Difference between revisions of "1998 AIME Problems/Problem 9"
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== Solution == | == Solution == | ||
=== Solution 1 === | === Solution 1 === | ||
− | Let the two mathematicians be <math>M_1</math> and <math>M_2</math>. Consider plotting the times that they are on break on a [[coordinate plane]] with one axis being the time <math>M_1</math> arrives and the second axis being the time <math>M_2</math> arrives (in minutes past 9 a.m.). The two mathematicians meet each other when <math>|M_1-M_2| \leq m</math>. Also because the mathematicians arrive between 9 and 10, <math>0 \leq M_1,M_2 \leq 60</math>. | + | Let the two mathematicians be <math>M_1</math> and <math>M_2</math>. Consider plotting the times that they are on break on a [[coordinate plane]] with one axis being the time <math>M_1</math> arrives and the second axis being the time <math>M_2</math> arrives (in minutes past 9 a.m.). The two mathematicians meet each other when <math>|M_1-M_2| \leq m</math>. Also because the mathematicians arrive between 9 and 10, <math>0 \leq M_1,M_2 \leq 60</math>. Therefore, <math>60\times 60</math> square represents the possible arrival times of the mathematicians, while the shaded region represents the arrival times where they meet. |
− | + | <asy> | |
− | + | import graph; | |
+ | size(180); | ||
+ | real m=60-12*sqrt(15); | ||
+ | draw((0,0)--(60,0)--(60,60)--(0,60)--cycle); | ||
+ | fill((m,0)--(60,60-m)--(60,60)--(60-m,60)--(0,m)--(0,0)--cycle,lightgray); | ||
+ | draw((m,0)--(60,60-m)--(60,60)--(60-m,60)--(0,m)--(0,0)--cycle); | ||
+ | xaxis("$M_1$",-10,80); | ||
+ | yaxis("$M_2$",-10,80); | ||
+ | label(rotate(45)*"$M_1-M_2\le m$",((m+60)/2,(60-m)/2),NW,fontsize(9)); | ||
+ | label(rotate(45)*"$M_1-M_2\ge -m$",((60-m)/2,(m+60)/2),SE,fontsize(9)); | ||
+ | label("$m$",(m,0),S); | ||
+ | label("$m$",(0,m),W); | ||
+ | label("$60$",(60,0),S); | ||
+ | label("$60$",(0,60),W); | ||
+ | </asy> | ||
+ | It's easier to compete the area of the unshaded region over the area of the total region, which is the probability that the mathematicians do not meet: | ||
<div style="text-align:center;"> | <div style="text-align:center;"> | ||
<math>\frac{(60-m)^2}{60^2} = .6</math><br /> | <math>\frac{(60-m)^2}{60^2} = .6</math><br /> |
Revision as of 21:06, 5 August 2015
Problem
Two mathematicians take a morning coffee break each day. They arrive at the cafeteria independently, at random times between 9 a.m. and 10 a.m., and stay for exactly minutes. The probability that either one arrives while the other is in the cafeteria is and where and are positive integers, and is not divisible by the square of any prime. Find
Solution
Solution 1
Let the two mathematicians be and . Consider plotting the times that they are on break on a coordinate plane with one axis being the time arrives and the second axis being the time arrives (in minutes past 9 a.m.). The two mathematicians meet each other when . Also because the mathematicians arrive between 9 and 10, . Therefore, square represents the possible arrival times of the mathematicians, while the shaded region represents the arrival times where they meet. It's easier to compete the area of the unshaded region over the area of the total region, which is the probability that the mathematicians do not meet:
So the answer is .
Solution 2
Case 1:
We draw a number line representing the time interval. If mathematician comes in at the center of the time period, then the two mathematicions will meet if comes in somewhere between minutes before and after comes (a total range of minutes). However, if comes into the cafeteria in the first or last minutes, then the range in which is reduced to somewhere in between and .
We know try to find the weighted average of the chance that the two meet. In the central minutes, and have to enter the cafeteria within minutes of each other; so if we fix point then has a probability of meeting.
In the first and last minutes, the probability that the two meet ranges from to , depending upon the location of with respect to the endpoints. Intuitively, the average probability will occur at .
So the weighted average is:
Solving this quadratic, we get two roots, . However, , so we discard the greater root; and thus our answer .
See also
1998 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
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.