Difference between revisions of "2016 AIME I Problems/Problem 13"
(→Solution) |
(→Solution) |
||
Line 2: | Line 2: | ||
Freddy the frog is jumping around the coordinate plane searching for a river, which lies on the horizontal line <math>y = 24</math>. A fence is located at the horizontal line <math>y = 0</math>. On each jump Freddy randomly chooses a direction parallel to one of the coordinate axes and moves one unit in that direction. When he is at a point where <math>y=0</math>, with equal likelihoods he chooses one of three directions where he either jumps parallel to the fence or jumps away from the fence, but he never chooses the direction that would have him cross over the fence to where <math>y < 0</math>. Freddy starts his search at the point <math>(0, 21)</math> and will stop once he reaches a point on the river. Find the expected number of jumps it will take Freddy to reach the river. | Freddy the frog is jumping around the coordinate plane searching for a river, which lies on the horizontal line <math>y = 24</math>. A fence is located at the horizontal line <math>y = 0</math>. On each jump Freddy randomly chooses a direction parallel to one of the coordinate axes and moves one unit in that direction. When he is at a point where <math>y=0</math>, with equal likelihoods he chooses one of three directions where he either jumps parallel to the fence or jumps away from the fence, but he never chooses the direction that would have him cross over the fence to where <math>y < 0</math>. Freddy starts his search at the point <math>(0, 21)</math> and will stop once he reaches a point on the river. Find the expected number of jumps it will take Freddy to reach the river. | ||
==Solution== | ==Solution== | ||
− | + | Clearly Freddy's <math>x</math>-coordinate is irrelevant, so we let <math>E(y)</math> be the expected value of the number of jumps it will take him to reach the river from a given <math>y</math>-coordinate. Observe that <math>E(24)=0</math>, and <cmath>E(y)=1+\frac{E(y+1)+E(y-1)+2E(y)}{4}</cmath> for all <math>y</math> such that <math>1\le y\le 23</math>. Also note that <math>E(0)=1+\frac{2E(0)+E(1)}{3}</math>. This gives <math>E(0)=E(1)+3</math>. Plugging this into the equation for <math>E(1)</math> gives that <cmath>E(1)=1+\frac{E(2)+3E(1)+3}{4},</cmath> or <math>E(1)=E(2)+7</math>. Iteratively plugging this in gives that <math>E(n)=E(n+1)+4n+3</math>. Thus <math>E(23)=E(24)+95</math>, <math>E(22)=E(23)+91=E(24)+186</math>, and <math>E(21)=E(22)+85=E(24)+273=\boxed{273}</math>. | |
== See also == | == See also == | ||
{{AIME box|year=2016|n=I|num-b=12|num-a=14}} | {{AIME box|year=2016|n=I|num-b=12|num-a=14}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 23:28, 5 March 2017
Problem
Freddy the frog is jumping around the coordinate plane searching for a river, which lies on the horizontal line . A fence is located at the horizontal line . On each jump Freddy randomly chooses a direction parallel to one of the coordinate axes and moves one unit in that direction. When he is at a point where , with equal likelihoods he chooses one of three directions where he either jumps parallel to the fence or jumps away from the fence, but he never chooses the direction that would have him cross over the fence to where . Freddy starts his search at the point and will stop once he reaches a point on the river. Find the expected number of jumps it will take Freddy to reach the river.
Solution
Clearly Freddy's -coordinate is irrelevant, so we let be the expected value of the number of jumps it will take him to reach the river from a given -coordinate. Observe that , and for all such that . Also note that . This gives . Plugging this into the equation for gives that or . Iteratively plugging this in gives that . Thus , , and .
See also
2016 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.