. Expanding (x + y + z)^2, and replacing by what we know from the instructions, we get that xy + xz + yz = 1. Thus, setting xyz = k, we can use Vieta's theorem to get that u^3 - sqrt(3)*u^2 + u - k = 0 has x, y, z as solutions. However, by computing the derivative of f(u) = u^3 - sqrt(3)*u^2 + u - k, we see that it has no extremum. Therefore, we know that the only possibility is to have all three solutions being equal (x = y = z since it is thus stritcly increasing), which means that f(u) = (u - x)^3 u^3 - 3u^2*x + 3ux^2 - x^3 = u^3 - sqrt(3)u^2 + u - k. Two polynomials are equal, if and only if their coefficients are equal, and those 4 equations leads us to get (without any contradiction) that x = sqrt(3)/3 and k = 1/sqrt(27). Since we had that x = y = z, we have that x = y = z = sqrt(3)/3. QED :D
I started like this and got xy + yz + zx = 1. Then using the identity (x - y)^2 + (y - z)^2 + (z - x)^2 = 2(x^2 + y^2 + z^2) - 2(xy + yz + zx), I got (x - y)^2 + (y - z)^2 + (z - x)^2 = 2 . 1 - 2 . 1 = 0, so that (x - y)^2 + (y - z)^2 + (z - x)^2 = 0. From there immediately one gets x=y=z and then easily one gets the solution x=y=z=sqrt(3)/3 from the equation x + y + z =1
I did the almost the same.
I haven't gone through all the 75 or so comments, so I don't know whether anyone has given the solution that I'm about to give.
It's given that x+y+z=√3 -- (i)
and x²+y²+z²=1 -- (ii)
Squaring (i), we get: x²+y²+z²+2xy+2yz+2zx = 3 -- (iii)
Multiplying (ii) throughout by 3, we get: 3x²+3y²+3z²=3 -- (iv)
Subtracting (iii) from (iv), we get:
(3x²+3y²+3z²)-(x²+y²+z²+2xy+2yz+2zx)=3-3=0
i.e, 2x²+2y²+2z²-2xy-2yz-2zx=0
The LHS can be written as (x-y)²+(y-z)²+(z-x)²=0 -- (v)
If x, y, and z are all real, each component of LHS must be real. In that case, because each component is also a perfect square, it must be the case that each component is >=0. But the RHS=0, and thus each component of LHS must be 0.
Thus (x-y)²=(y-z)²=(z-x)²=0
i.e., (x-y)=(y-z)=(z-x)=0
In other words, x=y=z
Using this in (i), we get: 3x=√3 ==> x = (√3)/3=1/√3
Thus x=y=z=1/√3
Very good! 🤩
@@SyberMath Thank you.
I like this solution. It's simple, yet elegant!!
@@mark75343416 Thank you.
I'm stunned to see this solution 🤩🤩
Ooh I love these types of questions
Very nice problem, and interesting solution .
I solved it in a shorter way I think.
First, I squared the first equation:
(x+y+z)^2=x^2+y^2+z^2+2(xy+xz+yz)
1+2(xy+xz+yz)=3
2(xy+xz+yz)=2
xy+xz+yz=1
Then, I created a new equation and made some math:
x^2+y^2+z^2=xy+xz+yz
x^2+y^2+z^2-xy-xz-yz=0
2*(x^2)+2*(y^2)+2*(z^2)-2xy-2xz-2yz=0
(x^2-2xy+y^2)+(x^2-2xz+z^2)+(y^2-2yz+z^2)=0
(x-y)^2+(x-z)^2+(y-z)^2=0
Each of the squares must be equal to zero, because that is the only way sum of square would be zero. Thus, we get x=y=z=sqrt(3)/3. The final step is simply a substitution in the first equation.
Negative square inside square root hence it must be zero. That was nice.🔥
Thank you! 😊
There might be a simple way.
x + y + z = \sqrt{3} is a plane, which pass (0, \sqrt{3}, 0), (\sqrt{3}, 0, 0) and (0, 0, \sqrt{3})
x ^ 2 + y ^ 2 + z ^ 3 = 1 is a sphere, whose origin is in (0, 0, 0) and the radius is 1.
Therefore if a plane cut a sphere, there can be three possibilities:
1) a circle
2) a tangent piont
3) they don't touch.
