Difference between revisions of "2003 AIME I Problems/Problem 11"
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− | The [[probability]] that <math>x</math> lies in this range is <math>\frac 1{45} \cdot \left(\frac 12 \arctan 2\right) = \frac{\arctan 2}{90}</math> so that <math>m = 2</math>, <math>n = 90</math> and our answer is <math> | + | The [[probability]] that <math>x</math> lies in this range is <math>\frac 1{45} \cdot \left(\frac 12 \arctan 2\right) = \frac{\arctan 2}{90}</math> so that <math>m = 2</math>, <math>n = 90</math> and our answer is <math>92</math>. |
== See also == | == See also == |
Revision as of 02:53, 6 November 2006
Problem
An angle is chosen at random from the interval
Let
be the probability that the numbers
and
are not the lengths of the sides of a triangle. Given that
where
is the number of degrees in
and
and
are positive integers with
find
Solution
Note that the three expressions are symmetric with respect to interchanging and
, and so the probability is symmetric around
. Thus, take
so that
. Then
is the largest of the three given expressions and those three lengths not forming a triangle is equivalent to a violation of the triangle inequality
This is equivalent to
and, using some of our trigonometric identities, we can re-write this as
and since we've chosen
this means
so
or
.
The probability that lies in this range is
so that
,
and our answer is
.