Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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).

share|improve this question
1  
Would you be alright with a coordinate geometry proof? –  J. M. Jan 1 '12 at 15:37
    
@J.M. Yes, that would be interesting. –  David Mitra Jan 1 '12 at 16:00
    
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. –  Henning Makholm Jan 1 '12 at 17:07
    
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). –  t.b. Jan 2 '12 at 5:28

1 Answer 1

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 $72^\circ$.

enter image description here




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.

enter image description here enter image description here

share|improve this answer
1  
My coordinate geometry route was essentially this, but you've managed to go about it purely synthetically. Nice! –  J. M. Jan 2 '12 at 0:45

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.