In this case it is trivial to guess and verify it is the scenario 2.
when you find xy+yz+zx = 1 = x^2+y^2+z^2
you can confirm x=y=z
I interpreted x+y+z as the dot product of (1,1,1) and (x,y,z). Given that (x,y,z) must be a unit vector and that the length of (1,1,1) is sqrt(3), (x,y,z) must be a (positive) multiple of (1,1,1) for the dot product to be sqrt(3)
At first blush, I see that you're intersecting a plane with a 2-sphere, so there is going to be a circle of solutions. With a little more playing you can also obtain the orientation of the sphere you're dealing with. The equation you obtain after getting rid of z is:
(x-\sqrt{3}/2)^2+(y-\sqrt{3}/2)^2-xy=2
write u=x-\sqrt{3}/2\\y-\sqrt{3}/2, and then you have the LHS as:
u^T*A*u=2, for some A
Then what you do is diagonalise A using standard linear algebra techniques and you're left with your circle of solutions.
It takes work, but it's easy work.
Wow! Good thinking!
An intersection between a sphere and a plane must be: a circle (infinite real solutions), a point or nothing (infinite complex solutions). If you are looking for a unique solution, it must be a point. By simetry, x, y and z must be equal. The rest is trivial
It also means that the sphere touches the plane, which I think is quite nice
Also, there could be infinite solutions, so it could be a circle... the question doesn't specifically ask for one solution.
We can use (x+y+z) ^2 = x^2+y^2+z^2+2xy+2yz+2zx, from that we get xy+yz+zx=1.
Now as we know that x^2+y^2+z^2-(xy+yz+zx) = 1/2[(x-y) ^2+(y-z) ^2+(z-x) ^2].
As LHS is zero, this means RHS must be zero and that can happen only when x=y=z.
Hence x+y+z=√3 will become 3x=√3 hence, x=1/√3.
Hence x=y=z=1/√3 as x=y=z.
From the second equation, (x,y,z) is a point on the unit sphere centered at the origin in 3D. And notice (1/sqrt(3), 1/sqrt(3), 1/sqrt(3)) is also on this unit sphere. Now, the first equation can be interpreted as the cosine of the angle between (x,y,z) and (1/sqrt(3), 1/sqrt(3), 1/sqrt(3)) is equal to 1. But cosine value equals 1 only when the angle is 0 modulo 2pi. Therefore, (x,y,z) can only be (1/sqrt(3), 1/sqrt(3), 1/sqrt(3)).
Nice!
very nice!
i will keep supporting so much1
congrats on 21 k subs!
Thank you so much! 💖
Hey, anyone have thoughts on this? Assign any real positive values to a and b. We find a' and b' using the iterative function (x'/y') = (x+yn)/(x+y). Upon iteration, a/b tends to sqrt(n). Ex. say x = 7, y = 4, n = 3. One iteration gets us 19/11 which is closer to sqrt(3). Another gets us 26/15, and another 71/41.
Applying this with n = 5 yields an interesting comparison to the Fibonacci series, of course.
If x² = y² = z² = ⅓, then they add to 1. Then x + y + z = 3*1/sqrt(3) = sqrt(3), so x = y = z = 1/sqrt(3) is one solution. Negative numbers won't work.
Now maximize x + y + z with the constraint that x² + y² + z² - 1 = 0, using Lagrange multipliers.
L = x + y + z - λ(x² + y² + z² - 1). 0 = ∂L/∂x = 1 - 2λx, and identical for y and z. That means that x = y = z maximizes their sum, and anything else gives a lower sum. So z = y = z = 1/sqrt(3) is the only solution.
Solving through, and substituting x=a+bi
Then answers are permutations of
(a+bi,0.5(sqrt(3)-a-sqrt(3)b-i(1-sqrt(3)a+b)),0.5(sqrt(3)-a+sqrt(3)b+i(1-sqrt(3)a-b)))
Interesting values:
(sqrt(3),i,-i)
(2sqrt(3)+i,-sqrt(3)+2i,-3i)
really nice I was about to tell the discriminate should be equal to zero. 👍👌
second equation shows the solution is a sphere centered at (0,0,0). but you do not even need to use that, since, by symmetry, first equation shows that x=y=z=sqrt(3)/3
You need to prove that the equation has only one solution (x = y = z = 1/√3). It's not very difficult but must be done.
That's right!
Nice problem Syber 🤩 though the solution was a bit tedious 😅. I hope you won't mind me saying like that 😁.
