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We are given the equilateral triangle A.
On each edge of the triangle we pick a point:

  • randomly (probability distribution is uniform)
  • independently of others

We construct new triangle B from randomized points.
Task is to find the chance of B containing the centre of circle circumscribed around triangle A.

I would appreciate hints and pointers. Thanks!

Edit
Suppose $P_0$ is chosen on the bottom line of triangle A. $P_O = (x_0,0), x_0 \in [0,1]$
Let's find geometrical place of points $P_1, P_2$.
If $P_1$ is on the upper left edge, that $P_1 = (x_1,\sqrt 3 x_1), x_1 \in [0, \frac 1 2]$.
Then $P_2$ is upper right edge, and $P_2 = (x_2, \sqrt 3 (1 - x_2)), x_2 \in [\frac 1 2, 1]$.

Centre point is $C = (\frac 1 2, \frac 1 {2 \sqrt 3 })$.
Now, we shall find the constraints for points.
For fixed $x_0$, $P_1$ and $P_2$ must lay below the line between $P_0$ and $C$.
But I wonder, if there is a way to describe constraint better?
Since this most trivial way suffers from situation when line $CP_0$ and one of the edges intersects outside of range $x_m$.
Edit 2
I have listened to the useful hints about advantages of angle-based view on the problem, and believe that now I have the geometry part figured:
As earlier, We fixed $P_0$ on the bottom edge and on the step 1 we pick $P_1$ on the left edge.
Consider $\angle \alpha$ line between bottom edge and line $CP_0$. Then $\alpha \in [\frac \pi 6, \frac \pi 2]$.
Then prohibited sector for left edge has angle $\frac \pi 2 - \alpha$.
Using same logic on the step 2 we obtain prohibited sector for point $P_2$.

And now it looks like I have to get down to all that integration, so your suggestions are welcome!

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  • $\begingroup$ Start with the geometry (ignoring the probability), in what circumstances is it true? That is, start with the point $P_1(x,0)$ on the base - what are $P_2$ and $P_3$ as a function of $x$? $\endgroup$ Commented Nov 27, 2014 at 0:18
  • $\begingroup$ @DaleM Would you be so kind to take a look on my edit? $\endgroup$ Commented Nov 27, 2014 at 23:49
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    $\begingroup$ It would be more common to have the distribution of points be uniform rather than Gaussian. In particular, a Gaussian distribution has infinite tails, while you problem limits the chosen points to the sides of the triangle. $\endgroup$ Commented Nov 28, 2014 at 0:10
  • $\begingroup$ @RossMillikan After your reply, I rechecked with task, and distribution turned out to be uniform actually. I misread it first time. Sorry for my mistake. $\endgroup$ Commented Nov 28, 2014 at 8:05
  • $\begingroup$ When the points are picked with an uniform distribution, the probability is $\frac{\log(4)}{3} \approx 0.4620981203733$. $\endgroup$ Commented Nov 29, 2014 at 8:11

5 Answers 5

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Let $a_1, a_2, a_3$ be the vertices of $A$. Choose a coordinate system such that $$a_1 = (0,0),\quad a_2 = (1,0),\quad a_3 = \left(\frac12,\frac{\sqrt{3}}{2}\right)$$ The centroid of triangle $A$ will be located at $$c = \frac{a_1 + a_2 + a_3}{3} = \left(\frac12, \frac{1}{2\sqrt{3}}\right)$$

Let $b_1, b_2, b_3$ be the vertices of $B$. By definition of triangle $B$, there exists three uniform random numbers $u_1, u_2, u_3$ taking values over $[0,1]$ such that $$b_1 = (1-u_1) a_1 + u_1 a_2,\quad b_2 = (1-u_2) a_2 + u_2 a_3,\quad b_3 = (1-u_3) a_3 + u_3 a_1$$

