PUMaC 2020 · 加试 · 第 3 题

Problem 3.4. Let C be a simple closed ant-path on the surface of P . It divides the surface S ( P ), and the vertices of V ( P ) consequently, into two parts. The sum of pointiness of vertices in each part is equal to 2 π . Proof 3.4. We will consider only one of the two parts and show the sum of the pointiness in that part is equal to 2 π . As the sum of all pointiness in 4 π , the sum in the other part must be 2 π as well. Let S denote one of the parts, and let M , M , . . . , M be the faces of that part (note 1 1 2 k that some of M may not be faces, but parts of faces of P intersected by C ). Let n denote i i the number of vertices of M , and let α , · · · , α be the angles of M . Finally, let C touch i i, 1 i,n i i l of M s. i The idea is to compute the sum of all the angles α in two different ways. Therefore, i,j ∑ ∑ k n i let S = α and note that: i,j i =1 j =1 n k k i ∑ ∑ ∑ S = α = ( n − 2) π i,j i i =1 j =1 i =1 On the other hand, we may group α by the vertex of S they touch. The sum of angles i,j 1 around points of S that lie on C is exactly π , because C is an ant-path. The sums of angles 1 around other vertices of S are 2 π − p ( v ), by the definition of the pointiness. Therefore, we 1 ∑ get S = lπ + 2 π − p ( v ). Equating the two expressions for S gives a way to compute v ∈ S 1 ∑ p ( v ) in the following way: v ∈ S 1 k ∑ ∑ ( n − 2) π = lπ + 2 π − p ( v ) i i =1 v ∈ S 1 k ∑ ∑ p ( v ) = π (2 V ( S ) − ( n − 2) + l ) 1 i v ∈ S i =1 1 ∑ k Therefore, it suffices to show that 2 V ( S ) − ( n − 2) + l = 2. We will be able 1 i i =1 show this using the Euler formula, applied on the graph we obtain by projecting S onto 1 the plane. The number of vertices of this graph is V ( S ) + l , the number of its edges is 1 ∑ k ∑ n + l i i =1 (here, we used the formula | f | = 2 E established earlier) and the number of f 2 its faces is k + 1 (because we count the outside face as well). Therefore, we have: ∑ k n + l i i =1 V ( S ) + l − + k + 1 = 2 1 2 ∑ k l ( n ) − 2 k i i =1 V ( S ) + − = 1 1 2 2 k ∑ 2 V ( S ) − ( n − 2) + l = 2 1 i i =1 This completes the proof. The previous problem now gives an even simpler solution to problem 2.5. As there is no way to find two vertices with pointiness adding up to 2 π (by the constraints of the problem), there are no simple closed ant-paths on the surface of the tetrahedron. Furthermore, the previous claim gives us a very strong tool when showing that a given polyhedron has no simple closed ant-paths - it is enough to check that no subset of its vertices has the sum of pointiness equal to 2 π . In some sense, this means that the most of polyhedra do not have simple closed ant-paths on their surfaces (making these concepts more precise would be out of scope of this power round). The inverse of the previous theorem does not hold, as we will show in the following problem.