HMMT 二月 2013 · 几何 · 第 3 题

英文解析

3. 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, asi +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
   (not to scale)A
   A A11
   2 6
   B B
   6 54
   1B
   ?B
   B B94
   2 3
   A A
   3 5
   4A
   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
   of triangle A A A and triangle A A A be k . Then, by similar triangles,2
   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 = =
   k A A A A2
   4 6 1 3
   √
   9 A B
   3 1
   2=
   k A A
   1 3
   Summing yields 6 /k = 1, so k = 36. To finish, the area of B B B B B B is equivalent to the area 2
   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.