The vertices of a triangle are three unifomly random points on a unit circle. The side lengths are, in random order, $a,b,c$.
There is a convoluted proof that $P(ab>c)=\frac12$. But since the probability is so simple, I am looking for an intuitive explanation, for example one that utilizes symmetry.
(If an intuitive explanation can be given, then I will remove this question from my list of probability questions that have answer $\frac12$ but resist intuitive explanation.)
One can use basic probability theory to prove that $$P(ab<kc)=\frac{2}{\pi}\arctan k,\qquad k>0.$$ Without loss of generality, the vertices opposite the sides $a,b,c$ are $$A=e^{2i\beta},\qquad B=e^{-2i\alpha},\qquad C=1,$$ where $\alpha,\beta\in(0,\pi)$ are independent random variables with uniform distribution. Then $$a=BC=2\sin\alpha,\qquad b=AC=2\sin\beta,\qquad c=AB=2|\sin(\alpha+\beta)|,$$ and hence $$ab<kc \quad\Longleftrightarrow\quad 2\sin\alpha\sin\beta<k|\sin(\alpha+\beta)| \quad\Longleftrightarrow\quad \frac{1}{k}<\left|\frac{\cot\alpha+\cot\beta}{2}\right|.$$ Now $\cot\alpha$ and $\cot\beta$ are independent random variables with standard Cauchy distribution, hence their average also has standard Cauchy distribution. So if $X\in(-\pi/2,\pi/2)$ is a random variable with uniform distribution, then $$P(ab<kc)=P\left(\frac{1}{k}<|\cot X|\right)=P(|\tan X|<k)=P(|X|<\arctan k)=\frac{2\arctan k}{\pi}.$$
The following argument is less or more the same as that of Iosif Pinelis, but with less computations and the symmetry is rather explicit. It may be explained without complex numbers, but the explanation I have in mind is rather longer.
We use complex coordinates on the unit circle. Let one vertex be 1, other two by $x$ and $y$, so a random triangle has "coordinates" $(x,y)$, coordinates are uniform on the circle and independent. Partition all triangles onto quadruples of the form $(x,y),(ix,-iy),(-ix,iy),(-x,-y)$. I claim that (in general position) exactly two out of these four triangles satisfy $ab>c$, and two satisfy $ab<c$, where $c=|x-y|$, $a=|1-x|$, $b=|1-y|$. The main combinatorial reason for that is the following
Observation. If $A,B,C,D$ are non-zero real numbers, and $AC<0$ or $BD<0$, then exactly two of the four numbers $AB,BC,CD,DA$ are positive.
We rewrite the inequality $ab<c$ as $$|x-y|^2>|1-x|^2\cdot |1-y|^2\Leftrightarrow -\frac{(x-y)^2}{xy}>\frac{(1-x)^2(1-y)^2}{xy} \Leftrightarrow \\ 0>\frac{(1+xy-x-y-i(x-y))(1+xy-x-y+i(x-y))}{xy}=:AB,$$ where I denote $xy=z^2$, $A=(1+xy-x(1+i)-y(1-i))/z$, $B=(1+xy-x(1-i)-y(1+i))/z$. Note that the numbers $A,B$ are real (and in general position non-zero), and for three other triangles analogous inequalities read as $0>BC$, $0>CD$, $0>DA$, where $C=(1+xy-x(-1-i)-y(-1+i))/z$ and $D=(1+xy-x(-1+i)-y(-1-i))/z$.
To apply the observation, it remains to prove that $AC<0$ or $BD<0$. Well, if $(x/z)(1+i)=u+iv$ (for real $u,v$), then $A,C=(1+xy)/z\pm 2u$ and $B,D=(1+xy)/z\pm 2v$. We have $u^2+v^2=2$, thus $\max(|u|,|v|)\geqslant 1$, and so the result.
Without loss of generality, the three points on the unit circle are $1,e^{is},e^{it}$, where $s$ and $t$ are uniformly and independently distributed on the interval $[0,2\pi]$, so that $a=|e^{is}-1|$, $b=|e^{it}-1|$, $c=|e^{is}-e^{it}|$.
Then the condition $ab>c$ can be rewritten as $$|u|<h(v):=\cos^{-1}g(v) \tag{10}\label{10}$$ for $$(u,v):=\Big(\frac{s-t}2,\frac{s+t}2\Big)\in[-\pi/2,\pi/2]\times[0,2\pi],$$ where $$g(v):=\frac{\cos v+\sqrt{2-\cos^2 v}}2;$$ see details on \eqref{10} at the end of this answer.
Here is the set of all pairs $(v,u)\in[0,2\pi]\times[-\pi/2,\pi/2]$ satisfying condition \eqref{10}:
Note the symmetries $$h(2\pi-v)=h(v)\quad\text{and especially}\quad h(v)+h(\pi-v)=\pi/2;\tag{20}\label{20}$$ see details on this at the end of the answer.
