Hello, billionj!

Determine the length of the two longest sides on an isosceles triangle

with the angles of 36°, 72°, 72°, and the base length of 550 feet. This is probably not an acceptable solution to this problem.

But it is a fascinating bit of mathematical trivia . . . hope you enjoy it.

Code:

*
/ \
/36°\
/ \
/ \ φ
/ \
/ \
/ 72° 72° \
* - - - - - - - *
1

This is a *very* *special* isosceles triangle.

The ratio of the equal side to the base is: .$\displaystyle \phi \:=\:\frac{1+\sqrt{5}}{2}$ . . . the Golden Ratio.

Hence, with this triangle:

Code:

*
/ \
/36°\
/ \
x / \
/ \
/ \
/ 72° 72° \
* - - - - - - - *
550

we have: .$\displaystyle \frac{x}{550} \:=\:\phi \quad\Rightarrow\quad x \;\approx\;890$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

The Fibonacci sequence: .$\displaystyle 1,1,2,3,5,8,13,21,34,55,89,144,233, 377, \hdots$

. . where each term is the sum of the preceding two terms.

The ratios of consecutive terms approach the Golden Mean.

. . $\displaystyle \begin{array}{ccc}

\phi & = & 1.618 033 989 \\ \hline \\[-3mm]

\dfrac{55}{34} &=& 1.617 647 059 \\ \\[-2mm]

\dfrac{233}{144} &=& 1.618 025 751 \\ \\[-2mm]

\dfrac{987}{610} &=& 1.618 032 787 \\ \\[-2mm]

\dfrac{2584}{1597} &=& 1.618 033 813 \\

\vdots && \vdots \end{array}$

Hence, any pair of consecutive terms of the Fibonacci sequence

. . will form a triangle *very close* to the 36°-72°-72° triangle.

Example:

Code:

A
*
/ \
/ \
/ \
377 / \ 377
/ \
/ \
/ \
B * - - - - - - - * C
233

The vertex angle is: .$\displaystyle A \;\approx\;36.0002^o$

The base angles are: .$\displaystyle B \:=\:C \;\approx\;71.9999^o$