Difference between revisions of "2019 AIME I Problems/Problem 5"
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==Solution== | ==Solution== | ||
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| + | We label a move from <math>(a,b)</math> to <math>(a,b-1)</math> as down (<math>D</math>), from <math>(a,b)</math> to <math>(a-1,b)</math> as left (<math>L</math>), and from <math>(a,b)</math> to <math>(a-1,b-1)</math> as slant (<math>S</math>). To arrive at <math>(0,0)</math> without arriving at an axis first, the particle must first go to <math>(1,1)</math> then do a slant move. The particle can arrive can be done through any permutation of the following 4 different cases: <math>SSS</math>, <math>SSDL</math>, <math>SDLDL</math>, and <math>DLDLDL</math>. | ||
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| + | There is only <math>1</math> permutation of <math>SSS</math>. Including the last move, there are <math>4</math> possible moves, making the probability of this move <math>\frac{1}{3^4}</math>. | ||
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| + | There are <math>\frac{4!}{2!} = 12</math> permutations of <math>SSDL</math>, as the ordering of the two slants do not matter. There are <math>5</math> possible moves, making the probability of this move <math>\frac{12}{3^5}</math>. | ||
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| + | There are <math>\frac{5!}{2! \cdot 2!} = 30</math> permutations of <math>SDLDL</math>, as the ordering of the pairs of do not matter. There are <math>6</math> possible moves, making the probability of this move <math>\frac{30}{3^6}</math>. | ||
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| + | There are <math>\frac{6!}{3! \cdot 3!} = 30</math> permutations of <math>DLDLDL</math>, as the ordering of the two slants do not matter. There are <math>7</math> possible moves, making the probability of this move <math>\frac{20}{3^7}</math>. | ||
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| + | Adding these, we get the total probability as <math>\frac{1}{3^4} + \frac{12}{3^5} + \frac{30}{3^6} + \frac{20}{3^7} = \frac{245}{3^7}</math>. Therefore, the answer is <math>245 + 7 = 342</math>. | ||
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==See Also== | ==See Also== | ||
{{AIME box|year=2019|n=I|num-b=4|num-a=6}} | {{AIME box|year=2019|n=I|num-b=4|num-a=6}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
IMO 1999 | IMO 1999 | ||
Revision as of 23:31, 14 March 2019
Problem 5
A moving particle starts at the point 
and moves until it hits one of the coordinate axes for the first time. When the particle is at the point 
, it moves at random to one of the points 
, 
, or 
, each with probability 
, independently of its previous moves. The probability that it will hit the coordinate axes at 
is 
, where 
and 
are positive integers. Find 
.
Solution
We label a move from to as down (
), from to as left (
), and from to as slant (
). To arrive at without arriving at an axis first, the particle must first go to 
then do a slant move. The particle can arrive can be done through any permutation of the following 4 different cases: 
, 
, 
, and 
.
There is only 
permutation of . Including the last move, there are 
possible moves, making the probability of this move 
.
There are 
permutations of , as the ordering of the two slants do not matter. There are 
possible moves, making the probability of this move 
.
There are 
permutations of , as the ordering of the pairs of do not matter. There are 
possible moves, making the probability of this move 
.
There are 
permutations of , as the ordering of the two slants do not matter. There are 
possible moves, making the probability of this move 
.
Adding these, we get the total probability as 
. Therefore, the answer is 
.
See Also
| 2019 AIME I (Problems • Answer Key • Resources) | ||
| Preceded by Problem 4 |
Followed by Problem 6 | |
| 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. 
IMO 1999
