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.

Solution

|Contact| |Front page| |Contents| |Geometry| |Store|

Copyright © 1996-2017 Alexander Bogomolny

Solution 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 (a, b, c) and (A, B, C) are proportional:

 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

  1. G. Polya, J. Kilpatrick, The Stanford Mathematics Problem Book, Dover, 2009
  2. G. Polya, Mathematical Discovery, John Wiley & Sons, 1981

|Contact| |Front page| |Contents| |Store| |Geometry|

Copyright © 1996-2017 Alexander Bogomolny

 62054133

Search by google: