PUMaC 2008 · 几何（B 组） · 第 4 题

PUMaC 2008 — Geometry (Division B) — Problem 4

(3 points) A cube is divided into 27 unit cubes. A sphere is inscribed in each of the corner unit cubes, and another sphere is placed tangent to these 8 spheres. What is the smallest possible value for the radius of the last sphere?

解析

Draw a 12-sided regular polygon. If the vertices going clockwise are A, B, C, D, E, F, etc, draw a line between A and F, B and G, C and H, etc. This will form a smaller 12-sided regular polygon in the center of the larger one. What is the area of the smaller one divided by the are of the larger one? √ ( ANS: 7 − 4 3. The diameter of a circle inscribed inside the inner dodecagon is equal to the side length of the outer dodecahedron. This can be seen from the parallel lines that are drawn from any two adjacent outer vertices that define opposite sides of the inner dodecagon. √ 2 The area of the inner dodecagon with diameter of an inscribed circle x is: 3(2 − 3) x The area √ 2 of the outer dodecagon with side length x is: 3(2 + 3) x √ √ √ 2 − 3 2 √ Thus the ratio of their areas is: = (2 − 3) = 7 − 4 3 2+ 3 The above formulas can be found using the right triangle with one angle of 15 degrees with hypotenuse length R that makes up 1/24th of the dodecagon. (1/2) x can be plugged in for a 2 corresponding side and used in the formula Area(dodecagon) = 3 R CB: AP)