Thanks everyone for the kind words--it is a dream to make these videos, but every creator has ups and downs and I appreciate the support :) A small but important clarification. At 6:08 I substitute a = b = c and state that produces the only solution (0, 0, 0). But it is easy to verify (R, R, R) is a solution for any constant R. So where did this go wrong?! If a = b = c = 0, then the two vectors in the cross-product are (0, 0, 0), so any vector is orthogonal to them. So the lesson is the case a = b = c = 0 should be treated separately. It ends up having several solutions, according to WolframAlpha: www.wolframalpha.com/input?i=x%5E2-yz%3D0%2Cy%5E2-xz%3D0%2Cz%5E2-xy%3D0 As I explained in the video, I didn't solve this problem. But I did do research on Math StackExchange, Topper, and I also asked the RUclips community page and Twitter. None of those explanations covered this corner case, and I missed it too. But of course someone pointed it out after I published :) Unfortunately that means the discussion did not get into the video. On the flip side, I think it's great these videos bring awareness to something new that was never discussed on the internet before. Thank you!
I have one question: if I treat the system of the equations from a geometric point of view it corresponds to a set of quadratic forms of the type as the one shown between round brackets (ax^2 + by^2 + cz^2 + dxy + eyz + fxz + gx + hy + jz + k = 0 omitting 'i' coefficient for clarity). Therefore, geometrically, we have three surfaces in the 3D space intersecting each other. The conclusion you pointed out is that the aforementioned surfaces always have for each real set of values (a,b,c) 2 shared points in common. For real numbers such that not at the same time a,b,c they're all equal to each other, am I right? Thanks for your reply.
Not only the case a=b=c=0 but the case when (c,a,b) and (b,c,a) are linearly dependent. If we are looking for complex solutions then (a,b,c) would be a multiple of (1,r,r^2) where r is a cubic root of unity. And I’ve been lazy to complete the solution afterwards…
I've come up with a different process Subtracting the given equations we obtain (x-y) (x+y+z) =-1 (y-z) (x+y+z) =-2 (x-z) (x+y+z) =-3 And by dividing these equations we obtain 2x+z=3y and substituting the above equation in place of y in the above 3 equations we get three equations in x, z 4x^2+z^2-5xz =27 3z^2-2x^2-xz=15 3x^2-2xz-z^2=6 now eliminate the xz term -7x^2+7z^2=24 now eliminate the constant term then we obtain 19x^2-8xz-11z^2=0 Divide by z^2 on both sides and substitute t=x/z 19t^2-8t-11=0 t=1 ❌ t=-11/19 x/z=-11/19 substitute the above x, z relation in -7x^2+7z^2=24 solve for x, y, z x=-11/√70, y=-1/√70 z=19/√70
I liked what you did when multiplying the three equations by the variable needed to cancel them all out when you add the equations together. That's some outside-the-box thinking that simplifies the math significantly. There seems to be a much simpler solution using only substitution and basic algebra from there, though. We are given: A) x² - yz = 2 B) y² - xz = 3 C) z² - xy = 5 As you showed, by multiplying A, B, and C by y, z, and x respectively, then adding the resulting equations, you can cancel out the squared terms. D) 0 = 5x + 2y + 3z As you also showed, this also works when you change the order to z, x, y: E) 0 = 3x + 5y + 2z Here's where I break from your solution. Subtract E from D and rearrange: F) z = 3y - 2x Substitute the right side of F for the z in either D or E. It doesn't matter which, either way it simplifies to: G) x = 11y Sub the right side of G for the x in F: H) z = -19y Sub the right side of G for x and the right side of H for z in either A, B, or C. Again, it doesn't matter which, it will simplify to: I) y² = 1 / 70 Or: J) y = ±1 / √70 Combining J with G and H gives you either: x = 11 / √70, y = 1 / √70, z = -19 / √70 OR x = -11 / √70, y = -1 / √70, z = 19 / √70 Either of which satisfy A, B, and C.
In general, a=b=c leads to a degeneracy, since the cross product is 0. It seems to have an inf # of solutions. Indeed, in the solution, this scenario leads to a 0/0 indeterminate form.
Your solution with dot and cross products is BEAUTIFUL, congrats! I want to share with you what I did, with elementary manipulation: Let's number the equations: x²-yz = 2 [1] y²-xz = 3 [2] z²-xy = 5 [3] Operating [1]-[2], [2]-[3] and [3]-[1], and factoring the difference of squares that appear, we get respectively: (x-y)(x+y+z) = -1 (y-z)(x+y+z) = -2 (z-x)(x+y+z) = 3 From these we see that x+y+z≠0 and also x,y,z are all different. Now we can isolate x+y+z from these and we get -1/(x-y) = -2/(y-z) = 3/(z-x) that simplifies to 2x-3y+z=0 (no matter which equality you choose you get the same thing). This means that z = -2x+3y. We substitute this back into [1], [2] and [3]: x²-y(-2x+3y) = 2 y²-x(-2x+3y) = 3 (-2x+3y)²-xy = 5 Expanding them we get x²+2xy-3y² = 2 2x²-3xy+y² = 3 4x²-13xy+9y² = 5 And now comes the trick! Let's divide all these by xy (for a moment suppose that xy≠0) and let r=x/y: x/y+2-3y/x = 2/(xy) ⇒ r+2-3/r = 2/(xy) 2x/y-3+y/x = 3/(xy) ⇒ 2r-3+1/r = 3/(xy) 4x/y-13+9y/x = 5/(xy) ⇒ 4r-13+9/r = 5(xy) and if we isolate 1/(xy) from the right hand side of each expression we get 1/2 (r+2-3/r) = 1/3 (2r-3+1/r) = 1/5 (4r-13+9/r) that simplifies to r²-12r+11=0, no matter which equality you choose. This has solutions r=1 and r=11. For r=1 you get x=y=z but we saw before that this is not possible. For r=11 you get x=11y so z=-19y and the solution is (x,y,z)=(11y,y,-19y). Getting back for example to equation [1]: 121y²+19y² = 2 ⇒ y = ±1/√70 ⇒ x = ±11/√70 ⇒ z = ∓19/√70. Finally it's easy to see that xy≠0. If it was, the equations [1],[2],[3] would yield contradictions.
it will be much simpler if u add twice of all three eq . 2x²-2yz+2y²-2xz+2z²-2xy=20 (x-y)²+(y-z)²+(z-x)²=20 as -1/(x-y)=-2/(y-z)=3/(z-x)=k(constant) then x-y=-1/k, y-z=-2/k,z-x=3/k. plugging all these above we can easily get k and then all variables
@@MindYourDecisions Can you share Why the explanations for a simple algebra problem were not making sense..do you have the links to share? Hope to hear from you..thanks for sharing..
@@sanketkulshrestha How is it a trick..indont see thst it is a trick at all..what Presh didis.more.of a trick..why would.anyone think to eliminate to eliminate squared term after all? Kind of.comes out of nowhere..
I'm a Math teacher from India. I watch your videos quite frequently. I've deep respect for your ability to generate new ideas , your skills and knowledge ; but above all, for your exemplary honesty and candidness. You maintain a strict standard of mathematical logic while explaining things. Very neat presentation , in all respects !!
@@johnsmithsu310 nah i much prefer harder problems cuz that way i wont have to deal with 2 braincell people in my college. And there are plenty of good institutes you can get in for anyone won't can't get to iits, not as good as iit but still good enough for them
@@bhaskarkhandewal3257 I know. I wrote xy instead of yz while solving in first equation. It ended up in a 4th degree one variable equation which I solved using graphing calculator.
Wow, just wow. Usually a 'trick' is an easy way to find what would be impossible the hard way without a computer. But in this case, even with the trick, which I think is really hard to discover if you haven't seen this kind of problem before, it's still a lot of work. And in the time constraints of a math contest, it's easy to make a mistake in all that algebra. This solution leaves me wondering if there is not, in fact, a more straight forward solution. When I rank Presh videos on a 1 to 5 scale meaning '1' is easy and '5' being very challenging, this is definitely a solid '5'. And I'm relieved to have saved time by watching the video than spending countless hours trying to come up with the trick to solve it. Trust me, all the conventional techniques for solving 3 equations in 3 unknowns don't work here.
I have to admit: I tried it, but wasn't able to solve it myself 🤯. Later I continued to watch the video- twice! The second time I stopped it between all the steps. Finally I understood the method, but it is hard and difficult. At the end I checked the values from the solution by using the pocket calculator application from my phone: works! 2, 3 and 5 coming out and confirm that solution. Perfect! Thanks for the video and the great explanation!
Hi there. Sorry to bother you, but I have a question to ask. I've looked for his email in the channel description in order to send him a math problem that's really hard for me to solve, could you say where specifically it's located, please?
Nice! but don't forget about the exceptions in some of the steps you apply: - Cross Product with the isn't quite valid; we ought to handle the a=b=c=0 case separately (and standard factorisation of each of the three original equations suffices to show that we can have x=y=z equal to any constant, not just 0) - likewise, we should handle the case where the two vectors are parallel,which in this case means a=b=c. In this case, our initial multiplying step indicate a whole plane of solutions, not just a vector. Not sure where we go from there. - dividing by a^3+b^3+c^3-3abc, we ought to handle the case where that is 0. By AM-GM inequality, that's only true if a=b=c, but we know that isn't true if our two vectors were different anyway. Finally, we should probably check our solutions as good practice (so easy to drop marks not doing that😢)
I am not yet done with the video (at 6:20) and stopped to see if someone had realized that there was a small mistake there. I also knew a = b = c = 0 x = y = z or x + y + z = 0, but you can't say that x = y = z = 0. I can confirm that you get a plane of solutions for a = b = c. The plane can be expressed as x + y + z = 0. Although I am sure you (@Nathan Brown) know that, because you pointed that out but if someone wants to know why, here is the reason: (1) x^2 - yz = a (2) y^2 -xz = b (3) z^2 - xy = c (1) - (2): a - b = x^2 -y^2 -zy + zx = (x+y)(x-y) + z*(x-y) = (x+y+z)*(x-y) (2) - (3): b - c = y^2 - z^2 - xz + xy = (y+z)*(y-z) + x*(y-z) = (x+y+z)*(y-z) (3) - (1): c - a = z^2 - x^2 - xy + yz = (z+x)(z-x) + y*(z-x) = (x+y+z)*(z-x) Now a = b = c => a - b = b - c = c - a = 0. so that means (x+y+z)*(x-y) = 0, (x+y+z)*(y-z) = 0, (x+y+z)*(z-x) = 0. say a = b = c = 0, then x^2 - yz = 0, y^2 - xz = 0, z^2 - xy = 0. Because all of the are zero, these equations are all equal: x^2 -yz = y^2 -xz = z^2 -xy. Adding xy + yz + zx on every equation we get x * (x+y+z) = y * (x+y+z) = z * (x+y+z). So for a = b = c = 0 there are two cases: x+y+z = 0 (what we want to show) or x = y = z. (BUT NOT NECESSARILY 0). Now to a = b = c ≠ 0: We had derived (x+y+z)*(x-y) = 0 (x+y+z)*(y-z) = 0 (x+y+z)*(z-x) = 0 earlier. These equations imply that either (x+y+z) = 0 or x - y = 0, y - z = 0 and z - x = 0 so we either have x+y+z = 0 what we want to show, or x = y = z. But because of the three equations (1), (2) and (3): x^2 -yz = a y^2 - xz = b z^2 - xy = c we can replace all the x and y by z: z^2 - z*z = a z^2 - z*z = b z^2 - z*z = c so a = b = c = 0. But we had claimed a = b = c ≠ 0, so the case that x = y = z can not happen and thus we have proven that when a = b = c all the tuples (x,y,z) for which x+y+z = 0 are in fact solutions to the given system of equations and when a = b = c = 0 there are even more solutions: x = y = z. Now because x + y + z = 0 can be written as 1*x + 1*y + 1*z = 0 which can be rewritten again as * = 0 which is the equation of a plane that is perpendicular to the Vector and goes through the origin, All the points on that plane are solutions for the original system of equations. I haven't seen the video fully, but I am looking forward for the third point you mentioned.
The math community is not just composed of professional mathematicians, it is MUCH bigger than this. It includes people who simply love math, who love to solve any kind of enigma in a complete and if possible in the clearest way possible. So there is no doubt that you are a MASSIVE part of the global math community ! Personally, I consider your channel to be the best when it comes to present crystal clear solutions to quite challenging problems. Keep it up ;)
The three combinations of two of the equations lead to this: (y-x) * (x+y+z) = 1 (z-y) * (x+y+z) = 2 (z-x) * (x+y+z) = 3 y = x + a z = x + 3a a * (3x+4a) = 1 x = (1/a - 4a)/3 = 1/3a - 4a/3 Then substitute the stuff in the first original equation.
Although quite challenging, I found that solving this problem with the fixed constants (2, 3, 5) algebraically is relatively easy and quite doable; see below. Solving the problem with generalized constants (a, b, c) while applying the same procedure is technically possible, but results in rather clumsy expressions: that's where a more insightful approach (such as in the video) would be welcome. x² - yz = 2 (eq. 1) y² - zx = 3 (eq. 2) z² - xy = 5 (eq. 3) It can be readily observed that if (x,y,z) = (p,q,r) is a solution, then (x,y,z) = (-p, -q, -r) is a solution too. Subtracting eq. 1 from eq. 2 and from eq. 3 , yields (after factoring): (y - x)(x + y + z) = 1 (eq. 4) (z - y)(x + y + z) = 2 (eq. 5) From these, we conclude that (x+y+z) ≠ 0 , and that (z-y) = 2(y-x) (eq. 6) Let u = (y-x) ==> x = y-u z-y = 2u ==> z = y+2u Now substitute x and z in equations (1), (2), (3) : (y-u)² - y(y+2u) = 2 y² - (y+2u)(y-u) = 3 (y+2u)² - (y-u)y = 5 After expanding the expressions, these three equations are reduced to: u² - 4uy = 2 (eq. 7) 2u² - uy = 3 (eq. 8) 4u² + 5uy = 5 (eq. 9) The sum of these three results gives us: 7u² = 10 ==> u² = 10/7 u = ±√(10/7) u = ±10/√(70) After entering this result into eq. 7 , we can solve for y: 10/7 - 4(±10/√(70))y = 2 -(±40/√(70))y = 4/7 (±70/√(70))y = -1 y = ∓1/√(70) and determine x and z : x = (y - u) ==> x = ∓11/(√70) z = (y+2u) ==> z = ±19/(√70) Check if the solutions satisfy the original equations: x² - yz = = (∓11/√70)² - (∓1/√70)(±19/√70) = 121/70 + 19/70 = 140/70 = 2 y² - zx = = (∓1/√70)² - (±19/√70)(∓11/√70) = 1/70 + 19*11/70 = 1/70 + 209/70 = 210/70 = 3 z² - xy = = (±19/√70)² - (∓11/√70)(∓1/√70) = 361/70 - 11/70 = 350/70 = 5 ==> all works out! So the solution is: (x, y, z) = ±( 11/√70 , 1/√70 , -19/√70 ) which can be rewritten as (x, y, z) = ±(11 , 1 , -19)/√70
This was my (not particularly elegant) solution: First label the equations as follows: x² - yz = 2 (1) y² - xz = 3 (2) z² - xy = 5 (3) Subtracting (1) from (2) gives: y^2 - x^2 - xz + yz = 1. We can then factor the LHS to get (y - x)(x + y + z) = 1. After doing this, it's not too difficult to see that if we do (3) - (2) and (3) - (1), we get the following similar looking equations: (z - y)(x + y + z) = 2 and (z - x)(x + y + z) = 3. Motivated by the common factor of x + y + z present in each equation, at this point I decided to let A = x + y + z. We can now rewrite (noting that A is not 0) and label the three equations we obtained as: y - x = 1/A (4) z - y = 2/A (5) z - x = 3/A (6) At this point I was just hoping that somehow I could calculate A explicitly, and from there go on two find x,y and z. I decided to substitute y = x + 1/A and z = x + 3/A into (1). This gave the following: 2 = x^2 - (x + 1/A)(x + 3/A) = x^2 - x^2 - 4x/A - 3/A^2 = -4x/A - 3/A^2. The x^2 terms cancelled, and I now had an expression for x in terms of A alone, namely x = -3/4A - A/2. I then used this expression, and the equations (4) and (5) to find y and z in terms of A. This gave y = 1/4A - A/2 and z = 9/4A - A/2. Promising? Then I remembered that A = x + y + z. I immediately substituted the expressions I had found for x, y and z to get x + y + z = 7/4A - 3A/2 . Now comes the magic: A = x + y + z = 7/4A - 3A/2, and so, on multiplying through by A, we obtain a quadratic equation for A: A^2 = 7/4 - 3A^2/2 , thus 5/2 * A^2 = 7/4 and hence A^2 = 7/10. Seemingly out of nowhere we have found the possible values of A, namely +sqrt(7/10) or -sqrt(7/10) , and with these we have solved the equations: The final solutions then, can be written as follows: x = -3/4A - A/2 y = 1/4A - A/2 z = 9/4A - A/2 where A is either +sqrt(7/10) or -sqrt(7/10). (We can check that these solutions do indeed work. You can also write out x,y and z explicitly, and manipulate them into the forms shown in the video if you like, however I cannot be bothered! ) I hope you enjoyed this solution as much as I did finding it. I was suprised that A could be calculated in a seemingly circular way, nonetheless it worked out!
My solution is quite simple. Since you have 2y+3z=-5x (video 3:28) and 5y+2z=-3x (video 4:06), we can easily calculate y=(1/11)x and z=(-19/9)x, replace them to x² - yz = 2, we have x = ±11/√70. That it！
I don't see how you get that value for x after replacing the values of y and z into x² - yz = 2. Oh, I see ... z = -(19/11)x and not -(19/9)x. Then you get x = ±11/√70. But, well done.
Thank you for the interesting and also beautiful solution. I've seen and failed to solve this kind of problem before. After watching this video, I know at least two ways to solve it (the other one from the screenshot of math stackexchange page shown in this video). One thing that I need to point out is that when a, b and c is 0 (as shown in the video, 6:15), the solution is x=y=z and not necessarily x=y=z=0 Keep posting, Presh. I have subscribed for a few year now. I really like how you find problems from various tests and competitions around the world.
I love the elegance of the general solution using vectors! This problem stumped me for a minute, but eventually I managed to brute force it using only elimination and substitution: [1] x^2-yz=2 [2] y^2-xz=3 [3] z^2-xy=5 [2]-[1] gives us: y^2-x^2+yz-xz=1 (y-x)(y+x)+z(y-x)=1 (y-x)(x+y+z)=1 x+y+z=1/(y-x) Similarly, [3]-[2] gives us: z^2-y^2+xz-xy=2 (z-y)(z+y)+x(z-y)=2 (z-y)(x+y+z)=2 x+y+z=2/(z-y) We now have two expressions both equal to (x+y+z), so they must be equal to each other: 1/(y-x)=2/(z-y) Cross multiplying produces: z-y=2y-2x Add y to both sides: [4] z=3y-2x Now substitute z back into [1]: x^2-y(3y-2x)=2 [5] x^2-3y^2+2xy=2 Substitute z back into [2]: y^2-x(3y-2x)=3 y^2+2x^2-3xy=3 [6] 2x^2+y^2-3xy=3 Combine [5]+3[6] to eliminate the y^2 term: 3(2x^2+y^2-3xy)=3(3) 6x^2+3y^2-9xy=9 x^2-3y^2+2xy=2 Adding these two lines we get: 7x^2-7xy=11 Now solve for y in terms of x: 7xy=7x^2-11 [7] y=x-11/(7x) Substitute y back into [4]: z=3(x-11/(7x))-2x z= 3x-33/(7x)-2x [8] z=x-33/(7x) We now have y and z in terms of x, substitute them back into [1]: x^2-(x-11/(7x))(x-33/(7x))=2 Distribute: x^2-(x^2-11/7-33/7+363/(49x^2))=2 Distribute the - and combine like terms: x^2-x^2+44/7-363/(49x^2)=2 x^2 terms cancel, subtract 44/7 from both sides and negate both sides: 363/(49x^2)=30/7 Prime factor and cross multiply: 7(363)=30(49x^2) 3(7)(11)(11)=2(3)(5)(7)(7)x^2 Divide both sides by 21: 121=70x^2 x^2=121/70 x=+or-11/sqrt(70) x=+or-11sqrt(70)/70 Plug x back into [7] to get y, use the positive value and then negate to get the other value: y=11sqrt(70)/70-11/7(sqrt(70)/11) y=11sqrt(70)/70-sqrt(70)/7 y=11sqrt(70)/70-10sqrt(70)/70 y=+or-sqrt(70)/70 Plug x back into [8] to get z, use the positive value and then negate to get the other value: z=11sqrt(70)/70-33/7(sqrt(70)/11 z=11sqrt(70)/70-3sqrt(70)/7 z=11sqrt(70)/70-30sqrt(70)/70 z=-or+19sqrt(70)/70 And that's the answer! :)
Well your answers are correct but in the video he shows how he got stumped by people doing a different method, that is also your method, of solving because of the cross-multiplication part and the solution in the video is the only one he understood.
Multiply first equation by y, second by z, and third by x. Add the resulting three equations to get 5x + 2y + 3z = 0 (i). Now multiply the first equation by z, the second by x, and the third by y. Add the resulting three equations to get 3x + 5y + 2z = 0 (ii). Use cross-multiplication method to get x/-11 = -y/1 = z/19 = k. So x = -11k, y = -k, z = 19k. (iii) Substitute these values back into the first equation (x^2 -yz = 2, or the second or third) and solve for k. (-11k)^2 - (-k)(19k) = 2 ==> k = +- 1/sqrt(70). Now substitute this back into the three equations of (iii) to get x = +- 11/sqrt(70), y = +- 1/sqrt(70), z = +- 19/sqrt(70). Hopefully, I didn't make any basic algebra mistakes. Now I can watch the video to see what the trick is. Addition: Your solution is the same method! Instead of using what is called the cross-multiplication method, you used the equivalent but more elegant method of using the determinant.
If a=b=c=0 there is the line of solutions x=y=z. I think you should also consider when (c,a,b) and (b,c,a) are multiples. Then b=ar, c=ar^2 where r is a cube root of unity…
You're right. Then by homogeneity and symmetry it is enough to solve the case a=1, b=r, c=r^2. For r^3=1. Case r=1 is easy. If r is not 1 then 1+r+r^2=0 so x**2+y**2+z**2-xy-yz-zx=0. What else? Also (x,y,z) perp to (1,r,r^2). I'm stuck, I don't want to start making substitutions...
