May the Product of Planes Be a Sphere?
Show that it is impossible to find (real or complex) numbers a, b, c, A, B, and C such that the equation
|x² + y² + z² = (ax + by + cz)(Ax + By + Cz)|
holds identically for independent variables x, y, z.
Let's expand the right-hand side of the hypothetical identity:
|aAx² + bBy² + cCz² + (bC + cB)yz + (cA + aC)xz + (aB + bA)xy.|
Now equating corresponding (i.e., by the same powers of the variables) coefficients on both sides of the identity, we obtain
|(1)||aA = bB = cC = 1|
|(2)||bC + cB = cA + aC = aB + bA = 0|
We derive from (2) that
|bC = -cB, cA =-aC, aB = -bA,|
and multiplying these three equations, we derive further that
|abcABC = -abcABC|
|abcABC = 0.|
Yet we derive from (1) that
|abcABC = 1.|
Consequently, the hypothetical identity is impossible.
If the identity at hand holds for any values of the three variables x, y, z, it is bound to hold for x, y, z that satisfy the system
|ax + by + cz = r|
|Ax + By + Cz = R|
for any choice of r and R. The two equations determine a plane each, thus, unless the planes are parallel, the triples of the solutions to the system lie on a straight line. While the left-hand side of the identity satisfies
|x² + y² + z² = rR.|
Since this is true for any selection of r and R, we may as well choose both positive, making the above a sphere with the radius √rR. We arrive at a contradiction by observing that no sphere contains a straight line.
In case where the two planes are parallel, the vectors
|α(a, b, c) = (A, B, C)|
for some α. Choosing again positive r and R makes α positive and the left-hand side of the hypothetical identity represent a sphere. Now, since a sphere does not contain a plane, the result is contradictory in this case also.
- G. Polya, J. Kilpatrick, The Stanford Mathematics Problem Book, Dover, 2009
- G. Polya, Mathematical Discovery, John Wiley & Sons, 1981