What is the probability that a "happy triangle" contains the centre of its circumcircle? Is it really $1/2$?

Question

On a circle, choose three uniformly random independent points $A,B,C$.
Inside the circle, choose a uniformly random independent point $D$.
Triangle $ABC$ is called "happy" if it contains $D$.

What is the probability that a happy triangle contains the centre of its circumcircle?

Simulations suggest that the probability is $1/2$.

Given the simplicity of this probability, I wonder if there is an intuitive explanation.

Context

This is a variation of the classic question, "What is the probability that a triangle, whose vertices are three random points on a circle, contains the centre of the circle?" The answer to that question is $1/4$, and there is an intuitive explanation.

My attempt

Assume the circle is a unit circle.

From circle triangle picking, the expected area of $\Delta ABC$ is $\dfrac{3}{2\pi}$.

$$\therefore P(\Delta ABC\text{ contains }D)=\dfrac{\frac{3}{2\pi}}{\text{Area of circle}}=\dfrac{3}{2\pi^2}$$

Call the centre of the circle $O$.

$$P\left(\text{$\Delta ABC$ contains $O$ | $\Delta ABC$ contains $D$}\right)$$

$$=\frac{P(\text{$\Delta ABC$ contains $O$ and $D$})}{P(\text{$\Delta ABC$ contains $D$})}$$

$$=\dfrac{2\pi^2}{3}P\left(\text{$\Delta ABC$ contains $O$ and $D$}\right)$$

I do not know how to calculate $P\left(\text{$\Delta ABC$ contains $O$ and $D$}\right)$.

I used this formula to make my simulations.

Possibly related question: "Probability that a random triangle with vertices on a circle contains an arbitrary point inside said circle"

Add this one to the list?

I might add this question to my list of probability questions that have answer $1/2$ but resist intuitive explanation.

Other elegant properties of happy triangles?

Simulations suggest that, for a happy triangle inscribed in a unit circle:

• The expected area is $\pi/4$. (Thus, a happy triangle divides the circle into four regions of equal expected area.)
• The expected shortest side length is $1$.
• The expected product of the three side lengths is $\pi$.
• The probability that it contains (another) random point in the circle is $1/4$.
• If the sides are, in random order, $a,b,c$, then $P(ab<c)=1/3$. (For not-necessarily-happy $\Delta ABC$, this probability is $1/2$).

Answer 1

Write $O$ and $D$ as shorthand for the events $O\in\triangle$ and $D\in\triangle$. By Bayes' theorem, $$P(O|D)P(D)=P(D|O)P(O)$$ There is a very simple descsription of $P(D)$: the expected proportional area of the triangle to the circle, $E[|\triangle|/\pi]$. Thus, we have $$P(O|D)E[ |\triangle|]=E[|\triangle||O]P(O)\tag{*}$$ We also know $P(O)=1/4$ by a simple argument. Imagine instead we first picked pairs of antipodal points $\pm A$, $\pm B$, $\pm C$, and then chose the triangle with vertices $(v_1,v_2,v_3)=(\pm A, \pm B, \pm C)$ for an independent choice of signs. There are $8$ possible triangles from this process, but opposite choices of signs give the same triangle $\triangle, -\triangle$ up to inversion, so we only need to consider $4$ choices by choosing representatives. We also know that $O\in\triangle$ exactly when $v_3$ lies in the minor arc between $-v_1$ and $-v_2$. There is only one such triangle among the sign choices for $v_1,v_2$, so WLOG, let this be $\triangle ABC$. We then choose $\triangle A'BC,\triangle AB'C,\triangle ABC'$ as the other three representatives, where $X'$ is shorthand for $-X$. $\triangle ABC$ is the only one containing $O$, so $P(O)=1/4$.

We can use the same ideas to find $E[|\triangle||O]$. In particular, $$|\triangle ABC|=|\triangle A'BC|+|\triangle AB'C|+|\triangle ABC'|\tag{**}$$ This follows from some simple geometry. Let $a,b,c$ be the distances from $O$ to the sides $BC, CA, AB$, respectively. $\triangle AB'C'$ and $\triangle ABC$ share the same base length $\overline{BC}$, but differ in height by $2a$. Thus, $$|\triangle ABC|-|\triangle A'BC|=|\triangle ABC|-|\triangle AB'C'|=a\overline{BC}$$ It follows that $$\begin{align*} 3|\triangle ABC|-|\triangle A'BC|-|\triangle AB'C|-|\triangle ABC'|=a\overline{BC}+b\overline{CA}+c\overline{AB} \\ = 2(|\triangle OBC|+|\triangle OCA|+|\triangle OAB|) \\ = 2|\triangle ABC| \end{align*}$$ from which $(**)$ follows.

