This problem from the 2012 AMC 10a gave me son a lot of trouble yesterday:
Problem #16 from the 2012 AMC 10A
You can see from the beginning that he’s really confused about how to even approach it:
So, after the introduction to the problem, we move to the whiteboard to begin solving the problem. His first idea is to try to look at the least common multiple of the speeds – this is a tricky and difficult approach, unfortunately.
Next he tries to find the time it takes for each runner to run a lap. This, again, is a difficult approach, but at least this approach uses the idea of speed.
Finally I ask him to think about what it means for two of the runners to meet.
Next we looked at how long it would take for two of the runners to meet. We find that these two particular runners meet after 2,500 seconds.
He’s still a little confused about how to describe the situation in the problem. For example, he initially thinks that one of the runners will have completed 5 laps in 2,500 seconds. But, it does feel as though a few of the ideas are beginning to sink in.
Now we come to the final step in the problem. We know that two of the runners meet for the first time after 2,500 second. What about the 3rd runner – when does that runner meet the other two?
For the last part of the problem we returned to the original problem to see if we could have learned anything at all from the choices. We decided to check if the three runners were together after 1,000 seconds:
So, this is a really challenging problem. Even in the last video – which was after talking about the problem in the morning and then again for 20 minutes at night – he’s still a little confused.
It was interesting to me that his initial thought on how to approach the problem revolved around least common multiples. The change that running around a track – and thus having differences that were multiples of 500m – made a huge difference in terms of level of difficulty of the problem. Feels like we are going to have to review a few more problems similar to this one to help the ideas from this problem sink in a little more.