I solved it in a geometrical/graphical approach.
In a 3D space, x+y+z=√3 represents a plane passing through (0,0,√3),(0,√3,0),(√3,0,0) and x²+y²+z²=1 represents a sphere centred at origin having a radius of 1 unit.
The intersection points of two curves represent the common solution to the equations corresponding to those curves respectively.
I imagined that sphere and the plane and found it to be symmetric about the line x=y=z.
When a sphere and a plane intersects each other it either gives a circle or a point (i.e. they just touch each other at that point) and all the solution(s) should lie on this circle (or a point).
Since, the whole thing here is symmetric about the line x=y=z the centre of the circle should be on the line x=y=z and obviously on the plane x+y+z=√3. So, the centre is (1/√3,1/√3,1/√3). Now, this point also lies on the sphere. So this point is the only solution.
Therefore, x=y=z=1/√3 .
X + Y + Z = 3^(1/2) symmetry applies so test from x == 3^(-1/2) ...
It is solvable geometrically. Polar distance of plane is 1, therefore this plane is tangent for sphere, then there is unique solution, where x=y=z because of symmetry
By Cauchy-Schwarz inequality,
x*1 + y*1 + z*1
very well done bro, thanks for sharing
My pleasure
Hi there ! thanks for this system. We can use a tricky way. We can easily notice that 1/sqrt(3) is a solution of this system. So we can then calculate (x-1/sqrt(3))² + (y-1/sqrt(3))² + (z-1/sqrt(3))². Using the 2 equations of the system we can easily find that this sum is equal to 0 ... as each member of the sum is a positive number then all the squares are equal to 0 ...
Thanks for the info!
I just looked at the equations and thought "the obvious solution is 1/√3". Chamchamal, if the Earth were a sphere.
Good evening! I apologize but to talk about that x=y=z, by symmetry, we have to show first that we have only one solution.
Actually when it's proven that there's only one solution, then automatically x = y = z because of symmetry. And there are various ways to do so - by geometry (distance between the sphere's centre and the plane) or by algebra (AM-GM).
No need to apologize
@@phamnguyenductin , when I solved as the distance form the plane to the sphere's center is 1. We have only one solution. But before say that x=y=z by simetry we have to prove that we have only one solution.
Squaring first equation , we get ,x^2+y^2+z^2+2xy+2yz+2zx=3.Now x^2+y^2+z^2=xy+yz+zx.It means x=y=z .From first equation x=y=z=1/sq.rt.3
Nice!
I think,there Is symetrie. x=y=z., we have 2eq And 3 var. (x+y+z)^2=x^2+y^2+z^2+2(xy+xz+yz). x==y=z=sqrt3/3.
GOOD!! I also same answer ^^
I appreciate your method of solving the two equations .God bless you .
Thank you! 💖
Thank you for this video.
My pleasure!
I tried using Vieta’s formula to solve this. I got the sum and xy+xz+yz, but I couldn’t find a way to get xyz (except for noticing that 3xyz = sum of cubes). This was a cool solution!
i got a better solution: (x+y+z)²=3 then after some replacements and simplifications, we get xy+xz+yz = 1, then x²+y²+z²=xy+xz+yz, then we multiply both sides by 2, and we have 2x²+2y²+2z²=2xy+2xz+2yz, then we substract the RHS from both sides, and we get 2x²+2y²+2z²-2xy-2xz-2yz=0, then x²+y²-2xy+x²+z²-2xz+y²+z²-2yz=0, which means (x-y)²+(x-z)²+(y-z)²=0, and we know that the square roots are always positive because x, y and z are real number, then since the sum of the square roots is zero, then x-y = 0 and x-z = 0 and y-z = 0, which give us x=y=z, and by replacing in the original equation, we get x=y=z=sqrt(3)/2
Good morning! I really liked so much this problem. x^2+y^2+z^2= R^2 and Ax+By+Cx=D. If d= |D|/SQR(A^2+B^2+C^2) < R we have a circumference as solution and if d = R we have just a a point. Otherwise we have no real solution.
In the case of a circumference, the center O=(xo,yo,zo)= D /SQR(A^2+^B^2+C^2)(A,B,C) and the ratio r= SQR(R^2-d^2).
