In this video of Veritasium, the following situation is sketched (rephrased in my own words):
You run two laps on a running track of length $x$. The first lap you run at an (average) speed $v_1$, and it can be as slowly as you want (as long as it's greater than zero). It takes you time $t_1$. The second lap you run at an (average) speed at such an (average) speed $v_2$ that your total average speed $v_\text{avg} = 2v_1$. This lap takes you time $t_2$. What's $v_2$ (expressed in $v_1$)?
My attempt to solve it was the following: $$v_\text{avg} = 2v_1$$ $$\frac{2x}{t_1+t_2} = 2\frac{x}{t_1}$$ $$t_1 + t_2 = t_1$$
Which would mean that $t_2 = 0$, so you would have to run $v_2 \to \infty$ to satisfy the condition, which of course is physically impossible.
Am I doing something wrong? Or would it be a trick question?
It's a trick question, or at least a subtle one.
You can verify your solution without any algebra. Suppose it took you a minute the first time around. To do two trips at double that rate (whatever rate it turns out to be) you would have to complete two rounds in a minute. But you've already used up a minute ...