1998 IMO Problems/Problem 2: Difference between revisions
mNo edit summary |
m latex |
||
| (7 intermediate revisions by 2 users not shown) | |||
| Line 1: | Line 1: | ||
In a competition, there are < | ==Problem== | ||
integer. Each judge rates each contestant as either “pass” or “fail”. Suppose < | |||
is a number such that, for any two judges, their ratings coincide for at most < | In a competition, there are <math>a</math> contestants and <math>b</math> judges, where <math>b\ge3</math> is an odd | ||
contestants. Prove that < | integer. Each judge rates each contestant as either “pass” or “fail”. Suppose <math>k</math> | ||
is a number such that, for any two judges, their ratings coincide for at most <math>k</math> | |||
contestants. Prove that <math>\frac{k}{a}\ge\frac{b-1}{2b}</math>. | |||
==Solution== | |||
Let <math>c_i</math> stand for the number of judges who pass the <math>i</math>th candidate. The number of pairs of judges who agree on the <math>i</math>th contestant is then given by | |||
\begin{align*} | |||
{c_i \choose 2} + {{b - c_i} \choose 2} &= \frac{1}{2}\left(c_i(c_i - 1) + (b - c_i)(b - c_i - 1) \right) \\ | |||
&= \frac{1}{2}\left( c_i^2 + (b - c_i)^2 - b \right) \\ | |||
&\geq \frac{1}{2}\left( \frac{b^2}{2} - b \right) \\ | |||
&= \frac{b^2 - 2b}{4} | |||
\end{align*} | |||
where the inequality follows from AM-QM. Since <math>b</math> is odd, <math>b^2 - 2b</math> is not divisible by <math>4</math> and we can strengthen the inequality to | |||
<cmath>{c_i \choose 2} + {{b - c_i} \choose 2} \geq \frac{b^2 - 2b + 1}{4} = \left(\frac{b - 1}{2}\right)^2.</cmath> | |||
Letting <math>N</math> stand for the number of instances where two judges agreed on a candidate, it follows that | |||
<cmath> N = \sum_{i = 1}^a {c_i \choose 2} + {{b - c_i} \choose 2} \geq a \cdot \left( \frac{b - 1}{2} \right)^2. </cmath> | |||
The given condition on <math>k</math> implies that | |||
<cmath> N \leq k \cdot {b \choose 2} = \frac{kb(b - 1)}{2}. </cmath> | |||
Therefore | |||
<cmath> a \cdot \left( \frac{b - 1}{2} \right)^2 \leq \frac{kb(b - 1)}{2}, </cmath> | |||
which simplifies to | |||
<cmath> \frac{k}{a} \geq \frac{b - 1}{2b}. </cmath> | |||
==See Also== | |||
{{IMO box|year=1998|num-b=1|num-a=3}} | |||
Latest revision as of 21:01, 4 July 2024
Problem
In a competition, there are 
contestants and 
judges, where 
is an odd
integer. Each judge rates each contestant as either “pass” or “fail”. Suppose 
is a number such that, for any two judges, their ratings coincide for at most
contestants. Prove that 
.
Solution
Let 
stand for the number of judges who pass the 
th candidate. The number of pairs of judges who agree on the th contestant is then given by
\begin{align*} {c_i \choose 2} + {{b - c_i} \choose 2} &= \frac{1}{2}\left(c_i(c_i - 1) + (b - c_i)(b - c_i - 1) \right) \\ &= \frac{1}{2}\left( c_i^2 + (b - c_i)^2 - b \right) \\ &\geq \frac{1}{2}\left( \frac{b^2}{2} - b \right) \\ &= \frac{b^2 - 2b}{4} \end{align*}
where the inequality follows from AM-QM. Since is odd, 
is not divisible by 
and we can strengthen the inequality to
![\[{c_i \choose 2} + {{b - c_i} \choose 2} \geq \frac{b^2 - 2b + 1}{4} = \left(\frac{b - 1}{2}\right)^2.\]](http://latex.artofproblemsolving.com/b/1/7/b171b18ac10a2c0f0a7bbd1eabf1410edeba864b.png)
Letting 
stand for the number of instances where two judges agreed on a candidate, it follows that
![\[N = \sum_{i = 1}^a {c_i \choose 2} + {{b - c_i} \choose 2} \geq a \cdot \left( \frac{b - 1}{2} \right)^2.\]](http://latex.artofproblemsolving.com/d/d/2/dd2364f4214f7010c2137216e277e6e54722e63e.png)
The given condition on implies that
![\[N \leq k \cdot {b \choose 2} = \frac{kb(b - 1)}{2}.\]](http://latex.artofproblemsolving.com/d/8/2/d822e4cde94139c63856f9ca61185a3157b3afb3.png)
Therefore
![\[a \cdot \left( \frac{b - 1}{2} \right)^2 \leq \frac{kb(b - 1)}{2},\]](http://latex.artofproblemsolving.com/c/6/6/c66c6465bc61dc52b9e264ec903870f9bd984f66.png)
which simplifies to
![\[\frac{k}{a} \geq \frac{b - 1}{2b}.\]](http://latex.artofproblemsolving.com/b/3/1/b3122f2a058fc60b564edde0125e319f117b2450.png)
See Also
| 1998 IMO (Problems) • Resources | ||
| Preceded by Problem 1 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 3 |
| All IMO Problems and Solutions | ||
