olymmath

RU-ALLRUSSIAN 1976 Problem 15

    S is a sphere with unit radius. P is a plane through the center. For any point x on the sphere f(x)\mathrm{f}(\mathbf{x}) is the perpendicular distance from x\mathbf{x} to P. Show that if x\mathbf{x} , y\mathbf{y} , z\mathbf{z} are the ends of three mutually perpendicular radii, then f(x)2+f(y)2+f(z)2=1\mathrm{f}(\mathbf{x})^2 + \mathrm{f}(\mathbf{y})^2 + \mathrm{f}(\mathbf{z})^2 = 1 (\). Now let g(x)\mathrm{g}(\mathbf{x}) be any function on the points of S taking non- negative real values and satisfying (\). Regard the intersection of P and S as the equator, the poles as the points with f(x)=1\mathrm{f}(\mathbf{x}) = 1 and lines of longitude as semicircles through both poles. (1) If x\mathbf{x} and y\mathbf{y} have the same longitude and both lie on the same side of the equator with x\mathbf{x} closer to the pole, show that g(x)>g(y)\mathrm{g}(\mathbf{x}) > \mathrm{g}(\mathbf{y}) . (2) Show that for any x\mathbf{x} , y\mathbf{y} on the same side of the equator with x\mathbf{x} closer to the pole than y\mathbf{y} we have g(x)>g(y)\mathrm{g}(\mathbf{x}) > \mathrm{g}(\mathbf{y}) . (3) Show that if x\mathbf{x} and y\mathbf{y} are the same distance from the pole then g(x)=g(y)\mathrm{g}(\mathbf{x}) = \mathrm{g}(\mathbf{y}) . (4) Show that g(x)=f(x)\mathrm{g}(\mathbf{x}) = \mathrm{f}(\mathbf{x}) for all x\mathbf{x} .

    Dataset text NuminaMath olympiads. Quoted with source.

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