Hi Presh, thanks for another great video! 😁👍 This one threw me more than I expected! 😳 Just a quick comment: at 6:19, a=b=c=0 doesn't imply that = in the given system, since = for any real number, n, will also result in a=b=c=0 (each equation becomes n^2 - n.n =0.
I started to contradict this, but actually, I do see you were looking at the system, not his equations. You're correct that x=y=z=n implies a=b=c=0, so I think a study on his assumptions would be reasonable to find something off in the equations he has on the right. Based on his equations on the right, his assertion seems pretty obviously true. He did multiply the original equations by x, y, and z to eventually get what he has on the right, and so this needs to be handled more carefully to watch out for multiplication by 0 that may knock out possible solutions like the ones you found. I don't think it changes his final answer, but I do think it possibly changes his general answer. I'll have to write it down and think about it more.
Thank you very much for the tip about cross multiplying and then adding all three to cancel out all the squares. I couldn't figure out how to simplify. But after that I decided to take another route. We have two resulting equations: 2z + 3x + 5y = 0 (i) 2y + 3z + 5x = 0 (ii) Multiply (i) by 3 and (ii) by 2 6z + 9x + 15y = 0 4y + 6z + 10x = 0 Subtract the second from the first - you cancel out z: 15y - 4y + 9x - 10x =0 x = 11y Plug that value of x into eq (ii): 2z + 3×11y + 5y = 0 2z = -38y => z = -19y Now plug the values of x and z into the second equation of the problem: y² - xz = 3 y² - 11y × (-19y) = 3 210y² = 3 y² = 3/210 = 1/70 y = +/- 1/sqrt 70 Plug the value of y into the x and the z equations and you get: x = +/- 11/sqrt 70 z = -/+ 19/ sqrt 70 😀
This made me feel like I was back in linear algebra and analytic geometry class, which I haven't done for 20 years and I never expected to remember it as fondly as I just did!
Very beautiful solution! But I think it has a small flaw. If a=b=c=0, (0,0,0) isn't the only solution. Any x=y=z are solutions. The expression that leads to the conclusion that (0,0,0) is the only solution in that case is invalid, because if a=b=c=0, the expressions later intepreted as a dot product would be 0=0 so the reasoning would die there.
What an elegant solution! I love it so much! Many thanks for sharing this beautiful solution and for the time and effort you put together in making this awesome video!
Hi, thanks for your video! I solved it in a different way. If you subtract the first equation from the 2nd and rearrange the terms you get (y-x)(x+y+z)=1. Similarly, if you subtract the 1st equation from the 3rd you get (z-x)(x+y+z) = 3. Now define S = x+y+z as an auxiliaty variable. From the above results you get y=1/S+x and z = 3/S+x. Substitute these equations in S=x+y+z and you get x = S/3-4/(3S). Finally take one of the original equations, substitute y=1/S+x, z = 3/S+x and x = S/3-4/(3S) and you get S =+sqrt(7/10) or S =-sqrt(7/10). Knowing S you also know x, y and z. No need of vectors or determinants.
Nice solution, however, there's a serious mistake at 6:19. If a=b=c=0, than every x=y=z numners are solution. Because if a=b=c, than the cross product method gives 0, so it doesn't work properly. If you discuss the resoult, you'll fint that the denominator turns to 0 in case of a=b=c If you remained some energy please make an appendix video about it.
Amazing solution! Thank you so much ❤ Instead of having to use vectors, we can instead set up a third equation in the three variables x, y, and z. To do so, we need the identity x³+y³+z³-3xyz=(x+y+z)(x²+y²+z²-xy-yz-zx). Adding the three given equations gives x²+y²+z²-xy-yz-zx=a+b+c; multiplying eqn. 1, 2, and 3 respectively by x, y, and z, and then adding them up gives x³+y³+z³-3xyz=ax+by+cz. Thus, from the above identity, we get ax+by+cz=(x+y+z)(a+b+c) => (b+c)x+(c+a)y+(a+b)z=0. Combining this with the two other equations derived in the video, we have the following system of equations : 1) bx+cy+az=0 2) cx+ay+bz=0 3) (b+c)x+(c+a)y+(a+b)z=0 Then, we can solve this system to get the final ans.
This seems way harder than the standard rearrange and plug in method. Just rearrange the first equation for x, second for y, third for z and then plug them into each other to solve for one of the variables and then work backwards to get all of them…..
Anyone who contends that Mind your Decisions is not part of the math community, well I'd say they need to revisit their definition of "community". To me, a community is where people get along and don't make arbitrary distinctions to kick people out because they don't like a video title. Presh, thank you for your commitment to the channel. I have found many great problems through you. I hope you know we appreciate you.
I found a much easier solution, y*fx + z*fy + x*fz = 0 and z*fx + y*fz + x*fy = 0, substitute values (fx = 2, fy = 3, fz = 5), solve, easy. much easier than your method. but this channel is awesome. I subbed & liked.
Sometimes your comments make me think you doubt your contributions, and I would give you a compliment borrowed from my father, but he was Navy and he spoke plainly always. To suffice, he called me a heartless bastard for showing him a math problem that he would not be able to sleep until he figured out. It was the highest sort of compliment. You show me these kinds of problems all the time. Don't ever stop. The world is a better place with this sort of content in it.
My thought on the first step: "Oh, that's a need trick. That will probably be very helpful! My thoughts after that: "I'm utterly lost." I've never learnt this vector notation and I've no idea how you know is orthogonal to the other vectors or what that even means.
Thank you for an elegant solution Presh! My solution is similar to @NestorAbad's solution. Let x² - yz = 2 (1) y² - xz = 3 (2) z² - xy = 5 (3) By subtracting (1) from (2) and factoring we get (y - x)(x + z + y) = 1 (4) By subtracting (2) from (3) and factoring we get (z - y)(x + z + y) = 2 (5) Since x + y + z ≠ 0, we can divide (4) by (5) and express z = 3y - 2x. By substituting z into (1) and factoring we get (x + 3y)(x - y) = 2 (6) By substituting z into (2) and factoring we get (2x - y)(x - y) = 3 (7) Since x - y ≠ 0, we can divide (6) by (7) and express x = 11y. By substituting x into (6) and simplifying we get 140y² = 2 (8) From (8) we get y = ±1/√70, from x = 11y we get x = ±11/√70, and from z = 3y - 2x we get z = ∓19/√70.
It can be solved using algebra.. Just need to manipulate add all three and manipulate them as (x-y) ²+(y-z) ²+(z-x) ²=20..(i) then find ratios of x-y, y-z & z-x by subtracting one pair.. The ratio will be -1: -2 : 3 let them be -1t, -2t, 3t.. Put t in equation (i) calculate t... Further you will also get that 3y= 2x+z....after few more steps you will ge tthe answer
Suppose that a,b,D are sides of triangle such that D^2=a^2+b^2-2*a*b*cos(60°) using cosine theorem Similary define E,F. Thus D,E,F are sides a triangle. This triangle is divide by 3 triangle.
Here's the normal solution method (since I don't see it in the comments anywhere, and since quadratics are pointless, much less VECTORS(!), when simple linear equations work fine): xx - yz = 2 yy - xz = 3 zz - xy = 5 Multiply each equation by the nonsquared variable so as to be able to add all three equations and have things cancel out (multiply 1st one by y, 2nd by z, 3rd by x): xxy - yyz = 2y yyz - xzz = 3z zzx - xxy = 5x Add all three: 0 = 2y + 3z + 5x Repeat that process, but with 1st one multiplied by z, 2nd by x, and 3rd by y, again giving complete cancellation when adding all three, yielding: 0 = 2z + 3x + 5y Now eliminate one variable so as to be able to express one of the other two in terms of the third (and then substitute to get things down to just one variable). Choose z to be eliminated, so multiply the 3z equation by 2 and the 2z equation by 3 so that 6z appears in both: 0 = 2y + 3z + 5x MULTIPLY BY 2 0 = 2z + 3x + 5y MULTIPLY BY 3 --> 0 = 4y + 6z + 10x 0 = 6z + 9x + 15y SUBTRACT EQUATIONS: --> 0 = x - 11y So x = 11y, and can do same thing to get z in terms of y: 0 = 2y + 3z + 5x MULTIPLY BY 3 0 = 2z + 3x + 5y MULTIPLY BY 5 --> 0 = 6y + 9z + 15x 0 = 10z + 15x + 25y SUBTRACT EQUATIONS: --> 0 = -z - 19y So z = -19y Substitute into the very first equation of the problem: xx - yz = 2 --> 11y * 11y - y(-19y) = 2 --> 121yy + 19yy = 2 ---> 140yy = 2 --> y = +/- sqrt(1/70) --> x = +/- 11(sqrt(1/70)) --> z = -/+ 19(sqrt(1/70)) So the only "completely unexpected solution" to this is that anyone would want to use vectors or anything else complicated! I didn't see the emails or websites that Presh Talwalkar did, but they must have been very strange if they lead him not to see this straightforward solution.
There is another (shorter) way to get the solution : First you see that : (x-y)(x+y+z)=(x^2-yz)-(y^2-xz)=2-3=-1 And (x-z)(x+y+z)=(x^2-yz)-(z^2-xy)=2-5=-3 And (y-z)(x+y+z)=(y^2-xz)-(z^2-xy)=3-5=-2 Then we can say : y=x+k and z=y+2k and z=x+3k because x+y+z=cte. Substituting in initial equations : x^2-yz=x^2-(x+k)(x+3k)=2 gives -k(4x+3k)=2 and y^2-xz=(x+k)^2-x(x+3k)=3 gives k(k-x)=3 Then it is easy to get from x and k : k=sqrt(10/7) x=-11/sqrt(70) y=-1/sqrt(70) z=19/sqrt(70)
I love your content. some ideas (python): import sympy as sp # Define symbolic variables x, y, z = sp.symbols('x y z') # Define the system of equations eq1 = sp.Eq(x**2 - y*z, 2) eq2 = sp.Eq(y**2 - x*z, 3) eq3 = sp.Eq(z**2 - x*y, 5) # Solve the system of equations sol = sp.solve((eq1, eq2, eq3), (x, y, z)) # Print the solutions print(sol) [(-11*sqrt(70)/70, -sqrt(70)/70, 19*sqrt(70)/70), (11*sqrt(70)/70, sqrt(70)/70, -19*sqrt(70)/70)] which is equivalent to the solutions provided in the video. it turns out sqrt(70)/70 == 1/sqrt(70) and it turns out that sqrt(a)/a == 1/sqrt(a) !!!! (why is nobody talking about this, too simple I guess?) > Math.sqrt(2)/2 0.7071067811865476 > 1/Math.sqrt(2) 0.7071067811865475 Which I don't even know why they didn't teach us in highschool. code is by chatgpt, using this prompt. "how can i use sympy to model the non linear system of equations: x^2 - yz = 2 , y^2-xz = 3, z^2 - xy = 5 and provide solutions to the system" I didn't write the code, I'm not a genius.
