Phi and the Pentagon Triangle

Earlier we saw that the triangle shown here occurs in the pentagon and decagon. If the shorter side is of unit length and we say the longer side has length P, we can calculate P, the ratio of the longer to shorter sides of this "sharply pointed" isosceles triangle (i.e. two sides of the triangle are equal and therefore two of its angles are also). We do this by introducing a point D on side AC. We choose it so that it makes BD of length 1 also, so BCD is isosceles too. So we can write in its angles (BDC = 72° also leaving 180o-72°-72o=36° for angle DBC). In other words

Triangle BCD is the same shape as triangle ABC since their angles are equal. We also see that BD halves the 72 degree angle ABC, so ABD has two angles equal and it too is isosceles. This means that sides AD and BD are equal too, so AD is of length 1 also.

Now we deduce that BD is of length P-1 since AC is of length P and AB is of length 1 All we have done is justify the numbers and angles on the diagram here.

Now to calculate P!

Since BCD is the same shape as ABC, their sides are in the same ratios. So the longer-side-to-shorter-side ratio in BCD is also P, i.e.

Refer back to the Fibonacci and Geometry section above where we solved this equation to get

P=(1+V5)/2 = 1-6180339... or P=(I-V5)/2 =-0-6180339..

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