Systems of Equations degree 2
$begingroup$
Given the system
$$begin{aligned}
x^2+y^2+frac{sqrt 3}{2}xy&=32\
x^2+z^2+frac{1}{2}xz&=16\
y^2+z^2&=16,end{aligned}$$
find the value of $(xy+sqrt{3}xz+2yz).$
The answer is $32,$ so I think there will be a really nice solution to this.
I tried completing the square for each pair $x$ and $y$ but I could not find anything, I tried adding all three equations but I got nothing.
contest-math systems-of-equations quadratics
$endgroup$
|
show 2 more comments
$begingroup$
Given the system
$$begin{aligned}
x^2+y^2+frac{sqrt 3}{2}xy&=32\
x^2+z^2+frac{1}{2}xz&=16\
y^2+z^2&=16,end{aligned}$$
find the value of $(xy+sqrt{3}xz+2yz).$
The answer is $32,$ so I think there will be a really nice solution to this.
I tried completing the square for each pair $x$ and $y$ but I could not find anything, I tried adding all three equations but I got nothing.
contest-math systems-of-equations quadratics
$endgroup$
1
$begingroup$
Are you sure that you have made no typo?
$endgroup$
– Dr. Sonnhard Graubner
Oct 31 '18 at 10:12
1
$begingroup$
Eliminating $y,z$ i have found for $x$: $$9,{x}^{8}-735,{x}^{6}+21328,{x}^{4}-253952,{x}^{2}+1048576=0$$
$endgroup$
– Dr. Sonnhard Graubner
Oct 31 '18 at 10:16
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Im sorry but there is no typo that i am sure of, ive been palying around with the figures but i still cant find it. this was supposedly solve in 120 seconds.
$endgroup$
– SuperMage1
Oct 31 '18 at 10:55
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Where does this problem come from?
$endgroup$
– Yuriy S
Oct 31 '18 at 12:10
$begingroup$
@SuperMage1, see the edit to my answer
$endgroup$
– Yuriy S
Oct 31 '18 at 12:18
|
show 2 more comments
$begingroup$
Given the system
$$begin{aligned}
x^2+y^2+frac{sqrt 3}{2}xy&=32\
x^2+z^2+frac{1}{2}xz&=16\
y^2+z^2&=16,end{aligned}$$
find the value of $(xy+sqrt{3}xz+2yz).$
The answer is $32,$ so I think there will be a really nice solution to this.
I tried completing the square for each pair $x$ and $y$ but I could not find anything, I tried adding all three equations but I got nothing.
contest-math systems-of-equations quadratics
$endgroup$
Given the system
$$begin{aligned}
x^2+y^2+frac{sqrt 3}{2}xy&=32\
x^2+z^2+frac{1}{2}xz&=16\
y^2+z^2&=16,end{aligned}$$
find the value of $(xy+sqrt{3}xz+2yz).$
The answer is $32,$ so I think there will be a really nice solution to this.
I tried completing the square for each pair $x$ and $y$ but I could not find anything, I tried adding all three equations but I got nothing.
contest-math systems-of-equations quadratics
contest-math systems-of-equations quadratics
edited Dec 9 '18 at 13:03
user10354138
7,4322925
7,4322925
asked Oct 31 '18 at 9:44
SuperMage1SuperMage1
897210
897210
1
$begingroup$
Are you sure that you have made no typo?
$endgroup$
– Dr. Sonnhard Graubner
Oct 31 '18 at 10:12
1
$begingroup$
Eliminating $y,z$ i have found for $x$: $$9,{x}^{8}-735,{x}^{6}+21328,{x}^{4}-253952,{x}^{2}+1048576=0$$
$endgroup$
– Dr. Sonnhard Graubner
Oct 31 '18 at 10:16
$begingroup$
Im sorry but there is no typo that i am sure of, ive been palying around with the figures but i still cant find it. this was supposedly solve in 120 seconds.
