In an isosceles right-angled triangle shaped billiards table , a ball starts moving from one of the vertices adjacent to hypotenuse. When it reaches to one side then it will reflect its path. Prove that if we reach to a vertex then it is not the vertex at initial position By Sam Nariman
Problem
Source: Iran TST 2007, Day 2
Tags: geometry, geometric transformation, reflection, analytic geometry, geometry proposed
07.05.2007 20:21
There is a straightforward proof by contradiction. But I suppose the easiest (and in my view most elegant) proof is the following: By successively reflecting the billiard by one of its sides we get a tilling of the plain. Identifying all the arising triangles in the right manner the laws of reflection imply that the path of the ball in the billiard can be represented as a straight line in the plain. Now we introduce coordinates in the plain such that the original billiard has vertices $(0,0)$, $(0,1)$ and $(1,1)$, where the initial position of the ball is the origin. By the above described identifaction this vertex corresponds to every point in the plain with even coordinates. Now the clain follows from the simple fact that any straight path from the origin to a point with even coordinates passes through another lattice point first. (I hope the idea is not obstructed by my inability to properly express myself in English... )
08.05.2007 01:58
We can prove easily that the line formed by the $4k+i$ move is parallel to the line formed by the $4(k-1)+i$ move, and that is impossible go to our initial vertice with lines parallels to our first 4 lines!
08.05.2007 16:19
e.lopes wrote: ... the line formed by the $4k+i$ move is parallel to the line formed by the $4(k-1)+i$ move ... Are you sure about that? Denote the sides of the billiard by $s_{1}$, $s_{2}$ and $l$, where $l$ is the longest side and the ball starts at the vertex between $l$ and $s_{1}$. Then your statement is true, if the ball first hits $s_{2}, l, s_{1}, l$ in that particular order. However we also could have $s_{2},l,s_{1},s_{2}$ or $s_{2},l,s_{2}, s_{1}$, and after that we don't seem to get something parallel.
08.05.2007 17:22
solyaris wrote: e.lopes wrote: ... the line formed by the $4k+i$ move is parallel to the line formed by the $4(k-1)+i$ move ... Are you sure about that? Denote the sides of the billiard by $s_{1}$, $s_{2}$ and $l$, where $l$ is the longest side and the ball starts at the vertex between $l$ and $s_{1}$. Then your statement is true, if the ball first hits $s_{2}, l, s_{1}, l$ in that particular order. However we also could have $s_{2},l,s_{1},s_{2}$ or $s_{2},l,s_{2}, s_{1}$, and after that we don't seem to get something parallel. You're right, my friend! But in this case, we will have basically the same thing. After many 'moves', our lines will be parallel!
24.07.2007 14:58
thank you for this beautiful solution, solyaris.
08.10.2019 23:34
04.01.2021 14:46