I wanted to calculate the radius of the inscribed circle of an ideal triangle.

and when i dat calculate it i came to $\ln( \sqrt {3}) \approx 0.54 $ (being arcos(sec (30^o)) but then at https://en.wikipedia.org/w/index.php?title=Ideal_triangle&oldid=668440011

it says that the equilateral triangle that is made by connecting the intersections of the ideal triangle and its [[inscribed circle]] has a side length of $ 4\ln \left( \frac{1+\sqrt 5}{2}\right) \approx 1.925 $

but this would mean that the radius is less than half the side which is impossible

so didI make a mistake, or is wikipedia wrong?

PS the wikipedia page is corrected now and gives the right values)

  • Your calculation of $\ln(\sqrt{3})$ looks right to me. Another way to get it is to use the ideal triangle in the upper half plane having vertices $0,1,\infty$. Its inscribed circle has center $\frac{1}{2} + \frac{\sqrt{3}}{2}i$, and that circle intersects the hyperbolic line with endpoints $0,1$ at the point $\frac{1}{2} + \frac{1}{2} i$. The distance between these two points is $\ln(\sqrt{3})$. – Lee Mosher Dec 8 '15 at 17:53
  • Thanks I did it in a similar way (only using the points $-1,1,\infty i $, Am I correct wikipedia is wrong and the sides of equilateral triangle that is made by connecting the intersections of the ideal triangle and its inscribed circle is $\ln ( (3+\sqrt{5}) / 2) $ also could you transform your comment as answer then I can accept it – Willemien Dec 8 '15 at 20:30
up vote 2 down vote accepted

I think the Beltrami-Klein model is particularly useful here since a hyperbolic line is just a straight chord there. For simplicity use a regular triangle as the inscribed ideal triangle. Its inscribed circle has an Euclidean radius which is half that of the fundamental circle. So the chord along the radius gets divided $1:1$ by the midpoint but $3:1$ by the point on the circle, therefore the radius is

$$\frac12\ln\left(\frac11\cdot\frac31\right)\approx0.54$$

just as you computed.

Now to the edge length of the inscribed triangle. A bit of messing around with the Pythagorean theorem will tell you that the ratio between the Euclidean edge length and the length of its supporting chord is $1:\sqrt5$. Which means one endpoint divides the chord in a ratio of $\sqrt5-1:\sqrt5+1$ and the other in the reciprocal ratio $\sqrt5+1:\sqrt5-1$ so the length between them is

$$\frac12\ln\left(\frac{\sqrt5+1}{\sqrt5-1}\cdot\frac{\sqrt5+1}{\sqrt5-1}\right) =\ln\frac{\sqrt5+1}{\sqrt5-1} \approx0.96$$

This looks like a factor of two mistake. Some extra computation shows that this is in fact the case, since

$$\frac{\sqrt5+1}{\sqrt5-1} = \frac{\sqrt5 + 3}2 = \left(\frac{\sqrt5 + 1}2\right)^2 = \varphi^2$$

I will edit Wikipedia.

  • thanks addeded wikipedia aswell am wondering is the link to golden ratio really meaningfull – Willemien Dec 10 '15 at 8:12
  • @Willemien: Well, it's nice in terms of en.wikipedia.org/wiki/Special:WhatLinksHere/Golden_ratio to see where this thing does crop up more or less unexpectedly. – MvG Dec 10 '15 at 10:37
  • curiously math.stackexchange.com/a/114231/88985 gives another value of the edge length of the inscribed triangle. – Willemien Jan 7 '16 at 21:24
  • @Willemien: There is no contradiction. Have a look at this construction (created with Cinderella). On the left you have the Beltrami-Klein model, on the right the Poincaré disk (for some reason without the inscribed circle). The thing you described is the line $EG$ between incenter contact points. That other question asks about the distance, i.e. requires the normal $EK$ with a right angle at $K$. That's a different line, leading to a different distance. And the numeric result from Cinderella supports Will Jagy's answer there. – MvG Jan 7 '16 at 21:45
  • @Willemien posted an answer here with radius of the circle and edge length of the inscribed triangle, written using the upper half plane model. – Will Jagy Jan 7 '16 at 23:17

In the upper half plane model, we take the ideal triangle as $x=-1,$ $x=1,$ and $x^2 + y^2 = 1.$ For this problem, you want the inscribed circle, which is $$ x^2 + (y-2)^2 = 1, $$ which meets the ideal triangle at $(-1,2),$ $(1,2),$ $(0,1).$

The diameter along the $y$ axis goes from $(0,1)$ to $(0,3).$ The geodesic center of the circle is at the geometric mean of $1$ and $3,$ therefore $(0, \sqrt 3).$ Recall from my 2012 answer that this geodesic is $(0,e^t).$ Therefore the radius is $\log {\sqrt 3} - \log 1$ = $(1/2) \log 3.$

One edge of the inscribed triangle passes through $(1,2),$ $(0,1).$ This segment is unit speed parametrized by $$ 2 + \sqrt 5 \tanh t, \sqrt 5 \operatorname{sech} t $$ and both ending $t$ values will be negative, because $0 < 2$ and $1 < 2.$ Let us just use the reciprocal of $\cosh t$ and be careful a bout signs.

One endpoint has $$ \frac{\sqrt 5}{ \cosh t} = 1 $$ and the other has $$ \frac{\sqrt 5}{ \cosh t} = 2, $$ so $$ \cosh t = \frac{\sqrt 5}{2} $$ and $$ \cosh t = \sqrt 5 $$

Looked it up, $$ \operatorname{arcosh} x = \log \left(x + \sqrt {x^2 - 1}\right) $$ so the absolute value of the $t$ difference is

$$ \log \left( \sqrt 5 + 2 \right) - \log \left( \frac{\sqrt 5 + 1}{2} \right) = \log (\varphi^2 + \varphi) - \log \varphi = \log (\varphi + 1) = \log (\varphi^2) = 2 \log \varphi \approx 0.96242365 $$

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