Compass-and-straightedge construction of two segments, given their arithmetic and geometric means

Question

Given two segments, construct, using compass and straightedge, two other segments such that the lengths of given segments represent arithmetic and geometric means of the lengths of constructed segments.

Answer 1

1. Set your compass to the length of the arithmetic mean ($m$) and draw a circle with radius $m$. Call the center $O$.

2. Draw a radius $OA$.

3. Draw a perpendicular radius $OB$.

4. Set your compass to the length of the geometric mean ($g$) and mark a point $C$ on the line $OB$, such that $\overline{OC}=g$.

5. Draw a line perpendicular to $OB$ through $C$. Call its intersection with the circle $D$.

   Since the circle has radius $m$, we can write $\overline{OC}^2 + \overline{CD}^2 = m^2$. Since $\overline{OC}=g$, we then have:

   $$ g^2 + \overline{CD}^2 = m^2 \\ \overline{CD} = \sqrt{m^2 - g^2} $$

6. Set your compass to the length of $CD$ and draw another circle centered on $O$. Call its intersections with $OA$ points $E$ and $F$.

We now have two segments of interest, $AE$ and $AF$. Their lengths are, respectively:

$$ \begin{align} \overline{AE} &= m - \sqrt{m^2 - g^2} \\ \overline{AF} &= m + \sqrt{m^2 - g^2} \\ \end{align} $$

Calculating their arithmetic and geometric means gives:

$$ \begin{align} \frac{\overline{AE}+\overline{AF}}{2} &= \frac{m - \sqrt{m^2 - g^2} + m + \sqrt{m^2 - g^2}}{2} \\ &= \frac{m + m}{2} \\ &= m \\ \sqrt{\left(\overline{AE}\right)\left(\overline{AF}\right)} &= \sqrt{\left(m - \sqrt{m^2 - g^2}\right)\left(m + \sqrt{m^2 - g^2}\right)} \\ &= \sqrt{m^2 - \left(\sqrt{m^2 - g^2}\right)^2} \\ &= \sqrt{m^2 - \left(m^2 - g^2\right)} \\ &= \sqrt{g^2} \\ &= g \end{align} $$

Thus, as desired, we have constructed two segments $AE$ and $AF$ whose arithmetic mean is $m$ and geometric mean is $g$. QED