$endgroup$
– SuperMage1
Oct 31 '18 at 10:55
$begingroup$
Where does this problem come from?
$endgroup$
– Yuriy S
Oct 31 '18 at 12:10
$begingroup$
@SuperMage1, see the edit to my answer
$endgroup$
– Yuriy S
Oct 31 '18 at 12:18
|
show 2 more comments
1
$begingroup$
Are you sure that you have made no typo?
$endgroup$
– Dr. Sonnhard Graubner
Oct 31 '18 at 10:12
1
$begingroup$
Eliminating $y,z$ i have found for $x$: $$9,{x}^{8}-735,{x}^{6}+21328,{x}^{4}-253952,{x}^{2}+1048576=0$$
$endgroup$
– Dr. Sonnhard Graubner
Oct 31 '18 at 10:16
$begingroup$
Im sorry but there is no typo that i am sure of, ive been palying around with the figures but i still cant find it. this was supposedly solve in 120 seconds.
$endgroup$
– SuperMage1
Oct 31 '18 at 10:55
$begingroup$
Where does this problem come from?
$endgroup$
– Yuriy S
Oct 31 '18 at 12:10
$begingroup$
@SuperMage1, see the edit to my answer
$endgroup$
– Yuriy S
Oct 31 '18 at 12:18
1
1
$begingroup$
Are you sure that you have made no typo?
$endgroup$
– Dr. Sonnhard Graubner
Oct 31 '18 at 10:12
$begingroup$
Are you sure that you have made no typo?
$endgroup$
– Dr. Sonnhard Graubner
Oct 31 '18 at 10:12
1
1
$begingroup$
Eliminating $y,z$ i have found for $x$: $$9,{x}^{8}-735,{x}^{6}+21328,{x}^{4}-253952,{x}^{2}+1048576=0$$
$endgroup$
– Dr. Sonnhard Graubner
Oct 31 '18 at 10:16
$begingroup$
Eliminating $y,z$ i have found for $x$: $$9,{x}^{8}-735,{x}^{6}+21328,{x}^{4}-253952,{x}^{2}+1048576=0$$
$endgroup$
– Dr. Sonnhard Graubner
Oct 31 '18 at 10:16
$begingroup$
Im sorry but there is no typo that i am sure of, ive been palying around with the figures but i still cant find it. this was supposedly solve in 120 seconds.
$endgroup$
– SuperMage1
Oct 31 '18 at 10:55
$begingroup$
Im sorry but there is no typo that i am sure of, ive been palying around with the figures but i still cant find it. this was supposedly solve in 120 seconds.
$endgroup$
– SuperMage1
Oct 31 '18 at 10:55
$begingroup$
Where does this problem come from?
$endgroup$
– Yuriy S
Oct 31 '18 at 12:10
$begingroup$
Where does this problem come from?
$endgroup$
– Yuriy S
Oct 31 '18 at 12:10
$begingroup$
@SuperMage1, see the edit to my answer
$endgroup$
– Yuriy S
Oct 31 '18 at 12:18
$begingroup$
@SuperMage1, see the edit to my answer
$endgroup$
– Yuriy S
Oct 31 '18 at 12:18
|
show 2 more comments
2 Answers
2
active
oldest
votes
$begingroup$
I guess the two factors (1/2) shouldn't be there, which mean we have instead
begin{align}
x^2+y^2+sqrt{3}xy&=32tag{A}\
x^2+z^2+xz&=16tag{B}\
y^2+z^2&=16tag{C}.
