Crossed Ladders Problem
Consider the following situation:
One could think of the two diagonal line segments as ladders leaning against two opposite walls of an alley. Two interesting questions may be asked:
Given the heights \(h_1, h_2\) of the two ladders above the ground, how high above the ground \((h)\) is the point where they meet?
If the lengths of the ladders \(\ell_1, \ell_2\) are given, as well as the height \(h\) of their intersection point, how wide is the alley?
Let's explore some basic cases:
- If \(h_1 = h_2\), how much is \(h?\)
- If \(h_1 \ll h_2\) \((\)i.e. \(h_1\) is much smaller than \(h_2),\) how much is \(h?\)
If \(h_1 = h_2\), the situation is perfectly symmetric. The intersection point lies precisely halfway between the two walls. Because the intersection point lies halfway on each ladder, its height is also half of each ladder's height:
\[h = \frac{h_1}2 = \frac{h_2}2.\]
If \(h_1 \ll h_2\), the intersection point lies very close to the left wall. Its height is only slightly less than the height of the first ladder:
\[h \approx h_1.\ _\square\]
If the ladders have the same length \(\ell_1 = \ell_2 = \ell\) and cross at height \(h\), what is the width \(w\) of the alley?
In this case, we are again dealing with the symmetric situation. From the discussion above we know that \(h_1 = h_2 = 2h\). Applying the Pythagorean theorem, we find
\[\ell^2 = (2h)^2 + w^2 \implies w = \sqrt{\ell^2 - 4h^2}.\ _\square \]
Height of the Intersection Point
The answer to question 1 is
\[\frac 1 h = \frac 1 {h_1} + \frac 1 {h_2}.\]
Define a coordinate system with the \(x\)-axis along the ground. Without loss of generality, the vertical walls have equations \(x = 0\) and \(x = 1\). The two ladders can be described by the following linear equations:
\[\begin{align} \text{ladder 1: }\ y &= h_1(1- x) \\ \text{ladder 2: }\ y &= h_2x. \end{align}\]
Equating the two, we obtain
\[h_1(1-x) = h_2 x \implies x = \frac{h_1}{h_1+h_2},\]
and for the vertical coordinate we find
\[h = y = h_2x = \frac{h_1h_2}{h_1+h_2}.\]
Taking reciprocals,
\[\frac 1 h = \frac{h_1+h_2}{h_1h_2} = \frac 1{h_1} + \frac 1{h_2}.\ _\square\]