Also, the Jacobian $J:=\frac{\partial(s,t)}{\partial(u,v)}$ is $2$.
The probability (say $p$) in question is $\dfrac A{(2\pi)^2}$, where $A$ is the area of the set of all $(s,t)\in[0,2\pi]^2$ satisfying the condition $ab>c$, so that $$A=2\times2\int_0^{2\pi}dv\,h(v) =8\int_0^\pi dv\,h(v)=8\int_0^\pi dv\,h(\pi-v) \\ =8\times\frac12 \int_0^\pi dv\,[h(v)+h(\pi-v)] =8\times\frac12 \,\pi\times\pi/2=2\pi^2. $$ (This result is "obvious" from the above picture, because $A$ is $J$ times the depicted area (that is, twice the depicted area), whereas the depicted area is half of the area $2\pi\times\pi$ of the rectangle enclosing the depicted area.)
So, $p=\dfrac{2\pi^2}{(2\pi)^2}=\dfrac12$. $\quad\Box$
Details on \eqref{10}: Note that $a^2=(e^{is}-1)(e^{-is}-1)=2(1-\cos s)$. Similarly, $b^2=2(1-\cos t)$ and $c^2=(e^{i(s-t)}-1)(e^{-i(s-t)}-1)=2(1-\cos(s-t))$. So, $ab>c$ can be rewritten as $2(1-\cos s)(1-\cos t)>1-\cos(s-t)$ or as $2(1-(\cos s+\cos t)+\cos s\,\cos t)>1-\cos(s-t)$. Now use the identities $\cos s+\cos t=2\cos u\cos v$ and $2\cos s\,\cos t=\cos2u+\cos2v$, $\cos 2x=2\cos^2x-1$, and $s-t=2u$ to rewrite $ab>c$ as $\cos u\ge g(v)$, which is equivalent to \eqref{10}, because $0\le g\le1$.
Details on the identity $h(v)+h(\pi-v)=\pi/2$ in \eqref{20}: Since $0\le g\le1$, \eqref{10} implies $0\le h\le\pi/2$. So, for $$f(v):=g(\pi-v)=\frac{-\cos v+\sqrt{2-\cos^2 v}}2$$ we have $$0\le\cos(\pi/2-h(\pi-v))=\sin(h(\pi-v)) \\ =\sin(\cos^{-1} f(v)) =\sqrt{1-f(v)^2}=g(v)=\cos h(v). $$ Using again the fact that $0\le h\le\pi/2$, we get $\pi/2-h(\pi-v)=h(v)$, as desired.
Responding to the comment that "$\sqrt{1-f(v)^2}=g(v)$ is true but not obviously true", here is a symmetry proof of that: Letting $y:=\cos v$, $x_+:=g(v)=\frac{y+\sqrt{2-y^2}}2$, and $x_-:=f(v)=\frac{-y+\sqrt{2-y^2}}2$, we have $x_+ +x_-=\sqrt{2-y^2}$ and $2x_+ x_-=1-y^2$, so that $g(v)^2+f(v)^2=x_+^2+x_-^2=(x_+ +x_-)^2-2x_+ x_-=(2-y^2)-(1-y^2)=1$. Therefore and because $g(v)\ge0$, we get $\sqrt{1-f(v)^2}=g(v)$.
On a unit circle with centre $O$, draw parallel chords $PQ$ and $P'Q'$ such that $PQ'\perp P'Q$. Chord $MN$ is parallel to $PQ$ and passes through $R$, the intersection of $PQ'$ and $P'Q$. Minor arc $MN$ is colored red. $\alpha=\angle POQ$ and $\beta=\angle P'OQ'$, where $\alpha$ and $\beta$ are convex.
We use the fact that, for a triangle with side lengths $a,b,c$ inscribed in a unit circle, $\text{area}=\frac{abc}{4}$.
$\therefore P(ab<c)=P\left(\frac{abc}{4}<\frac{c^2}{4}\right)=P\left(\text{area}<\frac{c^2}{4}\right)$
Let $P$ be the first chosen vertex of the triangle, and $Q$ the second chosen vertex. Let $PQ=c$. If $ab\color{red}{<}c$, the third chosen point must lie on the red arc. (This is because the inscribed triangle's area must be smaller than $\frac{c^2}{4}$, which is the area of $\triangle PQR$.)
So $P(ab\color{red}{<}c)$ is the proportion of the red arc length to the circumference of the circle, averaged as $\color{red}{\alpha}$ goes from $0^\circ$ to $180^\circ$.
