Difference between revisions of "1998 AIME Problems/Problem 9"

Revision as of 22:26, 30 October 2016

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, $0 \leq M_1,M_2 \leq 60$ . Therefore, $60\times 60$ 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:

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

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

Solution 2

Case 1:

Case 2:

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 $60-2m$ 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:

$\frac{\frac{m}{30}(60-2m) + \frac{m}{40}(2m)}{60} = \frac{40}{100}$

$0 = \frac{m^2}{60} - 2m + 24$

$0 = m^2 - 120m + 1440$

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

@@ Line 48: / Line 48: @@
 So the weighted average is:
 :<math>\frac{\frac{m}{30}(60-2m) + \frac{m}{40}(2m)}{60} = \frac{40}{100}</math>
-:<math>0 = \frac{m^2}{60} - 2m + \frac{2}{5}</math>
+:<math>0 = \frac{m^2}{60} - 2m + 24</math>
 :<math>0 = m^2 - 120m + 1440</math>

1998 AIME (Problems • Answer Key • Resources)
Preceded by Problem 8	Followed by Problem 10
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