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## 'Golden Thoughts' printed from http://nrich.maths.org/

Sue from Madras College, St Andrew's, Scotland
sent in a good solution to this problem.

Let $SX = a$ and $XR= b$ then $PQ=a+b$ and, since the areas of the
triangles $SPX, XYR$ and $PQY$ are equal, we know that $YR= ax/b$
and $QY = ax/(a+b)$. Because $PS = QR$ we now have $$
\frac{ax}{a+b} + \frac{ax}{b} = x$$ and simplifying this we get
$b^2 -ab - a^2 = 0$. So, writing $t$ for $\frac{b}{a}$, we get the
quadratic equation $$ t^2 - t - 1 = 0$$ which has roots $$ t =
\frac{1 \pm \sqrt 5}{2}$$ but we know that $t$ is positive so $$ t
=\frac {1 + \sqrt 5}{2}.$$ Also, writing the ratio $RY/YQ$ in terms
of $t$ we get $$ \frac{RY}{YQ} = \frac{a+b}{b} = \frac {1}{t} +
1.$$ We already know that $t=\frac{1}{t} + 1$ because $t^2 = 1 +
t$. So $$ \frac{RX}{XS} = \frac {RY}{YQ} = \frac {1 + \sqrt
5}{2}.$$