end{align}
The rationale being that it gives a much nicer set of coefficients when completing the square
begin{align}
(x+frac{sqrt3}2y)^2+frac14y^2&=32\
(x+frac12z)^2+frac34z^2&=16\
end{align}
and we have accidental cancellation when eliminating $y,z$, giving the biquartic $37x^4-896x^2+4096=0$, and hence a nice form of the four real solutions
$$
(x,y,z)=left(8 sqrt{frac{7-2sqrt{3}}{37}},
4(3 + sqrt3)sqrt{frac{7-2sqrt3}{111}},
4sqrt{frac{21 - 6sqrt3}{37}}right),text{ and similar}
$$
and it is easy to check $(xy+sqrt{3}xz+2yz)^2=2^{10}$. Indeed, it follows immediately from noting the LHS of
$$
mathrm{(A)^2+(B)^2+(C)^2-2times(A)times(B)-2times(B)times(C)-2times(C)times(A)}
$$
is $-(xy+sqrt{3}xz+2yz)^2$, which I suppose is the only way to solve this within the 120 seconds limit.
$endgroup$
add a comment |
$begingroup$
Possible hint:
Let's introduce spherical coordinates:
$$x=r cos a cos b \ y=r sin a cos b \ z= r sin b$$
Then we have:
$$r^2 cos^2 b left(1+frac{sqrt{3}}{2}sin a cos a right)=32 \ r^2 left(cos^2 a cos^2 b+sin^2 b+frac{1}{2}cos a sin b cos b right)=16 \ r^2 (sin^2 a cos^2 b+sin^2 b)=16$$
And we need to find:
$$r^2 left(sin a cos a cos^2 b+(sqrt{3} cos a +2 sin a)cos b sin b right)$$
Let's rewrite the last equation as: $$r^2 cos^2 b (sin^2 a-1)=16-r^2 \ cos^2 a cos^2 b=frac{r^2-16}{r^2}$$
So we need to have $r^2 >16$ for a non-trivial and real solution.
The trivial solution $r^2=16$ doesn't work, because then we have either $cos a=0$ or $cos b=0$, which, after substitution, don't satisfy the whole system of equations.
Edit
I have used Newton's method to solve the system numerically. With the initial guess $(1,2,3)$, I get:
$$x=3.12392804494377 \ y=3.55350700146376 \ z=1.83646072393286$$
And:
$$xy+sqrt{3} xz+2yz=34.0893777894208 neq 32$$
With another initial guess $(-1,2,3)$, I get:
$$x=-3.70737480877305 \ y=-2.95892185457043 \ z=2.69161317030243$$
And:
$$xy+sqrt{3} xz+2yz=-22.2425349920001 neq 32$$
I think those are the only real solutions, because for the polynomial of Dr. Sonnhard Graubner $$9,{x}^{8}-735,{x}^{6}+21328,{x}^{4}-253952,{x}^{2}+1048576=0$$
we have only $4$ real roots:
$$x_{1,2}= pm 3.12392804494377 \ x_{3,4}=pm 3.70737480877305$$
Here's the R code I used (you would need the matrixcalc package to find the inverse Jacobian matrix):
library(matrixcalc);
f1 <- function(x,y,z){y^2+z^2-16};
f2 <- function(x,y,z){x^2+z^2+x*z/2-16};
f3 <- function(x,y,z){x^2+y^2+sqrt(3)/2*x*y-32};
J <- function(x,y,z){matrix(c(0, 2*y, 2*z, 2*x+z/2, 0, 2*z+x/2, 2*x+sqrt(3)*y/2, 2*y+sqrt(3)*x/2, 0), nrow=3, byrow=TRUE)}
Nm <- 18;
r0 <- c(1,2,3);
n <- 0;
while(n < Nm){n <- n+1;
r0 <- r0 - matrix.inverse(J(r0[1],r0[2],r0[3])) %*% c(f1(r0[1],r0[2],r0[3]),f2(r0[1],r0[2],r0[3]),f3(r0[1],r0[2],r0[3]))};
x <- r0[1];
y <- r0[2];
z <- r0[3];
f <- function(x,y,z){x*y+sqrt(3)*x*z+2*y*z};
paste(r0)
paste(f(x,y,z))
f1(x,y,z)
f2(x,y,z)
f3(x,y,z)
$endgroup$
add a comment |
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2 Answers
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active
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2 Answers
2
active
oldest
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active
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active
oldest
votes
$begingroup$
I guess the two factors (1/2) shouldn't be there, which mean we have instead
begin{align}
x^2+y^2+sqrt{3}xy&=32tag{A}\
x^2+z^2+xz&=16tag{B}\
y^2+z^2&=16tag{C}.
