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HMMT 二月 2013 · 几何 · 第 3 题

HMMT February 2013 — Geometry — Problem 3

专题
Discrete Math / 离散数学
难度
L3
来源
HMMT

题目详情

  1. Let A A A A A A be a convex hexagon such that A A ‖ A A for i = 1 , 2 , 3 (we take 1 2 3 4 5 6 i i +2 i +3 i +5 A = A for each i ). Segment A A intersects segment A A at B , for 1 ≤ i ≤ 6, as i +6 i i i +2 i +1 i +3 i ∼ shown. Furthermore, suppose that △ A A A △ A A A . Given that [ A B B ] = 1, [ A B B ] = = 1 3 5 4 6 2 1 5 6 2 6 1 4, and [ A B B ] = 9 (by [ XY Z ] we mean the area of △ XY Z ), determine the area of hexagon 3 1 2 B B B B B B . 1 2 3 4 5 6 A (not to scale) 1 1 A A 2 6 B B 4 6 5 B 1 B ? 4 9 B B 2 3 A A 3 5 A 4
解析
  1. Let A A A A A A be a convex hexagon such that A A ‖ A A for i = 1 , 2 , 3 (we take 1 2 3 4 5 6 i i +2 i +3 i +5 A = A for each i ). Segment A A intersects segment A A at B , for 1 ≤ i ≤ 6, as i +6 i i i +2 i +1 i +3 i ∼ shown. Furthermore, suppose that △ A A A △ A A A . Given that [ A B B ] = 1, [ A B B ] = = 1 3 5 4 6 2 1 5 6 2 6 1 4, and [ A B B ] = 9 (by [ XY Z ] we mean the area of △ XY Z ), determine the area of hexagon 3 1 2 B B B B B B . 1 2 3 4 5 6 A (not to scale) 1 1 A A 2 6 B B 4 6 5 B 1 B ? 4 9 B B 2 3 A A 3 5 A 4 Answer: 22 Because B A B A and B A B A are parallelograms, B A = A B and A B = 6 3 3 6 1 4 4 1 6 3 6 3 1 1 A B . By the congruence of the large triangles A A A and A A A , A A = A A . Thus, B A + 4 4 1 3 5 2 4 6 1 3 4 6 6 3 A B − A A = A B + A B − A A , so B B = B B . Similarly, opposite sides of hexagon 1 1 1 3 6 3 4 4 4 6 6 1 3 4 B B B B B B are equal, and implying that the triangles opposite each other on the outside of this 1 2 3 4 5 6 hexagon are congruent. Furthermore, by definition B B ‖ A A , B B ‖ A A , B B ‖ A A and B B ‖ A A . Let the area 5 6 3 5 3 4 1 3 6 1 4 6 1 2 1 5 2 of triangle A A A and triangle A A A be k . Then, by similar triangles, 1 3 5 2 4 6 Geometry Test √ 1 A B 1 6 = 2 k A A 1 3 √ 4 B B B B 6 1 1 6 = = 2 k A A A A 4 6 1 3 √ 9 A B 3 1 = 2 k A A 1 3 2 Summing yields 6 /k = 1, so k = 36. To finish, the area of B B B B B B is equivalent to the area 1 2 3 4 5 6 of the triangle A A A minus the areas of the smaller triangles provided in the hypothesis. Thus, our 1 3 5 answer is 36 − 1 − 4 − 9 = 22.