Hyperbolic Triangle With Longer Sides But Smaller Area

 
THEOREM 3.1. There are two triangles $\Delta_A, \Delta_B$ in the hyperbolic (or Lobachevsky) plane such that $1^{\circ}$ the length of each side of $\Delta_A$ does not exceed the length of the corresponding side of $\Delta_B$, $2^{\circ} \Delta_B$ is non-obtuse, but \[ \text { area } \Delta_A>\text { area } \Delta_B \text {. } \] Proof. Let $\Delta_A$ be an equilateral triangle with each of its angles equal to $\pi / 6$; and let $\Delta_B$ be a right isosceles triangle, with the two sides adjacent to the right angle of the same length as a side of $\Delta_A$. First of all, as we know, in hyperbolic spaces the hypotenuse of a right triangle is the longest side, so we have $1^{\circ}$ the length of each side of $\Delta_A$ does not exceed the length of the corresponding side of $\Delta_B$. Next, since $\Delta_B$ is a right triangle, and because the sum of the interior angles is less than $\pi$ in any triangle, $\Delta_B$ can not have an obtuse angle, that is to say $2^{\circ} \Delta_B$ is non-obtuse. Finally, compare the areas of $\Delta_A$ and $\Delta_B$. As the defect of $\Delta_A$ is equal to $\pi-3\times\pi/6=\pi / 2$, we have \[ \text { area } \Delta_A=\frac{1}{|K|} \frac{\pi}{2} \text {, } \] where $K$ is the curvature of the hyperbolic plane. On the other hand, the sum of the interior angles of $\Delta_B$ is greater than $\pi / 2$, hence its defect is less than $\pi / 2$, and we have \[ \text { area } \Delta_B<\frac{1}{|K|} \frac{\pi}{2} \text {. } \] Comparing these, we obtain \[ \text { area } \Delta_A>\text { area } \Delta_B \text {, } \] and the proof of Theorem 3.1 is completed.