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.
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