Then we can change for Wt=(w1,w2,w3)=(x - xo, y- yo, z-zo) and for W we have the center of the circumference at (0,0,0)
Taking (A,B,0) and rotating the axis X to align with vector (A,B,0), we have to multiply W by M, and rotating axis Z to align with (A,B,C) multipying by N.
| cos(u) sen(u) 0 | | cos (v) 0 -sen(v) |
M= | -sen(u) cos (u) 0 | N= | 0 1 0 |
| 0 0 1 | | sen(v) 0 cos(v) |
cos(u) = A/SQR(A^2+B^2) and Xt=(x-xo, y-yo,z-zo)cos(v) = C/SQR (A^2+B^2+C^2) and P= M*N
so W= P*X. Is so easy to show that M and N are orthogonal, so P is also orthogonal and than P^-1= Pt
as the axis W3 is orthogonal to the plane and we forced the plane pass to (0,0,0) for W. We have that w3=0 and w1^2+w2^2=r^2.
So X= (x,y,z,)t = (xo,yo,zo) + Pt* (w1, w2, 0)t where w1^2+w2^2 = r^2 and we can find w1 and w2 and then return to (x,y,z).
if d= R we have only Xt= D /SQR(A^2+^B^2+C^2)(A,B,C).
E.g., x^2+y^2+z^2 = 25 and x+y+z= 4SQR(3)
O=4sqr(3)/3 (1,1,1) and r= 3; we have
| SQR(6)/6 -SQR(2)/2 SQR(3)/3 | | w1 |
X= 4sqr(3)/3 (1,1,1)t + | SQR(6)/6 SQR(2)/2 SQR(3)/3 | * | w2 |
| -SQR(6)/3 SQR(2)/2 SQR(3)/3 | | 0 |
Where w1^2 + w2^2 = 9
E.g. , for w1=3 and w2=0; x= 4 SQR(3)/3 + SQR(6)/2 ; y= 4 SQR(3)/3 + SQR(6)/2; z= 4 SQR(3)/3 - SQR(6)
w1=2 and w2=SQR(5) ; x= SQR(6)/3 - SQR(10)/3 + 4SQR(3)/3 ; y = SQR(6)/3 + SQR(10)/3 + 4SQR(3)/3; z= -2SQR(6)/3 + 4SQR(3)/3.
Believe it, I tested and it works.
Wow! Nice work!
RMS(root mean square) of x,y,z = sqrt[(x²+y²+z²)/3]
= sqrt(1/3) = 1/√3
AM of x,y,z = (x+y+z)/3
= √3/3 = 1/√3
So, RMS=AM
But, RMS>=AM with equality occurring if all elements are equal
=> x=y=z
But, x+y+z=√3
=> x=y=z= *√3/3*
Best regards. From Brazil. I solved by Linear Álgebra. We have a surface of a ball with center (0,0,0) and radius 1. And a plane orthogonal with the vector (1,1,1) that have the point (0;0;3^0;5).
So we have a circumference for solution if the distance d from the center of the ball were less than the radius, one point if d=R and no solution if d>R. For this case d=R=1. As if(a,b,c) was a solution, any permutation is also a solution. And as we have only one solution a=b=c=3^(1/2)/3. My batery is finishing, and I am not home. Tomorrow, I guess, I will post a solution for x^2+y^2+z^2= R^2 and Ax+By+Cz=D and |D|/(A^2+B^2+C^2)^0;5
Thank you, Pedro! 😊
Nice system !!! Good job
Thanks!
There is a faster solution. x+y+z=sqrt(3), then x^2+y^2+z^2+2(xy+yz+xz)=3. Therefore, xy+yz+xz=1=x^2+y^2+z^2. As a result, 2(x^2+y^2+z^2-xy-yz-xz)=0. Then (x-y)^2+(y-z)^2+(z-x)^2=0, so x=y=z.
Since the plane’s distance to O(0,0,0) is equal to the radius of the sphere, it will be much easier to solve the problem with geometric methods.
At first glance I thought it would be related to vietas formula for cubic polynomial
x+y+z=√3
x^2+y^2+z^2=1
At first glance We understand
x=y=z=√3/3 But we have to prove that x,y,z are real numbers.we take
z=>y=>x, z=x+b, y=x+a
above equation z must have positive
by order numbers x,y,z We have a, b are non negative real numbers.
