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Inside triangle E (the large red triangle) there are 2 smaller similar triangles. A ( the yellow triangle ) and B ( the blue triangle ), both of these smaller triangles have a base length that is 1 / 2 the base of E and a height length that is 1 / 2 the height of E . Also, their bases and heights are parallel to the base and height of E. A and B can move freely inside E and they must wholly stay in E. Also, there bases and heights must remain parallel with the corresponding side of E. (the diagram shows what I mean better).

What is the probability that a point uniformly picked in E also lies in the intersection of A and B (green area)?

Simulations suggest .1 but someone gives an exact answer of 13 / 120 while another gets 1 / 10 confirming the simulations. I am unable to tell which is correct, I need help.


I'll assume that by "moving freely", you mean that the apices of the small triangles are uniformly distributed over the area in which they can lie such that the small triangles are contained in the large triangle.

Let's introduce Cartesian coordinates with the origin at the apex of the big triangle and the other two vertices at $(1,0)$ and $(0,1)$ and characterise the two small triangles by the coordinates of their apices, $x_1,y_1$ and $x_2,y_2$.

There are two possible configurations: the one you've drawn, where without loss of generality $x_2\gt x_1$ and $y_2\gt y_1$, and another one where without loss of generality $x_2\gt x_1$ and $y_1\gt y_2$.

In the first case, the small triangles overlap in a fraction $\left(\frac12-(x_2-x_1)-(y_2-y_1)\right)^2$ of the big triangle.

In the second case, again without loss of generality, let $x_2-x_1\gt y_1-y_2$. Then the small triangles overlap in a fraction $\left(\frac12-(x_2-x_1)\right)^2$ of the big triangle.

With appropriate symmetry factors of $2$ in the first case and $4$ in the second case, this yields a mean overlap of

\begin{align} \left(\frac18\right)^{-2}\left(2\int_0^\frac12\mathrm dx_2\int_0^{\frac12-x_2}\mathrm dy_2\int_0^{x_2}\mathrm dx_1\int_0^{y_2}\mathrm dy_1\left(\frac12-(x_2-x_1)-(y_2-y_1)\right)^2\\ +4\int_0^\frac12\mathrm dx_2\int_0^{\frac12-x_2}\mathrm dy_2\int_0^{x_2}\mathrm dx_1\int_{y_2}^{y_2+x_2-x_1}\mathrm dy_1\left(\frac12-(x_2-x_1)\right)^2 \right)\;, \end{align}

where $\frac18$ is the area in which the apices of the small triangles can lie.

The integrals are readily evaluated in Sage, with the result $\frac1{10}$ confirming your simulations and the answer of a certain other.

  • $\begingroup$ you are correct about what I meant, thanks for correcting me +1 $\endgroup$ – bobbym Jun 14 '16 at 10:06

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