end{align}
The rationale being that it gives a much nicer set of coefficients when completing the square
begin{align}
(x+frac{sqrt3}2y)^2+frac14y^2&=32\
(x+frac12z)^2+frac34z^2&=16\
end{align}
and we have accidental cancellation when eliminating $y,z$, giving the biquartic $37x^4-896x^2+4096=0$, and hence a nice form of the four real solutions
$$
(x,y,z)=left(8 sqrt{frac{7-2sqrt{3}}{37}},
4(3 + sqrt3)sqrt{frac{7-2sqrt3}{111}},
4sqrt{frac{21 - 6sqrt3}{37}}right),text{ and similar}
$$
and it is easy to check $(xy+sqrt{3}xz+2yz)^2=2^{10}$. Indeed, it follows immediately from noting the LHS of
$$
mathrm{(A)^2+(B)^2+(C)^2-2times(A)times(B)-2times(B)times(C)-2times(C)times(A)}
$$
is $-(xy+sqrt{3}xz+2yz)^2$, which I suppose is the only way to solve this within the 120 seconds limit.
$endgroup$
add a comment |
$begingroup$
I guess the two factors (1/2) shouldn't be there, which mean we have instead
begin{align}
x^2+y^2+sqrt{3}xy&=32tag{A}\
x^2+z^2+xz&=16tag{B}\
y^2+z^2&=16tag{C}.
end{align}
The rationale being that it gives a much nicer set of coefficients when completing the square
begin{align}
(x+frac{sqrt3}2y)^2+frac14y^2&=32\
(x+frac12z)^2+frac34z^2&=16\
end{align}
and we have accidental cancellation when eliminating $y,z$, giving the biquartic $37x^4-896x^2+4096=0$, and hence a nice form of the four real solutions
$$
(x,y,z)=left(8 sqrt{frac{7-2sqrt{3}}{37}},
4(3 + sqrt3)sqrt{frac{7-2sqrt3}{111}},
4sqrt{frac{21 - 6sqrt3}{37}}right),text{ and similar}
$$
and it is easy to check $(xy+sqrt{3}xz+2yz)^2=2^{10}$. Indeed, it follows immediately from noting the LHS of
$$
mathrm{(A)^2+(B)^2+(C)^2-2times(A)times(B)-2times(B)times(C)-2times(C)times(A)}
$$
is $-(xy+sqrt{3}xz+2yz)^2$, which I suppose is the only way to solve this within the 120 seconds limit.
$endgroup$
add a comment |
$begingroup$
I guess the two factors (1/2) shouldn't be there, which mean we have instead
begin{align}
x^2+y^2+sqrt{3}xy&=32tag{A}\
x^2+z^2+xz&=16tag{B}\
y^2+z^2&=16tag{C}.
end{align}
The rationale being that it gives a much nicer set of coefficients when completing the square
begin{align}
(x+frac{sqrt3}2y)^2+frac14y^2&=32\
(x+frac12z)^2+frac34z^2&=16\
end{align}
and we have accidental cancellation when eliminating $y,z$, giving the biquartic $37x^4-896x^2+4096=0$, and hence a nice form of the four real solutions
$$
(x,y,z)=left(8 sqrt{frac{7-2sqrt{3}}{37}},
4(3 + sqrt3)sqrt{frac{7-2sqrt3}{111}},
4sqrt{frac{21 - 6sqrt3}{37}}right),text{ and similar}
$$
and it is easy to check $(xy+sqrt{3}xz+2yz)^2=2^{10}$. Indeed, it follows immediately from noting the LHS of
$$
mathrm{(A)^2+(B)^2+(C)^2-2times(A)times(B)-2times(B)times(C)-2times(C)times(A)}
$$
is $-(xy+sqrt{3}xz+2yz)^2$, which I suppose is the only way to solve this within the 120 seconds limit.
