HMMT 二月 2012 · 几何 · 第 7 题

英文解析

7. Let S be the set of the points ( x , x , . . . , x ) in 2012-dimensional space such that | x | + | x | + · · · +
   1 2 2012 1 2
   2012
   | x | ≤ 1. Let T be the set of points in 2012-dimensional space such that max | x | = 2. Let p be a
   2012 ii =1
   randomly chosen point on T . What is the probability that the closest point in S to p is a vertex of S ?
   Answer: Note that T is a hypercube in 2012-dimensional space, containing the rotated 1
   2011
   hyperoctahedron S . Let v be a particular vertex of S , and we will consider the set of points x on T2
   such that v is the closest point to x in S . Let w be another point of S and let ℓ be the line betweenv and w . Then in order for v to be the closest point to x in S , it must also be so on the region of ℓcontained in S . This condition is then equivalent to the projection of x lying past v on the line ℓ , oralternatively that v lies in the opposite halfspace of w defined by the hyperplane perpendicular to ℓand passing through v . This can be written algebraically as ( x − v ) · ( w − v ) ≤ 0. Therefore, v is theclosest point to x if and only if ( x − v ) · ( w − v ) ≤ 0 for all w in S .
   Note that these conditions do not depend on where w is on the line, so for each line intersecting Snontrivially, let us choose w such that w lies on the hyperplane H containing all the vertices of Sexcept for v and − v . We can further see that the conditions are linear in w , so them holding for all w
   Geometry Testin H is equivalent to them holding on the vertices of the region S ∩ H , which are simply the verticesof S except for v and − v . Let us now compute what these conditions look like.
   Without loss of generality, let v = (1 , 0 , . . . , 0) and w = (0 , 1 , 0 , . . . , 0). Then the equation is of theform ( x − v ) · ( − 1 , 1 , 0 , . . . , 0) ≤ 0, which we can rewrite as ( x − 1 , x , x , . . . , x 012) · ( − 1 , 1 , 0 , . . . , 0) =
   1 2 3 2
   1 − x + x ≤ 0. For the other choices of w , we get the similar conditions that 1 − x + x ≤ 0 and
   1 2 1 ialso 1 − x − x ≤ 0 for each i ∈ { 2 , . . . , 2012 } . Note that if any x ∈ { 2 , − 2 } for i 6 = 1, then one of
   1 i ithese conditions trivially fails, as it would require 3 − x ≤ 0. Therefore, the only face of T where xcan lie is the face defined by x = 2, which gives us the conditions that − 1 + x ≤ 0 and − 1 − x ≤ 0,1
   1 i iso x ∈ [ − 1 , 1] for all i ∈ { 2 , . . . , 2012 } . This defines a 2011-dimensional hypercube of side length 2 onthe face of T defined by x = 2, and we obtain similar regions on each of the other faces correspondingito the other vertices of S . Therefore, the volume of the set of x for which x is closest to a vertex is 1
   2011 2011
   2 · 2012 · 2 and the volume of all the choices of x is 2 · 2012 · 4 , so the desired probability is
   .1
   2011 2