lew@ihlpa.UUCP (Lew Mammel, Jr.) (02/12/86)
Here's my version of the justification of Dick Binder's pentagon construction. It has the merit that it doesn't resort to coordinate calculations. Lemma 1: The length of either of the two equal sides of the isoceles triangle having angles 36-36-108 has the golden ratio to the length of the base. Proof: Draw a line from the vertex dividing the triangle into a 36-36-108 and a 36-72-72 triangle. Observe that the sides of the 36-72-72 triangle are equal to the sides of the original 36-36-108 triangle and also equal to the base of the smaller 36-36-108 triangle. Then by similarity, side/base (small) = base/(base + side), which is by definition the golden ratio. ( I gleaned this from my high school geometry text's construction of a decagon, which of course also gives a pentagon. ) Note: The same goes for the base/side ratio of the 36-72-72 triangle. The pentagon is a veritable cornucopia of golden ratios. Just draw the diagonals, forming an inscribed pentagram and you will see them all over the place. Incidentally, it is my belief that the golden ration is the basis for the supposed mystical power of the pentagram, although I've never seen this in print. Lemma 2: On a given circle, half the side of a circumscribed pentagon has the golden ratio to the side of an inscribed pentagon. Proof: Construct a circumscribed pentagon having sides tangent at the vertices of an inscribed pentagon. The side of the inscribed pentagon forms the base of a 36-36-108 triangle having as its sides 1/2 the sides of the circumscribed pentagon. Lemma 3: The side of a pentagon inscribed in a given circle has the ratio to the radius of sqrt( 1 + g**2 ), where g is the golden ratio. Proof: In the construction for Lemma 2, draw a radius to a vertex of the inscribed pentagon and an extended radius to an adjacent vertex of the circumscribed pentagon, forming a right triangle with the third side formed by half the side of the circumscribed pentagon. With the radius defined as unit length, the extended radius has length 2*g (by similarity) so the Pythagorean Theorem gives: 1 + s**2 * g**2 = 4 * g**2 where s is the desired value for the length of the side. A little algebra using the defining equation for g proves the lemma. Lemma 3 justifies the construction. Note that line OD has the golden ratio to OB, so that BD has the required length. Lew Mammel, Jr. AT&T Bell Labs, Naperville IL.