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This seemingly trivial question inspired this one: Somewhere within a circle of unit radius is placed a circle of radius $r, 0 < r < 1$, such that this inner circle's center is uniformly distributed within $1-r$ of the outer circle's center. An equilateral triangle with uniformly distributed orientation is inscribed within the inner circle. (That is, all three vertices are on the inner circle's circumference.)

What is the probability (in terms of $r$) that the outer circle's center is inside the triangle?

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  • $\begingroup$ To choose a random circle then a random orientation for an inscribed triangle is just like to choose a random triangle (with a tailor-made definition of "random"). For any fixed orientation, it is not difficult to figure which triangles enclose the centre of the big circle. By symmetry orientation does not really matter, so we just need to compute the ratio between the area of an equilateral triangle and the area of a circle. $\endgroup$ Commented Aug 21, 2015 at 19:19

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Let's assign some labels. Let the outer circle have centre $O$, the inner circle have centre $P$, let the triangle be $T$ and let r.v. X be distance $\overline{OP},\;$ so $0\lt X\lt 1-r$. Also, let $E$ be the event that $O$ is inside $T$.

For $O$ to be inside $T$, it must also be inside the inner circle, so we want $X\lt r$.

Since the orientation of $T$ is uniformly distributed, the probability of $O$ being inside it equals the fraction of the circumference of a circle, centre $P$ with radius $X$, that is inside $T$. Referring to the diagram of the inner circle, this equals $\frac{6A}{2\pi} = \frac{3A}{\pi}$. Using the Sine Rule, we see:

$$\sin B = \dfrac{r\sin(\pi/6)}{X} = \dfrac{r}{2X}.$$ enter image description here

Angle $B$ is obtuse, so:

$$A = \pi-\dfrac{\pi}{6}-B = \dfrac{5\pi}{6}-\left(\pi-\sin^{-1}\left(\frac{r}{2X}\right)\right) = \sin^{-1}\left(\frac{r}{2X}\right) - \dfrac{\pi}{6}.$$

Note that if $X\lt r/2,\;$ then $O$ is always inside $T$. For one thing, this means that if $2/3 \lt r\lt 1$ then $P(E)=1,\;$ so we assume hereon that $0 \lt r\lt 2/3$.

Now, in the range $r/2\lt X\lt \min\{r,1-r\},\;$ we have $$P(E) = \dfrac{3A}{\pi} = \dfrac{3}{\pi} \sin^{-1}\left(\dfrac{r}{2X}\right) - \dfrac{1}{2}.$$

The distribution of $X$ is as follows. Since $P$ is uniformly distributed inside a circle (of radius $1-r$), $P(X=x)$ is proportional to the circumference of a circle with radius $X$, which means that $f_X(x) = cx$ for some constant $c$, which we find:

$$1=\int_{0}^{1-r}cx\;dx = \dfrac{c}{2}\left[x^2\right]_{0}^{1-r} = \dfrac{c}{2}(1-r)^2,\; \text{ so } c=\dfrac{2}{(1-r)^2},\; \text{ and } f_X(x) = \dfrac{2x}{(1-r)^2}.$$

So we have,

$$P(E) = P(X\lt r/2) + P(E\;\cap\; r/2 \lt X\lt \min\{r,1-r\}).$$

$$P(X\lt r/2) = \int_{0}^{r/2}f_X(x)\;dx = \dfrac{1}{(1-r)^2} \left[x^2\right]_{0}^{r/2} = \dfrac{r^2}{4(1-r)^2}.$$

For $0\lt r\lt 1/2,\; \min\{r,1-r\} = r,\;$ so in this range,

\begin{eqnarray*} P(E\;\cap\; r/2\lt X\lt r) &=& \int_{x=r/2}^r \left( \dfrac{3}{\pi} \sin^{-1}\left(\dfrac{r}{2x}\right) - \dfrac{1}{2} \right) \dfrac{2x}{(1-r)^2}\;dx \\ &=& \dfrac{1}{(1-r)^2} \left[\dfrac{3x}{4\pi} \left(r\sqrt{4-r^2/x^2} + 4\sin^{-1}\left(\dfrac{r}{2x}\right) \right) - \dfrac{x^2}{2} \right]_{x=r/2}^r \\ && \qquad\qquad\qquad\qquad\qquad\qquad\text{using Wolfram Alpha} \\ &=& \dfrac{r^2}{4\pi(1-r)^2} \left(3\sqrt{3} - \pi \right) \\ && \\ \therefore\quad P(E) &=& \dfrac{3\sqrt{3}r^2}{4\pi(1-r)^2}. \end{eqnarray*}

For $1/2 \lt r \lt 2/3,\;$ we just change the upper limit of integration, but it results in a rather ugly expression for $P(E)$:

\begin{eqnarray*} P(E\cap r/2\lt X\lt r) &=& \int_{x=r/2}^{1-r} \left( \dfrac{3}{\pi} \sin^{-1}\left(\dfrac{r}{2x}\right) - \dfrac{1}{2} \right) \dfrac{2x}{(1-r)^2}\;dx \\ && \\ \therefore\quad P(E) &=& \dfrac{1}{4\pi(1-r)^2} \left[ 3r(1-r)\sqrt{4-\dfrac{r^2}{(1-r)^2}} + 12(1-r)^2\sin^{-1}\left( \dfrac{r}{2(1-r)} \right) -2(1-r)^2\pi \right]. \end{eqnarray*}

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  • $\begingroup$ Hunh. I didn't expect that. Let me look at this when I get a chance; I'll accept if it passes muster. Looks reasonable at first glance, though. $\endgroup$
    – Brian Tung
    Commented Aug 22, 2015 at 17:01
  • $\begingroup$ @BrianTung Hi Brian. Yes, it would be good of you to check. I might well have a mistake or misunderstood something. $\endgroup$
    – Mick A
    Commented Aug 22, 2015 at 17:11
  • $\begingroup$ I've accepted this because it looks basically right, but for some reason I'm not getting the same value for $r=1/2$ in the two expressions. I haven't double checked yet, though. $\endgroup$
    – Brian Tung
    Commented Aug 26, 2015 at 1:10
  • $\begingroup$ @BrianTung Sorry, I made a mistake in the last expression. Fixed now. I replaced $4(1-r)\sin^{-1}\left( \dfrac{r}{2(1-r)} \right)$ with $12(1-r)^2\sin^{-1}\left( \dfrac{r}{2(1-r)} \right)$. I get now $\left(3\sqrt{3}\right)/\left(4\pi\right)$ for both expressions with $r=1/2$. $\endgroup$
    – Mick A
    Commented Aug 26, 2015 at 2:31

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