3
$\begingroup$

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.

$\endgroup$
4
  • 1
    $\begingroup$ 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. $\endgroup$ May 20, 2022 at 12:24
  • 2
    $\begingroup$ @JaapScherphuis You are right. As I couldn't find it myself I thought it would be ok to post. $\endgroup$
    – Simd
    May 20, 2022 at 12:26
  • $\begingroup$ I found the same question here. I can't see why it is interesting... $\endgroup$
    – xd y
    May 20, 2022 at 13:18
  • $\begingroup$ @xdy Maybe more interesting if you like ladders. It's just a simply stated puzzle that is non trivial to solve. $\endgroup$
    – Simd
    May 20, 2022 at 15:05

2 Answers 2

3
$\begingroup$

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

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

$\endgroup$
2
  • $\begingroup$ I really like how you have made the problem look almost easy to solve. $\endgroup$
    – Simd
    May 21, 2022 at 6:31
  • 1
    $\begingroup$ @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. $\endgroup$
    – loopy walt
    May 21, 2022 at 8:35
0
$\begingroup$

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.

$\endgroup$
1
  • 2
    $\begingroup$ There are clever ways to solve it that avoid the quartic, instead solving two successive quadratic equations. $\endgroup$ May 20, 2022 at 12:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.