The Geometer's Angle No. 4: From Pentagon to
A Discovery on the Generation of the Regular Heptagon from the
Equilateral Triangle and Pentagon.
Mill Valley, California 94941 USA
The New Year always begins with increasing Light.
In the last column, I mentioned
that we would visit the golden section, phi (from now
on we will use the symbol f to represent
the golden section in our demonstration). Now before you move
onto another article, I can assure you that we will be drawing
our information from the regular pentagon, and using the mathematical
standard: f = (Ö5
+ 1) ÷ 2. We won't be debating uses, approximations and
the myriad of other difficulties f
sometimes elicits in the world of scholarship and postmodern
thought. That said, I present the discovery of the first two
regular, odd-sided polygons -- equilateral triangle and pentagon
-- generating the next odd-sided polygon, the regular heptagon.
This construction achieves a heptagon that is within an incredibly
small percent deviation from the ideal. The heptagon, the side
of which subtends a rational angle number having the repeating
decimal expansion 51.428571428571...°, cannot be precisely
rendered with compasses and straightedge. The present construction,
however, comes about as close as possible. The relationship between
the incircle and excircle of the regular pentagon is the key
to this construction, and their ratio is 2 : f.
In other words, the golden section plays the critical role in
the establishment of this construction.
We will begin with a construction and a brief analysis of
the regular pentagon. This specific construction is credited
to Albrecht Dürer. There are a variety of procedures, but
I find this one to be of interest because of its utilization
of the vesica piscis and the circle.
DRAWING 1. Step-by-step
procedure for the construction of the pentagon from the vesica
piscis, after Albrecht Dürer
- Draw the horizontal line KM, a couple of inches in length,
as in Figure
- With the compasses opened to this length (KM), place the
pin in K, draw the arc, AMZ, and extend the arc beyond A and
Z, to P and G, as in Figure 1.2.
- Repeat this process with the pin in M, to draw arc, AKZ,
and extend the arc beyond A and Z,
to R and N.
- Draw the vertical line (perpendicular bisector to KM) AZ,
as in Figure 1.3.
- AKMZ is a vesica piscis with its two axes (AZ : KM
:: Ö3 : 1).
- Continuing with the same radius, KM, in Figure
1.4, with the pin in Z, draw
arc GKMN. (ZGKM and ZKMN are also both vesicas, and of
the same size as the first.)
- Locate the point Q where arc GKMN and the vertical AZ intersect.
- As shown in Figure 1-5.a, draw a straight line from G through
Q to intersect arc GMAP at P; continuing, draw a straight line
from N through Q to intersect arc NPAR at R.
- Using the pin of the compasses in points P and R, rotate
sides PK and RM to intersect at point J, as in Figure
- In Figure 1-6, points K,P, J, R and M are vertexes
of a regular pentagon.
- Draw the regular pentagon, KPJRM. (KP=PJ=JR=RM=MK.)
We will now look at this first drawing of the pentagon to
see some of the relationships within the five-sided figure, including
the golden section.
- As part of our subject on the golden section, we find that
where any two pentagonal chords (the straight lines connecting
every other vertex) intersect one another divides those chords
into, "extreme and mean ratio", the golden section.
- The pentagonal star - the pentagram - can be drawn inside
the pentagon by drawing the five chords. These chords will intersect
around the center of the pentagram and form a new, smaller pentagon.
This progression continues both micro- and macro-cosmically to
infinity. The sides of the pentagons and the arms of the pentagrams
will be in a golden section progression based on (Ö5
+ 1) ÷ 2 = f, or approximately
- The triangle formed from the center of the pentagon to the
base (cathetus), the half-base (base), and the line from the
center of the pentagon to the vertex (hypoteneuse) is a right,
scalene triangle, whose angles are 36°- 54° - 90°.
(Very rarely, one hears the triangle called, "The
Sacred MR Triangle of the Egyptians".)
- The cathetus's (vertical) length of this triangle forms the
radius of the incircle of the pentagon.
- The hypoteneuse of the triangle forms the radius of the pentagon's
circumcircle, or excircle.
- These two circles are in a 2 : f
(approximately 1.23606 : 1) relationship. This relationship is
the key to the entire drawing, as the spacing between the two
circles of the pentagon will be used for the smaller circle which
will enclose the equilateral triangle, and this spacing doubled
will enclose the heptagon. Hence, the spacing of the concentric
rings based those of the pentagon will be in the ratio 1 : 2.
