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1998 AIME Problems/Problem 9

Revision as of 15:15, 11 August 2012 by Fro116 (talk | contribs) (→‎Solution 1)

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 $m$ minutes. The probability that either one arrives while the other is in the cafeteria is $40 \%,$ and $m = a - b\sqrt {c},$ where $a, b,$ and $c$ are positive integers, and $c$ is not divisible by the square of any prime. Find $a + b + c.$

Solution

Solution 1

Let the two mathematicians be $M_1$ and $M_2$. Consider plotting the times that they are on break on a coordinate plane with one axis being the time $M_1$ arrives and the second axis being the time $M_2$ arrives (in minutes past 9 a.m.). The two mathematicians meet each other when $|M_1-M_2| \leq m$. Also because the mathematicians arrive between 9 and 10, $0 \leq M_1,M_2 \leq 60$.

We can use geometric probability to find the probability that the mathematicians do not meet:

$\frac{(60-m)^2}{60^2} = .6$
$(60-m)^2 = 36\cdot 60$
$60 - m = 12\sqrt{15}$
$\Rightarrow m = 60-12\sqrt{15}$

So the answer is $60 + 12 + 15 = 087$.

Solution 2

Case 1:

AIME 1998-9.png Case 2:

AIME 1998-9b.png

We draw a number line representing the time interval. If mathematician $M_1$ comes in at the center of the time period, then the two mathematicions will meet if $M_2$ comes in somewhere between $m$ minutes before and after $M_1$ comes (a total range of $2m$ minutes). However, if $M_1$ comes into the cafeteria in the first or last $m$ minutes, then the range in which $M_2$ is reduced to somewhere in between $m$ and $2m$.

We know try to find the weighted average of the chance that the two meet. In the central $\displaystyle 60-2m$ minutes, $M_1$ and $M_2$ have to enter the cafeteria within $m$ minutes of each other; so if we fix point $M_1$ then $M_2$ has a $\frac{2m}{60} = \frac{m}{30}$ probability of meeting.

In the first and last $2m$ minutes, the probability that the two meet ranges from $\frac{m}{60}$ to $\frac{2m}{60}$, depending upon the location of $M_1$ with respect to the endpoints. Intuitively, the average probability will occur at $\frac{\frac{3}{2}m}{60} = \frac{m}{40}$.

So the weighted average is:

$\frac{\frac{m}{30}(60-2m) + \frac{m}{40}(2m)}{60} = \frac{40}{100}$
$0 = \frac{m^2}{60} - 2m + \frac{2}{5}$
$0 = m^2 - 120m + 1440$

Solving this quadratic, we get two roots, $\displaystyle 60 \pm 12\sqrt{15}$. However, $m < 60$, so we discard the greater root; and thus our answer $60 + 12 + 15 = 087$.

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
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