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

(lines, wikify. Credit for 1st solution goes to joml88)
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== Problem ==
 
== Problem ==
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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 <math>m</math> mintues.  The [[probability]] that either one arrives while the other is in the cafeteria is <math>40 \%,</math> and <math>m = a - b\sqrt {c},</math> where <math>a, b,</math> and <math>c</math> are [[positive]] [[integer]]s, and <math>c</math> is not divisible by the square of any [[prime]].  Find <math>\displaystyle a + b + c.</math>
+
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 <math>m</math> minutes.  The [[probability]] that either one arrives while the other is in the cafeteria is <math>40 \%,</math> and <math>m = a - b\sqrt {c},</math> where <math>a, b,</math> and <math>c</math> are [[positive]] [[integer]]s, and <math>c</math> is not divisible by the square of any [[prime]].  Find <math>a + b + c.</math>
  
 
__TOC__
 
__TOC__
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== Solution ==
 
== Solution ==
 
=== Solution 1 ===
 
=== Solution 1 ===
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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]] and shading in the places where they would be there at the same time as such.
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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.
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+
<asy>
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We can count the area that we don't want in terms of <math>m</math> and solve:
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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);
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label("$60$",(60,0),S);
 +
label("$60$",(0,60),W);
 +
</asy>
 +
It's easier to compute 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 />
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=== Solution 2 ===
 
=== Solution 2 ===
−
[[Image:AIME_1998-9.png|350px]]
+
Case 1:
 +
 
 +
[[Image:AIME_1998-9.png]]
 +
Case 2:
  
−
[[Image:AIME_1998-9b.png|350px]]
+
[[Image:AIME_1998-9b.png]]
  
 
We draw a [[number line]] representing the time interval. If mathematician <math>M_1</math> comes in at the center of the time period, then the two mathematicions will meet if <math>M_2</math> comes in somewhere between <math>m</math> minutes before and after <math>M_1</math> comes (a total range of <math>2m</math> minutes). However, if <math>M_1</math> comes into the cafeteria in the first or last <math>m</math> minutes, then the range in which <math>M_2</math> is reduced to somewhere in between <math>m</math> and <math>2m</math>.
 
We draw a [[number line]] representing the time interval. If mathematician <math>M_1</math> comes in at the center of the time period, then the two mathematicions will meet if <math>M_2</math> comes in somewhere between <math>m</math> minutes before and after <math>M_1</math> comes (a total range of <math>2m</math> minutes). However, if <math>M_1</math> comes into the cafeteria in the first or last <math>m</math> minutes, then the range in which <math>M_2</math> is reduced to somewhere in between <math>m</math> and <math>2m</math>.
  
−
We know try to find the [[weighted average]] of the chance that the two meet. In the central <math>\displaystyle 60-2m</math> minutes, <math>M_1</math> and <math>M_2</math> have to enter the cafeteria within <math>m</math> minutes of each other; so if we fix point <math>M_1</math> then <math>M_2</math> has a <math>\frac{2m}{60} = \frac{m}{30}</math> probability of meeting.
+
We know try to find the [[weighted average]] of the chance that the two meet. In the central <math>60-2m</math> minutes, <math>M_1</math> and <math>M_2</math> have to enter the cafeteria within <math>m</math> minutes of each other; so if we fix point <math>M_1</math> then <math>M_2</math> has a <math>\frac{2m}{60} = \frac{m}{30}</math> probability of meeting.
  
 
In the first and last <math>2m</math> minutes, the probability that the two meet ranges from <math>\frac{m}{60}</math> to <math>\frac{2m}{60}</math>, depending upon the location of <math>M_1</math> with respect to the endpoints. Intuitively, the average probability will occur at <math>\frac{\frac{3}{2}m}{60} = \frac{m}{40}</math>.
 
In the first and last <math>2m</math> minutes, the probability that the two meet ranges from <math>\frac{m}{60}</math> to <math>\frac{2m}{60}</math>, depending upon the location of <math>M_1</math> with respect to the endpoints. Intuitively, the average probability will occur at <math>\frac{\frac{3}{2}m}{60} = \frac{m}{40}</math>.
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So the weighted average is:
 
So the weighted average is:
 
:<math>\frac{\frac{m}{30}(60-2m) + \frac{m}{40}(2m)}{60} = \frac{40}{100}</math>
 
:<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>
 
:<math>0 = m^2 - 120m + 1440</math>
  
−
Solving this [[quadratic equation|quadratic]], we get two roots, <math>\displaystyle 60 \pm 12\sqrt{15}</math>. However, <math>m < 60</math>, so we discard the greater root; and thus our answer <math>60 + 12 + 15 = 087</math>.
+
Solving this [[quadratic equation|quadratic]], we get two roots, <math>60 \pm 12\sqrt{15}</math>. However, <math>m < 60</math>, so we discard the greater root; and thus our answer <math>60 + 12 + 15 = 087</math>.
  
 
== See also ==
 
== See also ==
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[[Category:Intermediate Combinatorics Problems]]
 
[[Category:Intermediate Combinatorics Problems]]
 +
{{MAA Notice}}

Latest revision as of 18:34, 18 February 2018

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 compute 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:

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 $60-2m$ 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 $60 + 12 + 15 = 087$.

See also

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

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