Proving Existence of a Triangle with Defect > 14 Degrees in Hyperbolic Geometry

dancergirlie
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Homework Statement


How to show that there exists a triangle whose defect is greater than 14 degrees


Homework Equations





The Attempt at a Solution



No idea what to do here... something about the angle of parallelism
 
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You need to give us more information here. What do you mean by "defect?"

My best guess is that you're asking for an example of a triangle in hyperbolic geometry whose angles add up to less than 166 degrees.
 
I think the easiest way to show such a triangle exists is to explicitly write it down and compute its angle measures.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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