HMMT 二月 2008 · 几何 · 第 8 题
HMMT February 2008 — Geometry — Problem 8
题目详情
- [ 6 ] Let ABC be an equilateral triangle with side length 2, and let Γ be a circle with radius centered at 2 the center of the equilateral triangle. Determine the length of the shortest path that starts somewhere on Γ, visits all three sides of ABC , and ends somewhere on Γ (not necessarily at the starting point). √ Express your answer in the form of p − q , where p and q are rational numbers written as reduced fractions.
解析
- [ 6 ] Let ABC be an equilateral triangle with side length 2, and let Γ be a circle with radius centered at 2 the center of the equilateral triangle. Determine the length of the shortest path that starts somewhere on Γ, visits all three sides of ABC , and ends somewhere on Γ (not necessarily at the starting point). √ Express your answer in the form of p − q , where p and q are rational numbers written as reduced fractions. √ 28 Answer: − 1 Suppose that the path visits sides AB, BC, CA in this order. Construct points 3 ′ ′ ′ ′ ′ ′ ′ A , B , C so that C is the reflection of C across AB , A is the reflection of A across BC , and B is 1 ′ ′ ′ the reflection of B across A C . Finally, let Γ be the circle with radius centered at the center of 2 ′ ′ ′ ′ ′ ′ ′ A B C . Note that Γ is the image of Γ after the three reflections: AB, BC , C A . 3 ′ ′ B C A ′ B C A When the path hits AB , let us reflect the rest of the path across AB and follow this reflected path. ′ ′ When we hit BC , let us reflect the rest of the path across BC , and follow the new path. And when ′ ′ ′ ′ we hit A C , reflect the rest of the path across A C and follow the new path. We must eventually end ′ up at Γ . ′ It is easy to see that the shortest path connecting some point on Γ to some point on Γ lies on the line connecting the centers of the two circles. We can easily find the distance between the two centers √ √ ( ) 2 1 28 ′ 2 √ to be 3 + = . Therefore, the length of the shortest path connecting Γ to Γ has length 3 3 √ 28 − 1. By reflecting this path three times back into ABC , we get a path that satisfies our conditions. 3