Track / Stage 6 / 125 of 400 #1125 of 1964

Problem 1125

National olympiad, first round

Geometry Difficulty 6.1 Prove it

272. Two spheres $\alpha$ and $\beta$ touch the sphere $\omega$ at points $A$ and $B$ . On the sphere $\alpha$ , a point $M$ is taken, the line $M A$ intersects the sphere $\omega$ again at point $N$ , the line $N B$ intersects the sphere $\beta$ again at point $K$ . Find the geometric locus of such points $M$ , for which the line $M K$ is tangent to the sphere $\beta$ .

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

272. First, let us prove that if the line $M K$ is tangent to the sphere $\beta$ , then it is also tangent to the sphere $\alpha$ . Consider the section of the given spheres by the plane passing through the points $M, K, A, B$ , and $N$ (Fig. 55). $\widehat{M K B}$ is measured by half the arc $\overline{K B}$ enclosed within this angle, hence, $\bar{M} \widehat{K B}=\widehat{B A N}$ , since the angular measurements of the arcs $\overline{K B}$ and $\overline{B N}$ are equal (the arcs are taken on opposite sides of the line $K N$ if the tangency is external (Fig. 55, a), and on the same side if the tangency is internal (Fig. 55, b)). From this, it follows that $\widehat{A M K}=\widehat{A B N}$ or $\widehat{A M K}=180^{\circ}-\widehat{A B N}$ , which means that $\widehat{A M K}$ is measured by half of $A M$ , since the corresponding arcs $\overparen{A M}$ and $\widehat{A N}$ have equal angular measurements, i.e., $M K$ is tangent to the circle where the given section intersects the sphere $\alpha$ .

Now we can prove that the geometric locus of points $M$ is a circle.

Fig. 55.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.