HMMT 十一月 2013 · 冲刺赛 · 第 33 题
HMMT November 2013 — Guts Round — Problem 33
题目详情
- [ 17 ] On each side of a 6 by 8 rectangle, construct an equilateral triangle with that side as one edge such that the interior of the triangle intersects the interior of the rectangle. What is the total area of all regions that are contained in exactly 3 of the 4 equilateral triangles? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . HMMT NOVEMBER 2013, 9 NOVEMBER 2013 — GUTS ROUND
解析
- [ 17 ] On each side of a 6 by 8 rectangle, construct an equilateral triangle with that side as one edge such that the interior of the triangle intersects the interior of the rectangle. What is the total area of all regions that are contained in exactly 3 of the 4 equilateral triangles? √ √ √ 96 3 − 154 288 − 154 3 154 154 3 √ √ Answer: OR OR 96 − OR 96 − Let the rectangle be ABCD with 3 3 3 3 AB = 8 and BC = 6. Let the four equilateral triangles be ABP , BCP , CDP , and DAP (for 1 2 3 4 convenience, call them the P -, P -, P -, P - triangles). Let W = AP ∩ DP , X = AP ∩ DP , and 1 2 3 4 1 3 1 4 Y = DP ∩ CP . Reflect X, Y over the line P P (the line halfway between AB and DC ) to points 4 2 2 4 ′ ′ X , Y . √ 3 First we analyze the basic configuration of the diagram. Since AB = 8 < 2 · 6 , the P -, P - triangles 2 4 2 √ intersect. Furthermore, AP ⊥ BP , so if T = BP ∩ AP , then BP = 6 < 4 3 = BT . Therefore P 1 2 2 1 2 2 lies inside triangle P BA , and by symmetry, also triangle P DC . 1 3 It follows that the area we wish to compute is the union of two (congruent) concave hexagons, one of ′ ′ ′ which is W XY P Y X . (The other is its reflection over Y Y , the mid-line of AD and BC .) So we seek 2 ′ ′ ′ ′ 2[ W XY P Y X ] = 2([ W XP X ] − [ P Y P Y ]) . 2 4 2 4 √ √ 2 1 1 6 3 ′ ′ It’s easy to see that [ W XP X ] = [ ADP ] = = 3 3, since W XP X and its reflections over 4 4 4 3 3 4 ′ lines DW X and AW X partition 4 ADP . 4 ′ ′ It remains to consider P Y P Y , a rhombus with (perpendicular) diagonals P P and Y Y . If O denotes 2 4 2 4 √ √ 3 1 the intersection of these two diagonals (also the center of ABCD ), then OP is P B − AB = 3 3 − 4, 2 2 2 2 the difference between the lengths of the P -altitude in 4 CBP and the distance between the parallel 2 2 OP ′ 2 √ lines Y Y , CB . Easy angle chasing gives OY = , so 3 √ √ OP · OY 2 2 86 − 48 3 2 ′ 2 2 [ P Y P Y ] = 4 · = √ OP = √ (3 3 − 4) = √ , 2 4 2 2 3 3 3 and our desired area is √ √ √ 172 − 96 3 96 3 − 154 ′ ′ √ √ 2[ W XP X ] − 2[ P Y P Y ] = 6 3 − = , 4 2 4 3 3 √ 288 − 154 3 or . 3