Now instead let $P'$ be the first chosen vertex, and $Q'$ the second chosen vertex. Let $P'Q'=c$. If $ab\color{red}{>}c$, the third chosen point must lie on the red arc. (This is because the inscribed triangle's area must be greater than $\frac{c^2}{4}$, which is the area of $\triangle P'Q'R$.)
So $P(ab\color{red}{>}c)$ is the proportion of the red arc length to the circumference of the circle, averaged as $\color{red}{\beta}$ goes from $0$ to $180^\circ$, or by symmetry, as $\color{red}{\beta}$ goes from $180^\circ$ to $0^\circ$.
Obviously, $\alpha+\beta=180^\circ$ ($\angle POP'=2\times \angle PQ'P=90^\circ$).
Therefore the red arc length averaged as $\alpha$ goes from $0^\circ$ to $180^\circ$, is exactly the same as the red arc length averaged as $\beta$ goes from $180^\circ$ to $0^\circ$.
Therefore $P(ab<c)=P(ab>c)$. And since $P(ab<c)=1-P(ab>c)$, we have $P(ab>c)=\frac12.$
Denoting the angles corresponding to edges of side lengths $a,b,c$ with $A,B,C$ respectively, by the sine law:
$$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2.$$
Thus $c>ab$ amounts to $\sin(C)>2\sin(A)\sin(B)$. Replacing the LHS with $\sin(A+B)$ and the RHS with $\cos(A-B)-\cos(A+B)$, we can rephrase $c>ab$ as $\sin(A+B)+\cos(A+B)>\cos(A-B)$, or equivalently $\sqrt{2}\cos\left(A+B-\frac{\pi}{4}\right)>\cos(A-B)$. Denoting $A-B$ by $y$ and $A+B$ by $x$, they are subject only to $|y|<x<\pi$; and one should investigate the region $\sqrt{2}\cos\left(x-\frac{\pi}{4}\right)>\cos(y)$. Changing $y$ to $-y$ won't change the inequalities. So WLOG, we may assume that $0<y<x<\pi$. Notice that $\sqrt{2}\cos\left(x-\frac{\pi}{4}\right)>\cos(y)$ automatically holds when $x\in\left(0,\frac{\pi}{2}\right)$ because then the LHS is larger than $1$. As for the case of $x\in\left(\frac{\pi}{2},\pi\right)$, one should have $y>\cos^{-1}\left(\sqrt{2}\cos\left(x-\frac{\pi}{4}\right)\right)$. All in all, we have the picture below in which the ratio of the area of the yellow area to the area of the triangle $0<y<x<\pi$ is equal to $\Bbb{P}(c>ab)$.
We conclude that
$$\Bbb{P}(c>ab)=\frac{1}{2}\Leftrightarrow\int_{\frac{\pi}{2}}^{\pi}\cos^{-1}\left(\sqrt{2}\cos\left(x-\frac{\pi}{4}\right)\right){\rm{d}}x=\frac{\pi^2}{4}\approx 2.467401.$$
The value of this integral according to Wolfram Alpha:
Finally, I compute $\int_{\frac{\pi}{2}}^{\pi}\cos^{-1}\left(\sqrt{2}\cos\left(x-\frac{\pi}{4}\right)\right){\rm{d}}x =\int_{\frac{\pi}{2}}^{\pi}\cos^{-1}\left(\cos x+\sin x\right){\rm{d}}x$ using symmetry. Notice that with the change of variable $x\mapsto\frac{3\pi}{2}-x$:
$$ \begin{split} &\int_{\frac{\pi}{2}}^{\pi}\cos^{-1}\left(\cos x+\sin x\right){\rm{d}}x\\ &=\int_{\frac{\pi}{2}}^{\pi}\cos^{-1}\left(\cos\left(\frac{3\pi}{2}-x\right)+\sin\left(\frac{3\pi}{2}-x\right)\right){\rm{d}}x =\int_{\frac{\pi}{2}}^{\pi}\cos^{-1}\left(-\cos x-\sin x\right){\rm{d}}x\\ &=\int_{\frac{\pi}{2}}^{\pi}\left[\pi-\cos^{-1}\left(\cos x+\sin x\right)\right]{\rm{d}}x=\frac{\pi^2}{2}-\int_{\frac{\pi}{2}}^{\pi}\cos^{-1}\left(\cos x+\sin x\right){\rm{d}}x. \end{split} $$ Hence $\int_{\frac{\pi}{2}}^{\pi}\cos^{-1}\left(\cos x+\sin x\right){\rm{d}}x=\frac{\pi^2}{4}.$
Starting with a picture for the vertex $a$ and the angle $\alpha$
the event
$$a b > c$$ can be written as
$$|2 \sin(\alpha)| |2 \sin(\beta)|> |2 \sin(\alpha-\beta)|$$ the interval for $\alpha$ and $\beta$ is $(-\pi/2, \pi/2)$ and can be extended to a full period$(-\pi, \pi]$ without any change for the probability. The angles $\alpha$ and $\beta$ shall be the argument of two complex numbers $x$ and $y$ with $\alpha = \arg(x)$ and $\beta = \arg(y)$. $x$ as well as $y$ shall have a normal gaussian distribution for real and imaginary part. This ensures that the angles $\alpha$ and $\beta$ are uniform distributed. And gives equivalent events:
\begin{align} 2 \frac{|\Im(x)|}{|x|}\frac{|\Im(y)|}{|y|} &> \frac{|\Im(x y^*)|}{|x y^*|}\\ 2 |\Im(x)||\Im(y)| &> |\Im(x y^*)|=|\Im(x) \Re(y) - \Im(y)\Re(x)|\\ 1 &> \biggl| \frac{1}{2} \Biggl( {\frac{\Re(y)}{\Im(y)} + \frac{-\Re(x)}{\Im(x)}} \Biggr)\biggr| \end{align} Between the vertical bars is the sample mean of two standard Cauchy distributed quotients and this is again standard Cauchy distributed. Looking at another definition of Cauchy distribution, where a random straight line through $(0,1)$ in a cartesian system intersects with the x axis, then it is obvious that the probability for this event is $1/2$.
