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.
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Copyright © 1996-2018 Alexander BogomolnySolution 1
[Stanford Problem Book, #54.3, Mathematical Discovery, #2.66]
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 |
or
abcABC = 0. |
Yet we derive from (1) that
abcABC = 1. |
Consequently, the hypothetical identity is impossible.
Solution 2
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.
Reference
- G. Polya, J. Kilpatrick, The Stanford Mathematics Problem Book, Dover, 2009
- G. Polya, Mathematical Discovery, John Wiley & Sons, 1981
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