# $5^{\text {nd }}$ International Mathematical Olympiad 

12 - 24 July 2011 Amsterdam The Netherlands

![](https://cdn.mathpix.com/cropped/2024_04_17_481ddb3576adc2313a2eg-01.jpg?height=2070&width=1902&top_left_y=744&top_left_x=-3)

## 52nd International <br> Mathematical Olympiad <br> 12-24 July 2011 <br> Amsterdam <br> The Netherlands

Problem shortlist with solutions

## IMO regulation: <br> these shortlist problems have to be kept strictly confidential until IMO 2012.

The problem selection committee

Bart de Smit (chairman), Ilya Bogdanov, Johan Bosman,

Andries Brouwer, Gabriele Dalla Torre, Géza Kós,

Hendrik Lenstra, Charles Leytem, Ronald van Luijk,

Christian Reiher, Eckard Specht, Hans Sterk, Lenny Taelman

The committee gratefully acknowledges the receipt of 142 problem proposals by the following 46 countries:

Armenia, Australia, Austria, Belarus, Belgium, Bosnia and Herzegovina, Brazil, Bulgaria, Canada, Colombia, Cyprus, Denmark, Estonia, Finland, France, Germany, Greece, Hong Kong, Hungary, India, Islamic Republic of Iran, Ireland, Israel, Japan, Kazakhstan, Republic of Korea, Luxembourg, Malaysia, Mexico, Mongolia, Montenegro, Pakistan, Poland, Romania, Russian Federation, Saudi Arabia, Serbia, Slovakia, Slovenia, Sweden, Taiwan, Thailand, Turkey, Ukraine, United Kingdom, United States of America


## Algebra

A1. Find the roots of the quadratic equation \(x^2 - 4x + 4 = 0\).

Solution. The equation simplifies to \((x - 2)^2 = 0\), so the root is \(x = 2\).

A2. Determine the value of \(y\) in the equation \(2y + 3 = 7\).

Solution 1. Subtracting 3 from both sides gives \(2y = 4\).

Solution 2. Dividing both sides by 2 gives \(y = 2\).

## Combinatorics

C1. How many ways can 3 objects be arranged in a line?

Solution. The number of permutations of 3 objects is \(3! = 6\).