Copyright © University of Cambridge. All rights reserved.

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

Congratulations Joseph from Colyton Grammar School, Adam from the
University of Waterloo, Shaun from Nottingham High School and
Andrei from Tudor Vianu National College, Bucharest, Romania for
your solutions.

(1) If the area of the ellipse equals the area of the annulus then
$\pi ab = \pi b^2 - \pi a^2$ and so $ab = b^2 - a^2 $. Then,
dividing by $a^2$, $$ b/a = (b/a)^2 -1.$$ The ratio we want to find
is $b/a$, the ratio of the longer to the shorter axis of the
ellipse. So let $b/a = x$ then $$x^2 - x - 1 = 0.$$ Using quadratic
formula: $$x = {1\pm \sqrt 5 \over 2}$$ We choose the positive root
knowing that the ratio $b/a$ is positive so this ratio is equal to
the golden ratio.

(2) Note that $R$ appears itself in the nested root. Therefore we
can say $$R = \sqrt (1 + R)$$ and so $$ R^2 - R - 1 = 0.$$ We have
a quadratic of the same form as above. Hence we find $R$ to be the
golden ratio.