$endgroup$
I guess the two factors (1/2) shouldn't be there, which mean we have instead
begin{align}
x^2+y^2+sqrt{3}xy&=32tag{A}\
x^2+z^2+xz&=16tag{B}\
y^2+z^2&=16tag{C}.
end{align}
The rationale being that it gives a much nicer set of coefficients when completing the square
begin{align}
(x+frac{sqrt3}2y)^2+frac14y^2&=32\
(x+frac12z)^2+frac34z^2&=16\
end{align}
and we have accidental cancellation when eliminating $y,z$, giving the biquartic $37x^4-896x^2+4096=0$, and hence a nice form of the four real solutions
$$
(x,y,z)=left(8 sqrt{frac{7-2sqrt{3}}{37}},
4(3 + sqrt3)sqrt{frac{7-2sqrt3}{111}},
4sqrt{frac{21 - 6sqrt3}{37}}right),text{ and similar}
$$
and it is easy to check $(xy+sqrt{3}xz+2yz)^2=2^{10}$. Indeed, it follows immediately from noting the LHS of
$$
mathrm{(A)^2+(B)^2+(C)^2-2times(A)times(B)-2times(B)times(C)-2times(C)times(A)}
$$
is $-(xy+sqrt{3}xz+2yz)^2$, which I suppose is the only way to solve this within the 120 seconds limit.
edited Dec 9 '18 at 13:06
answered Dec 7 '18 at 10:19
user10354138user10354138
7,4322925
7,4322925
add a comment |
add a comment |
$begingroup$
Possible hint:
Let's introduce spherical coordinates:
$$x=r cos a cos b \ y=r sin a cos b \ z= r sin b$$
Then we have:
$$r^2 cos^2 b left(1+frac{sqrt{3}}{2}sin a cos a right)=32 \ r^2 left(cos^2 a cos^2 b+sin^2 b+frac{1}{2}cos a sin b cos b right)=16 \ r^2 (sin^2 a cos^2 b+sin^2 b)=16$$
And we need to find:
$$r^2 left(sin a cos a cos^2 b+(sqrt{3} cos a +2 sin a)cos b sin b right)$$
Let's rewrite the last equation as: $$r^2 cos^2 b (sin^2 a-1)=16-r^2 \ cos^2 a cos^2 b=frac{r^2-16}{r^2}$$
So we need to have $r^2 >16$ for a non-trivial and real solution.
The trivial solution $r^2=16$ doesn't work, because then we have either $cos a=0$ or $cos b=0$, which, after substitution, don't satisfy the whole system of equations.
