\documentclass{article}
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\title{December Exam 2}
\author{This is a 4 hour paper.}
\date{Various questions from Ukraine}
\begin{document}
\maketitle
\begin{enumerate}
\item Does there exist a function
$f : \mathbb R \rightarrow \mathbb R$ such that for all
$x,y \in \mathbb R$ the following equality holds:
\[ f(xy) = \max \{f(x),y\} + \min \{f(y),x\}?\] 
\noindent {\bf Solution}
{ \em Put $x = y =1$ to obtain that
\[ f(1) = \max \{ f(1), 1\} + \min \{ f(1), 1\} = f(1) + 1\]
which is absurd.}
\item Positive integers $a$ and $n$ are such that
$n$ divides $a^2 + 1$. Prove that there is a positive 
integer $b$ such that $n(n^2 +1)$ divides $b^2 +1$. \newline
\noindent {\bf Solution } {\em 
We have $a^2 \equiv -1 \mbox{ mod } n$ and 
$n^2 \equiv -1 \mbox{ mod } n^2 +1.$ Now
$n, n^2 +1$ are coprime (a common divisor would divide 1)
so we may apply the Chinese Remainder Theorem
to find an integer $b$ such that 
$b \equiv a \mbox{ mod } n$ and $b \equiv n \mbox{ mod }n^2+1$
so $b^2 \equiv a^2 \equiv -1 \mbox{ mod } n$ and
$b^2 \equiv n^2 \equiv -1 \mbox{ mod } n^2+1$.
Thus $b^2 \equiv -1 \mbox{ mod } n(n^2+1)$
and therefore $n(n^2+1) \mid b^2 + 1.$ Of course, if you are Tom Coker
or Paul Jefferys,
then you are at liberty to side-step the Chinese Remainder Theorem
and simply hurl a bolt of lightning by
letting $b = an^2 + a + n$ which visibly
has the appropriate divisibility properties!}

\item An acute angled triangle $ABC$ is such that $AC$ and $BC$
are of different lengths, and is inscribed in a circle $\omega$.
Let $N$ be the midpoint of the arc $AC$ (and $B$ does not lie
on this arc). Let $M$ be the midpoint of the arc $BC$ (and $A$
does not lie on this arc). Let $D$ be the point on the arc
$MN$ such that $DC \parallel NM$. Let $K$
be an arbitrary point on the arc $AB$ (and $C$ does not lie on this arc).
Let $O, O_1$ and $O_2$ be the incentres of the triangles
$ABC,\ ACK,\ CBK$ respectively. Suppose that $L$ is the intersection
of the line $DO$ and the circle $\omega$ such that $L \not = D$.
Prove that the points $K, O_1, O_2, L$ are concyclic. \newline
{\bf So far no solution has ben offered.}

\item Let $a,b,c$ and $\alpha, \beta, \gamma$ be positive
real numbers such that $\alpha + \beta + \gamma = 1$.
Prove the inequality
\[ \alpha a + \beta b + \gamma c 
+ 2 \sqrt{(\alpha \beta + \beta \gamma + \gamma \alpha)
(ab + bc + ca) } \leq a + b + c.\]
\noindent {\bf Solution} (Thanks to Tim Austin and Paul Jefferys)
{\em 
Replacing $a$ by $a/(a +b +c)$, $b$ by
$b/(a +b+c)$ and $c$ by $c/(a+b+c)$ we see that it suffices
to assume that $a + b + c = 1$ and to show that
\[ \alpha a + \beta b + \gamma c +
2 \sqrt{(\alpha \beta + \beta \gamma + \gamma \alpha)
( ab + bc + ca) } \leq 1.\]
Now by the Cauchy-Schwarz inequality,
\[ \alpha a + \beta b + \gamma c \leq \sqrt{ \alpha^2 + \beta^2 + 
\gamma^2}\sqrt{a^2 + b^2 + c^2}\ \ \ \ \ \  (\dagger)\]
Let $p = \alpha \beta + \beta \gamma + \gamma \alpha$
and $q = ab + bc + ca$. Now
$\alpha^2 + \beta^2 + \gamma^2 = 1 - 2p$ and 
$a^2 + b^2 + c^2 = 1 - 2q$. Now $(\dagger)$ implies that
\[ \alpha a + \beta b + \gamma c + 2 \sqrt{pq}
\leq \sqrt{1-2p} \sqrt{1-2q} + 2\sqrt{pq}\]
\[ \leq (1 - 2p + 1 - 2q)/2 + p + q = 1\]
using the GM-AM inequality.}
\end{enumerate}
\end{document}