A "Random" Triangle

If the first two sides are $x,y$ , chosen uniformly and independently from $(0,1)$ , then the third side $z$ must satisfy the inequalities $x \le y+z, y \le x+z, z \le x+y$ , and hence we must have $|x-y| \le z \le x+y$ Since the cosines of the angles $X,Y,Z$ are $\cos X \; = \; \frac{y^2+z^2-x^2}{2yz} \hspace{1cm} \cos Y \; = \; \frac{x^2 + z^2 - y^2}{2xz} \hspace{1cm} \cos Z \; = \; \frac{x^2 + y^2 - z^2}{2xy}$ the triangle will be acute provided that $x^2 \le y^2 + z^2, y^2 \le x^2 + z^2, z^2 \le x^2 + y^2$ , and so we must have $|x^2 - y^2| \; \le \; z^2 \; \le \; x^2 + y^2$ Since $|x^2 - y^2| \ge (x-y)^2$ and $x^2 + y^2 \le (x+y)^2$ , we deduce that the probability that the triangle is acute, given that the first two sides are $x,y$ , is $\frac{\sqrt{x^2+y^2} - \sqrt{|x^2 - y^2|}}{x+y - |x-y|}$ and hence the probability of an acute-angled triangle is therefore (putting $y = xu$ ) in the last line): $\begin{aligned} p & = \; \int_0^1\,dx \int_0^1\,dy \frac{\sqrt{x^2+y^2} - \sqrt{|x^2 - y^2|}}{x+y - |x-y|} \; = \; 2\int_0^1\,dx \int_0^x\,dy \frac{\sqrt{x^2 + y^2} - \sqrt{x^2 - y^2}}{2y} \\ & = \; 2\int_0^1\,dx \int_0^1\,du \frac{\sqrt{1+u^2} - \sqrt{1-u^2}}{2u} \times x \; =\; \tfrac12\int_0^1 \frac{\sqrt{1+u^2} - \sqrt{1-u^2}}{u}\,du \end{aligned}$ Thus (it is simplest to check this integral by differentiating back again) $p \; = \; \tfrac12\Big[-\sqrt{1-u^2} + \sqrt{1+u^2} + \ln\big(1 + \sqrt{1-u^2}\big) - \ln\big(1 + \sqrt{1+u^2}\big)\Big]_0^1 \; = \; \tfrac12\big(\sqrt{2} - \ln(1 + \sqrt{2})\big) \; = \; 0.26642$ making an obtuse-angled triangle more likely.

A quick Monte Carlo test in Excel, looking at 1000 randomly generated triangles at a time, consistently gives a proportion of roughly 26% of acute-angled triangles.