Consider following 4 events concerning the relative position of $c$ with respect to these 6 vertices. $$\begin{cases} \mathcal{E}_0 &\stackrel{def}{=} \text{Event}( c \in B = \triangle b_1 b_2 b_3 )\\ \mathcal{E}_1 &\stackrel{def}{=} \text{Event}( c \in \triangle a_1 b_1 b_3 )\\ \mathcal{E}_2 &\stackrel{def}{=} \text{Event}( c \in \triangle a_2 b_2 b_1 )\\ \mathcal{E}_3 &\stackrel{def}{=} \text{Event}( c \in \triangle a_3 b_3 b_2 )\\ \end{cases}$$ With probability one, these $4$ events are mutually disjoint. More precisely, $$\text{Prob}( \mathcal{E}_i \,\cap\, \mathcal{E}_j ) = 0\quad\text{ for } 0 \le i < j \le 3$$ Notice $\text{Prob}( \bigcup_{i=0}^3 \mathcal{E}_i ) = 1$ and by symmetry, $$\text{Prob}( \mathcal{E}_1 ) = \text{Prob}( \mathcal{E}_2 ) = \text{Prob}( \mathcal{E}_3 )$$ The probability we want is given by $$\text{Prob}( c \in B ) = \text{Prob}(\mathcal{E}_0) = 1 - 3\text{Prob}(\mathcal{E}_1)$$

Let $u = u_1$ and $v = 1-u_3$, we have $b_1 = (u,0), b_3 = \left(\frac{v}{2},\frac{v\sqrt{3}}{2}\right)$. The line passing through $b_1, b_3$ is given by $$\left|\begin{matrix} x & y & 1\\ u & 0 & 1\\ \frac{v}{2} & \frac{v\sqrt{3}}{2} & 1\\ \end{matrix}\right| = 0$$ Plugging the coordinates of $c$ into LHS and treat the resulting expression as a function of $u,v$. It is not hard to see the condition for $c \in \triangle a_1 b_1 b_3$ is simply

$$\left|\begin{matrix} \frac12 & \frac{1}{2\sqrt{3}} & 1\\ u & 0 & 1\\ \frac{v}{2} & \frac{v\sqrt{3}}{2} & 1\\ \end{matrix}\right| \ge 0 \quad\iff\quad 3uv - u - v \ge 0 \quad\iff\quad u \ge \frac12 \,\land\, v \ge \frac{u}{3u-1} $$ This leads to $$\text{Prob}(\mathcal{E}_1) = \int_{\frac12}^1 \left(1 - \frac{u}{3u-1}\right) du = \frac{3 - \log 4}{9}$$ and hence $$\text{Prob}( c \in B ) = 1 - 3\left(\frac{3 - \log 4}{9}\right) = \frac{\log 4}{3} \approx 0.4620981203733$$

As a double check, I have performed a numerical simulation of this problem. Out of $10^9$ copies of triangle $B$ generated, $462114483$ of them contains $c$. This leads to an estimate of the probability $c \in B$ at $0.462114 \pm 0.000016\;(\leftarrow 1 \sigma)$. This is consistent with what we have just derived.

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This question was also posed by Allen Gu in a puzzle-of-the-week post in FiveThirtyEight.com.

Allen Gu's solution is shown here.

The curator of the weekly puzzles, Zach Wissner-Gross also shared his own solution the following week

Other solutions presented:

Emma Knight's PDF

Josh Silverman's blog

Rajeev Pakalapati's tweet

Jonah's Mathematica Notebook

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This is only a hint, since the answer involves dependent probabilities which require (I believe) the use of a distribution function. The best I can do is a simulation which will only ever give an estimate...

Refer to this sketch: https://drive.google.com/file/d/0B-C4qnUqxtAVTnNnVElfLVRhSWMtS0EwRTVfUW5BT3lXZ0Vv/view?usp=sharing

I like Dale's hint. But my suspicion is that thinking in terms of angles might be easier. Picture a given side being intersected by a ray that comes from the center of the circle and sweeps a 120 degree sector, thus intersecting all points on that side (like an old-fashioned radar screen with the sweeping hand).

The set up the scenario in which we first randomly select a point on the bottom side (A), next we pick one on the left side (B), and finally pick one on the right (C). When measuring angles, I'm always sweeping counter-clockwise with the 0 angle being at the start of the side and 120 angle being at the end of the side.

The point on Side A can be selected by any angle from 0-120 degrees, because we are always able to construct a triangle that does not contain the center of the circle, regardless of where the point lies on side A.

Side B ("left" side) can be any angle from 60-120. Any angle between 0-60 puts the point in a position in which no "legal" triangle can be made.

To complete a "legal" triangle, Side C ("right" side) must be between 0 and (Angle B - 60).

Hope this is of some help. Very cool problem, by the way.

