Difference between revisions of "2005 AIME I Problems/Problem 13"
(sol by casework) |
m (→Solution 2) |
||
(14 intermediate revisions by 11 users not shown) | |||
Line 1: | Line 1: | ||
== Problem == | == Problem == | ||
− | A particle moves in the [[Cartesian | + | 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> | ||
Line 8: | Line 8: | ||
__TOC__ | __TOC__ | ||
+ | |||
== Solution == | == Solution == | ||
=== Solution 1 === | === Solution 1 === | ||
Line 13: | Line 14: | ||
*'''Case ''' <math>d = 1</math>: It is easy to see only <math>2</math> cases. | *'''Case ''' <math>d = 1</math>: It is easy to see only <math>2</math> cases. | ||
− | *'''Case ''' <math>d = 2</math>: There are two diagonals. We need to generate a string with <math>3</math> <math>R</math>'s, <math>3</math> <math>U</math>'s, and <math>2</math> <math>D</math>'s such that no two <math>R</math>'s or <math>U</math>'s are adjacent. The <math>D</math>'s split the string into three sections (<math>-D-D-</math>): by the [[Pigeonhole principle]] one of the two letters must | + | *'''Case ''' <math>d = 2</math>: There are two diagonals. We need to generate a string with <math>3</math> <math>R</math>'s, <math>3</math> <math>U</math>'s, and <math>2</math> <math>D</math>'s such that no two <math>R</math>'s or <math>U</math>'s are adjacent. The <math>D</math>'s split the string into three sections (<math>-D-D-</math>): 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 <math>R</math> and <math>U</math> stay together, then there are <math> | + | :If both <math>R</math> and <math>U</math> stay together, then there are <math>3 \cdot 2=6</math> ways. |
:If either <math>R</math> or <math>U</math> splits, then there are <math>3</math> places to put the letter that splits, which has <math>2</math> possibilities. The remaining letter must divide into <math>2</math> in one section and <math>1</math> in the next, giving <math>2</math> ways. This totals <math>6 + 3\cdot 2\cdot 2 = 18</math> ways. | :If either <math>R</math> or <math>U</math> splits, then there are <math>3</math> places to put the letter that splits, which has <math>2</math> possibilities. The remaining letter must divide into <math>2</math> in one section and <math>1</math> in the next, giving <math>2</math> ways. This totals <math>6 + 3\cdot 2\cdot 2 = 18</math> ways. | ||
− | *'''Case ''' <math>d = 3</math>: Now <math>2</math> <math>R</math>'s, <math>2</math> <math>U</math>'s, and <math>3</math> <math> | + | *'''Case ''' <math>d = 3</math>: Now <math>2</math> <math>R</math>'s, <math>2</math> <math>U</math>'s, and <math>3</math> <math>D</math>'s, so the string is divided into <math>4</math> partitions (<math>-D-D-D-</math>). |
− | :If the </math> | + | :If the <math>R</math>'s and <math>U</math>'s stay together, then there are <math>4 \cdot 3 = 12</math> places to put them. |
:If one of them splits and the other stays together, then there are <math>4 \cdot {3\choose 2}</math> places to put them, and <math>2</math> ways to pick which splits, giving <math>4 \cdot 3 \cdot 2 = 24</math> ways. | :If one of them splits and the other stays together, then there are <math>4 \cdot {3\choose 2}</math> places to put them, and <math>2</math> ways to pick which splits, giving <math>4 \cdot 3 \cdot 2 = 24</math> ways. | ||
:If both groups split, then there are <math>{4\choose 2}=6</math> ways to arrange them. These add up to <math>12 + 24 + 6 = 42</math> ways. | :If both groups split, then there are <math>{4\choose 2}=6</math> ways to arrange them. These add up to <math>12 + 24 + 6 = 42</math> ways. | ||
− | *'''Case ''' <math>d = 4</math>: Now <math>1</math> <math>R</math>, <math>1 U</math>, <math>4 D</math>'s (<math>-D-D-D-D-</math>). There are <math>5</math> places to put <math>R</math>, <math>4</math> places to put <math>U</math>, giving <math>20</math> ways. | + | *'''Case ''' <math>d = 4</math>: Now <math>1</math> <math>R</math>, <math>1</math> <math>U</math>, <math>4</math> <math>D</math>'s (<math>-D-D-D-D-</math>). There are <math>5</math> places to put <math>R</math>, <math>4</math> places to put <math>U</math>, giving <math>20</math> ways. |
*'''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. | ||
− | Together, these add up to <math>2 + 18 + 42 + 20 + 1 = \boxed{ | + | Together, these add up to <math>2 + 18 + 42 + 20 + 1 = \boxed{083}</math>. |
=== Solution 2 === | === Solution 2 === | ||
− | Another possibility is to use block-walking and [[recursion]]: for each vertex, the number of ways to reach it is the number of ways to reach the vertex to | + | Another possibility is to use block-walking and [[recursion]]: for each vertex, the number of ways to reach it is <math>a + b + c</math>, where <math>a</math> 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), <math>b</math> is the number of ways to reach the vertex from the vertex diagonally down and left, and <math>c</math> 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 <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}\ | ||
+ | b_{i,j} &= s(i-1,j-1) \ | ||
+ | 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); | ||
+ | |||
+ | 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 <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 == | ||
Line 34: | Line 56: | ||
[[Category:Intermediate Combinatorics Problems]] | [[Category:Intermediate Combinatorics Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 23:21, 28 July 2022
Problem
A particle moves in the Cartesian plane according to the following rules:
- From any lattice point the particle may only move to or
- There are no right angle turns in the particle's path.
How many different paths can the particle take from to ?
Contents
[hide]Solution
Solution 1
The length of the path (the number of times the particle moves) can range from to ; notice that gives the number of diagonals. Let represent a move to the right, represent a move upwards, and to be a move that is diagonal. Casework upon the number of diagonal moves:
- Case : It is easy to see only cases.
- Case : There are two diagonals. We need to generate a string with 's, 's, and 's such that no two 's or 's are adjacent. The 's split the string into three sections (): 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 and stay together, then there are ways.
- If either or splits, then there are places to put the letter that splits, which has possibilities. The remaining letter must divide into in one section and in the next, giving ways. This totals ways.
- Case : Now 's, 's, and 's, so the string is divided into partitions ().
- If the 's and 's stay together, then there are places to put them.
- If one of them splits and the other stays together, then there are places to put them, and ways to pick which splits, giving ways.
- If both groups split, then there are ways to arrange them. These add up to ways.
- Case : Now , , 's (). There are places to put , places to put , giving ways.
- Case : It is easy to see only case.
Together, these add up to .
Solution 2
Another possibility is to use block-walking and recursion: for each vertex, the number of ways to reach it is , where 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), is the number of ways to reach the vertex from the vertex diagonally down and left, and 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 the triplet . Let . Let all lattice points that contain exactly one negative coordinate be assigned to . This leaves the lattice points of the first quadrant, the positive parts of the and axes, and the origin unassigned. As a seed, assign to . (We will see how this correlates with the problem.) Then define for each lattice point its triplet thus: It is evident that is the number of ways to reach from . Therefore we compute vertex by vertex the triplets with . Finally, after simple but tedious calculations, we find that , so .
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.