# 2005 AIME I Problems/Problem 13

## Problem

A particle moves in the Cartesian Plane according to the following rules:

1. From any lattice point $(a,b),$ the particle may only move to $(a+1,b), (a,b+1),$ or $(a+1,b+1).$
2. There are no right angle turns in the particle's path.

How many different paths can the particle take from $(0,0)$ to $(5,5)$?

## Solution

### Solution 1

The length of the path (the number of times the particle moves) can range from $l = 5$ to $9$; notice that $d = 10-l$ gives the number of diagonals. Let $R$ represent a move to the right, $U$ represent a move upwards, and $D$ to be a move that is diagonal. Casework upon the number of diagonal moves:

• Case $d = 1$: It is easy to see only $2$ cases.
• Case $d = 2$: There are two diagonals. We need to generate a string with $3$ $R$'s, $3$ $U$'s, and $2$ $D$'s such that no two $R$'s or $U$'s are adjacent. The $D$'s split the string into three sections ($-D-D-$): by the Pigeonhole principle all of at least one of the two letters must be all together (i.e., stay in a row).
If both $R$ and $U$ stay together, then there are $3 \cdot 2=6$ ways.
If either $R$ or $U$ splits, then there are $3$ places to put the letter that splits, which has $2$ possibilities. The remaining letter must divide into $2$ in one section and $1$ in the next, giving $2$ ways. This totals $6 + 3\cdot 2\cdot 2 = 18$ ways.
• Case $d = 3$: Now $2$ $R$'s, $2$ $U$'s, and $3$ $D$'s, so the string is divided into $4$ partitions ($-D-D-D-$).
If the $R$'s and $U$'s stay together, then there are $4 \cdot 3 = 12$ places to put them.
If one of them splits and the other stays together, then there are $4 \cdot {3\choose 2}$ places to put them, and $2$ ways to pick which splits, giving $4 \cdot 3 \cdot 2 = 24$ ways.
If both groups split, then there are ${4\choose 2}=6$ ways to arrange them. These add up to $12 + 24 + 6 = 42$ ways.
• Case $d = 4$: Now $1$ $R$, $1$ $U$, $4$ $D$'s ($-D-D-D-D-$). There are $5$ places to put $R$, $4$ places to put $U$, giving $20$ ways.
• Case $d = 5$: It is easy to see only $1$ case.

Together, these add up to $2 + 18 + 42 + 20 + 1 = \boxed{83}$.

### Solution 2

Another possibility is to use block-walking and recursion: for each vertex, the number of ways to reach it is $a + b + c$, where $a$ is the number of ways to reach the vertex from the left (without having come to that vertex (the one on the left) from below), $b$ is the number of ways to reach the vertex from the vertex diagonally down and left, and $c$ is the number of ways to reach the vertex from below (without having come to that vertex (the one below) from the left).

Assign to each point $(i,j)$ the triplet $(a_{i,j}, b_{i,j}, c_{i,j})$. Let $s(i,j) = a_{i,j}+ b_{i,j}+ c_{i,j}$. Let all lattice points that contain exactly one negative coordinate be assigned to $(0,0,0)$. This leaves the lattice points of the first quadrant, the positive parts of the $x$ and $y$ axes, and the origin unassigned. As a seed, assign to $(0,1,0)$. (We will see how this correlates with the problem.) Then define for each lattice point $(i,j)$ its triplet thus:

$a_{i,j} = s(i-1,j) - c_{i-1,j}$

$b_{i,j} = s(i-1,j-1)$

$c_{i,j} = s(i,j-1) - a_{i,j-1}$

It is evident that $s(i,j)$ is the number of ways to reach $(i,j)$ from $(0,0)$. Therefore we compute vertex by vertex the triplets $(a_{i,j}, b_{i,j}, c_{i,j})$ with $0 \leq i, j \leq 5$. Finally, after simple but tedious calculations, we find that $(a_{5,5}, b_{5,5}, c_{5,5}) = (28,27,28)$, so $s(i,j)=28+27+28 = \boxed{83}$.

[Note:Someone may wish to add an image.] Template:Incomplete
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 90, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}