HMMT 二月 2003 · 冲刺赛 · 第 31 题

HMMT February 2003 — Guts Round — Problem 31

[10] A cylinder of base radius 1 is cut into two equal parts along a plane passing through the center of the cylinder and tangent to the two base circles. Suppose that each piece’s surface area is m times its volume. Find the greatest lower bound for all possible values of m as the height of the cylinder varies. 2 2 2

解析

A cylinder of base radius 1 is cut into two equal parts along a plane passing through the center of the cylinder and tangent to the two base circles. Suppose that each piece’s surface area is m times its volume. Find the greatest lower bound for all possible values of m as the height of the cylinder varies. Solution: 3 Let h be the height of the cylinder. Then the volume of each piece is half the volume 1 of the cylinder, so it is πh . The base of the piece has area π , and the ellipse formed 2 √ 2 h by the cut has area π · 1 · 1 + because its area is the product of the semiaxes times 4 π . The rest of the area of the piece is half the lateral area of the cylinder, so it is πh . Thus, the value of m is √ √ 2 2 π + π 1 + h / 4 + πh 2 + 2 h + 4 + h = πh/ 2 h √ 2 4 = + 2 + + 1 , 2 h h a decreasing function of h whose limit as h → ∞ is 3. Therefore the greatest lower bound of m is 3. 2 2 2