Berkeley Math Circle Monthly Contest 1

Solution:
There are three solutions:
$$1 = \frac{1}{3} + \frac{1}{3} + \frac{1}{3} = \frac{1}{2} + \frac{1}{4} + \frac{1}{4} = \frac{1}{2} + \frac{1}{3} + \frac{1}{6}$$
Now we must prove that these are the only solutions. If the fraction $1/2$ appears in the expression, the remaining fractions must add to $1/2$, so one of them is greater than or equal to $1/4$. If this fraction is $1/3$, we get the solution $1/2 + 1/3 + 1/6$, and if this fraction is $1/4$, we get the solution $1/2 + 1/4 + 1/4$. Thus, if the fraction $1/2$ is used, we cannot get any new solutions.
If the fraction $1/2$ does NOT appear in the expression, then all three fractions are at most $1/3$. Then their sum will certainly be less than $1$ unless they are all equal to $1/3$. Thus in this case, we only get the third solution, $1/3 + 1/3 + 1/3$.