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HMMT 二月 2009 · 几何 · 第 6 题

HMMT February 2009 — Geometry — Problem 6

专题
Discrete Math / 离散数学
难度
L3
来源
HMMT

题目详情

  1. [ 5 ] Let ABC be a triangle in the coordinate plane with vertices on lattice points and with AB = 1. Suppose the perimeter of ABC is less than 17. Find the largest possible value of 1 /r , where r is the inradius of ABC .
解析
  1. [ 4 ] Let ABC be a triangle in the coordinate plane with vertices on lattice points and with AB = 1. Suppose the perimeter of ABC is less than 17. Find the largest possible value of 1 /r , where r is the inradius of ABC . √ √ Answer: 1 + 5 2 + 65 Solution: Let a denote the area of the triangle, r the inradius, and p the perimeter. Then a = rp/ 2, so r = 2 a/p > 2 a/ 17. Notice that a = h/ 2 where h is the height of the triangle from C to AB , and h is an integer since the vertices are lattice points. Thus we first guess that the inradius is minimized when h = 1 and the area is 1 / 2. In this case, we can now assume WLOG that A = (0 , 0), B = (1 , 0), and √ √ 2 2 C = ( n + 1 , 1) for some nonnegative integer n . The perimeter of ABC is n + 2 n + 2 + n + 1 + 1. Since n = 8 yields a perimeter greater than 17, the required triangle has n = 7 and inradius r = 1 /p = √ √ 1 √ √ which yields the answer of 1 /r = 1 + 5 2 + 65. We can now verify that this is indeed 1+5 2+ 65 2 minimal over all h by noting that its perimeter is greater than 17 / 2, which is the upper bound in the case h ≥ 2.