Difference between revisions of "2005 Canadian MO Problems/Problem 1"
(Added a solution) |
|||
| Line 4: | Line 4: | ||
[[Image:CanMO_2005_1.png]] | [[Image:CanMO_2005_1.png]] | ||
==Solution== | ==Solution== | ||
| + | |||
| + | Define a last triangle of a row as the triangle in the row that the path visits last. From the last triangle in row <math>k</math>, the path must move down and then directly across to the last triangle in row <math>k+1</math>. Therefore, there is exactly one path through any given set of last triangles. For <math>1\le m\le n-1</math>, there are <math>m</math> possible last triangles for the <math>m</math>th row. The last triangle of the last row is always in the center. Therefore, <math>f(n)=(n-1)!</math>, and <math>f(2005)=2004!</math>. | ||
| + | |||
==See also== | ==See also== | ||
*[[2005 Canadian MO]] | *[[2005 Canadian MO]] | ||
Revision as of 23:34, 31 July 2006
Problem
Consider an equilateral triangle of side length 
, which is divided into unit triangles, as shown. Let 
be the number of paths from the triangle in the top row to the middle triangle in the bottom row, such that adjacent triangles in our path share a common edge and the path never travels up (from a lower row to a higher row) or revisits a triangle. An example of one such path is illustrated below for 
. Determine the value of 
.
Solution
Define a last triangle of a row as the triangle in the row that the path visits last. From the last triangle in row 
, the path must move down and then directly across to the last triangle in row 
. Therefore, there is exactly one path through any given set of last triangles. For 
, there are 
possible last triangles for the th row. The last triangle of the last row is always in the center. Therefore, 
, and 
.
