HMMT 二月 2011 · CALCGEO 赛 · 第 1 题
HMMT February 2011 — CALCGEO Round — Problem 1
题目详情
- Let ABC be a triangle such that AB = 7, and let the angle bisector of ∠ BAC intersect line BC at D . If there exist points E and F on sides AC and BC , respectively, such that lines AD and EF are parallel and divide triangle ABC into three parts of equal area, determine the number of possible integral values for BC .
解析
- Determine the smallest value of x for which every distortion of H is necessarily convex. Answer: 4 ′ A A X X 1 6 A A 2 5 ′ A A Y 3 Y 4 ′ Let H = A A A A A A be the hexagon, and for all 1 ≤ i ≤ 6, let points A be considered such that 1 2 3 4 5 6 i ′ ′ ′ ′ ′ ′ ′ ′ A A < 1. Let H = A A A A A A , and consider all indices modulo 6. For any point P in the plane, i i 1 2 3 4 5 6 ′ let D ( P ) denote the unit disk { Q | P Q < 1 } centered at P ; it follows that A ∈ D ( A ). i i ′ ′ Let X and X be points on line A A , and let Y and Y be points on line A A such that A X = 1 6 3 4 1 ′ ′ ′ ′ A X = A Y = A Y = 1 and X and X lie on opposite sides of A and Y and Y lie on opposite sides 1 3 3 1 ′ ′ ′ ′ of A . If X and Y lie on segments A A and A A , respectively, then segment A A lies between the 3 1 6 3 4 1 3 ′ ′ x lines XY and X Y . Note that is the distance from A to A A . 2 1 3 2 Calculus & Geometry Individual Test ′ A A X 1 X 6 A A 2 5 ′ Y A A 3 Y 4 x If ≥ 2, then C ( A ) cannot intersect line XY , since the distance from XY to A A is 1 and the 2 1 3 2 ′ ′ ′ distance from XY to A is at least 1. Therefore, A A separates A from the other 3 vertices of the 2 1 3 2 ′ hexagon. By analogous reasoning applied to the other vertices, we may conclude that H is convex. x ′ ′ If < 2, then C ( A ) intersects XY , so by choosing A = X and A = Y , we see that we may choose 2 1 3 2 ′ ′ A on the opposite side of XY , in which case H will be concave. Hence the answer is 4, as desired. 2