We get x^2+y^2+z^2=xy+yz+zx
x^2-xy+y^2-yz+z^2-zx=0
x(x-y)+y(y-z)+z(z-x)=0
x(-a)+(x+a)(a-b)+(x+b)b=0
-ax+ax-bx+a^2-ab+bx+b^2=0
a^2-ab+b^2=0
(a-b)^2=-ab
As a and b are real numbers, square of a-b is nonnegative,-ab also nonnegative.Then ab is less than or equal to zero , buy here a,b both are
non negative real numbers,only
option ab=0.
(a-b)^2=0
a-b=0, a=b=0
x=y=z=√3/3
I actually have a faster solution. I find xy+yz+xz thru (x+y+z)² - x²+y²+z², which is 1. therefore x²+y²+z²-xy-xz-yz= (x-y)²+(y-z)²+(x-z)²/2=1-1=0. So we have x=y=z. Substituting these back into the equation we have x=y=z= rt(3)/ 3.
Help me kindly check if this is correct. Thank you!
Equations #1^2-#2 -> xy+yz+zx=1 -> (x-y)^2+(y-x)^2+(z-x)^2=0 -> x=y=z.
Nice!
With algebraic symetry in the system: x=y=z. And the rest is very easy.
So you can solve any system like that?
What if you have
x+y+z=3
x^2+y^2+z^2=1?
i was able to guess the solution yahoo! 😂🙌
*Solve[{x + y + z == Sqrt[3], x^2 + y^2 + z^2 == 1}, {x, y, z}, Reals] // Simplify*
Hehe! Good job!
I did it a different way
First I squared first equ and expanded and got xy+yz+xz=1 We see that x^2+y^2+z^2=xy+yz+xz=1. 2x^2+2y^2+2z^2-2xy-2yz-2xz=0 (x-y)^2+(y-z)^2+(z-x)^2=0 So x=y=z Sub it in first equ we get x=y=z=√3/3
Nice!
At the first sight without any calculations is obvious that X=Y=Z, x,y,z=√3/3 = 1/√3
Inner Product ?
Vectors! 🤩
@@SyberMath Yup, Cauchy-Schwartz offers a quick solution to this (though this was a very fun video, especially the negative inside a radical at the end)
Awesome 😎😎😎😎
Very underrated
I like you so much I will keep supporting you 😊😌😊😊
Thank you! 💖
I answered this immediately before opening this video
Wow! 😁
Pearhaps more easy to consider we have an obvious solution and it is the unique ?
I tried this problem by squearing équation 1 but it did not work porperly....
Let's make it simpler:
We have the following system:
x + y + z = √3 (1)
x^2 + y^2 + z^2 = 1 (2)
⟹ (x+y+z)^2 - (x^2+y^2+z^2 ) = 2(xy+yz+zx) = 2
⟹ xy + yz + zx = 1 (3)
⟹ (x^2+y^2+z^2 ) - (xy+yz+zx) = [(x-y)^2+(y-z)^2+(z-x)^2] / 2 = 0
which implies :
(x-y)^2 + (y-z)^2 + (z-x)^2 = 0
The last equation is satisfied (for real x, y, z) if and only if:
x = y = z = √3/3 = 1/√3
which is the single real solution to the given system of equations.
easier way:
(x + y + z)^2 = x^2 + y^2 + z^2 + 2(xy + yz + xz) = 3
=> xy + yz + xz = 1
but x^2 + y^2 + z^2 always > or = xy + yz + xz, in this case is equal
=> x = y = z = 1/sqrt(3)
Good solution.
Thanks
I wonder if it can be proven that for a+b+c+... = sqrt(n) and a²+b²+c²+... = 1, where n is the amount of variables, the only solution is to for all variables be equal to sqrt(n)/n
Yes, either by RMS-AM inequality or Cauchy-Schwarz inequality (utilizing their equality conditions)
Yes it can, using the Cauchy-Schwartz inequality, which says that the dot product of two vectors must be less than or equal to the product of their magnitudes. Equality hold if and only if one of the two vectors is a scalar multiple of the other, ie their corresponding terms are in the same ratio.
Consider the vectors (a,b,c,...) and (1,1,1,...) of equal dimensions, and let a^2 + b^2 + c^2 +... = 1. The dot product of the two is a+b+c+..., and their magnitudes are, respectively, 1 and sqrt(n) (where n is the number of dimensions/quantities). Thus by C-S, a+b+c+...
Much easier solution: What if they're all the same value? Then they will be root 3 over 3, from the first equation. Try it in the second equation. Oh! It works. Done.
