XLII OM - I - Problem 12

For any natural number $ n $, consider the rectangular prism $ K_n $ with edge lengths $ 1,1,n $ and the set $ R_n $ consisting of $ 4n+1 $ points: the vertices of the rectangular prism $ K_n $ and the points dividing its longer edges into unit segments. We randomly select three different points from the set $ R_n $. Let $ p_n $ be the probability that the selected points are the vertices of an obtuse triangle. Calculate $ \lim_{n\to \infty} p_n $.
Note: Three collinear points do not form a triangle.
Let $ Q(n) $ denote the number of all three-element subsets of the set $ R_n $, $ T(n) $ - the number of non-degenerate triangles with vertices in the set $ R_n $, and $ P(n) $ - the number of obtuse triangles. The probability of interest is the fraction $ p_n = P(n)/Q(n) $. Notice that



(Justification of the second equality: from the number of all triples, we subtract the number of collinear triples of points - that is, triples of points lying on any of the four edges, each of which contains $ n +1 $ points of the set $ R_n $.)
After expanding the binomial coefficients, we obtain the formulas



Let us divide the set $ R_n $ into four-element "layers" $ H_0, H_1, \ldots, H_n $: the vertices of one of the square faces of the cube $ K_n $ form the layer $ H_0 $, the vertices of the other - the layer $ H_n $; the layer $ H_k $ consists of points of the set $ R_n $ at a distance of $ k $ units from the plane of the layer $ H_0 $.
Since we are interested in the limiting (as $ n \to \infty $) behavior of the quantities under consideration, we can assume that $ n \geq 4 $.
Fix a natural number $ r $, $ 3 \leq r \leq n-1 $, and choose any numbers $ k, m \in \{0,1,\ldots,n\} $ such that $ m - k = r + 1 $. There are thus $ r $ layers between layers $ H_k $ and $ H_m $. By choosing any points $ A \in H_k $, $ C \in H_m $, and $ B \in H_l $, where $ l \in \{k + 1,\ldots,m-1\} $, we obtain either an obtuse triangle or a collinear triple of points (since $ m-k \geq 4 $). We have $ 16 $ ways to position the pair of points $ A $ and $ C $ and $ 4r $ possible positions for point $ B $ - a total of $ 64r $ possibilities; $ 4r $ of them give collinear triples. The remaining $ 60r $ are obtuse triangles.
Since the numbers $ k $, $ m $ (with a given difference $ r+1 $) can be chosen in $ n-r $ ways, the number $ P(n) $ of all possible obtuse triangles is at least the sum



Thus, we have a two-sided estimate



The quantities $ Q(n) $, $ T(n) $, $ S(n) $ are third-degree polynomials (in $ n $). The coefficient of $ n^3 $ in $ Q(n) $ is $ 32/3 $, while in $ T(n) $ and $ S(n) $ the coefficient of $ n^3 $ is $ 10 $. Therefore,



and by the squeeze theorem,



