Human Pyramid Are Heavy

Note that the person who supported the most weight is the person in the middle of the ninth row. Give him a name, $S$ , for Strongest or Superman, whatever you like.

In the eighth row, there are 2 people whom weight is supported by $S$ . They provide a weight of $64$ , and there are $2$ ways for the weight to reach $S$ . Total weight, $64\times2=128$ .

In the seventh row, there are 3 people whom weight is supported by $S$ . They provide a weight of $32$ , and there are $4$ ways for the weight to reach $S$ . Total weight, $32\times4=128$ .

In the sixth row, there are 4 people whom weight is supported by $S$ . They provide a weight of $16$ , and there are $8$ ways for the weight to reach $S$ . Total weight, $16\times8=128$ .

Using the same logic, you will get an equation like this:

$64\times2+21\times4+16\times8+8\times16+4\times30+2\times50+1\times70+\frac{1}{2}\times70$

The tricky part is counting the ways for the weight to reach $S$ . Actually it's pretty simple, just add up the weight from the two people below them.

I'll write my general solution first. Suppose there are as many rows as desired, and let $n,k$ be any nonnegative integers with $0 \le k \le n$ . We consider the person in row $n$ , position $k$ (start with 0, like in Pascal's Triangle). Let $w_{n,k}$ represent the weight of the person in that position. Let $f(n,k)$ represent the weight supported by the person in that position.

Consider any two people $P$ and $Q$ , say in positions $(a,b)$ and $(n,k)$ respectively, such that $Q$ is supporting (part of) the weight of $P$ in the pyramid. Let $g(a,b,n,k)$ represent the proportion of the weight of the person in position $(a,b)$ that is supported by the person in position $(n,k)$ . The weight of $P$ is evenly distributed among the two people in positions $(a+1,b)$ and $(a+1,b+1)$ . Therefore (for example) $g(a,b,a+1,b)=g(a,b,a+1,b+1)=\frac{1}{2} w_{a,b}$ . Similarly, it can be shown by induction that $g(a,b,n,k) = \frac{1}{2^{n-a}} \binom{n-a}{k-b} w_{a,b}$ (the only exception: if $a = n$ and $b = k$ then $g(n,k,n,k) = 0$ since a person is assumed to not be supporting their own weight).

Now, let's find which of those values are nonzero for fixed $(n,k)$ . We must have $a \le n$ and $0 \le k-b \le n-a$ . Hence $0 \le b \le k$ and $b \le a \le n-k+b$ . Therefore, the total weight supported by person $Q$ is: $f(n,k) = -w_{n,k} + \sum_{b = 0}^k \sum_{a = b}^{n-k+b} \frac{1}{2^{n-a}} \binom{n-a}{k-b} w_{a,b}$

Note that we subtract $w_{n,k}$ because a person does not support their own weight.

If we assume that every person's weight is the same (say $w$ ), then we can simplify this formula a little by letting $s = k - b$ and $t = (n - a) - (k - b)$ :

$f(n,k) = w \left(-1 + \sum_{s = 0}^k \sum_{t = 0}^{n-k} \frac{1}{2^{s+t}} \binom{s+t}{s} \right)$

For this particular problem, we plug in $n = 8, k = 4$ and $w_{a,b} = 128$ for $0 \le b \le a$ , and sure enough the answer is $\boxed{837}$ .

Unfortunately I don't have a general solution, but for a pyramid of height 9, here we are: Notice first that the entire weight of the 36 people on upper rows will be distributed among people of the bottom row, with the heaviest load on the person in the centre. We now focus on this person: For a given person in the pyramid, we label them to be on row $k$ , starting from the bottom row $k=0$ , and lateral position $m$ , starting the count at $m=0$ for the leftmost person from whom there is a diagonal line connecting them to the centre person of the bottom row (so for example, the two central people in row $k=1$ will have positions 0 and 1, and everyone else in row $k=1$ has position $m<0$ or $m>1$ ). Then the person at $k, m$ contributes a weight of $\binom{k}{m} \times 128 \times 2^{-k}$ (we can see this by counting the number of ways this person has some chain of supporters leading down to the central person). Here we exclude $k=m=0$ , who contributes no weight to themselves. Then the total weight on the central person is the sum of each contribution from positions $m=0$ to $m=k$ in rows $k=1$ to $k=4$ , plus the contribution from positions $m=k-4$ to $m=4$ in rows $k=5$ to $k=8$ . The former sum is just 128 times four rows; the latter sum can be done by hand, though there's probably a cleverer way of doing it. The total weight on the central person is then $128 \times 4+325=837$ . (At this point, you should call some medical help for them.)

Am I the only person to brute force this? I did omit the side people and worked my way down centre. However, it still took me 3 tries.

There are some pretty good solutions here but one more doesn't hurt, right? The weight supported by the arms and knees of each person is half of the weight on each person being supported by them plus its own weight, so the $n-th$ person at layer $m$ (from top to bottom) supports a weight of:

$w_{m, n} = 0.5 \cdot (w_{m-1, n-1} + w_{m-1, n}) + 128 \\ w_{0, 0} = 128$

from that, it's easy to iterate over this to get that the weights on the arms and knees of each person in the last row are:

$[255.5, 506.0, 734.0, 902.0, 965.0, 902.0, 734.0, 506.0, 255.5]$

But since we want to know the weight that the person is supporting, not counting themselves, we can subtract their own weight which gives us $965 - 128 = 837$ .

To iterate over this I got lazy and simply wrote this python program:

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def weight(m):
    """ 
    Weight on the layer m
    """
    memo_table = [[0. for _ in range(m)] for _ in range(m)]
    memo_table[0][0] = 128.

    for layer in range(m):
        for person in range(layer+1):
            memo_table[layer][person] = 0.5*(memo_table[layer-1][person-1] + memo_table[layer-1][person]) + 128

    return memo_table[m-1]

assert sum(weight(9)) == 128*45
print(weight(9))