Maths Olympiad Prep

Library / /14 of 22

Algebra Difficulty 5.2 AIME, harder Prove it United States

Problem:
Find, with proof, all ways to write 11 as a sum of three fractions, each with numerator 11 and positive integer denominator. (The order of the fractions is irrelevant, so for instance 12+14+14\frac{1}{2}+\frac{1}{4}+\frac{1}{4} is the same as 14+14+12\frac{1}{4}+\frac{1}{4}+\frac{1}{2}.)

Solution

Solution:
There are three solutions:
1=13+13+13=12+14+14=12+13+16 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 1/21/2 appears in the expression, the remaining fractions must add to 1/21/2, so one of them is greater than or equal to 1/41/4. If this fraction is 1/31/3, we get the solution 1/2+1/3+1/61/2 + 1/3 + 1/6, and if this fraction is 1/41/4, we get the solution 1/2+1/4+1/41/2 + 1/4 + 1/4. Thus, if the fraction 1/21/2 is used, we cannot get any new solutions.
If the fraction 1/21/2 does NOT appear in the expression, then all three fractions are at most 1/31/3. Then their sum will certainly be less than 11 unless they are all equal to 1/31/3. Thus in this case, we only get the third solution, 1/3+1/3+1/31/3 + 1/3 + 1/3.

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.