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I'm a math teacher and this is a question for my students.

I'm certain I'm missing something very simple but I cant seem to get it. I made the question just through sketching pretty pattern, seeing the length in that case for $n=12$ was almost the golden ratio and it got me interested.

$x$ and $y$ are the lengths seen in the picture.

EDIT: changed the wording thanks to suggestions

The question

Lewis Winstanley
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    Please do not post questions in images. Instead use MathJax to write it directly in your post. Furthermore provide some context (why are you interested in the question, what things related to it do you already know) and show the work you have done trying to solve the problem yourself… – Jonas Linssen Mar 08 '22 at 11:33
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    What are $x,,y$? – J.G. Mar 08 '22 at 12:27
  • Are the triangles supposed to be $90-45-45$? You say isosceles, but do not say right. It would be better to specify that the triangles are each rotated by $\frac {2\pi}n$ from the previous one or something like that. – Ross Millikan Mar 08 '22 at 14:35
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    I don't see the point to work on this fan-like structure. You need only consider 2 neighbor triangles. – Jean Marie Mar 08 '22 at 17:35
  • @JeanMarie Although the triangles make the picture visually appealing, arguably they are also a pedagogical asset because they're mathematically unnecessary. :) I for one have developed the (bad?) habit of paring down problems to contain Exactly The Right Hypotheses, which deprives students the benefit and pleasure of seeing and extracting the essential information. – Andrew D. Hwang Mar 09 '22 at 15:18
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    @Andrew D. Hwang I fully agree, and I wouldn't have made this comment if the question comes from a student, but here it's the teacher who asks the question... – Jean Marie Mar 09 '22 at 17:46

1 Answers1

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For convenience with the unit circle, let's place the first (right isosceles) triangle with its base on the positive horizontal axis, so it has an acute angle at the origin and its right angle at $(1, 0)$. If each successive triangle is rotated $2\pi/n$ clockwise, the right angle of the $n$th triangle is at $(\cos(2\pi/n), \sin(2\pi/n))$. Consequently, we have $x = \sin(2\pi/n)$, $y = 1 - \cos(2\pi/n)$, and $$ \frac{x}{y} = \frac{\sin(2\pi/n)}{1 - \cos(2\pi/n)} = \frac{2\sin(\pi/n)\cos(\pi/n)}{1 - \cos^{2}(\pi/n) + \sin^{2}(\pi/n)} = \cot\frac{\pi}{n}. $$ The question is, what choice of $n > 1$ makes $x/y$ as close as possible to $\gamma = \frac{1}{2}(1 + \sqrt{5}) \approx 1.618034$.

Slopes approximating the golden ratio

The diagram shows the unit circle (centered at the origin) and the "final positions" of the right angle if we take $n = 2, \dots, 24$. The points $n = 5$ and $n = 6$ bracket the blue line (whose slope is $-\gamma$). The ratio $x/y$ is monotone in $n$, so one of these is optimal. We have $$ \frac{\sin(2\pi/5)}{1 - \cos(2\pi/5)} = \frac{\sqrt{2}\sqrt{5 + \sqrt{5}}}{\sqrt{5} - 1} \approx 1.37638,\qquad \frac{\sin(2\pi/6)}{1 - \cos(2\pi/6)} = \sqrt{3} \approx 1.73205. $$ Of these, $n = 6$ is closer.

The aspect ratio of the right triangles is immaterial, because the second acute angle (not at the origin) has no effect on the position of the right angle.

Andrew D. Hwang
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