PUMaC 2020 · 加试 · 第 2 题
PUMaC 2020 — Power Round — Problem 2
题目详情
Problem 2.4. Let C be a piecewise linear curve on the surface S ( P ) of the polyhedron, with breaking points T , T , ..., T . Prove that C is an ant-path if and only if it satisfies the 1 2 k following three properties: • All of the breaking points T , T , ..., T lie on the edges of P , 1 2 k • C contains no vertices in its interior, • If the breaking point T is contained in a short subsegment KL ( K and L being on i the different faces of P ) of C , and on the edge AB of P , then ∠ KT A = ∠ BT L . i i Remark: Formally, to solve this problem you need to prove two directions. However, you will find one of them easy due to the work we already did. Of special interest are ant-paths that do not have beginnings or ends: the closed ant- paths. Definition 2.E. A closed ant-path is a ant-path C whose beginning and end coincide. If this point is denoted by P , C needs to satisfy: there is a subsegment X Y of C which P P contains P , and on which C agrees with the shortest path between X and Y . A closed P P ant-path is simple if it has no self-intersections. Interestingly enough, the simple closed ant-paths are relatively rare. The following problem shows that generic tetrahedrons almost never contains simple closed ant-paths.
解析
Problem 2.5. Let A A A A be a tetrahedron, and let θ denote the sums of angles at 1 2 3 4 i the vertex A (in other words, θ = ∠ A A A + ∠ A A A + ∠ A A A , and similarly for i 1 2 1 3 2 1 4 4 1 3 the other indices). If θ + θ 6 = 2 π for all i 6 = j ∈ [4], then there is no simple closed ant-path i j with 3 or 4 segments on S ( A A A A ). 1 2 3 4 Hint: Try to draw such a path and show it cannot exist due to angle constraints given by the problem 2.4. Proof 2.5. (Solution for 3 or 4 breaking points.) Suppose the contrary, that is there is γ a closed simple and path with 3 or 4 breaking points. We first do the case that γ , the closed simple ant path, is abc , where a, b, c are points on three of the sides of the tetrahedron which share a vertex. Let v be the vertex that 1 all three of the edges contain, and let α be the pointiness at v . Note that the sum of 1 1 all face angles in a tetrahedron is 4 π (4 triangles with π each.) Then consider the sum of angles in the tetrahedron v abc . On the hand it is 4 π . However, since γ is an ant path, 1 we have shown the angle ∠ cav is equal to the exterior angle bav by problem 2.1.3, that is 1 1 ∠ cav + ∠ bav = π . From this, we get that the sum of all angles is π for each of the vertices 1 1 a, b, c , π for the triangle abc , and finally α for the sum of the face angles at v . Hence we 1 1 get 4 π = π + 3 π + α 1 Which is a contradiction as α > 0. 1 Now we do the case that γ has 4 breaking points, that is it is abcd , where the points are on four different sides of the tetrahedron (no 3 of which are on the same face). Then take v , v to be two vertices of the tetrahedron on one side of the path. Let α , α be the 1 2 1 2 pointiness of each of the vertices, respectively. Then let Σ be the total sum of angles in the quadrilaterals v abv , v cdv and triangles v ad and v bc . Clearly Σ = 6 π . On the other 1 2 2 1 1 2 hand, since γ is an ant-path, the sum of the angles ∠ dav + ∠ bav = π , since the angles 1 1 dav and the exterior angle bav are the same by problem 2.1.3. Similarly at each of the 1 1 vertices a, b, c, d the sum of the angles in the mentioned triangles and quadrilaterals is π . The other angles sum up to α + α . 1 2 Σ = 4 π + α + α 1 2 And hence α + α = 2 π , which is a contradiction with the assumption that no two 1 2 pointiness sum up to 2 π . The above result can be generalized to many other polyhedra. In some sense, almost no generic polyhedra have simple closed ant paths on their surfaces. However, in order to prove that, we will need to develop a stronger machinery. 3 More about polyhedra We will now recall some of the well-known properties of the polyhedrons, and also introduce some on the new ones. One of the most famous results describing 3-dimensional polyhedra is the celebrated Euler’s formula: