HMMT 二月 2008 · 几何 · 第 8 题

英文解析

8. [ 6 ] Let ABC be an equilateral triangle with side length 2, and let Γ be a circle with radius centered at the center of the equilateral triangle. Determine the length of the shortest path that starts somewhere 2
   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 reducedfractions.
   √
   Answer: − 1 Suppose that the path visits sides AB, BC, CA in this order. Construct points 28
   ′ ′ ′ ′ ′ ′ ′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
   When the path hits AB , let us reflect the rest of the path across AB and follow this reflected path.A
   ′ ′
   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
   √
   √
   ( )
   1 282
   ′
   √2
   to be 3 + = . Therefore, the length of the shortest path connecting Γ to Γ has length
   √33 − 1. By reflecting this path three times back into ABC , we get a path that satisfies our conditions.28 3