Puzzle o' the Day 404!

[captaino]

In a desperate attempt to bring life to a seemingly moribund social media platform (Should I get a Tumblr or something? Suggestions?), here's a classic puzzle

Comments

[copyright1983]

Let AB be the alley floor, AC and BD be the ladders, and E be the point where the ladders intersect. Let F be the point on AB directly below E. We're searching for the length of EF, given AD = 15 and BC = 10.

We know, from construction, that <|BFE ~ <|BAD and <|AFE ~ <|ABC. Therefore, EF/AD = BF/BA, and EF/BC = AF/AB. Adding these equations to each other and plugging in what we know, EF/10 + EF/15 = (BF + AF)/AB = 1. Thus EF = 6.

[captaino]

That's a lovely solution! I hadn't seen it done in precisely that way.

[wantonhalo]

Seconded! This makes it a little more clear to me why the width doesn't matter.

[wantonhalo]

Hm, this is interesting. Let h_1 and h_2 be the heights where the ladders reach the walls. Then using analytic geometry, I get h_1*h_2/(h_1+h_2) for the height where the ladders meet. In this case, they meet at height 6. I don't have a good intuition for the width not mattering, though the algebra plainly shows it doesn't.

[chaithedog]

Good to see you again, Nick! For Project Euler fans, there's this one as well, which requires you to program an algorithm:

https://projecteuler.net/problem=309

[captaino]

Hi Carl! Thanks for reminding me that the "Crossed Ladders Problem" has a harder variant; I'll definitely use it in an upcoming PotD.