These are the corrections to The IMO Compendium (Second Edition). Thanks to Alexander Bogomolny, Stijn Cambie, Jorgen Doenhardt, Jim Michael, Alberto Torregrosa, and Jozsef Pelikan. The list is updated on March 28, 2013.

- Page 27, problem 5: Change \(AMKD\) to \(AMCD\). Correct the page number from \(327\) to \(27\).
- Page 52, last footnote. Change
the word ``maximum’’ to ``maximal’’
- Page 235, problem 6, last line. Change
the first word ``integers’’ to ``integer’’
- Page 353, problem 3, first line. Change ``\(\cos 2x=1+\cos^2x\)’’ to ``\(\cos2x=2\cos^2x-1\)’’.
- Page 730, problem 2, second line. Change ``\(f(x_1)f(y)=f(\varphi(x_1))=f(\varphi(x_2))=\)’’ to ``\(f(x_1)f(y)=2f(\varphi(x_1))=2f(\varphi(x_2))=\)’’
- Page 764, problem 29. In line 5 change \((4x^2-1)\) to \((4x^2-1)^2\). In line 7 change \(16y^2\) to \(16y^4\).
In line 12 change \(2l+y\) to \(2x+l\).
- Page 781, problem 11. Change lines 8-21 to:
_quot_ Let us now prove the other direction, that the total perimeter is greater than or equal to \((m+1)2^{m+2}\). Denote by \(R\) the set of rectangles in partition that do not contain diagonal squares, and let \(Q\) be the sum of perimeters of rectangles from \(R\). Since the sum of the perimeters of diagonal squares is equal to \(4\cdot 2^m\) it suffices to prove that \(Q\geq m\cdot 2^{m+1}\). Assume that \(n=|Q|\). Let \(R_i\subseteq R\) be the set of those rectangles that contain at least one square from the \(i\)th row. Similarly, let \(C_i\subseteq R\) be the set of rectangles that contain the squares from the \(i\)th column. Clearly, \(C_i\cap R_i=\emptyset\) and
\[Q=2\left(\sum_{i=1}^{2^m}|R_i|+\sum_{i=1}^{2^m}|C_i|\right).\]
Let \(\mathcal F\) be the collection of all subsets of \(R\). Let \(\mathcal F_i\) denote the collection of those subsets \(S\) that satisfy
\(R_i\subseteq S\) and \(C_i\cap S=\emptyset\). Since \(\mathcal F_1\), \(\mathcal F_2\), \(\dots\), \(\mathcal F_{2^m}\) are disjoint sets and \(|\mathcal F_i|= 2^{n-|C_i|-|R_i|}\), using Jensen’s inequality applied to \(f(x)=2^{-x}\), we obtain \[2^{n}=|\mathcal F|\geq 2^{n}\sum_{i=1}^{2^m}2^{-|C_i|-|R_i|}\geq 2^{n}\cdot 2^m\cdot 2^{-\frac1{2^m}\sum(|C_i|+|R_i|)}=2^{n+m-\frac1{2^m}\cdot\frac Q2}.\] This yields \(Q\geq m\cdot 2^{m+1}\), which is the relation we wanted to prove._quot_
- Page 782, problem 12. Line 10: Before the sentence ``It is impossible that \(x_i+x_{i+2}\geq i\) \(\dots\),’’ add the following paragraph:
``First consider the case when \(x_i\geq 1\) for some \(i\), say \(i=1\). Then \(x_3 < 1\) and \(x_4 < 1\). If \(x_2+x_4\geq1\), Cinderella should empty the buckets \(1\) and \(2\). Otherwise, Cinderella should empty the buckets \(1\) and \(5\). The inequalities \(x_2+x_4\geq 1\) and \(x_3+x_5\geq 1\) cannot hold simultaneously.’’

- Page 782, problem 12. Line 16: Replace \(6\) by \(3\).