This problem could be considered a hard one compared to most of the other problems. The solution method is not that obvious at first glance. Are there other methods to solve it? Any ideas?
Very Good Video .I dont know for all real but maybe this is right for positive consider triangle abc with side length √19, √39, 7 by law of cosine, x=pa, y=pb, z=pb. and apb, apc, bpc are 120º degree by area of triangle using sine we get xy+xz+yz=31. sum all of the equation and get (x+y+z)²=100, x+y+z=10 substrack each of 2 equation (z-x)(x+y+z)=10, (y-z)(x+y+z)=20, (z-x)=1, (y-z)=2 then x+x+1+x+3=10, x=2, z=3, y=5 maybe you can get negative answer by x+y+z=-10 or using negative length
I solved it like this: x^2 + y^2 + xy = (x+y)^2 - xy = 39 so we have a square - a number getting 39 the first square bigger than 39 is 7^2 it can be written as 5 + 2, which is a valid setting for x and y, but since we don't know neither x nor y from 5 and 2 we will move to the 2nd equation, using the same reformation of the equation and choosing y to be the solution 5, we can see that z is 3. so we didn't even have to look at the 3rd equation. Thanks.
When you get the equation "3x^2-35xz+22z^2 = 0"(11:17) , you can just decompose the equation like this"(3x-2z)(x-11z) = 0",so we can easily konw the solution of the equation: 3x = 2z OR x = 11z.
Here is how I solved it, although I did end up using a calculator to get a discriminant at a step in the problem. If you take the difference of the three equations you get: (z-x)(x+y+z) = 10 (A) (y-x)(x+y+z) = 30 (y-z)(x+y+z) = 20 note that if x+y+z = 0 you get a contradiction, so x+y+z is not 0. Then you divide the equations against each other and get (y-z)/(z-x) = 2, which gives y = 3z-2x. Also if x=z then the first and second equation would be the same, so you get a contradiction. Similar for the others, since all the equations have different values. If you substitute your equation for y into the first equation you can simplify and get: x^2+3z^2-3zx = 13. You can subtract that equation from the third equation and simplify to get: z^2-2xz+3 = 0. This is a quadratic that yields z = x /pm sqrt(x^2-3), given this is over the reals x^2 /ge 3. Using our y equation we get that y = x /pm 3 sqrt(x^2-3) Conveniently, z-x = /pm sqrt(x^2-3) so if we substitute that into (A) you get what is essentially a quadratic in x^2. I kept careful track of the pluses and minuses, because it will be relevant in the final step, but for this part you don't care. Furthermore around this time I realized x^2 = 4 is a solution, but it'll be seen anyways if you just continue on. If I let t = x^2, then: 9(t^2 - 3t) = (22-4t)^2 You can rearrange and solve the quadratic, but I did use a calculator for the discriminant (not saying it was necessary). t = x^2 = 4, 11^2/7, so x = /pm 2, /pm 11/sqrt(7) So now you just need to acquire the triples (since both sides of the pluses and minuses do not work), from the last quadratic before I squared everything I kept that info and you just confirm that right options. (2,5,3), (-2,-5,-3),(-11/sqrt(7),19/sqrt(7), -17/sqrt(7)),(11/sqrt(7),-19/sqrt(7),1/sqrt(7)) At some point in the problem you could have argued that if (x,y,z) is a triple then (-x,-y,-z) is one as well since that is straightforward from the given information.
