Art of Problem Solving: 2013 AMC 12 A #24

Art of Problem Solving's Richard Rusczyk solves 2013 AMC 12 A #24.

[@ArtofProblemSolving]

If ever there was a problem to skip, this one was it. Even if you knew exactly what to do from the start, it would still take you a long time to do it, and you'd have a non-trivial chance of making an arithmetic error... The only way to get this one inside the time limit is to crush everything else so fast that you still have a lot of time for this one.

[@NintendoCollecting]

This question always stumped me. Thank you for showing your solution. You explained it quite well, but could have explained things a little further - especially the start and setting up the question with only half a 12-gon. I think there was no way I would have had enough time on the AMC even with your method in mind

[@planck8181]

Wow the last part is as good as the first part. It was clever to fully use the factoring in the arithmetic to preserve the factors of the fail#

[@Swiftclaw123]

On the test, how could we do this before the time limit!?

[@vishaalram7801]

You don't

[@aradhya9550]

I thought the question was saying that the 3 segments should form a triangle, not that they should be ABLE to form a triangle.....

[@vishaalram7801]

I thought the problem said that you have to choose 3 distinct segments so you cant have b,b,f

[@madhavgopakumar8597]

Vishaal Ram true

[@madhavgopakumar8597]

Vishaal Ram never mind, if you flip, richard's 12 gon, upside down, you will notice that all the segments, except f repeat

[@wontpower]

When he says b, he's referring to any segment that spans two tiny arc lengths. So there are actually 12 distinct bs in total.

[@M_Chen333]

Wouldn't a 12-gon be a dodecagon?

[@tychoelling5422]

I don't know...

[@anticorncob6]

Yes.

[@xiaoxia7874]

people might not know what a dodecagon is

[@penguinchess]

10000th view! Nice video :)