What if it does not work? 😁
10 sec ... Symetric , so x=y=z ... x= srqt(3)/3
x = y = z = 1/√3 ?
I have thought you will show, that the plane touches the sphere at exactly 1 point in 3d space.
I understood the the discriminant setting equal to zero approach, but you always had the option to take minus sign common from the parentheses in the radical to cancel out the minus sign outside, leaving you with flipped terms in parentheses which could be further evaluated after the square and radical cancelled, so what is wrong with this approach !
You cannot take -1 outside
@@SyberMath yeah, dumb question, I realized that after posting.......
@@SyberMath hey, you are quite expert maths, you might have heard of Photomath, right ?, Could you make videos on how to clear their test for teaching, will be highly appreciated, personally I cleared the knowledge part but failed the teaching skill section, don't know where did I go wrong, would really love any kind of advice or help, I am also enthusiastic when it comes to math......
Even mathway cant solve diophantine equation humans are way more better than computers.....
Haha, that's right! Don't let @leech hear this!
😂
@@SyberMath 😆😆
how did you not just see immediately that x=y=z=1/sqrt(3) is obviously the answer
But you have to prove that it's the only solution, just guessing won't work
@@anshumanagrawal346 That is the dumbest way to show that. Its a sphere and a plane. Show they're tangential at that point. There are so many other ways to show thats the only solultion
@@kouverbingham5997 The solution he shows is meant to be simple, using only some basic algebra, as most importantly, nothing "smart". His goal to is to present a solution everyone can understand
@@anshumanagrawal346 You have a very skewed sense of "simple." In no way is that solution simple.
@@kouverbingham5997 What if someone doesn't know about 3d Co-ordinate geometry
Thanks sir, before I watch this video I have try as I can and I got stuck. But Idk I really like your video sir
Glad to hear that
Good sir.
Many many thanks
Symmetric functions of the roots could work but one equation is missing
I think you could still use this method. To make things easier, I made the substitution x = u/sqrt(3), y = v/sqrt(3), z = w/sqrt(3). to get u+v+w = 3 and u^2+v^2+w^2 = 3. I get the polynomial P(x) = x^3 - 3x^2 + 3x + a_0. If you look at the graph and try various values of a_0 (or find where the derivative is 0), you can see that you get 3 real roots only when a_0 =-1, so P(x) = (x-1)^3, so u=v=w=1, meaning x=y=z=1/sqrt(3). If you don't restrict yourself to reals, you can get other answers by choosing different a_0. For example, if a_0 = 0, then { 0, (sqrt(3) +- i) / 2} satisfies the original equations.
awesome proof
Thank you!
i am strongest emotionally, mentally and physically.
this was a very simple geometry problem, no need for so much algebra...
余は英語はよくわからないが
x=y=z=(√3)/3は直観的に見える。
他の解があるのかないのか？の検討が問題である
Like this one a much
Forget what I say you, I have sea videos on complex nomber... 😉
Awesome.
I am learnig E.language , I hope that once a day chat whit you fluencly without wrong .
Yours sincerely.
Thanks! I hope so too but your English is fine!
Why do I exist
I don't know
Hocam,türkçe videolar da atsanız?Çok iyi anlatıyorsunuz da 😀
😁
Desde el principio se veía que la respuesta era raíz de tres todo tercios, saludoa
¡Bien!
I decided this for 5 sec, because x=y=z from the begin :)
Why?
We see that it's symmetric system - x, y and z can replace each other
answer : x=y=z= (ROOT 3) / 3 ^^
I got the correct answer, but my solution is much shorter (although there could be flaws in it).
What's your solution?
@@SyberMath I squared the first equation and found out while expanding the whole thing that xy+yz+xz=1, which I then set equal to the second equation. I was able to conclude y=x, z=y, and x=z. Thus, by the transitive property x=y=z. So then I plugged in one of the letters to have it equal to sqrt(3)/3, making the only ordered triple solution (sqrt(3)/3, sqrt(3)/3, sqrt(3)/3).
Where's is this problem useful? Everything we learn should be useful in life. So where can we apply this problem in real life?
magic！
:) 👍
so long
Nice logic of solution for 3 variables by just two equations. Straight away if we replace x=y=z=k also we get values
MURTHY RETIRED MATHS TEACHER SANGAREDDY 2O21O624TH