It's a bit hard to see because of a small arthimetic mistake, but all the solutions are: x = ±2, y = ±5, z = ±3 and x = ± 11/√7 , y = ± 19/√7 , z = ± 1/√7
This can be done without too much mathematical rigour and some magic assumptions. (x+y)^2=39+xy (y+z)^2=49+yz (x+z)^2=19+xz Then assume x, y, z as positive integers The next integer square after 39 is 49 so xy=10 For 49 is 64 so yz=15 For 19 is 25 so zx=6 Clearly x=2, y=5 and z=3 Sorry for cheating but I foud it cool. (
Without any mathematical rigour. I supposed that x,y and z were integers and that the squares on the left side were the next you can find from the first number on the right side. Luckily these two assumptilns were right.😉
Here is another easier/alternate way to reach out a solution to this problem. We assume that x, y and z are positive integers and there exists a solution to it. If we deduct one equation from the other we have three distinct equations as follows: (x+y+z)(y-z) = 20 .... (1), (x+y+z)(z-x) = 10 .... (2) and (x+y+z)(y-x) = 30 .... (3). And here lies the logic. (x+y+z)>>x ; (x+y+z)>>y ; also (x+y+z)>>z . And x+y+z is a common factor to the all the above deduced three equations. The right side values of the equations are 20, 10 and 30 and each value is the the product of two factors as per our equations. Therefore, x+y+z will be the HCF (Highest Common Factor) of 20, 10 and 30. In this case it is 10. In other words, x+y+z=10 ...(4). So that from the above three equations, we have y-z=2 ... (5), z-x=1 ... (6) and y-x=3 ... (7). Let us now solve for x, y and z. Consider (any) for z value. From eqn (6) x=z-1 and from (5) y=z+2. Putting these values in eqn (4), we have (z-1) + (z+2) + z = 10; from which we get z =3. Putting this value in equation(4) we get x+y=7 ...(8) Using eqn (7) and eqn (8), we get x = 2 and y=5. Therefore, the solutions sets are (x,y,z) =(2,5,3). This is a particular solution to the problem. For general solution, SyberMath is preferred.
For positive solution. Consider triangle with sides sqrt(39),7, sqrt(19), because x^2+x*y+y^2= x^2-2*cos(120°)=39. Note that ( a, b,c) is solution) Iff ( - a,-b,-c) is solution.
Thank you! 💖 I think the original problem had a, b, and c on the right hand side and I replaced them with numbers to make it easier. The original version would be quite complicated (If I remember correctly, the solution took like 4 or 5 pages). Maybe one day I can do it for fun in a long video (possibly in a live session)
Several of your answers are WRONG. Here are the correct answers: [x = 11/sqrt(7), y = -19/sqrt(7), z = 1/sqrt(7)], [x = -11/sqrt(7), y = 19/sqrt(7), z = -1/sqrt(7)] [x = 2, y = 5, z = 3], [x = -2, y = -5, z = -3]
Really interesting problem!! I solved for all real solutions ALERT: I screwed up. I checked the solutions I gave and none works except ±(2, 5, 3). I don't know where's the mistake. I will let the comment here. Let me know if you find the mistake. I got, like you, y = 3z - 2x (*) But instead of substituing this in the first equation I solved for z in terms of x using equation (3). That is, I solved z^2 + (x)z + (x^2 - 19) = 0 Then I got z = ( - x ± sqrt( 76 - 3x^2 ) )/2 From this I conclude that | x | ≤ sqrt( 76/3 ) Substituing z in (*) gives us y in terms of x So, all real solution have the form | x | ≤ sqrt( 76/3 ) z = ( - x ± sqrt( 76 - 3x^2 ) )/2 y = 3z - 2x It is easy to get all integer solution because sqrt( 76/3 ) is small Actually, sqrt( 76/3 ) ~ 5.033 So x belongs to {0, ±1, ±2, ±3, ±4, ±5} I checked all these (it is not complicated) And I got that the integers solutions are (2, 5, 3) (2, -19, -5) (3, 0, 2) (3, --21, -5) (5, -16, -2) (5, --19, -3) And also the negatives of all these 6 solution (viewing them as vectors) Maybe there is a linear algebra aproach using the fact that the three equations of the system are obtained using the quadratic form induced by the matrix ( 1 1) ( 0 1) But I don't know how to use this fact. Thanks for the video !!
minus the equation then factor it will get (y+z+x)(z-x)=10 (y-x)(x+y+z)=30 (y-z)(x+y+z)=20 thus 6z-6x=y-x=3y-3z this is another way to find the relationship of x y z
There is a big mistake. 1 + 11 + 121 = 133 not 132 (15:50) :D
It was good!
