Tetrahedral stack of spheres

[MarkFL]

Consider a tetrahedral stack of spheres as in this image:



Suppose there is a stack consisting of $n$ layers, and the spheres all have diameter $d$.

Showing your work:

a) Compute the number $N$ of spheres in the stack as a function of $n$. Then, express $N$ as a binomial coefficient.

b) Compute the height $h$ of the stack as a function of $d$ and $n$.

 

[mathworker]

b)My attempt on '\(\displaystyle b\)':
Tetrahedron formed is of side \(\displaystyle (n-1).d\) .The vertex \(\displaystyle A\)(center of upper sphere),centroid of triangle formed by lower spheres \(\displaystyle G\) and one of lower vertices B(or C or D) forms right angled triangle.
\(\displaystyle \begin{align*}\text{AG}=\text{height}&=\sqrt{(\text{AB})^2-(\text{BG})^2}\\&=\sqrt{(n-1)^{2}.d^{2}-\frac{(n-1)^{2}.{d}^2}{3}}\\&=\sqrt{\frac{2}{3}}(n-1)d\end{align*}\)

 

[MarkFL]

b)My attempt on 'b':
Tetrahedron formed is of side \(\displaystyle (n-1).d\) .The vertex A(center of upper sphere),centroid of triangle formed by lower spheres G and one of lower vertices B(or C or D) forms right angled triangle.
\(\displaystyle \begin{align*}\text{AG}=\text{height}&=\sqrt{(\text{AB})^2-(\text{BG})^2}\\&=\sqrt{(n-1)^{2}.d^{2}-\frac{(n-1)^{2}.{d}^2}{3}}\\&=\sqrt{\frac{2}{3}}(n-1)d\end{align*}\)

You are close...but you have left out something.

 

[mathworker]

\(\displaystyle \sqrt{\frac{2}{3}}(n-1)d+d\)?(adding the upper and lower part of spheres)

 

[MarkFL]

\(\displaystyle \sqrt{\frac{2}{3}}(n-1)d+d\)?(adding the upper and lower part of spheres)

Yes, that is correct...well done!

 

[mathworker]

a)counting from each layer,
total number of spheres is,
\(\displaystyle \sum_{n=1}^n \frac{n(n+1)}{2}=\frac{n(n+1)(n+2)}{6}=\binom{n+2}{3} \)

 

[MarkFL]

a)counting from each layer,
total number of spheres is,
\(\displaystyle \sum_{n=1}^n \frac{n(n+1)}{2}=\frac{n(n+1)(n+2)}{6}=\binom{n+2}{3} \)

I would avoid using the same variable for both the index of summation and the upper limit; I would write instead:

\(\displaystyle \sum_{k=1}^n \frac{k(k+1)}{2}=\frac{n(n+1)(n+2)}{6}=\binom{n+2}{3} \)

May I ask how you determined the partial sum?

 

[mathworker]

\(\displaystyle \begin{align*}\sum\frac{n(n+1)}{2}&=\frac{1}{2}(\sum_n^2)+\sum n)\\&=\frac{1}{2}\left(\left(\frac{(n(n+1)(2n+1)}{6}\right)+\left(\frac{n(n+1)}{2}\right)\right)\\&=\frac{n(n+1)}{4}.\left(\frac{2n+1}{3}+1\right)\\&=\frac{n(n+1)(n+2)}{6}\end{align*}\)

 

[MarkFL]

Here is my solution:

Spoiler

a) First, we want to compute the number of spheres $S_k$ in the $k$ layer (where $1\le k\le n$), counting from the top. Each layer is a triangular grid of spheres with $k$ rows, and we may use the difference equation:

\(\displaystyle S_{j}-S_{j-1}=j\) where \(\displaystyle S_{1}=1\) and \(\displaystyle 1\le j\le k\)

The homogeneous solution is:

\(\displaystyle h_j=c_1\)

and we should then expect to find a particular solution of the form:

\(\displaystyle p_j=Aj^2+Bj\)

Substituting this particular solution into the difference equation, we find:

\(\displaystyle \left(Aj^2+Bj \right)-\left(A(j-1)^2+B(j-1) \right)=j\)

\(\displaystyle 2Aj-(A-B)=j-0\)

Equating coefficients, we find:

\(\displaystyle 2A=1\,\therefore\,A=\frac{1}{2}\)

\(\displaystyle A-B=0\,\therefore\,B=A=\frac{1}{2}\)

Hence:

\(\displaystyle p_j=\frac{1}{2}j^2+\frac{1}{2}j=\frac{j(j+1)}{2}\)

Thus, the general solution is:

\(\displaystyle S_{j}=h_j+p_j=c_1+\frac{j(j+1)}{2}\)

Now we may determine the parameter:

\(\displaystyle S_{1}=c_1+\frac{1(1+1)}{2}=c_1+1=1\,\therefore\,c_1=0\)

Thus, the solution satisfying the given conditions is:

\(\displaystyle S_{j}=\frac{j(j+1)}{2}\)

This number is referred to as the $n$th triangular number, and may be written as a binomial coefficient:

\(\displaystyle t_n=\frac{n(n+1)}{2}=\frac{(n+1)!}{2!((n+1)-2)!}={n+1 \choose 2}\)

Now, we want to find the sum of the spheres in all the layers, thus we may use the difference equation:

\(\displaystyle T_{k}-T_{k-1}=\frac{k(k+1)}{2}\) where \(\displaystyle T_1=1\) and \(\displaystyle 1\le k\le n\)

The homogeneous solution is:

\(\displaystyle h_k=c_1\)

and we should expect a particular solution of the form:

\(\displaystyle p_k=Ak^3+Bk^2+Ck\)

Substituting this particular solution into the difference equation, we find:

\(\displaystyle \left(Ak^3+Bk^2+Ck \right)-\left(A(k-1)^3+B(k-1)^2+C(k-1) \right)=\frac{1}{2}k^2+\frac{1}{2}k\)

\(\displaystyle 3Ak^2+(-3A+2B)k+(A-B+C)=\frac{1}{2}k^2+\frac{1}{2}k+0\)

Equating coefficients, we find:

\(\displaystyle 3A=\frac{1}{2}\,\therefore\,A=\frac{1}{6}\)

\(\displaystyle -3A+2B=\frac{1}{2}\,\therefore\,B=\frac{1}{2}\)

\(\displaystyle A-B+C=0\,\therefore\,C=\frac{1}{3}\)

And so we have:

\(\displaystyle p_k=\frac{1}{6}k^3+\frac{1}{2}k^2+\frac{1}{3}k= \frac{k^3+3k^2+2k}{6}= \frac{k(k+1)(k+2)}{6}\)

Thus, the general solution is:

\(\displaystyle T_k=h_k+p_k=c_1+\frac{k(k+1)(k+2)}{6}\)

Now we may determine the parameter:

\(\displaystyle T_1=c_1+\frac{1(1+1)(1+2)}{6}=c_1+1=1\,\therefore\,c_1=0\)

Thus, the solution satisfying the given conditions is:

\(\displaystyle T_k=\frac{k(k+1)(k+2)}{6}\)

And so, we may state:

\(\displaystyle N=T_n=\frac{n(n+1)(n+2)}{6}=\frac{(n+2)!}{3!((n+2)-3)!}={n+2 \choose 3}\)

This number is referred to as the $n$th tetrahedral number, or triangular pyramidal number.

b) To find the height of the stack, consider connecting the centers of the 4 spheres at each vertex with a line segment, each of length $(n-1)d$.

Now, to find the height $h_T$ of the resulting tetrahedron, we may drop a vertical line segment from the apex to the base and label its length $h_T$. Next draw a line segment from the apex down one of its faces bisecting the face, and we find its length $\ell$ from \(\displaystyle \sin\left(60^{\circ} \right)=\frac{\ell}{(n-1)d}\,\therefore\,\ell=\frac{\sqrt{3}}{2}(n-1)d\).

Now, observing that the vertical line $h_T$ intersects the base in the middle (the point equidistant from the 3 vertices), we can find the length of the line segment $b$ joining $h_T$ and $\ell$ by dividing the equilateral base into 3 congruent isosceles triangles each of which has an area \(\displaystyle \frac{1}{3}\) that of the equilateral base:

\(\displaystyle \frac{1}{2}b(n-1)d=\frac{1}{3}\cdot\frac{1}{2}\sin\left(60^{\circ} \right)((n-1)d)^2\)

\(\displaystyle b=\frac{1}{3}\cdot\frac{\sqrt{3}}{2}(n-1)d=\frac{1}{2\sqrt{3}}(n-1)d\)

Now we have a right triangle whose legs are $h_T$ and $b$ and whose hypotenuse is $\ell$, thus from the Pythagorean theorem, we obtain:

\(\displaystyle h_T=\sqrt{\ell^2-b^2}=\sqrt{\left(\frac{\sqrt{3}}{2}(n-1)d \right)^2-\left(\frac{1}{2\sqrt{3}}(n-1)d \right)^2}\)

\(\displaystyle h_T=(n-1)d\sqrt{\frac{3}{4}-\frac{1}{12}}=(n-1)d\sqrt{\frac{2}{3}}\)

Now, observing that we need to add 1 more spherical diameter to account for the radii below and above the tetrahedron, we then may state:

\(\displaystyle h=h_T+d=(n-1)d\sqrt{\frac{2}{3}}+d=d\left((n-1)\sqrt{\frac{2}{3}}+1 \right)\)