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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.
Golden Eggs
Age 16 to 18
Challenge Level 
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.
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