Triangle Centroid Probability
Geometry, Probability, Calculus: Medium/Hard
Geometry-Probability Problem
This math question was sent to me by my friend and puzzle fiend Sandi Sandman who in turn got it from The Riddler:
You have an equilateral triangle. You pick a point on each of the three sides, each of the three points picked independently and uniformly randomly. The three random points you picked form a triangle. What’s the chance that the centroid of the equilateral triangle that you started with is inside this smaller triangle?
What makes this question interesting is that it is deceptively simple. One expects there to be an intuitively simple solution with the answer being a simple rational number. As it turns out, neither is the case. In fact, I cannot think of a solution that does not use calculus.
Solution preamble: Basic centroid segment relationship
Before we get to the solution there is a result we are going to need. This is easily derived based on the above figure.
Consider equilateral \( \triangle \) ABC of side \(=1 \) (without loss of generality) and a point \( X\) on side \( AB\) with \( l(AX) = x \leq 0.5\). \( G\) is the centroid and \(XG \) intersects side \(BC\) at \( H\) with \( l(BH) = l\). Let us find \( l\) in terms of \( x\).
Extend \(XH\) and \( AB\) to intersect at \( I\). The segment lengths shown in the figure can be calculated knowing \( \angle CAB \) and \(\angle CBA\) are \(60^o \). Now considering similar triangles \( \triangle\) XJI, \(\triangle\) GKI and \( \triangle\) HLI we get:
\[\displaystyle \frac{d+\displaystyle\frac{x}{2}}{\displaystyle\frac{\sqrt{3}}{2}x} \quad = \quad \frac{d+\displaystyle\frac{1}{2}}{\displaystyle\frac{1}{2\sqrt{3}}} \quad = \quad \displaystyle\frac{d+1-\displaystyle\frac{l}{2}}{\displaystyle\frac{\sqrt{3}l}{2}} \]
From which we get:
\[\bbox[yellow, 10px]{ \mathit{l = \frac{2x -1}{3x-2}} \qquad \qquad \qquad \tag{1} \label{eq1} } \]
Now we are ready to proceed to the actual solution:
Solution 1: Look at the conditions when XYZ contains the centroid
Consider as before, equilateral \( \triangle\) ABC of side =1 unit (without loss of generality) shown in the figure above.
- G is the centroid and X, Y, Z are randomly chosen points on the three sides with distances \( x, y, z \) measured as shown.
- Y’ is where XG interesects side AC. Similarly Z’ is where YG intersects BC.
- Let \(F = F(x,y,z) \) be the joint pdf for \( \triangle \) XYZ to contain centroid G.
We now evaluate the probability by realizing that relative to a given point X, the points Y and Z can only be on certain parts of their respective sides if XYZ has to contain G. e.g. it can be seen that for the X, Y, Z as shown in the figure, \(F(x,y,z) = 0 \)
The required probability is then:
\begin{align*}
&= \int\limits_{x=0}^{1} \int\limits_{y=0}^{1} \int\limits_{z=0}^{1} F \ dxdydz \\
&= 2 \ \int\limits_{x=0}^{0.5} \int\limits_{y=0}^{1} \int\limits_{z=0}^{1} F \ dxdydz \qquad ( \textit{ by symmetry, probability is same when } x \leq 0.5 \ or \ x \geq 0.5 ) \\ \\
& \textit {now evaluate based on Y being in one of the three segments: below Y’, between Y’ and N, between N and A } \\ \\