Edit
I have used Newton's method to solve the system numerically. With the initial guess $(1,2,3)$, I get:
$$x=3.12392804494377 \ y=3.55350700146376 \ z=1.83646072393286$$
And:
$$xy+sqrt{3} xz+2yz=34.0893777894208 neq 32$$
With another initial guess $(-1,2,3)$, I get:
$$x=-3.70737480877305 \ y=-2.95892185457043 \ z=2.69161317030243$$
And:
$$xy+sqrt{3} xz+2yz=-22.2425349920001 neq 32$$
I think those are the only real solutions, because for the polynomial of Dr. Sonnhard Graubner $$9,{x}^{8}-735,{x}^{6}+21328,{x}^{4}-253952,{x}^{2}+1048576=0$$
we have only $4$ real roots:
$$x_{1,2}= pm 3.12392804494377 \ x_{3,4}=pm 3.70737480877305$$
Here's the R code I used (you would need the matrixcalc package to find the inverse Jacobian matrix):
library(matrixcalc);
f1 <- function(x,y,z){y^2+z^2-16};
f2 <- function(x,y,z){x^2+z^2+x*z/2-16};
f3 <- function(x,y,z){x^2+y^2+sqrt(3)/2*x*y-32};
J <- function(x,y,z){matrix(c(0, 2*y, 2*z, 2*x+z/2, 0, 2*z+x/2, 2*x+sqrt(3)*y/2, 2*y+sqrt(3)*x/2, 0), nrow=3, byrow=TRUE)}
Nm <- 18;
r0 <- c(1,2,3);
n <- 0;
while(n < Nm){n <- n+1;
r0 <- r0 - matrix.inverse(J(r0[1],r0[2],r0[3])) %*% c(f1(r0[1],r0[2],r0[3]),f2(r0[1],r0[2],r0[3]),f3(r0[1],r0[2],r0[3]))};
x <- r0[1];
y <- r0[2];
z <- r0[3];
f <- function(x,y,z){x*y+sqrt(3)*x*z+2*y*z};
paste(r0)
paste(f(x,y,z))
f1(x,y,z)
f2(x,y,z)
f3(x,y,z)
$endgroup$
add a comment |
$begingroup$
Possible hint:
Let's introduce spherical coordinates:
$$x=r cos a cos b \ y=r sin a cos b \ z= r sin b$$
Then we have:
$$r^2 cos^2 b left(1+frac{sqrt{3}}{2}sin a cos a right)=32 \ r^2 left(cos^2 a cos^2 b+sin^2 b+frac{1}{2}cos a sin b cos b right)=16 \ r^2 (sin^2 a cos^2 b+sin^2 b)=16$$
And we need to find:
$$r^2 left(sin a cos a cos^2 b+(sqrt{3} cos a +2 sin a)cos b sin b right)$$
Let's rewrite the last equation as: $$r^2 cos^2 b (sin^2 a-1)=16-r^2 \ cos^2 a cos^2 b=frac{r^2-16}{r^2}$$
So we need to have $r^2 >16$ for a non-trivial and real solution.
The trivial solution $r^2=16$ doesn't work, because then we have either $cos a=0$ or $cos b=0$, which, after substitution, don't satisfy the whole system of equations.
Edit
I have used Newton's method to solve the system numerically. With the initial guess $(1,2,3)$, I get:
$$x=3.12392804494377 \ y=3.55350700146376 \ z=1.83646072393286$$
And:
$$xy+sqrt{3} xz+2yz=34.0893777894208 neq 32$$
With another initial guess $(-1,2,3)$, I get:
$$x=-3.70737480877305 \ y=-2.95892185457043 \ z=2.69161317030243$$
And:
$$xy+sqrt{3} xz+2yz=-22.2425349920001 neq 32$$
I think those are the only real solutions, because for the polynomial of Dr. Sonnhard Graubner $$9,{x}^{8}-735,{x}^{6}+21328,{x}^{4}-253952,{x}^{2}+1048576=0$$
we have only $4$ real roots:
$$x_{1,2}= pm 3.12392804494377 \ x_{3,4}=pm 3.70737480877305$$
Here's the R code I used (you would need the matrixcalc package to find the inverse Jacobian matrix):
library(matrixcalc);
f1 <- function(x,y,z){y^2+z^2-16};
f2 <- function(x,y,z){x^2+z^2+x*z/2-16};
f3 <- function(x,y,z){x^2+y^2+sqrt(3)/2*x*y-32};
