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Difference between revisions of "2005 AIME I Problems/Problem 13"
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== Problem ==
 
== Problem ==
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A particle moves in the [[Cartesian Plane]] according to the following rules:
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A particle moves in the [[Cartesian plane]] according to the following rules:
  
 
# From any lattice point <math> (a,b), </math> the particle may only move to <math> (a+1,b), (a,b+1), </math> or <math>(a+1,b+1). </math>
 
# From any lattice point <math> (a,b), </math> the particle may only move to <math> (a+1,b), (a,b+1), </math> or <math>(a+1,b+1). </math>
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__TOC__
 
__TOC__
 +
 
== Solution ==
 
== Solution ==
 
=== Solution 1 ===
 
=== Solution 1 ===
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*'''Case ''' <math>d = 5</math>: It is easy to see only <math>1</math> case.
 
*'''Case ''' <math>d = 5</math>: It is easy to see only <math>1</math> case.
  
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Together, these add up to <math>2 + 18 + 42 + 20 + 1 = \boxed{83}</math>.
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Together, these add up to <math>2 + 18 + 42 + 20 + 1 = \boxed{083}</math>.
  
 
=== Solution 2 ===
 
=== Solution 2 ===
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Assign to each point <math>(i,j)</math> the triplet <math>(a_{i,j}, b_{i,j}, c_{i,j})</math>. Let <math>s(i,j) = a_{i,j}+ b_{i,j}+ c_{i,j}</math>. Let all lattice points that contain exactly one negative coordinate be assigned to <math>(0,0,0)</math>. This leaves the lattice points of the first quadrant, the positive parts of the <math>x</math> and <math>y</math> axes, and the origin unassigned. As a seed, assign to <math>(0,1,0)</math>. (We will see how this correlates with the problem.) Then define for each lattice point <math>(i,j)</math> its triplet thus:
 
Assign to each point <math>(i,j)</math> the triplet <math>(a_{i,j}, b_{i,j}, c_{i,j})</math>. Let <math>s(i,j) = a_{i,j}+ b_{i,j}+ c_{i,j}</math>. Let all lattice points that contain exactly one negative coordinate be assigned to <math>(0,0,0)</math>. This leaves the lattice points of the first quadrant, the positive parts of the <math>x</math> and <math>y</math> axes, and the origin unassigned. As a seed, assign to <math>(0,1,0)</math>. (We will see how this correlates with the problem.) Then define for each lattice point <math>(i,j)</math> its triplet thus:
 +
<cmath>\begin{align*}
 +
a_{i,j} &= s(i-1,j) - c_{i-1,j}\\
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b_{i,j} &= s(i-1,j-1) \\
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c_{i,j} &= s(i,j-1) - a_{i,j-1}.
 +
\end{align*}</cmath>
 +
It is evident that <math>s(i,j)</math> is the number of ways to reach <math>(i,j)</math> from <math>(0,0)</math>. Therefore we compute vertex by vertex the triplets <math>(a_{i,j}, b_{i,j}, c_{i,j})</math> with <math>0 \leq i, j \leq 5</math>.
 +
<asy>
 +
defaultpen(fontsize(8)+0.8+heavyblue); size(250);
  
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<math>a_{i,j} = s(i-1,j) - c_{i-1,j}</math>
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for(int i = 0; i<6; ++i) { draw((0,i)--(5,i)^^(i,0)--(i,5), gray+0.25); }
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label("$(0,0,0)$", (0,0), N);
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<math>b_{i,j} = s(i-1,j-1)</math>
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for(int i = 1; i<6; ++i) { label("$(0,0,1)$", (0,i), N); label("$(1,0,0)$", (i,0), N); label("$({"+string(i-1)+"},1,0)$", (i,1), N); label("$(0,1,{"+string(i-1)+"})$", (1,i), N);}
−
 
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real[] val={1,2,4,7};
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<math>c_{i,j} = s(i,j-1) - a_{i,j-1}</math>
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for(int i = 2; i<6; ++i) { label("$(1,{"+string(i-1)+"}, {"+string(val[i-2])+"})$", (2,i), N); label("$({"+string(val[i-2])+"},{"+string(i-1)+"}, 1)$", (i,2), N);}
−
 
 
−
It is evident that <math>s(i,j)</math> is the number of ways to reach <math>(i,j)</math> from <math>(0,0)</math>. Therefore we compute vertex by vertex the triplets <math>(a_{i,j}, b_{i,j}, c_{i,j})</math> with <math>0 \leq i, j \leq 5</math>. Finally, after simple but tedious calculations, we find that <math>(a_{5,5}, b_{5,5}, c_{5,5}) = (28,27,28)</math>, so <math>s(i,j)=28+27+28 = \boxed{83}</math>.
 
  
−
[Note:Someone may wish to add an image.]
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label("$(3,3,3)$", (3,3), N); label("$(9,9,9)$", (4,4), N); label("$(28,27,28)$", (5,5), N);
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{{incomplete|solution}}
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label("$(4,5,6)$", (3,4), N); label("$(6,5,4)$", (4,3), N);
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label("$(5,8,11)$", (3,5), N); label("$(11,8,5)$", (5,3), N);
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label("$(13,15,18)$", (4,5), N); label("$(18,15,13)$", (5,4), N);
 +
</asy>
 +
Finally, after simple but tedious calculations, we find that <math>(a_{5,5}, b_{5,5}, c_{5,5}) = (28,27,28)</math>, so <math>s(i,j)=28+27+28 = \boxed{083}</math>.
  
 
== See also ==
 
== See also ==

Latest revision as of 23:21, 28 July 2022

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 $2 + 18 + 42 + 20 + 1 = \boxed{083}$.

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: \begin{align*} 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}. \end{align*} 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 $0 \leq i, j \leq 5$. [asy] defaultpen(fontsize(8)+0.8+heavyblue); size(250); for(int i = 0; i<6; ++i) { draw((0,i)--(5,i)^^(i,0)--(i,5), gray+0.25); } label("$(0,0,0)$", (0,0), N); for(int i = 1; i<6; ++i) { label("$(0,0,1)$", (0,i), N); label("$(1,0,0)$", (i,0), N); label("$({"+string(i-1)+"},1,0)$", (i,1), N); label("$(0,1,{"+string(i-1)+"})$", (1,i), N);} real[] val={1,2,4,7}; for(int i = 2; i<6; ++i) { label("$(1,{"+string(i-1)+"}, {"+string(val[i-2])+"})$", (2,i), N); label("$({"+string(val[i-2])+"},{"+string(i-1)+"}, 1)$", (i,2), N);} label("$(3,3,3)$", (3,3), N); label("$(9,9,9)$", (4,4), N); label("$(28,27,28)$", (5,5), N); label("$(4,5,6)$", (3,4), N); label("$(6,5,4)$", (4,3), N); label("$(5,8,11)$", (3,5), N); label("$(11,8,5)$", (5,3), N); label("$(13,15,18)$", (4,5), N); label("$(18,15,13)$", (5,4), N); [/asy] Finally, after simple but tedious calculations, we find that $(a_{5,5}, b_{5,5}, c_{5,5}) = (28,27,28)$, so $s(i,j)=28+27+28 = \boxed{083}$.

See also

2005 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 12 		Followed by
Problem 14
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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