& = 2 \ \int\limits_{x=0}^{0.5} \ \left[\underbrace{ \ \int\limits_{y=0}^{CY} \int\limits_{z=0}^{1} F \ dydz}_{\textit{this is 0 because XYZ will be ‘under’ G}} \quad + \qquad
\underbrace{\int\limits_{y=CY}^{0.5} \int\limits_{z=0}^{1} F \ dydz}_{\textit{this is 1: XYZ will always contain G}} \qquad + \qquad \int\limits_{y=0.5}^{1} \ \int\limits_{z=0}^{1} F \ dydz \right ]dx \\ \\
& = 2 \ \int\limits_{x=0}^{0.5} \ \left[ \ \int\limits_{y=\frac{2x-1}{3x-2}}^{0.5} \ dy \quad + \qquad
\int\limits_{y=0.5}^{1} \int\limits_{z=0}^{1} F \ dydz \right ]dx \qquad ( \textit{ using result } \eqref{eq1} \textit{ to get CY in terms of} \ x \ ) \\ \\
& = 2 \ \int\limits_{x=0}^{0.5} \ \left[ \ \frac{x}{4-x} \quad + \qquad
\int\limits_{y=0.5}^{1} \int\limits_{z=0}^{1} F \ dydz \right ]dx \\ \\
& (\textit{now evaluate based on Z being between C and Z’ or Z’ and B }) \\ \\
& = 2 \ \int\limits_{x=0}^{0.5} \ \left[ \ \frac{x}{4-x} \quad + \qquad
\int\limits_{y=0.5}^{1} \ \left( \ \underbrace{\int\limits_{z=0}^{CZ’} F \ dz}_{\textit{this is 1 since XYZ will contain G in this range}} \ + \ \underbrace{\int\limits_{z=0}^{CZ’} F \ dz}_{\textit{this is 0}} \ \ \right )dy \right ]dx \\ \\
& \textit{now we can use } \eqref{eq1} \textit{ to solve for CZ’ in terms of y when y} \geq 0.5 \\ \\
& 1 – CZ’ = \frac{2(1-y) -1}{3(1-y) -2} \\
& \rightarrow \ CZ’ = \frac{y}{3y – 1} \\ \\
& \textit{continuing, our required probability…} \\ \\
& = 2 \ \int\limits_{x=0}^{0.5} \ \left[ \ \frac{x}{4-x} \quad + \quad
\int\limits_{y=0.5}^{1} \frac{y}{3y-1}dy \right ]dx \\ \\
& = 2 \ \int\limits_{x=0}^{0.5} \ \left[ \ \frac{x}{4-x} \quad + \quad \frac{1}{6} \quad + \quad \ln\frac{4}{9} \right ]dx \\ \\
&= \bbox[yellow, 10px]{ \ln\frac{16}{6} \approx 0.4621 \ \ \ \textit{or} \ \ \ 46.21 \% } \\ \\
\end{align*}
Solution 2: Look at when XYZ doesn't contain G
This is an easier, shorter solution provided by the esteemed Sandi Sandman.
Consider as before, equilateral \( \triangle\) ABC of side =1 unit.
- G is the centroid and X, Y, Z are randomly chosen points on the three sides with distances \( x, y \) measured as shown.
- Y’ is where XG interesects side AC. CY’ has length \(l\).
\( \triangle\) XYZ splits the original triangle into 4 triangles, 3 of which share exactly one vertex each with ( \triangle\) ABC. Without loss of generality, let us calculate the probability that G falls into one of these triangles, say \( \triangle\) AXY.
The only way \( \triangle\) AXY contains G is if \(x \leq 0.5 \) and \(y \leq l \).
So the probability that \( \triangle\) AXY contains G for a given \(x\) using \(\eqref{eq1} \) is:
\[ l = \frac{2x-1}{3x-2}\]
The total probability that \( \triangle\) AXY contains G is then:
\begin{align*}
&= \int\limits_{x=0}^{0.5} \frac{2x-1}{3x-2} \ dx \\ \\
& = \frac{2}{3} \left. \left[x \ \ + \ \ \frac{1}{6} \ln(2-3x) \right] \right\rvert_{0}^{0.5} \\ \\
& = \frac{1}{3} – \frac{1}{9}\ln4 \\ \\
\end{align*}
By symmetry, this has to be the same probability for having the other two such triangles contain G. So the final probability that \( \triangle\) XYZ contains G is
\begin{align*}
& = 1 – 3 \left[ \ \frac{1}{3} – \frac{1}{9}\ln4 \ \right ] \\ \\
&= \bbox[yellow, 10px]{ \ln\frac{16}{6} \approx 0.4621 \ \ \ \textit{or} \ \ \ 46.21 \% } \\ \\
\end{align*}
same as before.