Another solution without vector can be possible. X^2-yz =2 a y^2-xz =3 b z^2-xy =5 c equation Then a-b and we can get (x-y)(x+y+z)=-1 so x+y+z=-1/x-y Likewise, x+y+z=-1/x-y=-2/y-z=3/z-x So we can solve the problem
I just wanted to throw out a general statement for the various solutions others have gotten. You may have gotten similar answers, but you REALLY need to be careful when you multiply or divide variable expressions. If you're dividing, you're adding an assumption that that expression can't be 0. If you're multiplying, you may lose solutions if that expression is 0. Presh even made that assumption where he multiplied equations by x, y, and z but then claimed that they had to all be 0 if a, b, and c were 0. That would be a massive loss of solutions for a=b=c=0. I haven't fully thought through if that changes his final answer, but I do think it changes his answer to the general problem. This isn't a linear system, so there's no reason the homogeneous case has to have the same number of solutions as the non-homogeneous case.
7:38 At this point you really need to set aside the case a=b=c=1, or you end up dividing by 0. In fact, any case that involves a=b=c is extremely troublesome and has no solution except for the case a=b=c=x=y=z=0.
I did z(x^2 - yz) = 2z x^2z - z^2y = 2z z^2 - xy = 5 z^2 = 5 + xy x^2z - y(5 + xy) = 2z x^2z - 5y - xy^2 = 2z y^2 - zx = 3 y^2 = 3 + zx x^2z - 5y - x(3 + zx) = 2z x^2z - 5y - 3x - x^2z = 2z - 5y - 3x = 2z 5y + 3x + 2z = 0 Do the same process except multiply with y at the start (y(x^2 - yz) = 2y) and you get the following: 3z + 5x +2y = 0 Use these two and one of the starting equations to make the problem much easier to solve.
Sir please solve this problem I am really stuck with this : Find the possible pairs for m>n such that gcd(m+1,n+1) = gcd(m+2,n+2)= gcd(m,2m-n) =1 for m+n
If i had seen this video 5 months ago i would have gotten a bronze medal in my country’s maths olympiad But to be fair i didn’t know much about linear algebra 😅😅, great video.
I cannot do the vector solution. I was trying to use the concept 2+3=5; therefor x^2 -yx + y^2 -xz = z^2 -xy. I tried to combine and factor terms but had no luck. Thank you for the problem and solution.
Hey in 4:18 , you found two non coplanar planes intersecting each other, which should give the equation of a line and thus we should have infinite sets of values of (x,y,z) isn't it? May be I'm wrong somewhere... can you explain?
no, you have 3 equations in 3 unknowns with quadratic terms, one would expect 2 solutions. He does find a line (parameterized by t) that lies on the intersection of the two planes. He then plugs it into the original equation to find t. The point on the line which is the solution. Note he showed that the solution lies on the line, he didn't show that every point of the line was the solution.
I managed to get one of the solution sets just by messing around with the equations. I thought I had to be wrong though because these videos always seem to have neater looking solutions.
This is the method i had solved the question. For me the solution looked hard to understand as i am still unaware of vectors(just finished class 10th from India) so this was my approach By the method in vid we get 2y+ 3z+5x= 0 Also x²-yz = y²-zx -1 x²-y² +z(x-y)=-1 (x-y)(x+y) +z(x-y)= -1 ( x-y)(x+y+z) = -1 Similarly (y-z)(x+y+z) /2 = -1 So (x-y)2 = y-z 3y-2x=z gives 2y + 9y - 6x + 5x = 0 11y = x Giving -19y = z We have z² - xy = 5 361y² - 11y² = 5 350 y² = 5 y = ±1/√70 Giving x = ±11/√70 z = ±19/√70 for - when y is positive and + when y is negative. But it can be solved in other ways. This is another solution i did which only works 2+3 = 5 X² + y² -(3y-2x)(x+y) = (3y-2x)² -xy x² + 11y² -12xy=0 X = 12y ± √100y²/2 X= 12±10y/2 X= y(rejected as then x=y=x and 0= 5 would come) or 11y Gives z = -19y So then 361y² - 11y² = 5 350 y² = 5 y = ±1/√70 Giving x = ±11/√70 z = ±19/√70 for - when y is positive and + when y is negative.
It doesn't take me back that far! I didn't learn the meaning of dot or cross products until freshman university , linear algebra, a lot of what I've forgotten. I remember the formula for a dot product but I couldn't remember how to compute a cross product, so even if I knew that was the trick to solve these equations, I *still* would have had to look something up.
This solution is smart and awesome, but I wonder why you were not satisfied with the solution at 1:13, though the numbers are different there. If you set a = 1/(x + y + z) and x - y = -a, y - z = -2a, then x = z - 3a, y = z - 2a, so a = 1/(3z - 5a), which means z = (1/3)(5a + 1/a), y = (1/3)(-a + 1/a), x = (1/3)(-4a + 1/a) With this result and x^2 - yz = 1, we can get a = ± sqrt(10/7) so we can compute x, y and z.
I struggled to folow the solution mainly because have forgotten everything I ever knew about vectors and matrices. However, I tried subtracting one equation form another with a higher value, eg if the x squared one is subtracted fromthe y squared one you get x + y+ z = 1/y +x and similar totals for x + y + z for the other subtractions. (3/z + x and 2/z + y). If you do the algebra using these equalities (which I have nnot done) that may be an easier solution..
Bro cramed in physics and thought we wouldn't notice and this is the type of problem everyone underestimates but as the lore goes on so does the self confidence
I haven't seen the solution yet, but just a guess, maybe it's sonething to do with the fibonacci sequence or something similar? It just kind of looks like that with the 2, 3, and 5; and the kind of rotating variables.
My wife and I, scratching our heads: "Have you ever heard the term mutually orthoganal? Me neither." "What? Where did i, j, and k come from?" "This is so far over our heads that we can't even understand the explanation".
How would anyone say you aren’t part of the math community? You provide math content at a decently high level (up to calculus I know) so it’s quite odd anyone would claim you aren’t part of the math community.
Mind Your Decisions is "Recreational Mathematics", interesting problems which often bring powerful but obscure techniques to light. That in addition to being just plain fun. Anyone who claims MYD is *not* mathematics either doesn't understand the full breadth of the math universe or is just a stuffed shirt, fully eligible for being completely ignored. IMHO & FWIW.
This is another Wiles Moment, where showing how big the Fermat Margin should have been, is slightly anticlimactic... I really hoped for a slick solution for this specific case! (I suffered through the general solution at university) It would be interesting to see where the 70 in the root comes from/changes based on the initial values...
This is why I love hate math. "Use this trick to solve" means you have to know the trick in the first place, when and where to use it, then try it and hope it works. But it's fun to try to solve, and feels great if/when you do.
I didn't get a solution, but I did prove that X, Y, and Z were not integers and at least one was negative. If X is even, then by the first equation YZ is an even number, thus either Y or Z is even. If Y is even, then by the second equation, XZ must be odd. But since we said X was even, this is a contradition, thus Y is odd. But if XY is even, then Z must be odd since Z^2 is odd, thus YZ is odd, and thus requires X to be odd, contradicting the original statement.
I'm in 9th grade and can't do vectors yet so taking a bit of help from the video (in doing the first step) and solving my way I ended up with x=y=z=0 haha, I knew there was a problem with my solution and in fact there was lol
If you were able to lift my skull right now, you would probably see some tumbleweed passing by 😂 I often tend to call a mathmatician a mathmagician. Obviously with great respect and admiration!
better explanation: go back to the determinant with i j k vectors. you can replace first column with the sum of the 3 columns. that will give you vector , a+b+c, a+b+c in that column. If you assume a+b+c=0, then you calculate it by first column you get * something * t = so that would mean x=y=z whatever that "something*t" is. from there you go back to the system and left sides simply to 0. So in fact a+b+c=0 implies a=b=c=0
im not sure it is solvable apart from (x,y,z) = (0,0,0), because at least in real numbers anything squared is automatically positive number and hance it will always be greater than zero, unless it's the aformentioned solution. You must have made some mistake simplifying or doing something similar. You can of course look for imaginary solutions if you want
@@hydro63 ohh.....ok i understand.....thank you so much mate😁😁......after watching all these I would rather say that I am satisfied with myself having solved the problem this far😅😂😂
6:15 Why is x=y=z=0 the only solution? Shouldn't it be possible for all values of x,y,z if x=y=z? Because 1² - 1*1 = 0 and the same is true for all other values. Or did I miss something?
its not a solution of the system on the left, look at the vector equality at 6:03, then substitute a=b=c=0. It's the only solution of THAT equation, what you are saying is actually the consequence of that, we are looking for possible values of t, If a=b=c=0 then t can be any real number so we have infinite may solutions parameterized by t, and just so happen that the solution is {a=t,b=t,c=t | where t in R}
Thanks everyone for the kind words--it is a dream to make these videos, but every creator has ups and downs and I appreciate the support :) A small but important clarification. At 6:08 I substitute a = b = c and state that produces the only solution (0, 0, 0). But it is easy to verify (R, R, R) is a solution for any constant R. So where did this go wrong?! If a = b = c = 0, then the two vectors in the cross-product are (0, 0, 0), so any vector is orthogonal to them. So the lesson is the case a = b = c = 0 should be treated separately. It ends up having several solutions, according to WolframAlpha:
www.wolframalpha.com/input?i=x%5E2-yz%3D0%2Cy%5E2-xz%3D0%2Cz%5E2-xy%3D0
As I explained in the video, I didn't solve this problem. But I did do research on Math StackExchange, Topper, and I also asked the RUclips community page and Twitter. None of those explanations covered this corner case, and I missed it too. But of course someone pointed it out after I published :) Unfortunately that means the discussion did not get into the video. On the flip side, I think it's great these videos bring awareness to something new that was never discussed on the internet before. Thank you!
I have one question: if I treat the system of the equations from a geometric point of view it corresponds to a set of quadratic forms of the type as the one shown between round brackets (ax^2 + by^2 + cz^2 + dxy + eyz + fxz + gx + hy + jz + k = 0 omitting 'i' coefficient for clarity). Therefore, geometrically, we have three surfaces in the 3D space intersecting each other. The conclusion you pointed out is that the aforementioned surfaces always have for each real set of values (a,b,c) 2 shared points in common. For real numbers such that not at the same time a,b,c they're all equal to each other, am I right?
Thanks for your reply.
Not only the case a=b=c=0 but the case when (c,a,b) and (b,c,a) are linearly dependent. If we are looking for complex solutions then (a,b,c) would be a multiple of (1,r,r^2) where r is a cubic root of unity. And I’ve been lazy to complete the solution afterwards…
I've come up with a different process
Subtracting the given equations we obtain
(x-y) (x+y+z) =-1
(y-z) (x+y+z) =-2
(x-z) (x+y+z) =-3
And by dividing these equations we obtain
2x+z=3y
and substituting the above equation in place of y in the above 3 equations we get three equations in x, z
4x^2+z^2-5xz =27
3z^2-2x^2-xz=15
3x^2-2xz-z^2=6
now eliminate the xz term
-7x^2+7z^2=24
now eliminate the constant term then we obtain
19x^2-8xz-11z^2=0
Divide by z^2 on both sides and substitute t=x/z
19t^2-8t-11=0
t=1 ❌
t=-11/19 x/z=-11/19
substitute the above x, z relation in -7x^2+7z^2=24 solve for x, y, z
x=-11/√70,
y=-1/√70
z=19/√70
I liked what you did when multiplying the three equations by the variable needed to cancel them all out when you add the equations together. That's some outside-the-box thinking that simplifies the math significantly. There seems to be a much simpler solution using only substitution and basic algebra from there, though.
