RMO Training: a question from Nordic Mathematical Contest: 1988


Two spheres with the same centre have radii r and R, where r<R. From the surface of the bigger sphere we will try to select three points A, B and C such that all sides of the triangle ABC coincide with the surface of the smaller sphere. Prove that this selection is possible if and only if R<2r.


Assume A, B and C lie on the surface \Gamma of a sphere of a radius R and centre O, and AB, BC, and CA touch the surface \gamma of a sphere of radius r and centre O. The circumscribed and inscribed circles of ABC then are the intersections of the plane ABC with \Gamma and \gamma, respectively. The centres of these circles both are the foot D of the perpendicular dropped from O to the plane ABC. This point lies both on the angle bisectors of the triangle ABC and on the perpendicular bisectors of its sides. So, these lines are the same, which means that the triangle ABC is equilateral, and the centre of the circles is the common point of intersection of the medians of ABC. This again implies that the radii of the two circles are 2r_{1} and r_{1} for some real number r_{1}. Let OD=d. Then, 2r_{1}=\sqrt{R^{2}-d^{2}} and r_{1}=\sqrt{r^{2}-d^{2}}. Squaring, we get R^{2}-d^{2}=4r^{2}-4d^{2}, 4r^{2}-R^{2}=3d^{2} \geq 0, and 2r\geq R.

On the other hand, assume 2r \geq R. Consider a plane at the distance d=\sqrt{\frac{4r^{2}-R^{2}}{3}} from the common centre of the two spheres. The plane cuts the surfaces of the spheres along concentric circles of radii

r_{1}=\sqrt{\frac{R^{2}-r^{2}}{3}} and R_{1}=2\sqrt{\frac{R^{2}-r^{2}}{3}}.

The points A, B, and C can now be chosen on the latter circle in such a way that ABC is equilateral.

Reference: Nordic Mathematical Contest, 1987-2009.


Nalin Pithwa.

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