A Regular Pentagon and Powers of Phi +/Chapter 7

 There will be other things you find, but here are some things Barbara and Jenny, 9th graders at the time, found: A smaller pentagon FGHIJ is formed inside, but points in the opposite direction. The sum of the interior angles of a regular polygon is        (n-2) x 180° One interior angle of a regular pentagon is         (5-2) x 180° /5 = 108°.   There are only 3 different angles in the pentagon, 36°, 72° (2 x 36°), and 108° (3 x36°) in the figure.

 There are only 2 different shape triangles. The obtuse isosceles triangles like HCD (36°, 36°,108°) of only two sizes, and similar. The other shape triangles are the acute isosceles triangles (72°, 72°, 36°) of three sizes, and these are similar triangles also.

We found also that  sin 72°/ sin 36° = ø = 1.6..., and sin 108°/ sin 36° = ø, and

sin108°/ sin 36° = 1.

 Don proved to himself that GC = CH = ø, with trig functions:

 The acute triangles at the right taken out of the pentagon, all have the same angles 36°, 72°, and 72°, therefore are similar triangles (do the eye test). They are isosceles triangles (2 sides equal). It turns out they are golden triangles, because when we measure CH and GH, the ratio of the longest side to the shortest side CH/GH  =  the golden mean ø =~ 1.6. Which we also proved with trig, above. We used the symbol ø (the Greek letter phi) to represent it. So we let CH=ø and GH = 1.

Since triangle HBC is similar to triangle GCH, the sides are proportional, therefore they

said the following proportion must be true:             and

From that,  ø2 = ø+1. ø+1 is also the length of the side of the pentagon. Triangle ACE is

also similar to triangle GCH so the following proportion is true:

and therefore ø3 = 2ø+1. Find these lengths on the pentagon.

Extending the lines, Barbara and Jenny found ø4 = 3ø+2 and ø5 = 5ø+3.

They saw a pattern in the powers of ø and wrote a computer program which printed out:

... AND YOU SHALL MEET A HORRIBLE FATE... YOU SHALL SPEND ALL ETERNITY FINDING POWERS OF PHI ...

Barbara and Jenny solved this quadratic equation ø2 - ø -1 = 0 that they got above, using the quadratic formula, obtaining ø1 = (1+Sqrt 5)/2 = 1.61803 and ø2 = (1-Sqrt 5)/2 = -0.61803. They realized that ø1 * ø2 = -1  and ø1 + ø2 = 1.

What happens if you draw the diagonals of the smaller pentagon? Make up other questions.

See Cristobal Vila's beautiful short movie "Nature by Numbers" at

See Xah Lee's site on curves- especially the equiangular spiral

See also Golden Triangle, Pentagon, Sunflower head , and The Golden Angle on the MAP

Geoffrey graphs the ratios of the Fibonacci numbers and writes a program to get the infinite sequence which has a limit.

Tara writes a story "A Quest For The Sacred Golden Pineapple, Pine Cone and Artichoke"

Jamie an 8th grader, solved the quadratic equation x2 - x - 1 = 0,  by iteration to get Phi, and he and Don wrote a program to do this on a TI-84 Plus.

Comparing the spirals of fibonacci numbers, the Nautilus shell, and The equal-tempered Musical scale

See also Dr. Ron Knott's great webite for much more on the Fibonacci numbers.