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Given $u=(0,0)$, $v=(\frac12,0)$, $w=(\frac12,\frac12)$. Find the hyperbolic area of the hyperbolic triangle $[u,v,w]$ in $\mathbb{H}^2$.

My approach:

One of the angles, say $\alpha$, is a right angle, so we have that $\alpha=\frac{\pi}2$, so that $[u,v,w]$ is a right triangle. Now, this triangle is also isosceles, so two of its sides, say $b$ and $c$, are equal.

Now, $$b=c=\cosh^{-1}\left(\frac53\right)$$

Also, the other side, say $a$, is determined as follows:

$$a=\cosh^{-1}\left(\frac73\right)$$

Now, we get

$$\frac73 = \cosh b=\frac{\cos\beta}{\sin\alpha}=\cot\beta$$ $$\implies \beta = \cot^{-1}\left(\frac73\right)=\alpha$$ $$\implies A=\frac{\pi}2-2\cot^{-1}\left(\frac73\right)$$

I'd appreciate if someone would let me know whether or not my reasoning is correct in this solution.

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  • $\begingroup$ Which model of hyperbolic plane are you using? $\endgroup$ – Somos Jul 24 '17 at 3:42
  • $\begingroup$ @Somos Poincaré disc. $\endgroup$ – sequence Jul 24 '17 at 3:50
  • $\begingroup$ In that case, the triangle is a Euclidean right triangle and I do not think it is a hyperbolic right triangle. Why do you think that it is? $\endgroup$ – Somos Jul 24 '17 at 4:17
  • $\begingroup$ @Somos It is a hyperbolic triangle. $\endgroup$ – sequence Jul 24 '17 at 13:45
  • $\begingroup$ No definitely no right angle at any of the points know that hyperbolic lines not through the centre u are circles $\endgroup$ – Willemien Jul 25 '17 at 18:24
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The area of a hyperbolic triangle is equal to the angle deficit, i.e. to the difference between the hyperbolic sum of interior angles and the Euclidean sum of $\pi$.

Poincaré half plane

Draw the triangle, and it will look like this:

Half plane triangle

This triangle has two ideal points. I'm not sure I'd call it isosceles, since two legs have infinite length, but you might as well. It has two angles of $0$ and one of $\frac\pi2$, so the sum is $\frac\pi2$, which is $\frac\pi2$ less than the Euclidean sum of a triangle. Thus the area is $\frac\pi2\approx1.5708$.

Poincaré disk

Draw the triangle, and it will look like this:

Disk triangle

No right angle here, nor is it obviously isosceles. The segment $vw$ belongs to a circle with center at $\left(\frac54,\frac14\right)$. You can verify that by checking that inverting $v$ or $w$ results in a point on that circle, too. You can use that to compute the lines connecting that center to $v$ resp. $w$, then have the tangents in these points orthogonal to that, and finally compute the angle between tangent and the line to the origin. Sum the angles, subtract it from $\pi$ and you have the area $\pi-2\arctan3\approx0.6435$.

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