We are given:
A) x² - yz = 2
B) y² - xz = 3
C) z² - xy = 5
As you showed, by multiplying A, B, and C by y, z, and x respectively, then adding the resulting equations, you can cancel out the squared terms.
D) 0 = 5x + 2y + 3z
As you also showed, this also works when you change the order to z, x, y:
E) 0 = 3x + 5y + 2z
Here's where I break from your solution. Subtract E from D and rearrange:
F) z = 3y - 2x
Substitute the right side of F for the z in either D or E. It doesn't matter which, either way it simplifies to:
G) x = 11y
Sub the right side of G for the x in F:
H) z = -19y
Sub the right side of G for x and the right side of H for z in either A, B, or C. Again, it doesn't matter which, it will simplify to:
I) y² = 1 / 70
Or:
J) y = ±1 / √70
Combining J with G and H gives you either:
x = 11 / √70, y = 1 / √70, z = -19 / √70
OR
x = -11 / √70, y = -1 / √70, z = 19 / √70
Either of which satisfy A, B, and C.
In general, a=b=c leads to a degeneracy, since the cross product is 0. It seems to have an inf # of solutions. Indeed, in the solution, this scenario leads to a 0/0 indeterminate form.
Your solution with dot and cross products is BEAUTIFUL, congrats!
I want to share with you what I did, with elementary manipulation:
Let's number the equations:
x²-yz = 2 [1]
y²-xz = 3 [2]
z²-xy = 5 [3]
Operating [1]-[2], [2]-[3] and [3]-[1], and factoring the difference of squares that appear, we get respectively:
(x-y)(x+y+z) = -1
(y-z)(x+y+z) = -2
(z-x)(x+y+z) = 3
From these we see that x+y+z≠0 and also x,y,z are all different. Now we can isolate x+y+z from these and we get
-1/(x-y) = -2/(y-z) = 3/(z-x)
that simplifies to 2x-3y+z=0 (no matter which equality you choose you get the same thing). This means that z = -2x+3y. We substitute this back into [1], [2] and [3]:
x²-y(-2x+3y) = 2
y²-x(-2x+3y) = 3
(-2x+3y)²-xy = 5
Expanding them we get
x²+2xy-3y² = 2
2x²-3xy+y² = 3
4x²-13xy+9y² = 5
And now comes the trick! Let's divide all these by xy (for a moment suppose that xy≠0) and let r=x/y:
x/y+2-3y/x = 2/(xy) ⇒ r+2-3/r = 2/(xy)
2x/y-3+y/x = 3/(xy) ⇒ 2r-3+1/r = 3/(xy)
4x/y-13+9y/x = 5/(xy) ⇒ 4r-13+9/r = 5(xy)
and if we isolate 1/(xy) from the right hand side of each expression we get
1/2 (r+2-3/r) = 1/3 (2r-3+1/r) = 1/5 (4r-13+9/r)
that simplifies to r²-12r+11=0, no matter which equality you choose. This has solutions r=1 and r=11.
For r=1 you get x=y=z but we saw before that this is not possible.
For r=11 you get x=11y so z=-19y and the solution is (x,y,z)=(11y,y,-19y). Getting back for example to equation [1]:
121y²+19y² = 2 ⇒ y = ±1/√70 ⇒ x = ±11/√70 ⇒ z = ∓19/√70.
Finally it's easy to see that xy≠0. If it was, the equations [1],[2],[3] would yield contradictions.
A good approach but a bit of trick....😊
it will be much simpler if u add twice of all three eq .
2x²-2yz+2y²-2xz+2z²-2xy=20
(x-y)²+(y-z)²+(z-x)²=20
as
-1/(x-y)=-2/(y-z)=3/(z-x)=k(constant)
then x-y=-1/k, y-z=-2/k,z-x=3/k.
plugging all these above we can easily get k and then all variables
Thanks! It is always good to know multiple methods to solve problems, and we appreciate your detailed explanation!
@@MindYourDecisions Can you share Why the explanations for a simple algebra problem were not making sense..do you have the links to share? Hope to hear from you..thanks for sharing..
@@sanketkulshrestha How is it a trick..indont see thst it is a trick at all..what Presh didis.more.of a trick..why would.anyone think to eliminate to eliminate squared term after all? Kind of.comes out of nowhere..
I'm a Math teacher from India. I watch your videos quite frequently. I've deep respect for your ability to generate new ideas , your skills and knowledge ; but above all, for your exemplary honesty and candidness. You maintain a strict standard of mathematical logic while explaining things. Very neat presentation , in all respects !!
Greetings from Turkey
Please don't give this kind of problem to your students for the exam 💀💀💀
@@johnsmithsu310 nah i much prefer harder problems cuz that way i wont have to deal with 2 braincell people in my college. And there are plenty of good institutes you can get in for anyone won't can't get to iits, not as good as iit but still good enough for them
I was hoping the answer would be X=2 or something fun. No way I’m getting to vectors 😂
ditto
X = -1
Y = 1
Z = 2
Check it out
@@HackedPC satisfies two of the equations,second and third respectively
@@bhaskarkhandewal3257 I know. I wrote xy instead of yz while solving in first equation. It ended up in a 4th degree one variable equation which I solved using graphing calculator.
This was fun too
I appreciate your decisions
You are a part of the math community. You’re videos are often a nice warm up. You get me thinking. That’s essential!
Wow, just wow. Usually a 'trick' is an easy way to find what would be impossible the hard way without a computer. But in this case, even with the trick, which I think is really hard to discover if you haven't seen this kind of problem before, it's still a lot of work. And in the time constraints of a math contest, it's easy to make a mistake in all that algebra. This solution leaves me wondering if there is not, in fact, a more straight forward solution.
When I rank Presh videos on a 1 to 5 scale meaning '1' is easy and '5' being very challenging, this is definitely a solid '5'. And I'm relieved to have saved time by watching the video than spending countless hours trying to come up with the trick to solve it. Trust me, all the conventional techniques for solving 3 equations in 3 unknowns don't work here.
I have to admit: I tried it, but wasn't able to solve it myself 🤯. Later I continued to watch the video- twice! The second time I stopped it between all the steps. Finally I understood the method, but it is hard and difficult.
At the end I checked the values from the solution by using the pocket calculator application from my phone: works! 2, 3 and 5 coming out and confirm that solution.
Perfect! Thanks for the video and the great explanation!
Watching this felt like those going deeper into liminal spaces videos
Exactly
Hi there. Sorry to bother you, but I have a question to ask. I've looked for his email in the channel description in order to send him a math problem that's really hard for me to solve, could you say where specifically it's located, please?
@@allayar7 i found it in his website’s sidebar on the left
@@safwan6363 wow, thank you so much, bro)
X = -1
Y = 1
Z = 2
Check it out
Nice! but don't forget about the exceptions in some of the steps you apply:
- Cross Product with the isn't quite valid; we ought to handle the a=b=c=0 case separately (and standard factorisation of each of the three original equations suffices to show that we can have x=y=z equal to any constant, not just 0)
- likewise, we should handle the case where the two vectors are parallel,which in this case means a=b=c. In this case, our initial multiplying step indicate a whole plane of solutions, not just a vector. Not sure where we go from there.
- dividing by a^3+b^3+c^3-3abc, we ought to handle the case where that is 0. By AM-GM inequality, that's only true if a=b=c, but we know that isn't true if our two vectors were different anyway.
Finally, we should probably check our solutions as good practice (so easy to drop marks not doing that😢)
I am not yet done with the video (at 6:20) and stopped to see if someone had realized that there was a small mistake there. I also knew a = b = c = 0 x = y = z or x + y + z = 0, but you can't say that x = y = z = 0.
I can confirm that you get a plane of solutions for a = b = c. The plane can be expressed as x + y + z = 0. Although I am sure you (@Nathan Brown) know that, because you pointed that out but if someone wants to know why, here is the reason:
(1) x^2 - yz = a
(2) y^2 -xz = b
(3) z^2 - xy = c
(1) - (2):
a - b = x^2 -y^2 -zy + zx
= (x+y)(x-y) + z*(x-y)
= (x+y+z)*(x-y)
(2) - (3):
b - c = y^2 - z^2 - xz + xy
= (y+z)*(y-z) + x*(y-z)
= (x+y+z)*(y-z)
(3) - (1):
c - a = z^2 - x^2 - xy + yz
= (z+x)(z-x) + y*(z-x)
= (x+y+z)*(z-x)
Now a = b = c
=> a - b = b - c = c - a = 0.
so that means
(x+y+z)*(x-y) = 0,
(x+y+z)*(y-z) = 0,
(x+y+z)*(z-x) = 0.
say a = b = c = 0,
then x^2 - yz = 0,
y^2 - xz = 0,
z^2 - xy = 0.
Because all of the are zero, these equations are all equal:
x^2 -yz = y^2 -xz = z^2 -xy.
Adding xy + yz + zx on every equation we get
x * (x+y+z) = y * (x+y+z) = z * (x+y+z).
So for a = b = c = 0 there are two cases:
x+y+z = 0 (what we want to show)
or
x = y = z. (BUT NOT NECESSARILY 0).
Now to a = b = c ≠ 0:
We had derived
(x+y+z)*(x-y) = 0
(x+y+z)*(y-z) = 0
(x+y+z)*(z-x) = 0
earlier. These equations imply that either (x+y+z) = 0 or x - y = 0, y - z = 0 and z - x = 0
so we either have x+y+z = 0 what we want to show, or x = y = z.
But because of the three equations (1), (2) and (3):
x^2 -yz = a
y^2 - xz = b
z^2 - xy = c
we can replace all the x and y by z:
z^2 - z*z = a
z^2 - z*z = b
z^2 - z*z = c
so a = b = c = 0. But we had claimed a = b = c ≠ 0, so the case
that x = y = z can not happen and thus we have proven that when a = b = c
all the tuples (x,y,z) for which x+y+z = 0 are in fact solutions to the given system of equations
and when a = b = c = 0 there are even more solutions: x = y = z.
Now because x + y + z = 0 can be written as 1*x + 1*y + 1*z = 0
which can be rewritten again as * = 0 which is the equation
of a plane that is perpendicular to the Vector and goes through the origin,
All the points on that plane are solutions for the original system of equations.
I haven't seen the video fully, but I am looking forward for the third point you mentioned.
The math community is not just composed of professional mathematicians, it is MUCH bigger than this.
It includes people who simply love math, who love to solve any kind of enigma in a complete and if possible in the clearest way possible.
So there is no doubt that you are a MASSIVE part of the global math community !
Personally, I consider your channel to be the best when it comes to present crystal clear solutions to quite challenging problems.
Keep it up ;)
I didn't expect vectors to come into play here, but this solution was amazing!
this is the beauty of mathematics, you don't know what topic will come in the steps you are doing.
Really, a staggering approach! So simple, yet so not obvious! Thank you for sharing this problem and its solution with the community!
This was a 'mind-blown' solution. Thanks, Presh, for your exciting videos. I'm thrilled that I found your channel and am a proud subscriber.
The three combinations of two of the equations lead to this:
(y-x) * (x+y+z) = 1
(z-y) * (x+y+z) = 2
(z-x) * (x+y+z) = 3
y = x + a
z = x + 3a
a * (3x+4a) = 1
x = (1/a - 4a)/3 = 1/3a - 4a/3
Then substitute the stuff in the first original equation.