Oh, man! And 133 is divisible by 19! I should've checked that!
@@SyberMath a computer is good for that! :D
This problem could be considered a hard one compared to most of the other problems. The solution method is not that obvious at first glance. Are there other methods to solve it? Any ideas?
Very Good Video .I dont know for all real but maybe this is right for positive
consider triangle abc with side length √19, √39, 7
by law of cosine, x=pa, y=pb, z=pb. and apb, apc, bpc are 120º degree
by area of triangle using sine we get xy+xz+yz=31. sum all of the equation and get (x+y+z)²=100, x+y+z=10 substrack each of 2 equation
(z-x)(x+y+z)=10, (y-z)(x+y+z)=20,
(z-x)=1, (y-z)=2
then x+x+1+x+3=10, x=2, z=3, y=5
maybe you can get negative answer by x+y+z=-10 or using negative length
@@webtoon1121 Very nice!!!
I solved it like this:
x^2 + y^2 + xy = (x+y)^2 - xy = 39
so we have a square - a number getting 39
the first square bigger than 39 is 7^2
it can be written as 5 + 2, which is a valid setting for x and y, but since we don't know neither x nor y from 5 and 2 we will move to the 2nd equation, using the same reformation of the equation and choosing y to be the solution 5, we can see that z is 3.
so we didn't even have to look at the 3rd equation.
Thanks.
When you get the equation "3x^2-35xz+22z^2 = 0"(11:17) , you can just decompose the equation like this"(3x-2z)(x-11z) = 0",so we can easily konw the solution of the equation: 3x = 2z OR x = 11z.
I really like it, I know English little, but I understanded you. Very interesting. Good luck. Sorry for my English, i will study.
Thank you! Don't worry about your English! 😃
Here is how I solved it, although I did end up using a calculator to get a discriminant at a step in the problem.
If you take the difference of the three equations you get:
(z-x)(x+y+z) = 10 (A)
(y-x)(x+y+z) = 30
(y-z)(x+y+z) = 20
note that if x+y+z = 0 you get a contradiction, so x+y+z is not 0. Then you divide the equations against each other and get (y-z)/(z-x) = 2, which gives y = 3z-2x. Also if x=z then the first and second equation would be the same, so you get a contradiction. Similar for the others, since all the equations have different values.
If you substitute your equation for y into the first equation you can simplify and get: x^2+3z^2-3zx = 13. You can subtract that equation from the third equation and simplify to get: z^2-2xz+3 = 0.
This is a quadratic that yields z = x /pm sqrt(x^2-3), given this is over the reals x^2 /ge 3. Using our y equation we get that y = x /pm 3 sqrt(x^2-3)
Conveniently, z-x = /pm sqrt(x^2-3) so if we substitute that into (A) you get what is essentially a quadratic in x^2. I kept careful track of the pluses and minuses, because it will be relevant in the final step, but for this part you don't care. Furthermore around this time I realized x^2 = 4 is a solution, but it'll be seen anyways if you just continue on. If I let t = x^2, then:
9(t^2 - 3t) = (22-4t)^2
You can rearrange and solve the quadratic, but I did use a calculator for the discriminant (not saying it was necessary). t = x^2 = 4, 11^2/7, so x = /pm 2, /pm 11/sqrt(7)
So now you just need to acquire the triples (since both sides of the pluses and minuses do not work), from the last quadratic before I squared everything I kept that info and you just confirm that right options. (2,5,3), (-2,-5,-3),(-11/sqrt(7),19/sqrt(7), -17/sqrt(7)),(11/sqrt(7),-19/sqrt(7),1/sqrt(7))
At some point in the problem you could have argued that if (x,y,z) is a triple then (-x,-y,-z) is one as well since that is straightforward from the given information.
