Properties of the pentagon and dodecahedron
The regular pentagon has the property that a diagonal between any two vertices is φ times a
side, where φ is the Golden Ratio (1 + √5 )/2 = 1.618033989.. which was known to the
Greeks.
DIAGRAM 1
The Ratio can be easily found by the construction shown where sides of length “c” are
extended and a line drawn through to their intersection, the centre of a circumscribed circle
of the pentagon, the midpoint of a side and the opposite vertex of the pentagon. Using Euclid
proposition 32 Book I gives the sum of the equal exterior angles is 2π so each angle is 2π/5
which is identical to the five central angles around O. Taking the triangle Δ AOC the angle
AOC is 2 by 2π/5 and as angle OAC = OCA and the sum of the internal angles of a triangle
are π, angle OAC = (1/2) *( π – 2( 2π/5 )) =π/10. Taking triangle ΔOAE the angle AOF
=2π/10 so angle AEO = (π/2 - 2π/10) = 3π/10. Then angle EAC = π/10 + 3π/10 =2π/5 =
angle FAE so ΔAF F’ is congruent with ΔACF’ and sides AF = AC.
There are now two similar Δ’s; ΔAF F’ and ΔBF F’’ giving b: c/2 :: b + c : b/2 or b² = c
(b+c) which in quadratic form is b² – cb – c² =0
and has the solution b = ( c + √(c² + 4c² ) )/2 = c (1+√5 )/2 = φ c .
This solution can be arrived at more simply by Ptolemy’s theorem for cyclic quadrilaterals
but then the other sides are more difficult to calculate.
The Renaissance mathematician Piero della Francesca may not have known trigonometry so
the following approach uses only the theorem of Pythagoras.
Using Δ’s; ΔODF’’ and ΔCDF’’ on the diagram, we get by Pythagoras:R² = (b/2) ² + (R -CF’’) ² …..I and
c² = (b/2) ² + (CF’’) ² ……..II
and b = c φ …….III
Then eliminating (b/2) ² from equations I and II, we get R² – (R - CF’’) ² = c² - (CF’’) ²
gives :R² - (R² – 2R (CF’’) + (CF’’) ² = c² – (CF’’) ² which is simplified to :CF’’ = c²/2R then using equation II gives:-
c² = (c φ/2) ² + (c²/2R) ² which rearranges to:R = c √( 5 + √5 )/√10
By trigonometry:- R= (c/2) / Cos[3π/10] using Mathematica gives the same result.
DIAGRAM 2
Using triangle ΔAOF’ R²= y² + (c/2) ² giving y = c √ (2.5 + √5)/√10
The area of a pentagon is then five triangles:5 (1/2) y c = c² (5/2) √((5/2) + √5)/√10 = 1.72047740059.. c²
Piero della Francesca is attributed as the first to calculate the volume of the dodecahedron
with its 12 sides as regular pentagons. He had a clear mind for looking at three dimensional
shapes so he may have used the following approach.
The diagram shows the plane through a dodecahedron that bisects it including two sides and
4 altitudes (y + R) of pentagons. Surprisingly the bisection plane can be bisected two more
times to give a right angle at the centre.
The volume can be calculated as 12 x h x area (of a pentagon) following Euclid.
The height “h” joins the centre of the circumscribing circle radius ( = R) of a pentagon
surface to the centre of the solid.
R1 is the radius of the circumscribing sphere of the solid dodecahedron.
Then again using the theorem of Pythagoras:R1² = (c/2)² + a² …………………….I
R1²= R² + h² …………………………II
(y + R)² = a² + (√(y² + h² ) –(c/2) )² ....III
Then eliminate R1 from I and II and solving for a² gives:a² = R² + h² – (c/2) ²…………………..IV
Then substitute in III and expand gives:y² + R² + 2yR = R² + h² – (c/2)² + y² + h² + (c/2)² -2 (c/2) √(y² + h² )
which can be simplified to:2(h²- yR) = c √(y² + h² )
Substitute h² = H gives:2(H – yR) = c √( y² + H )
Square both sides gives :4( H² - 2 yRH + y²R² ) = c² ( y² + H)
This can be rewritten as a quadratic equation with H as the variable because y and R are
constants:4 H² - (8 yR + c²) H + (4 y² R²- c² y² ) = 0
The above values of y and R can now be substituted and the quadratic equation and solved to
give:H= c² (25 + 11 √5 ) 40
h = √H = 1.11351636442.. c
The volume is then:V = (12/3) x (1.11351636442.. c ) x (1.72047740059.. c² ) = 7.663118960625..c³
The volume of an icosahedron which was also calculated by Piero della Francesca responds
to a similar analysis using y = (c/6) √3 and R = (1/3) √3 for the surface isosceles triangles.
The bisecting surface can again be divided into four right angled elements.
The area of an isosceles triangle is:- A = (c²/2) √3
Using the same analysis as above, the height “h” is:h = (1/2) √(7 + 3√5 )/6 = 0.75576131407..c
The volume is then:V = (20/3) h A = 2.181694990.. c³