The Golden Ratio: Pentagonal Diagonals

Here is the previous post concerning the Golden Ratio. For a collection of all the posts concerning the Golden Ratio, click #GoldenRatio below.

We all know that the diagonal of a unit square (a square with sidelength 1) is 2\sqrt{2}. We could easily figure this out using the Pythagorean theorem. Then I ask you, what is the length of the diagonal of a regular unit pentagon? Try to solve this before moving on.

As you may have guessed, it is the golden ratio! But how did we figure it out? Well, take the pentagon pictured above. There are a lot of ways to solve this problem, but some require more algebra than others. Other techniques used could be the Law of Cosines or Ptolmey's Theorem. Here, we will just use good old geometry. You may want to draw a picture for yourself to follow what's happening.

Note that since we have a regular pentagon, AB=BC=CD=DE=EAAB = BC = CD = DE = EA. Also, because of its symmetry, all of its diagonals are equal, i.e. AC=BD=CE=DA=EBAC = BD = CE = DA = EB. Define point FF as the intersection of the lines ACAC and BDBD and draw ACAC and BDBD. Now, let's do some angle chasing.

  1. We know, or could easily calculate, that the interior angles of a pentagon are ABC=108\angle ABC = 108^{\circ}
  2. ΔABC\Delta ABC is isoceles because AB=BCAB = BC
  3. BAC=BCA=36\angle BAC = \angle BCA = 36^{\circ} because ΔABC\Delta ABC is isoceles
  4. Using the same logic on ΔBCD\Delta BCD, DBC=FBC=36\angle DBC = \angle FBC = 36^{\circ}
  5. By the AA theorem, ΔABCΔBFC\Delta ABC \sim \Delta BFC
  6. CFBC=BCAC\frac{CF}{BC} = \frac{BC}{AC} due to similarity. *
  7. ABF=10836=72\angle ABF = 108^{\circ}-36^{\circ} = 72^{\circ}
  8. BFA=72\angle BFA = 72^{\circ}
  9. ΔABF\Delta ABF is isoceles because ABF=BFA\angle ABF = \angle BFA
  10. AB=AFAB = AF *

Especially note steps 6 and 10. From there,

ACAB=AF+CFAB=AFAB+CFAB=AFAB+CFBC=1+BCAC=1+ABAC\frac{AC}{AB} = \frac{AF + CF}{AB} = \frac{AF}{AB} + \frac{CF}{AB} = \frac{AF}{AB} + \frac{CF}{BC} = 1 + \frac{BC}{AC} = 1 + \frac{AB}{AC}

This should start to look familiar. If not, let x=ACABx = \frac{AC}{AB}. Then x=1+1xx = 1 + \frac{1}{x}. Then xx is the golden ratio! Thus, for a regular unit pentagon, the diagonals have length ϕ\phi.

Can you supply your own proof of this property of the pentagon? Don't be afraid to use algebra. What else is special about this pentagon?

Here is the next note.

Note by Bob Krueger
7 years ago

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