Leaning tower

Relevant wiki: Torque - Equilibrium

A single brick placed on the edge is not tipping over, as long its center of mass (indicated by the dot) is supported by the wall underneath. Therefore, the maximal overhang $d_1 = \frac{1}{2} l$ is equal to the center position for one brick.

Assume you have already build a stable tower with $n$ bricks (at least for $n = 1$ ), so that the overhang is equal to its center of gravity: $d_n = \frac{1}{n \cdot m} \sum_{k = 1}^n m x_k = \frac{1}{n} \sum_{k = 1}^n x_k$ Now you place an additional brick underneath. The old tower is supported by the brick below, because the old center of mass aligns with the edge of the new brick. The position $x_{n+1}$ of new brick must be chosen, so that the combined center of mass lies on the edge of the wall.

Therefore $\begin{aligned} d_{n +1} &= \frac{1}{n+1} \sum_{k = 1}^{n+1} x_k = \frac{1}{n+1} \left(\sum_{k = 1}^{n} x_k + x_{n+1} \right) \\ &= \frac{1}{n+1} \left(n d_n + d_n + \frac{l}{2} \right) = d_n + \frac{l}{2(n+1)} \end{aligned}$ Induction results $\boxed{d_n = \frac{l}{2} \sum_{k=1}^n \frac{1}{k} }$ so that the overhang is given by a harmonic sum. The first values ​​of the sequence result to $d_1= \frac{1}{2} l, \quad d_2 = \frac{3}{4} l, \quad d_3 = \frac{11}{12} l, \quad d_4 = \frac{25}{24} l$ Therefore, four bricks are just enough to achieve an overhang $d > l$

For large values of $n$ , the harmonic sum can be approximated by an integral $\sum_{k = 1}^n \frac{1}{k} \gtrsim \int_1^{n+1} \frac{dx}{x} = \ln (n+1) \quad \Rightarrow \quad n \lesssim \exp\left( \frac{2 d}{l} \right) - 1$ In theory, an abitrary large overhang $d$ can be achieved, but the number $n$ of required bricks growns exponentially.

Edit: As some of you already noticed, there is a note in the problem, which states that there is only one brick per layer allowed. This note was added after I posted my solution.

I also got 4, but I placed the first brick half over the edge and then put the 2nd and 3rd next to each other with a slide gap between 2 and 3, so that 2 reaches over more than one brick length. The 4th Brick was put on top of 2 and 3 to make the construction stable. The 4th one is needed because of the gap between 2 and 3.

Relevant wiki: Torque - Equilibrium

Number the bricks $i = 0$ at the top, etc. Let $x_i$ be the position of the left side of a brick; set $x_0 = 0$ .

Let $u_n$ be the position of the center of mass of the first $n$ bricks. By definition, $u_n = \frac\ell 2 + \frac 1 n \sum_{i=0}^{n-1} x_i.$ Stability requires that the next brick (numbered $n$ ) should be place somewhere below this center of mass. In the extreme case, $x_n = u_n = \frac\ell 2 + \frac 1 n \sum_{i=0}^{n-1} x_i.$ We have $nx_n - (n-1)x_{n-1} = (n-(n-1))\frac \ell 2 + \left(\sum_{i=0}^{n-1} x_i - \sum_{i=0}^{n-2} x_i\right) = \frac\ell 2 + x_{n-1},$ $nx_n = \frac\ell 2 + nx_{n-1},$ $x_n = x_{n-1} + \frac 1 n \frac{\ell} 2 = \frac{\ell}2 \sum_{i = 1}^n \frac 1 i.$ Thus the first few terms are $x_0 = 0, \\ x_1 = 1 \cdot \frac \ell 2 = \tfrac12\ell, \\ x_2 = (1 + \tfrac12)\cdot \frac \ell 2 = \tfrac34\ell, \\ x_3 = (1 + \tfrac12 + \tfrac13)\cdot \frac \ell 2 = \tfrac{11}{12}\ell, \\ x_4 = (1 + \tfrac12 + \tfrac13 + \tfrac14)\cdot \frac \ell 2 = \tfrac{25}{24}\ell.$ After $\boxed{\text{four}}$ bricks the total overhang would therefore be greater than $\ell$ .

Consider a reference point A at a distance of l/2 from the edge of the table. Placing the first brick at an offset x off the table moves the center of mass of the structure to be offset by x from our reference point. Note for example that if x> l/2 the brick would fall off the table, If we add a second brick to the structure, the center of mass will move another x/2 to a total of x + x/2 form our reference point. It is easy to see that after n bricks are added the center of mass would have moved by:

x + x/2 + x/3 + ... + x/n

from A.

To solve our problem we need to find n such that:

nx = l

and

x + x/2 + x/3 + ... + x/n < l/2

which leads to

1 + 1/2 + ... + 1/n < n/2

Leading to the solution n = 4.