Difference between revisions of "1991 AIME Problems/Problem 13"

(Solution)
(Solution)
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P=\frac{r(r-1)}{(r+b)(r+b-1)}+\frac{b(b-1)}{(r+b)(r+b-1)}=\frac{r(r-1)+(t-r)(t-r-1)}{t(t-1)}=\frac{1}{2}.
 
P=\frac{r(r-1)}{(r+b)(r+b-1)}+\frac{b(b-1)}{(r+b)(r+b-1)}=\frac{r(r-1)+(t-r)(t-r-1)}{t(t-1)}=\frac{1}{2}.
 
</math>
 
</math>
Solving the quadratic equation for <math>r</math> in terms of <math>t</math>, one obtains
+
 
 +
Solving the resulting quadratic equation <math>r^{2}-rt+t(t-1)/4=0</math> for <math>r</math> in terms of <math>t</math>, one obtains that
 +
 
 +
<math>
 +
r=\frac{t\pm\sqrt{t}}{2}
 +
</math>
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=1991|num-b=12|num-a=14}}</math>
 
{{AIME box|year=1991|num-b=12|num-a=14}}</math>

Revision as of 18:22, 18 April 2007

Problem

A drawer contains a mixture of red socks and blue socks, at most 1991 in all. It so happens that, when two socks are selected randomly without replacement, there is a probability of exactly $\displaystyle \frac{1}{2}$ that both are red or both are blue. What is the largest possible number of red socks in the drawer that is consistent with this data?

Solution

Let $r$ and $b$ denote the number of red and blue socks, respectively. Also, let $t=r+b$.

The probability $P$ that when two socks are drawn without replacement, both are red or both are blue is given by

$P=\frac{r(r-1)}{(r+b)(r+b-1)}+\frac{b(b-1)}{(r+b)(r+b-1)}=\frac{r(r-1)+(t-r)(t-r-1)}{t(t-1)}=\frac{1}{2}.$

Solving the resulting quadratic equation $r^{2}-rt+t(t-1)/4=0$ for $r$ in terms of $t$, one obtains that

$r=\frac{t\pm\sqrt{t}}{2}$

See also

1991 AIME (ProblemsAnswer KeyResources)
Preceded by
Problem 12
Followed by
Problem 14
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

</math>

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