Although quite challenging, I found that solving this problem with the fixed constants (2, 3, 5) algebraically is relatively easy and quite doable; see below. Solving the problem with generalized constants (a, b, c) while applying the same procedure is technically possible, but results in rather clumsy expressions: that's where a more insightful approach (such as in the video) would be welcome.
x² - yz = 2 (eq. 1)
y² - zx = 3 (eq. 2)
z² - xy = 5 (eq. 3)
It can be readily observed that if (x,y,z) = (p,q,r) is a solution, then (x,y,z) = (-p, -q, -r) is a solution too.
Subtracting eq. 1 from eq. 2 and from eq. 3 , yields (after factoring):
(y - x)(x + y + z) = 1 (eq. 4)
(z - y)(x + y + z) = 2 (eq. 5)
From these, we conclude that (x+y+z) ≠ 0 , and that
(z-y) = 2(y-x) (eq. 6)
Let u = (y-x) ==> x = y-u
z-y = 2u ==> z = y+2u
Now substitute x and z in equations (1), (2), (3) :
(y-u)² - y(y+2u) = 2
y² - (y+2u)(y-u) = 3
(y+2u)² - (y-u)y = 5
After expanding the expressions, these three equations are reduced to:
u² - 4uy = 2 (eq. 7)
2u² - uy = 3 (eq. 8)
4u² + 5uy = 5 (eq. 9)
The sum of these three results gives us:
7u² = 10
==>
u² = 10/7
u = ±√(10/7)
u = ±10/√(70)
After entering this result into eq. 7 , we can solve for y:
10/7 - 4(±10/√(70))y = 2
-(±40/√(70))y = 4/7
(±70/√(70))y = -1
y = ∓1/√(70)
and determine x and z :
x = (y - u) ==> x = ∓11/(√70)
z = (y+2u) ==> z = ±19/(√70)
Check if the solutions satisfy the original equations:
x² - yz =
= (∓11/√70)² - (∓1/√70)(±19/√70) = 121/70 + 19/70 = 140/70
= 2
y² - zx =
= (∓1/√70)² - (±19/√70)(∓11/√70) = 1/70 + 19*11/70 = 1/70 + 209/70 = 210/70
= 3
z² - xy =
= (±19/√70)² - (∓11/√70)(∓1/√70) = 361/70 - 11/70 = 350/70
= 5
==> all works out!
So the solution is:
(x, y, z) = ±( 11/√70 , 1/√70 , -19/√70 )
which can be rewritten as
(x, y, z) = ±(11 , 1 , -19)/√70
This was my (not particularly elegant) solution:
First label the equations as follows:
x² - yz = 2 (1)
y² - xz = 3 (2)
z² - xy = 5 (3)
Subtracting (1) from (2) gives: y^2 - x^2 - xz + yz = 1. We can then factor the LHS to get (y - x)(x + y + z) = 1.
After doing this, it's not too difficult to see that if we do (3) - (2) and (3) - (1), we get the following similar looking equations:
(z - y)(x + y + z) = 2 and (z - x)(x + y + z) = 3. Motivated by the common factor of x + y + z present in each equation, at this point I decided to let A = x + y + z.
We can now rewrite (noting that A is not 0) and label the three equations we obtained as:
y - x = 1/A (4)
z - y = 2/A (5)
z - x = 3/A (6)
At this point I was just hoping that somehow I could calculate A explicitly, and from there go on two find x,y and z.
I decided to substitute y = x + 1/A and z = x + 3/A into (1). This gave the following:
2 = x^2 - (x + 1/A)(x + 3/A) = x^2 - x^2 - 4x/A - 3/A^2 = -4x/A - 3/A^2.
The x^2 terms cancelled, and I now had an expression for x in terms of A alone, namely x = -3/4A - A/2.
I then used this expression, and the equations (4) and (5) to find y and z in terms of A. This gave y = 1/4A - A/2 and z = 9/4A - A/2. Promising?
Then I remembered that A = x + y + z. I immediately substituted the expressions I had found for x, y and z to get x + y + z = 7/4A - 3A/2 .
Now comes the magic: A = x + y + z = 7/4A - 3A/2, and so, on multiplying through by A, we obtain a quadratic equation for A:
A^2 = 7/4 - 3A^2/2 , thus 5/2 * A^2 = 7/4 and hence A^2 = 7/10.
Seemingly out of nowhere we have found the possible values of A, namely +sqrt(7/10) or -sqrt(7/10) , and with these we have solved the equations:
The final solutions then, can be written as follows:
x = -3/4A - A/2
y = 1/4A - A/2
z = 9/4A - A/2
where A is either +sqrt(7/10) or -sqrt(7/10).
(We can check that these solutions do indeed work. You can also write out x,y and z explicitly, and manipulate them into the forms shown in the video if you like, however I cannot be bothered! )
I hope you enjoyed this solution as much as I did finding it. I was suprised that A could be calculated in a seemingly circular way, nonetheless it worked out!
My solution is quite simple. Since you have 2y+3z=-5x (video 3:28) and 5y+2z=-3x (video 4:06), we can easily calculate y=(1/11)x and z=(-19/9)x, replace them to x² - yz = 2, we have x = ±11/√70. That it！
I don't see how you get that value for x after replacing the values of y and z into x² - yz = 2.
Oh, I see ... z = -(19/11)x and not -(19/9)x.
Then you get x = ±11/√70.
But, well done.
@@davidbrisbane7206 you r right, i wrote it wrong.
Thank you for the interesting and also beautiful solution. I've seen and failed to solve this kind of problem before. After watching this video, I know at least two ways to solve it (the other one from the screenshot of math stackexchange page shown in this video).
One thing that I need to point out is that when a, b and c is 0 (as shown in the video, 6:15), the solution is x=y=z and not necessarily x=y=z=0
Keep posting, Presh. I have subscribed for a few year now. I really like how you find problems from various tests and competitions around the world.
Great!!! Thank you for this amazing question with the detail explanation!!!
I love the elegance of the general solution using vectors!
This problem stumped me for a minute, but eventually I managed to brute force it using only elimination and substitution:
[1] x^2-yz=2
[2] y^2-xz=3
[3] z^2-xy=5
[2]-[1] gives us:
y^2-x^2+yz-xz=1
(y-x)(y+x)+z(y-x)=1
(y-x)(x+y+z)=1
x+y+z=1/(y-x)
Similarly, [3]-[2] gives us:
z^2-y^2+xz-xy=2
(z-y)(z+y)+x(z-y)=2
(z-y)(x+y+z)=2
x+y+z=2/(z-y)
We now have two expressions both equal to (x+y+z), so they must be equal to each other:
1/(y-x)=2/(z-y)
Cross multiplying produces:
z-y=2y-2x
Add y to both sides:
[4] z=3y-2x
Now substitute z back into [1]:
x^2-y(3y-2x)=2
[5] x^2-3y^2+2xy=2
Substitute z back into [2]:
y^2-x(3y-2x)=3
y^2+2x^2-3xy=3
[6] 2x^2+y^2-3xy=3
Combine [5]+3[6] to eliminate the y^2 term:
3(2x^2+y^2-3xy)=3(3)
6x^2+3y^2-9xy=9
x^2-3y^2+2xy=2
Adding these two lines we get:
7x^2-7xy=11
Now solve for y in terms of x:
7xy=7x^2-11
[7] y=x-11/(7x)
Substitute y back into [4]:
z=3(x-11/(7x))-2x
z= 3x-33/(7x)-2x
[8] z=x-33/(7x)
We now have y and z in terms of x, substitute them back into [1]:
x^2-(x-11/(7x))(x-33/(7x))=2
Distribute:
x^2-(x^2-11/7-33/7+363/(49x^2))=2
Distribute the - and combine like terms:
x^2-x^2+44/7-363/(49x^2)=2
x^2 terms cancel, subtract 44/7 from both sides and negate both sides:
363/(49x^2)=30/7
Prime factor and cross multiply:
7(363)=30(49x^2)
3(7)(11)(11)=2(3)(5)(7)(7)x^2
Divide both sides by 21:
121=70x^2
x^2=121/70
x=+or-11/sqrt(70)
x=+or-11sqrt(70)/70
Plug x back into [7] to get y, use the positive value and then negate to get the other value:
y=11sqrt(70)/70-11/7(sqrt(70)/11)
y=11sqrt(70)/70-sqrt(70)/7
y=11sqrt(70)/70-10sqrt(70)/70
y=+or-sqrt(70)/70
Plug x back into [8] to get z, use the positive value and then negate to get the other value:
z=11sqrt(70)/70-33/7(sqrt(70)/11
z=11sqrt(70)/70-3sqrt(70)/7
z=11sqrt(70)/70-30sqrt(70)/70
z=-or+19sqrt(70)/70
And that's the answer! :)
Well your answers are correct but in the video he shows how he got stumped by people doing a different method, that is also your method, of solving because of the cross-multiplication part and the solution in the video is the only one he understood.
Thanks for the intro! I learned a lot about how you approach these videos!
Multiply first equation by y, second by z, and third by x. Add the resulting three equations to get 5x + 2y + 3z = 0 (i). Now multiply the first equation by z, the second by x, and the third by y. Add the resulting three equations to get 3x + 5y + 2z = 0 (ii). Use cross-multiplication method to get x/-11 = -y/1 = z/19 = k. So x = -11k, y = -k, z = 19k. (iii) Substitute these values back into the first equation (x^2 -yz = 2, or the second or third) and solve for k. (-11k)^2 - (-k)(19k) = 2 ==> k = +- 1/sqrt(70). Now substitute this back into the three equations of (iii) to get x = +- 11/sqrt(70), y = +- 1/sqrt(70), z = +- 19/sqrt(70). Hopefully, I didn't make any basic algebra mistakes. Now I can watch the video to see what the trick is.
Addition: Your solution is the same method! Instead of using what is called the cross-multiplication method, you used the equivalent but more elegant method of using the determinant.
If a=b=c=0 there is the line of solutions x=y=z. I think you should also consider when (c,a,b) and (b,c,a) are multiples. Then b=ar, c=ar^2 where r is a cube root of unity…
You're right. Then by homogeneity and symmetry it is enough to solve the case a=1, b=r, c=r^2. For r^3=1. Case r=1 is easy. If r is not 1 then 1+r+r^2=0 so x**2+y**2+z**2-xy-yz-zx=0. What else? Also (x,y,z) perp to (1,r,r^2). I'm stuck, I don't want to start making substitutions...
Hi Presh, thanks for another great video! 😁👍 This one threw me more than I expected! 😳
Just a quick comment: at 6:19, a=b=c=0 doesn't imply that = in the given system, since = for any real number, n, will also result in a=b=c=0 (each equation becomes n^2 - n.n =0.
I started to contradict this, but actually, I do see you were looking at the system, not his equations. You're correct that x=y=z=n implies a=b=c=0, so I think a study on his assumptions would be reasonable to find something off in the equations he has on the right. Based on his equations on the right, his assertion seems pretty obviously true. He did multiply the original equations by x, y, and z to eventually get what he has on the right, and so this needs to be handled more carefully to watch out for multiplication by 0 that may knock out possible solutions like the ones you found. I don't think it changes his final answer, but I do think it possibly changes his general answer. I'll have to write it down and think about it more.