A beautiful relationship also exists within the pentagon in
that the chord of the pentagon is the geometric mean between
the extremes of the height of the pentagon and the diameter of
the pentagon's excircle. Following is the procedure for establishing
DRAWING 2. The Pentagon's Geometric
- Given: pentagon, AKRMZ (Figure
- With center, G, circumscribe a circle tangent to the vertexes
of the pentagon.
- Draw the diameter, PN, of the circle.
- Draw the chord of the pentagon, KM.
- Drop the vertical, RB, from the vertex, R, passing through
the center, G, perpendicular to the chord, KM, the base, AZ,
and (in this drawing) the diameter, PN.
- RB intersects AZ at B, and is not a diameter.
We can now use the height of the pentagon, RB, to be the first
extreme (A), and the diameter of the circumscribing circle, PN,
as the second extreme (C ). We now apply the formula for finding
the geometric mean (B) between two extremes:
B = ÖAC
The geometric mean is, KM, the chord of the pentagon.
While studying this construction, my attention was drawn to
the small difference between the two extremes in our geometric
mean relationship. Taking their half-lengths, GN and GB, we can
generate two circles; that is, the circumscribing circle commonly
termed the excircle, and a second circle that is inscribed
within the pentagon and tangent to the sides internally, called
the incircle. Curiosity led me to experiment with this
length along with other concentric circles using this small difference
between the two circles. I did not know the delight I was to
find in my search!
360° ÷ 7 = 51. 428571428571
It is important now to discuss the mechanism within the pentagon
that will enable us to come extremely close to constructing a
regular heptagon. The angle subtended by 1/7 of a circle required
in generating the heptagon is,
As there is no construction known to generate this exact angle,
we can only approximate the 7 angles and sides to be "equal".
The protractor's calibrations do not have decimals such as these,
and AutoCad, like the protractor, cannot be accepted in the classical
requirements of using only compasses and straight edge for any
geometric constructions. In this drawing, however, we are given
a relatively simple and easy task, and with some care, the heptagon
should be drawn without great effort. Careful measure with the
compasses and accurate drafting will show just how close this
drawing truly is.
The Procedural Steps for the Generation of the Heptagon from
the Pentagon and the Equilateral Triangle, and Starting with
the Regular Pentagon.
- Draw a circle, O, with a radius of a few inches. When we
do the proof, I will assign a specific scale of measures to the
necessary geometric lengths for understanding the relationships.
For now, the drawing need only be a comfortable size for the
tools being used. (If you've done the other constructions, proceed
to the two primary circles of the pentagon.) Given: pentagon
- From the center of this circle, mark off 5 radii, OK, OP,
OJ, OR, and OM, spaced at every 72°. (For convenience, clarity,
and symmetry, place the radius, OJ, pointing up vertically to
the "12 o'clock" position. The drawing is now bilaterally
- Inscribe the sides of the pentagon within the circle where
the radii intersect the circle, at K, P, J, R, and M. This circle
is the circumcircle or excircle. A radius of this
circle, OM for example, is labeled r1.
- Inscribe a circle with the radius, r2, inside the pentagon
and tangent to each of the five mid-sides. This inscribed circle
is the incircle.
- In Figure 3.2, note well the small length of line,
a, the difference between the two radii, r1 and r2. These two
radii are in a ratio of 2 : f. This
length is extremely difficult to measure precisely with an inch
rule or meter stick, so use compasses or dividers as this small
line segment "a" is the key to the entire construction,
and its measurement must be as precise as possible.
Without it, we will not be able to draw either the circle
for the equilateral triangle or the circle for the heptagon.
shows the two primary circles of the pentagon O1 and O2, for
- Using "a", subtract it from the radius,
r2, the incircle of the pentagon, to obtain r3. Using r3, inscribe
the equilateral triangle FLE with one of the three vertexes in
the vertical "12 o'clock" position. (Figure 3.4.)
- From the radius r1 of the excircle of the pentagon, now mark
off two "a" lengths (labeled "b")
to make r4, the largest concentric circle, as in Figure
- Extend the base of the euilateral triangle FE to the left
and right to intersect this largest circle at points I and H,
also in Figure 3.5. (Note:
That the base vertexes of the equilateral triangle align precisely
with two of the seven vertexes of the heptagon is, by itself,
an extraordinary factor in this extraordinary construction. Without
the alignment, the construction fails.)
