You may also like

problem icon

Gold Again

Without using a calculator, computer or tables find the exact values of cos36cos72 and also cos36 - cos72.

problem icon

Pythagorean Golden Means

Show that the arithmetic mean, geometric mean and harmonic mean of a and b can be the lengths of the sides of a right-angles triangle if and only if a = bx^3, where x is the Golden Ratio.

problem icon

Golden Triangle

Three triangles ABC, CBD and ABD (where D is a point on AC) are all isosceles. Find all the angles. Prove that the ratio of AB to BC is equal to the golden ratio.

Golden Eggs

Stage: 5 Challenge Level: Challenge Level:2 Challenge Level:2

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.