Waste less time on Facebook — follow Brilliant.

Proofathon Contest 7: Combinatorics!

The Proofathon Spring Competition was a huge success thanks to you guys! This time we had two perfect scorers and the largest turn out we've ever had! The April Monthly: Combinatorics is already underway, so visit the Proofathon Contests Page for your chance to compete. If you are new to Proofathon, sign up at Proofathon.org and be sure to check out our Facebook for updates and news. Here's a problem from our last competition:

Find all polynomials \(p(x)\) such that for all real numbers \(a,b,c \neq 0\) satisfying \(\frac{1}{a}+\frac{1}{b} = \frac{1}{c}\), \[\frac{1}{p(a)}+\frac{1}{p(b)} = \frac{1}{p(c)}\]

Good luck!

Note by Logan Dymond
3 years, 4 months ago

No vote yet
1 vote


Sort by:

Top Newest

When i took the contest, my solution was plug in a=b, then rearrange to 2f(x)=f(2x), implying that f(x) is linear and so p = ax. Then plug in and show it works. Maybe my proof was too simple and had a flaw though.... Aayush Gupta · 3 years, 4 months ago

Log in to reply

Okay... I'm not sure whether this is super easy or really tough. I'm going to assume it's easy.

We can prove that for any class of polynomials \(p_n(x)=\frac{x}{n}\) that the conditions will be satisfied. Multiplying \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\) by \(n\), we achieve \(\frac{1}{\frac{1}{n}a}+\frac{1}{\frac{1}{n}b}+\frac{1}{\frac{1}{n}c}\), which is in the form given as long as \(p(x)=\frac{x}{n}\) which we've already stated. That's all I can think of. There's probably something I'm missing, though. @mention me if you have a cool proof! :D Finn Hulse · 3 years, 4 months ago

Log in to reply


Problem Loading...

Note Loading...

Set Loading...