perplexus.info :: Just Math : Golden Ratio

Golden Ratio (Posted on 2003-11-26)

The ancient Egyptians found a particular ratio very pleasing to the eye. Their architecture is full of examples of this ratio. And you can see it even in a golden rectangle.

A golden rectangle is one from which, if you remove a square from one end (with side equal to the shorter side of the rectangle), what remains is a rectangle that is similar (has identical proportions) to the original rectangle.

What is the ratio of the longer side, to the shorter side (in the golden rectangle), and how did you determine it?
_____________________

By the way, I realize many people are familiar with this ratio (in which case this is a very easy problem), but for those who haven't, do them a favor, and please don't post the solution.

Submitted by SilverKnight

Rating: 3.0000 (6 votes)

Solution: (Hide)

Let the longer side (of the golden rectangle) be x and the shorter side, y.
Then we are asking for the ratio x/y (the two sides of the original rectangle).
After we remove the y by y square the remaining rectangle has longer side, y and shorter side x-y.
It's ratio would be expressed as y/(x-y).
Since these two ratios must be equal we have:
x/y = y/(x-y)
Here are two solutions:
________________________
(1)
When y=1, then x/y = x. So, let's ASSUME y=1, and solve for x
x/1 = 1/(x-1)
x² - x = 1
x² - x - 1 = 0
--- quadratic formula ---
x = [1 ± √( (-1)² - 4(1)(-1) ) ] / 2(1)
x = (1 ± √5) / 2
--- take the positive solution ---
x = x/y = 1.618034 (because y is one)...
___________________________________
(2)
If you wish to do this more generally, then substitute z = x/y and solve for z
originally we had:
x/y = y/(x-y)
y/x = (x-y)/y
y/x = x/y - y/y
y/x = x/y - 1
1/z = z - 1
1 = z² - z
0 = z² - z - 1
now solve for z (as in the first solution), and we get the same answer.

Comments: ( You must be logged in to post comments.)

Subject	Author	Date
Puzzle Thoughts	K Sengupta	2024-04-18 13:03:28
re: Fibonacci series	TomM	2003-11-30 11:14:25
Fibonacci series	Lorne Hrynkiw	2003-11-30 03:59:00
Golden Proportion	Hal9000	2003-11-29 17:43:19
re: Golden Section - an interesting number	Gamer	2003-11-27 07:50:33
Golden Section - an interesting number	retiarius	2003-11-27 02:29:23
Da answer	Zuninga!	2003-11-27 01:07:58
Donald in MathMagicLand	brianjn	2003-11-26 23:16:09
A way to solve this	Gamer	2003-11-26 16:20:12
yay i am the first to post a solution	drew	2003-11-26 14:59:22

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