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  • $\begingroup$ I agree, use of angles probably makes it easier. So thank you for hit. But even your own sketch shows flaw in your logic. There are such values of $P_0$ (point on bottom side) which are able to construct valid triangle with point $P_1$ from anywhere from left side, yours "illegal" zone included. Point $x_0$ on your picture illustrates that. But still thanks, and I will write here about my progress. $\endgroup$ Commented Nov 27, 2014 at 23:17
  • $\begingroup$ Glad I could be of some help with the angles approach. You are correct, though, that my sketch shows all triangles you could construct such that they do NOT contain the center point, rather than those that DO contain the center point -- big difference and NOT what we are seeking. $\endgroup$ Commented Nov 28, 2014 at 3:18
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Without loss of generality, consider the triangle $\Delta ABC$ with vertexes $A(-1,0)$, $B(1,0)$, $C(0,\sqrt3)$ and center $O(0,\frac{1}{\sqrt3})$.

Consider the points $D(d,0)$ with $d\in [-1,0]$ (because the problem is symmetric we only need to consider 1 half and the double the result - this simplifies the process), $E(e\frac{1}{2}-1,e\frac{\sqrt3}{2})$ with $e\in [0,2]$ and $F(1-f\frac{1}{2},\frac{\sqrt3}{2}(1-f))$ with $f\in[0,2]$.

For any $d$ and $e$, legal values of $f$ require that $EF$ is "above" $O$ and $DF$ is "below" $O$. That is slope $EF\ge EO$ and slope $DF\le DO$. So

$$\begin{align} \frac{e\frac{\sqrt3}{2}-\frac{\sqrt3}{2}(1-f)}{\left(e\frac{1}{2}-1\right)-\left(1-f\frac{1}{2}\right)}&\ge\frac{e\frac{\sqrt3}{2}-\frac{1}{\sqrt3}}{\left(e\frac{1}{2}-1\right)-\left(0\right)}\\ \sqrt3\frac{e+f-1}{e+f-4}&\ge\frac{3e-2}{e-2}\\ \sqrt3(e+f-1)(e-2)&\ge(3e-2)(e+f-4)\\ \sqrt3e^2-2\sqrt3e+\sqrt3ef-2\sqrt3f-e+2&\ge3e^2+3ef-14e-2f+8\\ ((\sqrt3-3)e+2-2\sqrt3)f&\ge(3-\sqrt3)e^2+(2\sqrt3-13)e+6\\ f&\ge\frac{(3-\sqrt3)e^2+(2\sqrt3-13)e+6}{(\sqrt3-3)e+2-2\sqrt3}\\ \end{align}$$

And

$$\begin{align} \frac{0-f\frac{\sqrt3}{2}}{d-\left(1-f\frac{1}{2}\right)}&\le\frac{0-\frac{1}{\sqrt3}}{(d)-\left(0\right)}\\ \frac{\sqrt3f}{2d-2+f}&\le\frac{2}{\sqrt3d}\\ 3df&\le{4d-4+2f}\\ f&\le\frac{4d-4}{3d-2}\\ \end{align}$$

So

$$\frac{4d-4}{3d-2}\ge f\ge\frac{(3-\sqrt3)e^2+(2\sqrt3-13)e+6}{(\sqrt3-3)e+2-2\sqrt3}\\$$

NOTE: The algebra breaks down if the initial denominators are 0 but this only happens at the end points - make them open intervals if this bothers you. Also, I haven't checked it so I could have stuffed it up.

Now, what are the valid values and how does this relate to your probability distribution?

Note this is incredibly ugly even for a uniform distribution.

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  • $\begingroup$ Thanks for answer. I tried same approach with line equations. And I found it to monstrous as well. Maybe it could be easier with angles, like it's written in my 2nd edit? By the way, I made an edit, distributions actually are uniform. Sorry for that mistake. $\endgroup$ Commented Nov 28, 2014 at 9:17
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I may be mistaken but I dont believe there is a solution to this problem. It depends entirely on how you construct the triangle. Possibly. Have you heard of Bertrand's Paradox? Its a paradox in probability theory that asks a similar question: Take an equilateral triangle inscribed in a circle, what is the probability that a random chord of the circle is longer than the side of the triangle? Its simple enough to understand, and easy enough to answer, but it can be shown that you get drastically distinct answers depending on how you construct the chord.

Maybe Im wrong though. Maybe this is nothing like Bertrand's Paradox.

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