## 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.

|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

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

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

Copyright © 1996-2017 Alexander Bogomolny61198387 |