HMMT 十一月 2019 · 冲刺赛 · 第 12 题
HMMT November 2019 — Guts Round — Problem 12
题目详情
- [8] Four players stand at distinct vertices of a square. They each independently choose a vertex of the square (which might be the vertex they are standing on). Then, they each, at the same time, begin running in a straight line to their chosen vertex at 10mph, stopping when they reach the vertex. If at any time two players, whether moving or not, occupy the same space (whether a vertex or a point inside the square), they collide and fall over. How many different ways are there for the players to choose vertices to go to so that none of them fall over? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . HMMT November 2019, November 9, 2019 — GUTS ROUND Organization Team Team ID#
解析
- [8] Four players stand at distinct vertices of a square. They each independently choose a vertex of the square (which might be the vertex they are standing on). Then, they each, at the same time, begin running in a straight line to their chosen vertex at 10mph, stopping when they reach the vertex. If at any time two players, whether moving or not, occupy the same space (whether a vertex or a point inside the square), they collide and fall over. How many different ways are there for the players to choose vertices to go to so that none of them fall over? Proposed by: Carl Schildkraut Answer: 11 Observe that no two players can choose the same vertex, and no two players can choose each others vertices. Thus, if two players choose their own vertices, then the remaining two also must choose their own vertices (because they cant choose each others vertices), thus all 4 players must choose their own vertices. There is 1 way to choose the vertices in this case. Name the players top left, top right, bottom left, and bottom right, based on their initial positions. Assume exactly one player (without loss of generality, say the top left) chooses their own vertex. Then, the remaining 3 players have to form a triangle (recall no two player can choose each others vertices). There are 4 ways to choose which player chooses their own vertex, and 2 ways to choose which direction the players move in the triangle, thus there are 8 ways to choose the vertices in this case. Lastly, assume no one chooses their own vertex. We will first prove that no player can choose the vertex across them. Assume the contrary, without loss of generality, let the top left player chooses the bottom right vertex. Then, neither of the bottom left and the top right players can choose the others vertex, because they would meet the top left player at the center of the square. As they cant choose bottom right (it is chosen by the top left player), and cant choose their own vertex (by assumption), they both have to choose the top left vertex, which is an immediate contradiction. Now, the top left player has to choose either the top right vertex or the bottom left. Without loss of generality, let the player choose the top right vertex. Then, the top right player has to choose the bottom right vertex (as they can neither go across nor back to top left), the bottom right player has to choose the bottom left vertex, and the bottom left player has to choose the top left vertex, and all the choices are determined by the first players choice. There are 2 ways to choose where the first player goes, thus there are 2 ways to choose the vertices in this case. In total, there are 1 + 8 + 2 = 11 ways to choose the vertices.