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.

  • $\begingroup$ How are the two given segments defined? Just as lengths? So we can position them as we like? Or with coordinates? $\endgroup$ – Ivo Beckers Nov 13 '15 at 9:31
  • $\begingroup$ The segments are defined as segments, geometrically. Let me know if this sounds unclear. $\endgroup$ – user3900460 Nov 13 '15 at 9:42

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  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

  • $\begingroup$ Great! Another solution (assuming the given segments aren't equal, i.e. the case isn't trivial) is 1) constructing a right triangle with one cathetus equal to the shorter segment and hypotenuse equal to the longer segment, then 2) extending the hypotenuse beyond the vertex not shared with mentioned cathetus, and finally 3) drawing the circle centered on the mentioned vertex with radius equal to the cathetus that shares this vertex with hypotenuse. The circle marks on hypotenuse and its extension ends of two desired segments, both of which begin from the other end of hypotenuse. $\endgroup$ – user3900460 Nov 13 '15 at 12:58
  • $\begingroup$ The construction can be explained by the fact that distance to the horizon of a sphere is the geometric mean of the distance to the closest point of the sphere and the distance to the farthest point of the sphere. Whereas distance to the center of a sphere is the arithmetic mean of the same distances. $\endgroup$ – user3900460 Nov 13 '15 at 13:06

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