Difference between revisions of "2017 AMC 12A Problems/Problem 22"
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We let <math>c, e,</math> and <math>m</math> be the probability of reaching a corner before an edge when starting at an "inside corner" (e.g. <math>(1, 1)</math>), an "inside edge" (e.g. <math>(1, 0)</math>), and the middle respectively. | We let <math>c, e,</math> and <math>m</math> be the probability of reaching a corner before an edge when starting at an "inside corner" (e.g. <math>(1, 1)</math>), an "inside edge" (e.g. <math>(1, 0)</math>), and the middle respectively. | ||
| − | Starting in the middle, there is a <math> | + | Starting in the middle, there is a <math>\frac{4}{8}</math> chance of moving to an inside edge and a <math>\frac{4}{8}</math> chance of moving to an inside corner, so |
<cmath>m = \frac{1}{2}e + \frac{1}{2}c.</cmath> | <cmath>m = \frac{1}{2}e + \frac{1}{2}c.</cmath> | ||
| − | Starting at an inside edge, there is a <math> | + | Starting at an inside edge, there is a <math>\frac{2}{8}</math> chance of moving to another inside edge, a <math>\frac{2}{8}</math> chance of moving to an inside corner, a <math>\frac{1}{8}</math> chance of moving into the middle, and a <math>\frac{3}{8}</math> chance of reaching an outside edge and stopping. Therefore, |
| − | <cmath>e = \frac{1}{4}e + \frac{1}{4}c + \frac{1}{8}m + \frac{3}{8}0 = \frac{1}{4}e + \frac{1}{4}c + \frac{1}{8}m.</cmath> | + | <cmath>e = \frac{1}{4}e + \frac{1}{4}c + \frac{1}{8}m + \frac{3}{8}*0 = \frac{1}{4}e + \frac{1}{4}c + \frac{1}{8}m.</cmath> |
| − | Starting at an inside corner, there is a <math> | + | Starting at an inside corner, there is a <math>\frac{2}{8}</math> chance of moving to an inside edge, a <math>\frac{1}{8}</math> chance of moving into the middle, a <math>\frac{4}{8}</math> chance of moving to an outside edge and stopping, and finally a <math>\frac{1}{8}</math> chance of reaching that elusive outside corner. This gives |
| − | <cmath>c = \frac{1}{4}e + \frac{1}{8}m + \frac{1}{2}0 + \frac{1}{8}1 = \frac{1}{4}e + \frac{1}{8}m + \frac{1}{8}.</cmath> | + | <cmath>c = \frac{1}{4}e + \frac{1}{8}m + \frac{1}{2}0 + \frac{1}{8}*1 = \frac{1}{4}e + \frac{1}{8}m + \frac{1}{8}.</cmath> |
Solving this system of equations gives | Solving this system of equations gives | ||
Revision as of 00:10, 9 February 2017
Problem
A square is drawn in the Cartesian coordinate plane with vertices at 
, 
, 
, 
. A particle starts at 
. Every second it moves with equal probability to one of the eight lattice points (points with integer coordinates) closest to its current position, independently of its previous moves. In other words, the probability is 
that the particle will move from 
to each of 
, 
, 
, 
, 
, 
, 
, or 
. The particle will eventually hit the square for the first time, either at one of the 4 corners of the square or at one of the 12 lattice points in the interior of one of the sides of the square. The probability that it will hit at a corner rather than at an interior point of a side is 
, where 
and 
are relatively prime positive integers. What is 
?
Solution
We let 
and be the probability of reaching a corner before an edge when starting at an "inside corner" (e.g. 
), an "inside edge" (e.g. 
), and the middle respectively.
Starting in the middle, there is a 
chance of moving to an inside edge and a chance of moving to an inside corner, so
![\[m = \frac{1}{2}e + \frac{1}{2}c.\]](http://latex.artofproblemsolving.com/0/9/c/09c7e9598e237027f630d0902905a6fcedb167b6.png)
Starting at an inside edge, there is a 
chance of moving to another inside edge, a chance of moving to an inside corner, a 
chance of moving into the middle, and a 
chance of reaching an outside edge and stopping. Therefore,
![\[e = \frac{1}{4}e + \frac{1}{4}c + \frac{1}{8}m + \frac{3}{8}*0 = \frac{1}{4}e + \frac{1}{4}c + \frac{1}{8}m.\]](http://latex.artofproblemsolving.com/a/4/9/a49a41c7be57f048397713957790a19a0193f6f7.png)
Starting at an inside corner, there is a chance of moving to an inside edge, a chance of moving into the middle, a chance of moving to an outside edge and stopping, and finally a chance of reaching that elusive outside corner. This gives
![\[c = \frac{1}{4}e + \frac{1}{8}m + \frac{1}{2}0 + \frac{1}{8}*1 = \frac{1}{4}e + \frac{1}{8}m + \frac{1}{8}.\]](http://latex.artofproblemsolving.com/c/9/b/c9b1f9c9971c284cdbe08211d2529b7e7e7072e6.png)
Solving this system of equations gives
![\[m = \frac{4}{35},\]](http://latex.artofproblemsolving.com/6/c/7/6c7320d977b41ce17e21dd42138d560e104569ad.png)
![\[e = \frac{1}{14},\]](http://latex.artofproblemsolving.com/7/6/3/763d1d62d2d1de18ef658dc60ed70c272bf43963.png)
![\[c = \frac{11}{70}.\]](http://latex.artofproblemsolving.com/1/d/f/1df8ce7dea94a05220817b43cf3de8181554f7b2.png)
Since the particle starts at 
it is we are looking for, so the final answer is
![\[4 + 35 = \textbf{(E) }39.\]](http://latex.artofproblemsolving.com/8/5/a/85aaa96841e11969ebc63b96133a206a4315ca1c.png)
See Also
| 2017 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 21 |
Followed by Problem 23 |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
| All AMC 12 Problems and Solutions | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
