# Find hyperbolic area of hyperbolic triangle

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.

• Which model of hyperbolic plane are you using? – Somos Jul 24 '17 at 3:42
• @Somos Poincaré disc. – sequence Jul 24 '17 at 3:50
• 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? – Somos Jul 24 '17 at 4:17
• @Somos It is a hyperbolic triangle. – sequence Jul 24 '17 at 13:45
• No definitely no right angle at any of the points know that hyperbolic lines not through the centre u are circles – Willemien Jul 25 '17 at 18:24

## 1 Answer

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: 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: 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 $\pi-2\arctan3\approx0.6435$.