olymmath

Putnam 1940 B4

  • Session B

Prove that the locus of the point of intersection of three mutually perpendicular planes tangent to the surface ax2+by2+cz2=1    (where  abc≠0)ax^2 + by^2 +cz^2 =1\;\; (\text{where}\; abc \ne 0) is the sphere x2+y2+z2=1a+1b+1c.x^2 +y^2 +z^2 =\frac{1}{a}+\frac{1}{b}+\frac{1}{c}.

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