Thank you very much for the tip about cross multiplying and then adding all three to cancel out all the squares. I couldn't figure out how to simplify. But after that I decided to take another route.
We have two resulting equations:
2z + 3x + 5y = 0 (i)
2y + 3z + 5x = 0 (ii)
Multiply (i) by 3 and (ii) by 2
6z + 9x + 15y = 0
4y + 6z + 10x = 0
Subtract the second from the first - you cancel out z:
15y - 4y + 9x - 10x =0
x = 11y
Plug that value of x into eq (ii):
2z + 3×11y + 5y = 0
2z = -38y => z = -19y
Now plug the values of x and z into the second equation of the problem:
y² - xz = 3
y² - 11y × (-19y) = 3
210y² = 3
y² = 3/210 = 1/70
y = +/- 1/sqrt 70
Plug the value of y into the x and the z equations and you get:
x = +/- 11/sqrt 70
z = -/+ 19/ sqrt 70
😀
This made me feel like I was back in linear algebra and analytic geometry class, which I haven't done for 20 years and I never expected to remember it as fondly as I just did!
Very beautiful solution! But I think it has a small flaw. If a=b=c=0, (0,0,0) isn't the only solution. Any x=y=z are solutions.
The expression that leads to the conclusion that (0,0,0) is the only solution in that case is invalid, because if a=b=c=0, the expressions later intepreted as a dot product would be 0=0 so the reasoning would die there.
What an elegant solution! I love it so much! Many thanks for sharing this beautiful solution and for the time and effort you put together in making this awesome video!
Hi, thanks for your video! I solved it in a different way. If you subtract the first equation from the 2nd and rearrange the terms you get (y-x)(x+y+z)=1. Similarly, if you subtract the 1st equation from the 3rd you get (z-x)(x+y+z) = 3. Now define S = x+y+z as an auxiliaty variable. From the above results you get y=1/S+x and z = 3/S+x. Substitute these equations in S=x+y+z and you get x = S/3-4/(3S). Finally take one of the original equations, substitute y=1/S+x, z = 3/S+x and x = S/3-4/(3S) and you get S =+sqrt(7/10) or S =-sqrt(7/10). Knowing S you also know x, y and z. No need of vectors or determinants.
Nice solution, however, there's a serious mistake at 6:19.
If a=b=c=0, than every x=y=z numners are solution.
Because if a=b=c, than the cross product method gives 0, so it doesn't work properly. If you discuss the resoult, you'll fint that the denominator turns to 0 in case of a=b=c
If you remained some energy please make an appendix video about it.
Amazing solution! Thank you so much ❤
Instead of having to use vectors, we can instead set up a third equation in the three variables x, y, and z. To do so, we need the identity x³+y³+z³-3xyz=(x+y+z)(x²+y²+z²-xy-yz-zx).
Adding the three given equations gives x²+y²+z²-xy-yz-zx=a+b+c; multiplying eqn. 1, 2, and 3 respectively by x, y, and z, and then adding them up gives x³+y³+z³-3xyz=ax+by+cz. Thus, from the above identity, we get ax+by+cz=(x+y+z)(a+b+c) => (b+c)x+(c+a)y+(a+b)z=0. Combining this with the two other equations derived in the video, we have the following system of equations :
1) bx+cy+az=0
2) cx+ay+bz=0
3) (b+c)x+(c+a)y+(a+b)z=0
Then, we can solve this system to get the final ans.
This seems way harder than the standard rearrange and plug in method. Just rearrange the first equation for x, second for y, third for z and then plug them into each other to solve for one of the variables and then work backwards to get all of them…..
saying and doing it is two completely different things
Anyone who contends that Mind your Decisions is not part of the math community, well I'd say they need to revisit their definition of "community". To me, a community is where people get along and don't make arbitrary distinctions to kick people out because they don't like a video title.
Presh, thank you for your commitment to the channel. I have found many great problems through you. I hope you know we appreciate you.
I found a much easier solution, y*fx + z*fy + x*fz = 0 and z*fx + y*fz + x*fy = 0, substitute values (fx = 2, fy = 3, fz = 5), solve, easy. much easier than your method. but this channel is awesome. I subbed & liked.
Sometimes your comments make me think you doubt your contributions, and I would give you a compliment borrowed from my father, but he was Navy and he spoke plainly always. To suffice, he called me a heartless bastard for showing him a math problem that he would not be able to sleep until he figured out. It was the highest sort of compliment. You show me these kinds of problems all the time. Don't ever stop. The world is a better place with this sort of content in it.
My thought on the first step: "Oh, that's a need trick. That will probably be very helpful!
My thoughts after that: "I'm utterly lost."
I've never learnt this vector notation and I've no idea how you know is orthogonal to the other vectors or what that even means.
This problem has a very interesting story and I think the time you spent on this problem is worth it 'cause I perfectly understood the problem.
Thank you for an elegant solution Presh! My solution is similar to @NestorAbad's solution.
Let
x² - yz = 2 (1)
y² - xz = 3 (2)
z² - xy = 5 (3)
By subtracting (1) from (2) and factoring we get
(y - x)(x + z + y) = 1 (4)
By subtracting (2) from (3) and factoring we get
(z - y)(x + z + y) = 2 (5)
Since x + y + z ≠ 0, we can divide (4) by (5) and express z = 3y - 2x.
By substituting z into (1) and factoring we get
(x + 3y)(x - y) = 2 (6)
By substituting z into (2) and factoring we get
(2x - y)(x - y) = 3 (7)
Since x - y ≠ 0, we can divide (6) by (7) and express x = 11y.
By substituting x into (6) and simplifying we get
140y² = 2 (8)
From (8) we get y = ±1/√70, from x = 11y we get x = ±11/√70, and from z = 3y - 2x we get z = ∓19/√70.
what about
X = -1
Y = 1
Z = 2
Check it out
I solved exactly same way! Idea with dot and cross products certainly great, but looks like overcomplication in that concrete task.
@@HackedPC 😢🎉m fr nr
It can be solved using algebra.. Just need to manipulate add all three and manipulate them as (x-y) ²+(y-z) ²+(z-x) ²=20..(i) then find ratios of x-y, y-z & z-x by subtracting one pair.. The ratio will be -1: -2 : 3 let them be -1t, -2t, 3t.. Put t in equation (i) calculate t... Further you will also get that 3y= 2x+z....after few more steps you will ge tthe answer
Suppose that a,b,D are sides of triangle such that D^2=a^2+b^2-2*a*b*cos(60°) using cosine theorem Similary define E,F. Thus D,E,F are sides a triangle. This triangle is divide by 3 triangle.
this is actually a very clever solution, nice job, worth the time you spent on it :)
Here's the normal solution method (since I don't see it in the comments anywhere, and since quadratics are pointless, much less VECTORS(!), when simple linear equations work fine):
xx - yz = 2
yy - xz = 3
zz - xy = 5
Multiply each equation by the nonsquared variable so as to be able to add all three equations and have things cancel out (multiply 1st one by y, 2nd by z, 3rd by x):
xxy - yyz = 2y
yyz - xzz = 3z
zzx - xxy = 5x
Add all three:
0 = 2y + 3z + 5x
Repeat that process, but with 1st one multiplied by z, 2nd by x, and 3rd by y, again giving complete cancellation when adding all three, yielding:
0 = 2z + 3x + 5y
Now eliminate one variable so as to be able to express one of the other two in terms of the third (and then substitute to get things down to just one variable). Choose z to be eliminated, so multiply the 3z equation by 2 and the 2z equation by 3 so that 6z appears in both:
0 = 2y + 3z + 5x MULTIPLY BY 2
0 = 2z + 3x + 5y MULTIPLY BY 3
-->
0 = 4y + 6z + 10x
0 = 6z + 9x + 15y SUBTRACT EQUATIONS:
--> 0 = x - 11y
So x = 11y, and can do same thing to get z in terms of y:
0 = 2y + 3z + 5x MULTIPLY BY 3
0 = 2z + 3x + 5y MULTIPLY BY 5
-->
0 = 6y + 9z + 15x
0 = 10z + 15x + 25y SUBTRACT EQUATIONS:
--> 0 = -z - 19y
So z = -19y
Substitute into the very first equation of the problem:
xx - yz = 2 --> 11y * 11y - y(-19y) = 2 --> 121yy + 19yy = 2 ---> 140yy = 2
--> y = +/- sqrt(1/70)
--> x = +/- 11(sqrt(1/70))
--> z = -/+ 19(sqrt(1/70))
So the only "completely unexpected solution" to this is that anyone would want to use vectors or anything else complicated! I didn't see the emails or websites that Presh Talwalkar did, but they must have been very strange if they lead him not to see this straightforward solution.
There is another (shorter) way to get the solution :
First you see that :
(x-y)(x+y+z)=(x^2-yz)-(y^2-xz)=2-3=-1
And
(x-z)(x+y+z)=(x^2-yz)-(z^2-xy)=2-5=-3
And
(y-z)(x+y+z)=(y^2-xz)-(z^2-xy)=3-5=-2
Then we can say :
y=x+k and z=y+2k and z=x+3k
because x+y+z=cte.
Substituting in initial equations :
x^2-yz=x^2-(x+k)(x+3k)=2
gives -k(4x+3k)=2
and
y^2-xz=(x+k)^2-x(x+3k)=3
gives k(k-x)=3
Then it is easy to get from x and k :
k=sqrt(10/7)
x=-11/sqrt(70)
y=-1/sqrt(70)
z=19/sqrt(70)
I love your content.
some ideas (python):
import sympy as sp
# Define symbolic variables
x, y, z = sp.symbols('x y z')
# Define the system of equations
eq1 = sp.Eq(x**2 - y*z, 2)
eq2 = sp.Eq(y**2 - x*z, 3)
eq3 = sp.Eq(z**2 - x*y, 5)
# Solve the system of equations
sol = sp.solve((eq1, eq2, eq3), (x, y, z))
# Print the solutions
print(sol)
[(-11*sqrt(70)/70, -sqrt(70)/70, 19*sqrt(70)/70), (11*sqrt(70)/70, sqrt(70)/70, -19*sqrt(70)/70)]
which is equivalent to the solutions provided in the video.
it turns out sqrt(70)/70 == 1/sqrt(70)
and it turns out that sqrt(a)/a == 1/sqrt(a) !!!! (why is nobody talking about this, too simple I guess?)
> Math.sqrt(2)/2
0.7071067811865476
> 1/Math.sqrt(2)
0.7071067811865475
Which I don't even know why they didn't teach us in highschool.
code is by chatgpt, using this prompt.
"how can i use sympy to model the non linear system of equations: x^2 - yz = 2 , y^2-xz = 3, z^2 - xy = 5 and provide solutions to the system"
I didn't write the code, I'm not a genius.
I find this to be more interesting than the video itself!
Very nice explanation!
Another solution without vector can be possible.
X^2-yz =2 a
y^2-xz =3 b
z^2-xy =5 c equation
Then a-b and we can get
(x-y)(x+y+z)=-1
so x+y+z=-1/x-y
Likewise,
x+y+z=-1/x-y=-2/y-z=3/z-x
So we can solve the problem
I just wanted to throw out a general statement for the various solutions others have gotten. You may have gotten similar answers, but you REALLY need to be careful when you multiply or divide variable expressions. If you're dividing, you're adding an assumption that that expression can't be 0. If you're multiplying, you may lose solutions if that expression is 0. Presh even made that assumption where he multiplied equations by x, y, and z but then claimed that they had to all be 0 if a, b, and c were 0. That would be a massive loss of solutions for a=b=c=0. I haven't fully thought through if that changes his final answer, but I do think it changes his answer to the general problem. This isn't a linear system, so there's no reason the homogeneous case has to have the same number of solutions as the non-homogeneous case.
