PUMaC 2020 · 加试 · 第 5 题
PUMaC 2020 — Power Round — Problem 5
题目详情
Problem 5.3. Give an example showing the previous claim (i.e. find some d, m, n such that the segment from ( d, 0) to ( d + m, n ) cannot be produced from the ant-path using the above procedure). ′ Therefore, our goal is to find which segments from X = ( d, 0) to X = ( d + m, n ) correspond to actual closed ant-paths on the cube. There is one easy way to determine this for any d, m, n . Start by labeling the vertices of the unit square by a, b, c, d and by unrolling ′ the cube over the segment XX . When a vertex V of the cube falls onto the plane, label the ′ ′ ′ ′ ′ point by v ( V ∈ { A, B, C, D, A , B , C , D } ). If X ends up on the horizontal edge whose left label is a and whose right label is b (the left label corresponding to the point ( m, n ), and the right on to the point ( m + 1 , n )), then the cube have made a full rotation and we have an ant-path. ′ Note that the segment XX contains no points with integer coordinates. From now on, whenever we consider segments of this type, we will exclude the segments containing the integer points. In general, given some fixed m, n , and d ∈ [0 , 1], the labels of the points ( m, n ) and ( m + 1 , n ) could depend on d . However, we have the following claim that eliminates this possibility in case the points ( m, n ) and ( m + 1 , n ) were labeled a and b , as the following problem suggests.
解析
Problem 5.2. Prove that the point X will have the coordinates of the form ( d + m, n ), for some positive integers m, n ∈ Z . Further, prove that the length of the planar line
0 ′ segment XX is equal to the length of C . Proof 5.2. Since ant-paths of the cube cannot cross any vertex of the cube, the planar line segment corresponding to an ant-path must not touch any integer lattice points. This means that one can find a continuum of ant-paths that give the same sequence of unrolling moves; the corresponding line segments in the plane have the same slope but slightly different start points. This means we may assume d is irrational, since if it were rational we could simply consider nudging it to a nearby irrational number. Consider what would happen if we kept unrolling the ant-path forever, obtaining a line in the plane rather than a segment. The sequence of unrolling steps (rolling right or up) repeats, so clearly this line, and therefore the original line segment, must have rational slope. ′ X must have one coordinate an integer and be distance d from a lattice point, meaning it is of the form ( m + d, n ), ( m − d, n ), ( m, n + d ), or ( m, n − d ) for some nonnegative integers n m and n . But the slopes of the second through fourth possibilities are respectively , m − 2 d n + d n − d ′ , and , which are irrational for irrational d . Therefore X is of the form ( m + d, n ). m − d m − d ′ Suppose now that X = ( m + d, n ). Then the ant-path (a broken line) consists of precisely ′ m + n segments whose lengths are precisely the lengths of XX ∩ [ a, a + 1] × [ b, b + 1] for ′ some integers a and b . In other words, if you restrict the segment XX to a the unit squares in the integer lattice, you will get m + n smaller segments whose lengths match up exactly with the lengths of the segments of the original ant-path. Then the length of the ant-path is the same as the sum of the lengths of these planar segments, which is clearly just the ′ length of XX . However, not every segment going from ( d, 0) to ( d + m, n ) corresponds to an ant-path.