- Bisect (by p2) the angle q (BOH), subtended by this intersection
at H and the vertex J. This angle is very nearly 102.8571°,
and bisected, it yields, very nearly, 51.4285°, one of the
seven angles of the heptagon. (We will discuss these angles in
the proof.) Do the same bisection on the other side of the circle,
at IOB. (Figure 3-6.)
- With four of the seven sides of the heptagon completed by
the two bisections, and using the lengths of each of the four
completed sides of the heptagon to obtain the final three sides
by measuring them with the compasses or dividers, cut the rest
of the circle of the heptagon.
11. Draw the heptagon, UIXBWHY, as in
Proof for the Construction
(Please note that Figure 4.1 represents an enlargement of the essential
parts of the construction.)
- AOM is the 36°- 54°- 90° right, scalene triangle
in the regular pentagon.
- OA (r2) is the radius of the incircle (the cathetus) of the
- OM (r1) is the radius of the excircle (the hypoteneuse) of
- It is a fact that the ratio of OM : OA = 2 : f.
Without loss of generality, we may choose the scale such that
OM = 2, and OA = f.
- The smaller concentric circle for the equilateral triangle
has the radius, OT (r3). It has the same spacing "a"
from the incircle as the spacing between the pentagon's incircle
and excircle. By our choice of scale, a = 2 - f.
Therefore OT= OM -2a = 2 - 2 (2 - f)
= 2f - 2.
- In the equilateral triangle (FLE), it is a fact that the
side of the equilateral triangle bisects the radius OT at Q.
Therefore, OQ = OT/2 = (2f -2)/2 =
f - 1 = 1/f.
- Now the radius for the heptagon, OH (r4), is established
by doubling the spacing a out from the excircle
of the pentagon (this new spacing is labeled "b").
So, the radius of the heptagon, r4 = OH = 2 + 2 (2 - f)
= 6 - 2f.
- Next, we want to solve for angle QHO, which we label a. We can draw line ON parallel to line
QH, so that HON = a as well. Note,
angle NOB = 90°. In our approximate construction of the heptagon,
angle BOH is twice the central angle of the approximate hexagon.
Clearly, angle BOH = a + 90 degrees,
so it remains to find angle a. Using
right triangle trigonometry, sin a
= OQ/OH, i.e., a = sin-1 (OQ/OH).
Using our values for OH = 6 - 2 f
, and OQ =1/f, we obtain a
» 12. 92096638°, or, 12°
37' 15". Therefore, angle BOH »102.9209664°,
which we may cut in half to obtain the central angle of our approximate
heptagon, approximately 51.4604832°.
We can easily compute the difference between the actual and
approximate values for the central angle of the regular heptagon:
51. 4604832° Approximate
- 51. 4285714° Actual
» 0. 0319117° Difference;
thus the percent deviation is:
0.0319117° ÷ 51.4285714°
» 0. 0006205
= 0. 06205% high.
It should be noted that this error is extremely small, imperceptible
in even the most accurately executed pencil-and-paper constructions.
What should be pointed out is that f
has been used:
1.) in its organic form - to grow something; and,
2.) in its mathematical exactitude, as it maintains its integrity
within the pentagon. There is no question of approximation or
an issue of speculation.
This is an ideal situation. We have f
in both its major roles: mathematical and organic. In all the
discussions of f's use / nonuse /
misuse, I don't believe enough has been said in a positive mode
to advance the application of the number organically in architecture
and the arts. Too much can be made of its "cut", and
not enough on its overall qualities in the growth of structures,
systems, and patterns. It could be that f
is more ubiquitous than is commonly held, and that it does at
times exist beneath the surface if things. Hopefully, this construction
demonstrates that research can and should be continued on the
Next issue, we'll consider harmonic progressions, and a special
presentation of a general formula for harmonic progressions.
I will be showing how one harmonic progression can generate both
1- and 2-point linear perspective grids.
Have a great summer. Go on a construction search with f or o or h, or even e
! You never know
gratitude to mathematicians Stephen Wassell and Michael Schneider
for their continuing patience, assistance, and indulgence with
The correct citation for
this article is:
Reynolds, "From Pentagon to Heptagon: A Discovery on the
Generation of the Regular Heptagon from the Equilateral Triangle
and Pentagon.", Nexus Network Journal, vol. 3, no.
3 (Summer 2001), http://www.nexusjournal.com/GA-3.3.html
Copyright ©2005 Kim Williams
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