7:38 At this point you really need to set aside the case a=b=c=1, or you end up dividing by 0. In fact, any case that involves a=b=c is extremely troublesome and has no solution except for the case a=b=c=x=y=z=0.
I did
z(x^2 - yz) = 2z
x^2z - z^2y = 2z
z^2 - xy = 5
z^2 = 5 + xy
x^2z - y(5 + xy) = 2z
x^2z - 5y - xy^2 = 2z
y^2 - zx = 3
y^2 = 3 + zx
x^2z - 5y - x(3 + zx) = 2z
x^2z - 5y - 3x - x^2z = 2z
- 5y - 3x = 2z
5y + 3x + 2z = 0
Do the same process except multiply with y at the start (y(x^2 - yz) = 2y) and you get the following:
3z + 5x +2y = 0
Use these two and one of the starting equations to make the problem much easier to solve.
Aaaaaand my brain melted. Frigging vectors.
I was visualizing the problem in 3d space and felt like it had potential, but I never studied vector math so I didn't know what to do next.
Sir please solve this problem
I am really stuck with this :
Find the possible pairs for m>n such that gcd(m+1,n+1) = gcd(m+2,n+2)= gcd(m,2m-n) =1 for m+n
great video i have enjoyed your videos for several years now: always interesting problems.
If i had seen this video 5 months ago i would have gotten a bronze medal in my country’s maths olympiad
But to be fair i didn’t know much about linear algebra 😅😅, great video.
Yikes
Which country
I cannot do the vector solution. I was trying to use the concept 2+3=5; therefor x^2 -yx + y^2 -xz = z^2 -xy. I tried to combine and factor terms but had no luck. Thank you for the problem and solution.
Hey in 4:18 , you found two non coplanar planes intersecting each other, which should give the equation of a line and thus we should have infinite sets of values of (x,y,z) isn't it? May be I'm wrong somewhere... can you explain?
no, you have 3 equations in 3 unknowns with quadratic terms, one would expect 2 solutions. He does find a line (parameterized by t) that lies on the intersection of the two planes. He then plugs it into the original equation to find t. The point on the line which is the solution. Note he showed that the solution lies on the line, he didn't show that every point of the line was the solution.
@@shohamsen8986 Oh I got it! Thank you so much
Vector bashing a system of equations is a 10,000 IQ play.
I managed to get one of the solution sets just by messing around with the equations. I thought I had to be wrong though because these videos always seem to have neater looking solutions.
This is the method i had solved the question. For me the solution looked hard to understand as i am still unaware of vectors(just finished class 10th from India) so this was my approach
By the method in vid we get 2y+ 3z+5x= 0
Also x²-yz = y²-zx -1
x²-y² +z(x-y)=-1
(x-y)(x+y) +z(x-y)= -1
( x-y)(x+y+z) = -1
Similarly (y-z)(x+y+z) /2 = -1
So (x-y)2 = y-z
3y-2x=z gives
2y + 9y - 6x + 5x = 0
11y = x
Giving
-19y = z
We have z² - xy = 5
361y² - 11y² = 5
350 y² = 5
y = ±1/√70
Giving x = ±11/√70
z = ±19/√70 for - when y is positive and + when y is negative.
But it can be solved in other ways.
This is another solution i did which only works 2+3 = 5
X² + y² -(3y-2x)(x+y) = (3y-2x)² -xy
x² + 11y² -12xy=0
X = 12y ± √100y²/2
X= 12±10y/2
X= y(rejected as then x=y=x and 0= 5 would come) or 11y
Gives z = -19y
So then
361y² - 11y² = 5
350 y² = 5
y = ±1/√70
Giving x = ±11/√70
z = ±19/√70 for - when y is positive and + when y is negative.
Good backstory on the process
If I wouldn't have watched math memes, I wouldn't have gonna understand this cuz most things used here are learnt by me from memes
Umm Like your method in vector format, I kinda solve it in algebraic method and my solution matched your solutions, appreciate it😉
this takes me back to my second year in high school , that was brilliant and genious by the way Dr presh !
It doesn't take me back that far! I didn't learn the meaning of dot or cross products until freshman university , linear algebra, a lot of what I've forgotten. I remember the formula for a dot product but I couldn't remember how to compute a cross product, so even if I knew that was the trick to solve these equations, I *still* would have had to look something up.
For (a,b,c) = (1,4,4) and (5,5,6), the solutions are rational. Are there any such (a,b,c) with all of them different?
This solution is smart and awesome, but I wonder why you were not satisfied with the solution at 1:13, though the numbers are different there.
If you set a = 1/(x + y + z) and x - y = -a, y - z = -2a, then x = z - 3a, y = z - 2a, so a = 1/(3z - 5a), which means z = (1/3)(5a + 1/a), y = (1/3)(-a + 1/a), x = (1/3)(-4a + 1/a)
With this result and x^2 - yz = 1, we can get a = ± sqrt(10/7) so we can compute x, y and z.
I struggled to folow the solution mainly because have forgotten everything I ever knew about vectors and matrices. However, I tried subtracting one equation form another with a higher value, eg if the x squared one is subtracted fromthe y squared one you get x + y+ z = 1/y +x and similar totals for x + y + z for the other subtractions. (3/z + x and 2/z + y). If you do the algebra using these equalities (which I have nnot done) that may be an easier solution..
Sorry, small correction: You need to work with 1/y-x, etc. Now you see why I didn't struggle with the algebra.
Bro cramed in physics and thought we wouldn't notice and this is the type of problem everyone underestimates but as the lore goes on so does the self confidence
I haven't seen the solution yet, but just a guess, maybe it's sonething to do with the fibonacci sequence or something similar? It just kind of looks like that with the 2, 3, and 5; and the kind of rotating variables.
My wife and I, scratching our heads:
"Have you ever heard the term mutually orthoganal? Me neither."
"What? Where did i, j, and k come from?"
"This is so far over our heads that we can't even understand the explanation".
Thank you, Professor
For a = b = c = 1, the formula is invalid. (1, 1, -2)/sqrt(3) is a solution.
That must've been a really hard problem, it took you two years to solve it!
its not like he was always focusing on that problem, although i kinda see that ur joking but just in case im putting it
How would anyone say you aren’t part of the math community? You provide math content at a decently high level (up to calculus I know) so it’s quite odd anyone would claim you aren’t part of the math community.
Mind Your Decisions is "Recreational Mathematics", interesting problems which often bring powerful but obscure techniques to light. That in addition to being just plain fun. Anyone who claims MYD is *not* mathematics either doesn't understand the full breadth of the math universe or is just a stuffed shirt, fully eligible for being completely ignored. IMHO & FWIW.
Yup Math isn't just highschool math exercises.
Jesus.
This was very difficult. I had to get a pencil and paper and then I got stuck just before the vector portion of the problem.
This is another Wiles Moment, where showing how big the Fermat Margin should have been, is slightly anticlimactic...
I really hoped for a slick solution for this specific case! (I suffered through the general solution at university)
It would be interesting to see where the 70 in the root comes from/changes based on the initial values...
"70 in the root" being:
The cubic thing in the denominator that looks like its (a-b)², but cubed instead, but isn't.
5：50, here I cannot follow as my knowledge level is too low to realize the principle. I have to re-learn the course if time is free!
This is why I love hate math. "Use this trick to solve" means you have to know the trick in the first place, when and where to use it, then try it and hope it works. But it's fun to try to solve, and feels great if/when you do.
I didn't get a solution, but I did prove that X, Y, and Z were not integers and at least one was negative. If X is even, then by the first equation YZ is an even number, thus either Y or Z is even. If Y is even, then by the second equation, XZ must be odd. But since we said X was even, this is a contradition, thus Y is odd. But if XY is even, then Z must be odd since Z^2 is odd, thus YZ is odd, and thus requires X to be odd, contradicting the original statement.
Elegance!
My brain explosive 🤯 #😂😂😂
I'm in 9th grade and can't do vectors yet so taking a bit of help from the video (in doing the first step) and solving my way I ended up with x=y=z=0 haha, I knew there was a problem with my solution and in fact there was lol
If you were able to lift my skull right now, you would probably see some tumbleweed passing by 😂
I often tend to call a mathmatician a mathmagician. Obviously with great respect and admiration!
What happens if a+b+c=0? Because that would mean t=1/0 even though a, b and c are not equal to zero
in that case the equation at 7:33 have zero solutions, so there are no vectors perpendicular to both and so there are no solutions to the system.
better explanation: go back to the determinant with i j k vectors. you can replace first column with the sum of the 3 columns. that will give you vector , a+b+c, a+b+c in that column. If you assume a+b+c=0, then you calculate it by first column you get * something * t = so that would mean x=y=z whatever that "something*t" is. from there you go back to the system and left sides simply to 0.
So in fact a+b+c=0 implies a=b=c=0
Tried auditing linear algebra once. Never made it very far. So, i knew what determinates and cross products are, just could not follow this problem.
X = -1
Y = 1
Z = 2
Check it out
Damn the haters Presh. Keep up the good work.
If a= b =c =0, then we also have a solution where x = y = z =1
Its definitely harder when you don’t know about the cross product
So elegant.
Nice. Thank you.
X and y gets confused itself
Bro it just kept on going and going. I was already lost at vectors and he just didn’t stop😭
Its really beautiful!!
you arent in the math community, you ARE the math community 😁😁
Difficulty level in a 0 - 10 range: 70
that was insane, wow
I used plain algebra and I am getting an equation like this :
(x + y) ^2 + (y + z) ^2 + (z + x) ^2 = 0
What to do after this step??
im not sure it is solvable apart from (x,y,z) = (0,0,0), because at least in real numbers anything squared is automatically positive number and hance it will always be greater than zero, unless it's the aformentioned solution. You must have made some mistake simplifying or doing something similar. You can of course look for imaginary solutions if you want
@@hydro63 ohh.....ok i understand.....thank you so much mate😁😁......after watching all these I would rather say that I am satisfied with myself having solved the problem this far😅😂😂
Don't you mean
(x - y)^2 + (y - z)^2 + (z - x)^2 = 20
perhaps?
Or else you probably made some error along the way.
@@yurenchu ......probably.....i am wrong.....whatever it is.....thank you for helping 😁😁
Well explained! Math splaining? 😊
6:15 Why is x=y=z=0 the only solution? Shouldn't it be possible for all values of x,y,z if x=y=z? Because 1² - 1*1 = 0 and the same is true for all other values. Or did I miss something?
Yes,you are right.
Yes, you're right. If a=b=c=0 then the whole reasoning before isn't valid because both and are the 0 vector.
Totally agree!
its not a solution of the system on the left, look at the vector equality at 6:03, then substitute a=b=c=0. It's the only solution of THAT equation, what you are saying is actually the consequence of that, we are looking for possible values of t, If a=b=c=0 then t can be any real number so we have infinite may solutions parameterized by t, and just so happen that the solution is {a=t,b=t,c=t | where t in R}
absolutely awesome.