[Guest]

The following polynomial is constructed of 3 variables (A,B,C): 41A²+18B²+27C²+16AB+18AC-12BC

Make a change of variables which will transform this polynomial into one which has no cross product terms. That is, find functions F, G, and H of X, Y, and Z such that when A is replaced by F(X,Y,Z), B is replaced by G(X,Y,Z), and C is replaced by H(X,Y,Z) in the above polynomial, it is transformed into a polynomial of the form:

X²+Y²+Z²

Factor it into the form of:

(x₁A+y₁B+z₁C)²+j(x₂A+y₂B+z₂C)²+k(x₃A+y₃B+z₃C)²

There is only one way to factor it. All coefficients are integers...but some may be zero

[Guest]

the F, G, H equations are kind of messy and i would accept the factored form of the above expression as a good enough solution to this puzzle

Edit: will post solution in a day or two if needed

Edited July 16, 2010 by klose

[bonanova]

Let {A, B, C} be orthogonal vectors in 3-space.

Perform a 3-dimensional rotation to obtain orthogonal vectors {X, Y, Z}.

Solve for the three rotation angles by setting each term of 16AB+18AC-12BC separately to zero.

Or, wait two days for the answer to be posted.

[Guest]

x₁²+x₂²+x₃²=41

y₁²+y₂²+y₃²=18

z₁²+z₂²+z₃²=27

x₁y₁+x₂y₂+x₃y₃=8

x₁z₁+x₂z₂+x₃z₃=9

y₁z₁+y₂z₂+y₃z₃=-6

start with the squares

[superprismatic]

F(X,Y,Z)=(Z-Y)/8

G(X,Y,Z)=(Z+7y-8X)/40

H(X,Y,Z)=(9Z+23Y+8X)/120

[Guest]

F(X,Y,Z)=(Z-Y)/8

G(X,Y,Z)=(Z+7y-8X)/40

H(X,Y,Z)=(9Z+23Y+8X)/120

im pretty sure this doesnt work. none of the coefficients are 0 so all equations have to be dependent on X, Y, and Z.

(A+6B+2C)²+(4A+B+C)²+(-3A+3B-3C)²

what was yours?

[superprismatic]

im pretty sure this doesnt work. none of the coefficients are 0 so all equations have to be dependent on X, Y, and Z.

(A+6B+2C)²+(4A+B+C)²+(-3A+3B-3C)²

what was yours?

(-A-4B+3C)²+(-2A+B+3C)²+(6A+B+3C)²

I think there are 48 distinct solutions.

[superprismatic]

Hey, your factored form gives

26A²+46B²+14C²+2AB+30AC+8BC

when expanded. It solves a different problem than

the one in the OP!

[Guest]

Hey, your factored form gives

26A²+46B²+14C²+2AB+30AC+8BC

when expanded. It solves a different problem than

the one in the OP!

darn! i accidently put the coefficients in the wrong spot. I put x1, x2, x3 into x1, y1, z1 instead of their rightful place.

it should be: (A+4B-3C)²+(6A+B+3C)²+(2A+B-3C)²

i attribute my failures to being a noob at puzzle creation.

[superprismatic]

darn! i accidently put the coefficients in the wrong spot. I put x1, x2, x3 into x1, y1, z1 instead of their rightful place.

it should be: (A+4B-3C)²+(6A+B+3C)²+(2A+B-3C)²

i attribute my failures to being a noob at puzzle creation.

That doesn't work either -- it expands to

41A^2+18B^2+27C^2+24AB+18AC+24BC; the coefficients

of AB and BC are wrong.

Here is a list of all 48 possible solutions. There are

really only 8 because order amongst the 3 terms is

irrelevant:

-1-4 3 -2 1 3 6 1 3 means (-A-4B+3C)^2+(-2A+B+3C)^2+(6A+B+3C)^2
1 4-3 -2 1 3 6 1 3
-1-4 3 2-1-3 6 1 3
1 4-3 2-1-3 6 1 3
-1-4 3 -2 1 3 -6-1-3
1 4-3 -2 1 3 -6-1-3
-1-4 3 2-1-3 -6-1-3
1 4-3 2-1-3 -6-1-3
-1-4 3 6 1 3 -2 1 3
1 4-3 6 1 3 -2 1 3
-1-4 3 -6-1-3 -2 1 3
1 4-3 -6-1-3 -2 1 3
-1-4 3 6 1 3 2-1-3
1 4-3 6 1 3 2-1-3
-1-4 3 -6-1-3 2-1-3
1 4-3 -6-1-3 2-1-3
-2 1 3 -1-4 3 6 1 3
2-1-3 -1-4 3 6 1 3
-2 1 3 1 4-3 6 1 3
2-1-3 1 4-3 6 1 3
-2 1 3 -1-4 3 -6-1-3
2-1-3 -1-4 3 -6-1-3
-2 1 3 1 4-3 -6-1-3
2-1-3 1 4-3 -6-1-3
-2 1 3 6 1 3 -1-4 3
2-1-3 6 1 3 -1-4 3
-2 1 3 -6-1-3 -1-4 3
2-1-3 -6-1-3 -1-4 3
-2 1 3 6 1 3 1 4-3
2-1-3 6 1 3 1 4-3
-2 1 3 -6-1-3 1 4-3
2-1-3 -6-1-3 1 4-3
6 1 3 -1-4 3 -2 1 3
-6-1-3 -1-4 3 -2 1 3
6 1 3 1 4-3 -2 1 3
-6-1-3 1 4-3 -2 1 3
6 1 3 -1-4 3 2-1-3
-6-1-3 -1-4 3 2-1-3
6 1 3 1 4-3 2-1-3
-6-1-3 1 4-3 2-1-3
6 1 3 -2 1 3 -1-4 3
-6-1-3 -2 1 3 -1-4 3
6 1 3 2-1-3 -1-4 3
-6-1-3 2-1-3 -1-4 3
6 1 3 -2 1 3 1 4-3
-6-1-3 -2 1 3 1 4-3
6 1 3 2-1-3 1 4-3
-6-1-3 2-1-3 1 4-3

Edited July 18, 2010 by superprismatic