Among triangles of perimeter $3a$ and a side $a$, what is the probability of selecting an acute/right/obtuse/scalene/isosceles triangle?

Question

If I consider all possible triangles with perimeter equal to $3a$ and one side length equal to $a$, what is the probability of selecting an acute angled triangle? a right triangle? an obtuse angled triangle? a scalene triangle? an isosceles triangle?

How would I approach such a problem mathematically? (I know writing a simulation on a computer is quite easy.)

Please share if you have a reference that discusses how to solve problems like this. I am a math hobbyist at best, not a mathematician, so would appreciate something readable. Thank you.

Answer 1

Let's denote the triangle $ABC$, in a way that $\bar{BC}=a$, and $\bar{AB}+\bar{AC}=2a$. First of all, by triangle inequality, we have \begin{align*} \frac{a}{2} \leq \bar{AB} \leq \frac{3a}{2},\\ \frac{a}{2} \leq \bar{AC} \leq \frac{3a}{2}. \end{align*}

I assumed that the probability that you are considering is such that the probability of $\bar{AB}=b$ is uniform over all feasible values, i.e., for $\frac{a}{2} \leq b \leq \frac{3a}{2}$.

I consider the question of an acute angled triangle. Consider the case where it becomes right angled triangle. Then, either \begin{align*} \bar{AB}=&\frac{3a}{4}\\ \bar{Ac}=&\frac{5a}{4}, \end{align*} or inverse. Hence, the region of the acute triangles is as following:

\begin{align*} \frac{3a}{4} \leq \bar{AB} \leq \frac{5a}{4},\\ \frac{3a}{4} \leq \bar{AC} \leq \frac{5a}{4}. \end{align*}

Now, the probability is simply the proportion of line length of two regions, \begin{align*} \frac{\sqrt{2}\frac{a}{2}}{\sqrt{2}\frac{a}{2}}=\frac{1}{2}. \end{align*}

Answer 2

Given the side of length $a$, the locus of the remaining vertex is the following ellipse

Depending on how you weight the red vs the indigo arcs of the ellipse, you will get different probabilities. Weighting by arclength will require Elliptic Integrals.

The perimeter of the entire ellipse is $$ a\int_{-1}^1\sqrt{\frac{4-x^2}{1-x^2}}\,\,\mathrm{d}x=4aE\!\left(1;\tfrac12\right) $$ and the probability of being acute is $$ \frac{\int_0^{1/2}\sqrt{\frac{4-x^2}{1-x^2}}\,\,\mathrm{d}x}{\int_0^1\sqrt{\frac{4-x^2}{1-x^2}}\,\,\mathrm{d}x}=\frac{E\!\left(\frac12;\frac12\right)}{E\!\left(1;\frac12\right)}=0.35290989543987 $$ where $E(x;k)$ is the Incomplete Elliptic Integral of the Second Kind.

Answer 3

Another option is to look at the path of all such triangles through the "map of Trilandia".

Let $\rho,r,R$ denote the semiperimeter, inradius and the circumradius of a general triangle. Its shape is uniquely defined by the pair $v=\rho/R, v=r/R$.

Consider all possible shapes of triangles inscribed into the circle with $R=1$. Any particular shape is represented by the point $(v,u)$ on the map:

For $v\in[0,\tfrac12]$ the boundaries are defined by two curves, $u_{\min}$ (blue) and $u_{\max}$ (red)

\begin{align} u_{\min}&=\sqrt{27-(5-v)^2-2\sqrt{(1-2\,v)^3}} \tag{1}\label{1} ,\\ u_{\max}&=\sqrt{27-(5-v)^2+2\sqrt{(1-2\,v)^3}} \tag{2}\label{2} \end{align}

The point $E$ corresponds to the equilateral shape.

The orange line $u_{90}=v+2$ corresponds to all triangles with $90^\circ$ angle, and separates the areas of acute (top) and obtuse (bottom) kingdoms, so the top border represents exclusively acute isosceles shapes, while the bottom border line ($u_{\min}$) is split by the line $u_{90}=v+2$ at the check-point $(\sqrt2-1,\sqrt2+1)$ between the lower obtuse isosceles part and the short upper isosceles part. Note that part $BD$ of the orange line $u_{90}$ escapes the Trilandia, that means that for $v>\sqrt2-1$ no valid right triangle can be constructed.

The green curve represent all possible triangular shapes inscribed in a unit circle, for which one of the side lengths is $2\rho/3$.

To find the equation of the green line, recall that the three side length of the triangle, inscribed into unit circle are the roots of cubic equation \begin{align} x^3-2u\,x^2+(u^2+v^2+4v)\,x-4\,u\,v&=0 \tag{3}\label{3} , \end{align}

given that one side length is equal $\tfrac23\,u$, as a reminder of \eqref{3} divided by $x-\tfrac23\,u$, we have a condition \begin{align} u^2+9\,v^2-18\,v&=0 \tag{4}\label{4} , \end{align}

so the equation of the sought line is

\begin{align} u(v)&=3\sqrt{2v-v^2} \tag{5}\label{5} ,\\ u'(v)&= \frac{3(1-v)}{\sqrt{v\,(2-v)}} . \end{align}

It crosses the obtuse/acute border at the point $(\tfrac25,\tfrac{12}5)$.

So, if we assume that probabilities are proportional to the length of the curve, we have the total length of the curve of interest as

\begin{align} L&= \int_0^{1/2} \sqrt{1+(u'(v))^2} \, dv \approx 2.670 , \end{align}

and the length of the obtuse part

\begin{align} L_o&= \int_0^{2/5} \sqrt{1+(u'(v))^2} \, dv \approx 2.449 , \end{align}

so the probability of choosing an obtuse shape in this case is \begin{align} P_o&=\frac{L_o}{L} \approx 91.7\% , \end{align}

which leaves a pity $8.3\%$ for acute shapes.