OK, the difficult part is the proof for the sum of Cauchy distributions....
With this solution the problem can get another formulation:
A random straight line has distance 1 to the intersection of two other random straight lines, all in one plane. Find the probability for the event that the three lines enclose an area smaller than 1?
Here is an elementary proof using basic trigonometry for the more general formula $p(ab<rc)=\frac{2 \arctan(r)}{\pi}.$
Fixing one point on a circle of radius $r$ and placing the other two at angles $a$ and $b$ consecutively the condition becomes $4r^2\sin{a}\sin{b}>2r\sin{(b+a)}$ or $2r\sin{a}\sin{b}>|\sin{(b+a)}|$ with $a,b\in[0,\pi]$. Expanding we find that $2r\tan a\tan b>|\tan a+\tan b|$ or equivalently $2r>|\cot a+\cot b|$ and hence $2r>|\tan (a^{'})+\tan (b^{'})|$ where $a^{'},b^{'}\in[-\pi/2,\pi/2]$. Plotting the region, illustrated here for $r=1$, we have:
and focussing on the curve:
We see that we can compute the probability $p$ of point lying in the region by taking half the area of rectangle OABC assuming that we can show the curve is suitably symmetric.
This is easy to do by setting $s(x,y)=\tan(\frac{x+y}{2})+\tan(\frac{y-x}{2})$ and noting that the region $OABC$ is mapped to $[0,\pi]\times[0,2\arctan(r)]$ and the curve to $s(x,y)=2r$. The plot becomes
Using the tangent addition formula it is easy to prove that $$s(x,y)=2r \iff S(\pi-x,2 \arctan(r)-y)=2r$$ with $x\in[0,\pi]$,$y\in[0,\pi/2]$.
Hence probability $p=$ area OABC$/d^2$ with $d=\frac{\sqrt{2}\pi}{2}$ and $|OA|=|(\arctan(r),\arctan(r))|=\sqrt{2}\arctan(r)$.
Hence we obtain $$p= \frac{d \sqrt{2}\arctan(r)}{d^2}=\frac{\sqrt{2}\arctan(r)}{ \frac{\sqrt{2}\pi}{2}}= \frac{2 \arctan(r)}{\pi}.$$
We use the formula (from en.wikipedia.org/wiki/Circumcircle#Other_properties ):
diameter = a * b * c / ( 2 * area )
In our case diameter = 2 and we get:
4 * area = a * b * c
If h is the height of the triangle over the base c then area = h * c / 2 and we get:
2 * h = a * b
We are then interested in the probability h < c / 2.
If the side of the triangle giving the length c is fixed then the set of points giving a triangle satisfying the inequality is an "open" circle segment. All "open" circle segments can be obtained in this way from unique starting triangle side.
Now consider two triangle sides giving complementary circle segments. With probability one an extra point makes exactly one of the two completed triangles satisfy the inequality. And I think it should be easy to finish the argument 🙂
Instead of writing a bunch of formulas that I honestly feel are just tedious I give you a picture
The black circle segment is the part giving triangles satisfying the inequality (th red line is the side giving with length c). The roundish things are supposed to be circles Note that you can make a larger circle at the top giving the complementary segment (using this symmetry there are many ways to exploit symmetry to get the 1/2 probability, I can expand but if already this part is not understandable I again do not really see the point). Hope this gives enough intuition for the problem. For me the entire point of this question was to avoid doing lots of technical computations :)