Very good! Thank you!
It's a bit hard to see because of a small arthimetic mistake, but all the solutions are:
x = ±2, y = ±5, z = ±3 and
x = ± 11/√7 , y = ± 19/√7 , z = ± 1/√7
Yes. Sorry
@@SyberMath
Good video 👍
y=(-/+)19/√7 not y=(+/-)19/7.
This can be done without too much mathematical rigour and some magic assumptions.
(x+y)^2=39+xy
(y+z)^2=49+yz
(x+z)^2=19+xz
Then assume x, y, z as positive integers
The next integer square after 39 is 49 so
xy=10
For 49 is 64 so
yz=15
For 19 is 25 so
zx=6
Clearly x=2, y=5 and z=3
Sorry for cheating but I foud it cool.
(
Looks good! 😁
how did you find it? :((
Without any mathematical rigour. I supposed that x,y and z were integers and that the squares on the left side were the next you can find from the first number on the right side.
Luckily these two assumptilns were right.😉
Brilliant. How did you think about this method? It's genius.
Will be solution [11/srt(7),-19/sqrt(7),1/sqrt(7)] and [-11/sqrt(7),19/sqrt(7),-1/sqrt(7)]?
To get rid of complicated square roots, we can factor 3x^2 -35xz +22z^2 =0 (x-11z)(3x-2z)=0
Good!
Dear friend: A mistake is made in solving the equation for z2=19/132. It should be z2=19/133=1/7
Here is another easier/alternate way to reach out a solution to this problem. We assume that x, y and z are positive integers and there exists a solution to it. If we deduct one equation from the other we have three distinct equations as follows: (x+y+z)(y-z) = 20 .... (1), (x+y+z)(z-x) = 10 .... (2) and (x+y+z)(y-x) = 30 .... (3). And here lies the logic. (x+y+z)>>x ; (x+y+z)>>y ; also (x+y+z)>>z . And x+y+z is a common factor to the all the above deduced three equations. The right side values of the equations are 20, 10 and 30 and each value is the the product of two factors as per our equations. Therefore, x+y+z will be the HCF (Highest Common Factor) of 20, 10 and 30. In this case it is 10. In other words, x+y+z=10 ...(4). So that from the above three equations, we have y-z=2 ... (5), z-x=1 ... (6) and y-x=3 ... (7). Let us now solve for x, y and z. Consider (any) for z value. From eqn (6) x=z-1 and from (5) y=z+2. Putting these values in eqn (4), we have (z-1) + (z+2) + z = 10; from which we get z =3. Putting this value in equation(4) we get x+y=7 ...(8) Using eqn (7) and eqn (8), we get x = 2 and y=5. Therefore, the solutions sets are (x,y,z) =(2,5,3). This is a particular solution to the problem. For general solution, SyberMath is preferred.
That urge to ffwd to the end to see if the solutions are integers
computer finds four solutions, nice:
*Solve[{x^2 + x y + y^2 == 39, y^2 + y z + z^2 == 49, z^2 + z x + x^2 == 19}, {x, y, z}]*
The usual time tested method of addition, subtraction works very well. Silution 2,5,3
Assuming x,y,z are integer
x=2, y=5, z=3 are solutions.
For positive solution. Consider triangle with sides sqrt(39),7, sqrt(19), because x^2+x*y+y^2= x^2-2*cos(120°)=39. Note that ( a, b,c) is solution) Iff ( - a,-b,-c) is solution.
There are more solutions if you let x=y, y=z, or z=x.
For instance, x=y -> x = ±√13, y = ±√13, z = (±√37-√13)/2
GREAT SIR, KEEP EXPLAINING THESE TYPE OF TEDIOUS QUESTIONS
Thank you, I will
Very good thanks for this vidéo
Thank you too!
