返回题库

HMMT 二月 2006 · GEN1 赛 · 第 8 题

HMMT February 2006 — GEN1 Round — Problem 8

专题
Discrete Math / 离散数学
难度
L3
来源
HMMT

题目详情

  1. A dot is marked at each vertex of a triangle ABC . Then, 2, 3, and 7 more dots are marked on the sides AB , BC , and CA , respectively. How many triangles have their vertices at these dots?
解析
  1. A dot is marked at each vertex of a triangle ABC . Then, 2, 3, and 7 more dots are marked on the sides AB , BC , and CA , respectively. How many triangles have their vertices at these dots? Answer: 357 ( ) 15 Solution: Altogether there are 3 + 2 + 3 + 7 = 15 dots, and thus = 455 3 ( ) ( ) ( ) 2+2 2+3 2+7 combinations of 3 dots. Of these combinations, + + = 4 + 10 + 84 = 98 3 3 3 do not give triangles because they are collinear (the rest do give triangles). Thus 455 − 98 = 357 different triangles can be formed.