J <- function(x,y,z){matrix(c(0, 2*y, 2*z, 2*x+z/2, 0, 2*z+x/2, 2*x+sqrt(3)*y/2, 2*y+sqrt(3)*x/2, 0), nrow=3, byrow=TRUE)}
Nm <- 18;
r0 <- c(1,2,3);
n <- 0;
while(n < Nm){n <- n+1;
r0 <- r0 - matrix.inverse(J(r0[1],r0[2],r0[3])) %*% c(f1(r0[1],r0[2],r0[3]),f2(r0[1],r0[2],r0[3]),f3(r0[1],r0[2],r0[3]))};
x <- r0[1];
y <- r0[2];
z <- r0[3];
f <- function(x,y,z){x*y+sqrt(3)*x*z+2*y*z};
paste(r0)
paste(f(x,y,z))
f1(x,y,z)
f2(x,y,z)
f3(x,y,z)
$endgroup$
add a comment |
$begingroup$
Possible hint:
Let's introduce spherical coordinates:
$$x=r cos a cos b \ y=r sin a cos b \ z= r sin b$$
Then we have:
$$r^2 cos^2 b left(1+frac{sqrt{3}}{2}sin a cos a right)=32 \ r^2 left(cos^2 a cos^2 b+sin^2 b+frac{1}{2}cos a sin b cos b right)=16 \ r^2 (sin^2 a cos^2 b+sin^2 b)=16$$
And we need to find:
$$r^2 left(sin a cos a cos^2 b+(sqrt{3} cos a +2 sin a)cos b sin b right)$$
Let's rewrite the last equation as: $$r^2 cos^2 b (sin^2 a-1)=16-r^2 \ cos^2 a cos^2 b=frac{r^2-16}{r^2}$$
So we need to have $r^2 >16$ for a non-trivial and real solution.
The trivial solution $r^2=16$ doesn't work, because then we have either $cos a=0$ or $cos b=0$, which, after substitution, don't satisfy the whole system of equations.
Edit
I have used Newton's method to solve the system numerically. With the initial guess $(1,2,3)$, I get:
$$x=3.12392804494377 \ y=3.55350700146376 \ z=1.83646072393286$$
And:
$$xy+sqrt{3} xz+2yz=34.0893777894208 neq 32$$
With another initial guess $(-1,2,3)$, I get:
$$x=-3.70737480877305 \ y=-2.95892185457043 \ z=2.69161317030243$$
And:
$$xy+sqrt{3} xz+2yz=-22.2425349920001 neq 32$$
I think those are the only real solutions, because for the polynomial of Dr. Sonnhard Graubner $$9,{x}^{8}-735,{x}^{6}+21328,{x}^{4}-253952,{x}^{2}+1048576=0$$
we have only $4$ real roots:
$$x_{1,2}= pm 3.12392804494377 \ x_{3,4}=pm 3.70737480877305$$
Here's the R code I used (you would need the matrixcalc package to find the inverse Jacobian matrix):
library(matrixcalc);
f1 <- function(x,y,z){y^2+z^2-16};
f2 <- function(x,y,z){x^2+z^2+x*z/2-16};
f3 <- function(x,y,z){x^2+y^2+sqrt(3)/2*x*y-32};
J <- function(x,y,z){matrix(c(0, 2*y, 2*z, 2*x+z/2, 0, 2*z+x/2, 2*x+sqrt(3)*y/2, 2*y+sqrt(3)*x/2, 0), nrow=3, byrow=TRUE)}
Nm <- 18;
r0 <- c(1,2,3);
n <- 0;
while(n < Nm){n <- n+1;
r0 <- r0 - matrix.inverse(J(r0[1],r0[2],r0[3])) %*% c(f1(r0[1],r0[2],r0[3]),f2(r0[1],r0[2],r0[3]),f3(r0[1],r0[2],r0[3]))};
x <- r0[1];
y <- r0[2];
z <- r0[3];
f <- function(x,y,z){x*y+sqrt(3)*x*z+2*y*z};
paste(r0)
paste(f(x,y,z))
f1(x,y,z)
f2(x,y,z)
f3(x,y,z)
$endgroup$
Possible hint:
Let's introduce spherical coordinates:
$$x=r cos a cos b \ y=r sin a cos b \ z= r sin b$$
Then we have:
$$r^2 cos^2 b left(1+frac{sqrt{3}}{2}sin a cos a right)=32 \ r^2 left(cos^2 a cos^2 b+sin^2 b+frac{1}{2}cos a sin b cos b right)=16 \ r^2 (sin^2 a cos^2 b+sin^2 b)=16$$
And we need to find:
$$r^2 left(sin a cos a cos^2 b+(sqrt{3} cos a +2 sin a)cos b sin b right)$$
Let's rewrite the last equation as: $$r^2 cos^2 b (sin^2 a-1)=16-r^2 \ cos^2 a cos^2 b=frac{r^2-16}{r^2}$$
So we need to have $r^2 >16$ for a non-trivial and real solution.
