<html><head><meta name="color-scheme" content="light dark"></head><body><pre style="word-wrap: break-word; white-space: pre-wrap;">\nopagenumbers
\noindent {\bf 15th IMO 1973}
\vskip 25pt
\noindent {\bf A1}. $OP_1,OP_2,\ldots,OP_{2n+1}$ are unit vectors in a plane. $P_1,P_2,\ldots,P_{2n+1}$ all lie on the same side of a line through $O$. Prove that $|OP_1+\ldots+OP_{2n+1}|\ge1$.
\vskip 12pt
\noindent {\bf A2}. Can we find a finite set of non-coplanar points, such that given any two points, $A$ and $B$, there are two others $C$ and $D$, with the lines $AB$ and $CD$ parallel and distinct?
\vskip 12pt
\noindent {\bf A3}. $a$ and $b$ are real numbers for which the equation $x^4+ax^3+bx^2+ax+1=0$ has at least one real solution. Find the least possible value of $a^2+b^2$.
\vskip 12pt
\noindent {\bf B1}. A soldier needs to sweep a region with the shape of an equilateral triangle for mines. The detector has an effective radius equal to half the altitude of the triangle. He starts at a vertex of the triangle. What path should he follow in order to travel the least distance and still sweep the whole region?
\vskip 12pt
\noindent {\bf B2}. $G$ is a set of non-constant functions $f$. Each $f$ is defined on the real line and has the form $f(x)=ax+b$ for some real $a,b$. If $f$ and $g$ are in $G$, then so is $fg$, where $fg$ is defined by $fg(x)=f(g(x))$. If $f$ is in $G$, then so is the inverse $f^{-1}$. If $f(x)=ax+b$, then $f^{-1}(x)={x-b\over a}$. Every $f$ in $G$ has a fixed point (in other words we can find $x_f$ such that $f(x_f)=x_f$. Prove that all the functions in $G$ have a common fixed point.
\vskip 12pt
\noindent {\bf B3}. $a_1,a_2,\ldots,a_n$ are positive reals, and $q$ satisfies $0&lt;q&lt;1$. Find $b_1,b_2,\ldots,b_n$, such that:

(a) $a_i&lt;b_i$ for $i=1,2,\ldots,n$,

(b) $q&lt;{b_{i+1}\over b_i}&lt;{1\over q}$ for $i=1,2,\ldots,n-1$,

(c) $b_1+b_2+\ldots+b_n&lt;(a_1+a_2+\ldots+a_n){1+q\over1-q}$.
\vskip 20pt
\noindent \copyright John Scholes

\noindent jscholes@kalva.demon.co.uk

\noindent 19 August 2003

\bye
</pre></body></html>