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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.

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 Construction

Age 16 to 18 Challenge Level:

In this problem you start with a square, construct a golden rectangle and calculate the value of the golden ratio.

When you cut a square off a golden rectangle you are left with another rectangle whose sides are in the same ratio. In the diagram below, if you remove the orange square from rectangle $AEFD$, you are left with the yellow rectangle whose sides are in the same ratio as $AEFD$.

(1) Follow the instructions for drawing the rectangle $AEFD$. You can make the most accurate drawing by using a ruler and compasses. Draw a square $ABCD$ of side length 10 cm. Bisect $AB$ at $M$ and draw an arc of radius $MC$ to meet $AB$ produced at $E$. If you prefer you can just measure $MC$ and mark $E$ on $AB$ so that $ME=MC$. Draw $EF$ perpendicular to $AB$ to meet $DC$ produced at $F$.

Measure $AE$ and $BE$. From your measurements calculate the ratios $AE$/$AD$ and $BC/BE$. What do you notice?

(2) Calculate the exact lengths of $MC$, $AE$ and $BE$. Calculate the exact value of the ratios $AE$/$AD$ and $BC$/$BE$ and prove that they are equal. [Note: To get exact values you must work with surds and you will not be able to use a calculator.]

(3) Suppose this ratio is denoted by $\phi$ and take 10 cm as 1 unit then $AE$ is $\phi$ units. Show that $BE$ is $1/\phi$ units and hence $$\phi = 1 + {1\over \phi}\quad (1)$$

Explain just from equation (1), and without solving the equation, why the equation must have a solution between 1 and 2.

(4) Draw the graphs of $y=x$, $y=1/x$ and $y=1+1/x$ on the same axes and use your graph to find an approximate value for $\phi$.

(5) Solve equation (1) to find the value of $\phi$.