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In Robert Dixon's Mathographics, a regular pentagon is constructed with straightedge and compass only. It is the pentagon $ABCDE$ pictured below.

I am having trouble seeing why the central angles are all $72^\circ$, though. Can anyone provide the proof?

Also, does anyone know who this construction is due to? I haven't seen it anywhere, other than in Dixon's book; is it Dixon's result?

enter image description here

The result appears much more impressive without all the labels (which are slightly misplaced, please excuse this); however, I provided those so that answering would be easier. Also, it makes it easy to describe the steps in the construction:

1) Draw a circle (the red one) with center $h$.

2) Draw the perpendicular lines $\ell_1$ and $\ell_2$ through $h$. Locate the points of intersection $f$, $B$, and $g$ with the red circle.

3) Bisect the line segment $gh$. Denote the center by $a$.

4) Draw the green circle with center $a$ and radius $ah$.

5) Draw the other green circle (as in (3) and 4)).

6) Draw the line segment through $f$ and $a$.

7) Locate the points of intersection $b$ and $c$ of the line segment with the circle constructed in step 4).

8) Draw the blue arcs (both have center at $f$ and the radii are $fb$ and $fc$).

9) Locate the points of intersection $A$, $C$, $D$, and $E$.


I actually have the solution to the first question, and will post it unless a more elegant explanation is provided (which is probably likely). However, I find this construction particularly beautiful, and would like to know who it is attributed to (Dixon doesn't say, explicitly).

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    $\begingroup$ Would you be alright with a coordinate geometry proof? $\endgroup$ – J. M. is a poor mathematician Jan 1 '12 at 15:37
  • $\begingroup$ @J.M. Yes, that would be interesting. $\endgroup$ – David Mitra Jan 1 '12 at 16:00
  • $\begingroup$ The idea of the construction seems to me to be that the side (Ef) of a regular decagon has a certain nice(ish) relation to the corner radius of the decagon. This side is then being constructed using the Pythagorean triangle ahf. $\endgroup$ – Henning Makholm Jan 1 '12 at 17:07
  • $\begingroup$ On MathOpenRef there's an animated and detailed version of this construction. Unfortunately, they don't provide any references or justifications. On cut the knot there's a closely related construction attributed to Y. Hirano (19th century). $\endgroup$ – t.b. Jan 2 '12 at 5:28
  • $\begingroup$ I admire this question, but I had great difficulty seeing the colors you mention. I therefore reproduced your diagram as closely as I could and emphasized the colors. I hope you don't mind, but if you do mind go ahead and put your own graphic back. +1 $\endgroup$ – Rory Daulton Jul 5 '15 at 23:08
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Here is my explanation:

Let the radius of the circle drawn in step 1) be $2$ (units). Then the radius of the larger arc (through $A$ and $C$) drawn in step 8) is $1+\sqrt 5$ and the radius of the smaller arc is $ \sqrt5-1$ .

Triangle $\color{darkgreen}{\triangle hfD}$, thus, has side lengths $2$, $2$, and $\sqrt 5-1$; and so, is a golden triangle (see below) with angles $36^\circ$-$72^\circ$-$72^\circ$.
Triangle $\color{red}{\triangle fhC}$ has side lengths $2$, $2$, and $\sqrt 5+1$; and thus, is a golden triangle with angles $36^\circ$-$108^\circ$-$36^\circ$.

From this, one can deduce that the angles $\angle ChD$ and $\angle ChB$ have measures $72^\circ$ (note the angle formed by $\ell_1$ and the segment $\overline{hC}$ is $18^\circ$).

By symmetry, the angles $\angle BhA$ and $\angle AhE$ have measure $72^\circ$. The remaining angle, $\angle EhD$ must then have measure $72^\circ$.

Image 1


On golden triangles:

The golden ratio is $\tau={1+\sqrt 5\over 2}$; a golden triangle is an isosceles triangle with two sides in the golden ratio.

By considering similar triangles in the diagrams below, one can show that the golden triangles are the $36^\circ$-$72^\circ$-$72^\circ$ and the $36^\circ$-$108^\circ$-$36^\circ$ triangles.

Image 2 Image 3

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    $\begingroup$ My coordinate geometry route was essentially this, but you've managed to go about it purely synthetically. Nice! $\endgroup$ – J. M. is a poor mathematician Jan 2 '12 at 0:45

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