# How high does the ladder reach up the wall?

A ladder of length $$l$$ rests against a vertical wall. Suppose that there is a rung on the ladder which has the same distance $$d$$ from both the wall and the (horizontal) ground. Find explicitly, in terms of $$l$$ and $$d$$, the height $$h$$ from the ground that the ladder reaches up the wall.

• This is a classic maths puzzle, the ladder and box problem. It is so classic that I thought it must have been posted before, but I haven't found it. May 20, 2022 at 12:24
• @JaapScherphuis You are right. As I couldn't find it myself I thought it would be ok to post.
– Simd
May 20, 2022 at 12:26
• I found the same question here. I can't see why it is interesting...
– xd y
May 20, 2022 at 13:18
• @xdy Maybe more interesting if you like ladders. It's just a simply stated puzzle that is non trivial to solve.
– Simd
May 20, 2022 at 15:05

Let h,w be the height and width of the triangle formed by the ladder with the wall and the ground.

First let us divide through with d: $$H:=h/d, W:=w/d, L:=l/d, D:=d/d=1$$ Then

$$L^2=H^2+W^2$$ (1).

Because the inscribed unit square cuts the triangle into two similar triangles using for example the top one of the two we have

$$H/W = H-1$$ or $$H+W = HW$$ (2).

Taken together these yield a quadratic

$$L^2=(HW)^2-2HW$$ (3).

with positive solution in terms of $$P:=HW=H+W$$

$$P = 1 + \sqrt{1+L^2} (4).$$

Solving for H:

$$H=P-P/H$$ or $$H^2-HP=-P$$ or $$H=\frac{P \pm \sqrt{P^2-4P}}2$$.

Resubstituting we finally get

$$h = \frac{d + \sqrt{d^2+l^2} \pm \sqrt{l^2-2d^2-2\sqrt{d^2+l^2}}}2$$

• I really like how you have made the problem look almost easy to solve.
– Simd
May 21, 2022 at 6:31
• @graffe Thanks! Unlike Jaap I didn't know this problem. It's kind of neat and the little trick (2/3) satisfying to find, so thanks for posting it. May 21, 2022 at 8:35

To rephrase the problem, we wish to find the legs of a right triangle whose hypotenuse is $$l$$ and which contains an inscribed square of side $$d$$ which contains its right angle. One of these legs will be $$h$$ and the other will be something else - $$w$$ for width, say - and we can't actually tell which is which from the given information. To begin,

use the square to divide the triangle into two similar triangles - the legs of these are $$h-d$$ and $$d$$ and $$d$$ and $$w-d$$, respectively.
Scaling the former triangle by \frac{w}{d} and comparing to the original, we get that $$h = \frac{hw}{d}-w$$, or $$w = \frac{hd}{h-d}$$. Substituting this into $$h^2+w^2=l^2$$ and rearranging gives us a quartic in $$h$$ which can be solved by the usual methods.

• There are clever ways to solve it that avoid the quartic, instead solving two successive quadratic equations. May 20, 2022 at 12:28