The ladder and box problem - a classic challenge!
Why Mathematics Stack Exchange recommends it
Mathematics Stack Exchange users link it in answers to the ladder-and-box question because the video itself is based on that question and its answers, and it shows the solution worked out visually.
Recommended by 1 people across 1 different months, 2015 to 2021.
What it is
A MindYourDecisions video by Presh Talwalkar that solves the classic ladder and box puzzle: a 4-unit ladder touches the corner of a 1-unit cube against a wall, and the question is the distance between the top of the box and the top of the ladder. It walks through the algebraic solution step by step.
What people said
Recommends Sean Codie Verne · 2020-06 · Mathematics Stack Exchange · on “The position of a ladder leaning against a wall and touching a box under it”
Here are two solutions, the first is purely algebraic, the second uses some basic trigonometry. For an overview of the history of this problem see The “Ladder and Box” Problem: From Curves to Calculators [LINK] by P. Baggett and A. Ehrenfeucht [LINK] . This reference was provided by Presh Talwalkar to his video Ladder And Box Puzzle [this video] , which goes through the solution given in another answer [LINK] to this question. First solution With $c = |AB|$ the height of the box, $x = |CE|$ , $y = |BF|$ the lengths of the legs of the right triangle $AEF$ minus the height of the box, and $l = |EF|$ the length of the ladder: $\left(x+c\right)^2 + \left(y+c\right)^2 = l^2$ \left(theorem of Pythagora
Cites Blue · 2015-06 · Mathematics Stack Exchange · on “The position of a ladder leaning against a wall and touching a box under it”
EDIT (5.5 years later!) Had I known my solution would be featured in the video "Ladder and Box Problem" [this video] on Presh Talwalkar's "Mind Your Decisions" YouTube channel (thanks, Presh!), I would've made another pass. Shunting the original solution to this answer's edit history [LINK] , here's something of a streamlined presentation. From a length- $b$ ladder resting against a side- $r$ box, we can make the following diagram: [LINK] Then we have: $$\left(\;\color{green}{p + q + r} \;\right)^2 = \color{blue}{b}^2 + \color{red}{r}^2 \qquad\to\qquad p + q + r = \sqrt{b^2 + r^2} \tag{1}$$ (discarding the negative root). The semicircle recalls the classical construction of the geometric mean [LI
Cites nick012000 · 2021-10 · Mathematics Stack Exchange · on “The position of a ladder leaning against a wall and touching a box under it”
I just saw Presh Talwalker's video [this video] linking here, and I think that both he and everyone here are overcomplicating things. As far as I can tell, this problem can be solved with just Pythagoras's Theorem, similar triangles, and simultaneous equations. We start off by labeling the triangles the way that Presh Talwalker does; the length along the ground between the box and the ladder is equal to x; the length along the wall between the box is equal to y. From Pythagoras's Theorem, we can see that $(x+1)^2+(y+1)^2 = 4^2$ . We can also see that we have a pair of similar triangles formed between the box, the ladder, and the floor, and the wall, the box, and the ladder. Because they're simila