When I was preparing earlier this year for math competitions, I pooled a bunch of problems from different resources to practice with. I didn't complete all of them, and I decided to go back to them during my spring break, which recently started. On a one very fun problem, I made some progress but then got stuck and was wondering if someone could help me figure it out.
Here it is:
"What are the real-number solutions to the following system of equations?
$$ \begin{cases} (2m - \frac{1}{m})np = n+p \\ (3n - \frac{1}{n})mp = p+m \\ (2020p - \frac{1}{p})mn = m+n \\ \end{cases} $$ My approach:
- I noticed some things (i.e. the symmetry).
- I added the equations and then multiplied both sides by mnp.
- I got $$2025(mnp)^2-(m^2n^2 + m^2p^2 + n^2p^2) = 2m^2np + 2mn^2p + 2mnp^2$$
- I thought of Vieta's formulas.
- I realized that I can rewrite this as $2025(mnp)^2=(mn+mp+np)^2$.
- I rewrote this as $45|mnp| = |mn+mp+np|$.
- I separated this into 2 cases: $45mnp = mn+mp+np$ and $-45mnp = mn+mp+np$.
- I thought of using Vieta's formulas by taking a polynomial $P(x) = x^3 + ax^2 + bx + c$. Using Vieta's, this would give $b = 45c$ or $b = -45c$.
Then I got stuck. I would really appreciate any help with this! I have spent a while trying to figure this problem out, and I can't get it off my mind.
I will only talk about the case when all three of them are nonzero. The remaining case (i.e. at least one of them is zero) is left to you.
Assume that $m, n, p$ are all nonzero. We rewrite the first equation as: $$2m - 1/m = (n + p)/np = 1/n + 1/p,$$ and therefore $$1/m + 1/n + 1/p = 2m.$$
Similarly, the other two equations give $$1/m + 1/n + 1/p = 3n,$$$$1/m + 1/n + 1/p = 2020p.$$
From here it should be clear how to proceed.