From $(**)$ it follows immediately that $E[|\triangle||O]=2E[|\triangle|]$. Combining with $P(O)=1/4$ in $(*)$, we get $P(O|D)=1/2$ as desired.

Answer 2

Here I offer an intuitive proof of the fact that, in the diagram below (in which letters represent areas), $A+B+C+D+E+F=a+b+c+d+e+f$. This fact is used in the solutions found in @Kevin P. Costello's comments to the OP, and @Jacob's answer.

(For this proof, don't think in terms of letters; rather, think in term of the shapes in the diagram. I'm using letters just to communicate the idea.)

We have $A+a=b+c$. This is obvious when you consider the triangle with area $A+a+b+c$.

$\therefore A=b+c-a$

Similarly, we have:

$B=a+f-b$
$C=d+e-c$
$D=c+b-d$
$E=f+a-e$
$F=e+d-f$

Add the equations. On the RHS, each small letter appears twice as positive and once as negative (you don't have to count them; it's obvious from symmetry), so altogether exactly once.

$\therefore A+B+C+D+E+F=a+b+c+d+e+f$.

Answer 3

In the $xy$-plane, let

• $A,B,C$ be three points chosen uniformly at random from points on the standard unit circle.$\\[4pt]$
• $D$ be a point chosen uniformly at random from points in the interior of the standard unit disk.

and where all choices are mutually independent.

Using the letters "$D$","$O$" in a context dependent way, let

• $D$ be the event that the interior of $\Delta ABC$ contains the point $D$.$\\[4pt]$
• $O$ be the event that the interior of $\Delta ABC$ contains the point $O=(0,0)$.

Claim:$\;P(O{\,|}D)={\large{\frac{1}{2}}}$.

Proof:

We will make the following assumptions . . .

• $A,B,C$ are distinct.$\\[4pt]$
• $A,B,C$ are in counterclockwise order.$\\[4pt]$
• $A=(1,0)$.

The above assumptions will not affect the values for any of the results obtained below.

Based on these assumptions, we can write \begin{align*} B&=(\cos(u),\sin(u))\\[4pt] C&=(\cos(v),\sin(v))\\[4pt] \end{align*} where$\;0 < u < v < 2\pi$.

From the Shoelace formula, we get $$ \text{area}(\Delta ABC) = {\small{\frac{1}{2}}} \Bigl( (\sin(u)-\sin(v)-\sin(u)\cos(v)+\sin(v)\cos(u) \Bigr) $$ Note that the event $O$ occurs if and only if$\;0 < u < \pi < v < u + \pi$.

It follows that $ E\Bigl(\text{area}(\Delta ABC)\,{\large{|}}\,O\Bigr) $ can be expresses as $$ \frac { {\displaystyle{ \;\frac{1}{2} \int_0^\pi\int_\pi^{u+\pi} \Bigl( \sin(u)-\sin(v)-\sin(u)\cos(v)+\sin(v)\cos(u) \Bigr) \,dv\;du\;\; }} } { {\displaystyle{ \int_0^\pi\int_\pi^{u+\pi} \,dv\;du }} } $$ which evaluates to ${\Large{\frac{3}{\pi}}}$.

Then we get $$ P(D{\,|}O) = \frac { \,E\Bigl(\text{area}(\Delta ABC)\,{\large{|}}\,O\Bigr)\; } { \text{area}(\text{the unit disk}) } = \frac{3}{\pi^2} $$ As you've noted, we have

• $P(O)={\large{\frac{1}{4}}}$ $\\[9pt]$
• $P(D)={\large{\frac{3}{2\pi^2}}}$

hence by Bayes' formula, we get \begin{align*} P(O{\,|}D) &= P(D{\,|}O) \,{\cdot} \frac{P(O)}{P(D)} \\[4pt] &= \frac{3}{\pi^2} \,{\cdot} \frac{1}{4} \,{\cdot} \frac{2\pi^2}{3} \\[6pt] &= \frac{1}{2} \\[4pt] \end{align*} as was to be shown.