Difference between revisions of "1998 AIME Problems/Problem 9"
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== Problem == | == 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 <math>m</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> |
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Revision as of 20:55, 15 July 2011
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 and shading in the places where they would be there at the same time as such.
We can count the area that we don't want in terms of and solve:
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 |