From equation 1&2 We get
z is greater than x
( z+ x)^2=19+ zx
(z+ x)^2=25
zx=6
z=3, x=2.
y=5
Another ingenious solution!
The big question is who and how invented this problem.
Thank you! 💖
I think the original problem had a, b, and c on the right hand side and I replaced them with numbers to make it easier. The original version would be quite complicated (If I remember correctly, the solution took like 4 or 5 pages). Maybe one day I can do it for fun in a long video (possibly in a live session)
Several of your answers are WRONG. Here are the correct answers:
[x = 11/sqrt(7), y = -19/sqrt(7), z = 1/sqrt(7)],
[x = -11/sqrt(7), y = 19/sqrt(7), z = -1/sqrt(7)]
[x = 2, y = 5, z = 3],
[x = -2, y = -5, z = -3]
Awesome methode !!!!!!
Glad you think so!
that's y i did not plug y into the last equation
You are Great !! , Very helpful to me ~~
Glad to hear that! 💖
I think z2 = 19/133 instead of 19/132
Why?
It was a mistake, it should be 19/133
133 Z×Z=19,So, Z=Root7
Hey great video!
I have a question tho
Where are you from?
You have and interesting accent
TR
@@SyberMath yes a good accent
11:30 That is genius! Why don't they teach us that at school? :(
I have no idea! School curriculum is unfortunately very dry and repetitive. Teachers are often ill-prepared and receive very little support.
I understood the whole thing. Followed every step. Yup.
Nice!
good job
Thank you!
Really interesting problem!!
I solved for all real solutions
ALERT: I screwed up. I checked the solutions I gave and none works except ±(2, 5, 3). I don't know where's the mistake. I will let the comment here. Let me know if you find the mistake.
I got, like you, y = 3z - 2x (*)
But instead of substituing this in the first equation I solved for z in terms of x using equation (3).
That is, I solved z^2 + (x)z + (x^2 - 19) = 0
Then I got z = ( - x ± sqrt( 76 - 3x^2 ) )/2
From this I conclude that | x | ≤ sqrt( 76/3 )
Substituing z in (*) gives us y in terms of x
So, all real solution have the form
| x | ≤ sqrt( 76/3 )
z = ( - x ± sqrt( 76 - 3x^2 ) )/2
y = 3z - 2x
It is easy to get all integer solution because sqrt( 76/3 ) is small
Actually, sqrt( 76/3 ) ~ 5.033
So x belongs to {0, ±1, ±2, ±3, ±4, ±5}
I checked all these (it is not complicated)
And I got that the integers solutions are
(2, 5, 3)
(2, -19, -5)
(3, 0, 2)
(3, --21, -5)
(5, -16, -2)
(5, --19, -3)
And also the negatives of all these 6 solution (viewing them as vectors)
Maybe there is a linear algebra aproach using the fact that the three equations of the system are obtained using the quadratic form induced by the matrix
( 1 1)
( 0 1)
But I don't know how to use this fact.
Thanks for the video !!
Nice! Thanks
& 19/133=1/7
Dude, you’re freaking amazing! Thanks a bunch!
Thank you! and you're welcome! 😊
Good job
Thanks!
@@SyberMath wlcm
It's 133 and not 132
I got x=2, y=5 and z=3. I solved the values of x,y, and z fast. I was right.
How?
I did it in my head.
A little mistake, but very interesting problem. Thanks.
Yes! Thank you!
good verry good
Thanks
nice
Thanks
Great
Thank you!
minus the equation then factor it will get
(y+z+x)(z-x)=10 (y-x)(x+y+z)=30 (y-z)(x+y+z)=20
thus 6z-6x=y-x=3y-3z
this is another way to find the
relationship of x y z
Absolutely! Good thinking!
9:15 there is a easier method to solve that system
What is it?
@@SyberMath hydrocloric acid, I assume ;)
Anyone tried using a matrix? And then using Gaussian elimination?
The equations are not linear
I can solve this problem.
Cool! What is your method?