PUMaC 2010 · 几何（A 组） · 第 8 题

PUMaC 2010 — Geometry (Division A) — Problem 8

There is a point source of light in an empty universe. What is the minimum number of solid balls (of any size) one must place in space so that any light ray emanating from the light source intersects at least one ball? 2

解析

There is a point light source in an empty universe. What is the minimal number of solid balls (of any size) that one must place in the universe so that any light ray emanating from the light source intersects at least one ball? [Answer] 4 [Solution] It is easy to see that 3 is not enough. Since suppose we regard the plane passing through light source and parallel to the triangle formed by centers of three balls as equatorial plane, then all high-lattitude light rays cannot be blocked. To show 4 is enough, consider a regular tetrahedron ABCD with light source O at center. There exist four infinite cones containing tetrahedra OABC, OABD, OACD, OBCD. Any inscribed ball in a cone will block all light rays in that cone. Choose four non-intersecting balls inscribed in the four cones will give us the solution. The non-intersecting condition can be met since the four cones are infinite, and we can stipulate the radii of 4 balls to be drastically different. To be precise, for an inscribed ball with radius R , there exists α and β such that the ball lies entirely in the interior of the spherical shell centered at origin with inner radius αR and outer − 1 2 − 2 3 − 3 radius βR . Then we choose the fall balls with radius R, αβ R, α β R, α β R . This way, the four balls will not intersect since they lie in four different spherical shell regions. 4