Problem 1373

1. Determine the number of layers in the tetrahedral pile:

Each layer in the tetrahedral pile has a triangular number of balls. The $n$-th layer from the top has $\frac{n(n+1)}{2}$ balls. The total number of balls in the pile is given by the sum of the first $n$ triangular numbers:
$$\sum_{k=1}^{n} \frac{k(k+1)}{2} = 8436$$
Simplifying the sum, we get:
$$\sum_{k=1}^{n} \frac{k(k+1)}{2} = \frac{1}{2} \sum_{k=1}^{n} k(k+1) = \frac{1}{2} \left( \sum_{k=1}^{n} k^2 + \sum_{k=1}^{n} k \right)$$
Using the formulas for the sum of the first $n$ squares and the sum of the first $n$ integers:
$$\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}, \quad \sum_{k=1}^{n} k = \frac{n(n+1)}{2}$$
We get:
$$\frac{1}{2} \left( \frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2} \right) = 8436$$
Simplifying further:
$$\frac{1}{2} \left( \frac{n(n+1)(2n+1) + 3n(n+1)}{6} \right) = 8436$$
$$\frac{n(n+1)(2n+4)}{12} = 8436$$
$$\frac{n(n+1)(n+2)}{6} = 8436$$
$$n(n+1)(n+2) = 50616$$

2. **Solve for $n$:**

We need to find $n$ such that $n(n+1)(n+2) = 50616$. Testing values, we find:
$$n = 36 \quad \text{(since \( 36 \cdot 37 \cdot 38 = 50616 \))}$$

3. Calculate the height of the pile:

The height of the pile is determined by the distance between the layers of balls. Each layer is separated by a distance $D$, and there are $35$ such distances between the $36$ layers. Additionally, we need to account for the radius of the balls at the top and bottom of the pile.

The distance $D$ between the centers of the balls in adjacent layers can be found by considering the geometry of the tetrahedron formed by four balls. The side length of this tetrahedron is $2$ cm (twice the radius of the balls).

Let $A, B, C, D$ be the vertices of the tetrahedron, with $D$ at the top and $A, B, C$ forming the base. The height $D$ of this tetrahedron can be found using the Pythagorean theorem in the triangle formed by the centroid of the base and the top vertex.

The height of the tetrahedron is:
$$D = \frac{2\sqrt{6}}{3}$$

Therefore, the total height of the pile is:
$$35D + 2 = 35 \left( \frac{2\sqrt{6}}{3} \right) + 2 = \frac{70\sqrt{6}}{3} + 2$$

Simplifying, we get:
$$\frac{70\sqrt{6} + 6}{3}$$