The trivial solution $r^2=16$ doesn't work, because then we have either $cos a=0$ or $cos b=0$, which, after substitution, don't satisfy the whole system of equations.
Edit
I have used Newton's method to solve the system numerically. With the initial guess $(1,2,3)$, I get:
$$x=3.12392804494377 \ y=3.55350700146376 \ z=1.83646072393286$$
And:
$$xy+sqrt{3} xz+2yz=34.0893777894208 neq 32$$
With another initial guess $(-1,2,3)$, I get:
$$x=-3.70737480877305 \ y=-2.95892185457043 \ z=2.69161317030243$$
And:
$$xy+sqrt{3} xz+2yz=-22.2425349920001 neq 32$$
I think those are the only real solutions, because for the polynomial of Dr. Sonnhard Graubner $$9,{x}^{8}-735,{x}^{6}+21328,{x}^{4}-253952,{x}^{2}+1048576=0$$
we have only $4$ real roots:
$$x_{1,2}= pm 3.12392804494377 \ x_{3,4}=pm 3.70737480877305$$
Here's the R code I used (you would need the matrixcalc package to find the inverse Jacobian matrix):
library(matrixcalc);
f1 <- function(x,y,z){y^2+z^2-16};
f2 <- function(x,y,z){x^2+z^2+x*z/2-16};
f3 <- function(x,y,z){x^2+y^2+sqrt(3)/2*x*y-32};
J <- function(x,y,z){matrix(c(0, 2*y, 2*z, 2*x+z/2, 0, 2*z+x/2, 2*x+sqrt(3)*y/2, 2*y+sqrt(3)*x/2, 0), nrow=3, byrow=TRUE)}
Nm <- 18;
r0 <- c(1,2,3);
n <- 0;
while(n < Nm){n <- n+1;
r0 <- r0 - matrix.inverse(J(r0[1],r0[2],r0[3])) %*% c(f1(r0[1],r0[2],r0[3]),f2(r0[1],r0[2],r0[3]),f3(r0[1],r0[2],r0[3]))};
x <- r0[1];
y <- r0[2];
z <- r0[3];
f <- function(x,y,z){x*y+sqrt(3)*x*z+2*y*z};
paste(r0)
paste(f(x,y,z))
f1(x,y,z)
f2(x,y,z)
f3(x,y,z)
edited Oct 31 '18 at 13:05
answered Oct 31 '18 at 10:23
Yuriy SYuriy S
15.8k433118
15.8k433118
add a comment |
add a comment |
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1
$begingroup$
Are you sure that you have made no typo?
$endgroup$
– Dr. Sonnhard Graubner
Oct 31 '18 at 10:12
1
$begingroup$
Eliminating $y,z$ i have found for $x$: $$9,{x}^{8}-735,{x}^{6}+21328,{x}^{4}-253952,{x}^{2}+1048576=0$$
$endgroup$
– Dr. Sonnhard Graubner
Oct 31 '18 at 10:16
$begingroup$
Im sorry but there is no typo that i am sure of, ive been palying around with the figures but i still cant find it. this was supposedly solve in 120 seconds.
$endgroup$
– SuperMage1
Oct 31 '18 at 10:55
$begingroup$
Where does this problem come from?
$endgroup$
– Yuriy S
Oct 31 '18 at 12:10
$begingroup$
@SuperMage1, see the edit to my answer
$endgroup$
– Yuriy S
Oct 31 '18 at 12:18