# How Do You Solve A Problem Like Bulgaria?

“Which is larger: $\left(\frac{6}{5}\right)^{\sqrt{3}}$ or $\left( \frac{5}{4}\right)^{\sqrt{2}}$?

Bulgarian Mathematical Olympiad 2018, question 3 ((I think. I’ve lost the precise reference.))

The immediate reaction of most mathematics seeing this question is revulsion. Irrational powers? what foolishness is this?

Digging deeper - or making judicious use of calculating tools - reveals that the two numbers are surprisingly close together, so making use of the traditional tools for such problems (approximate the roots as rationals and show that the lower bound for one exceeds the upper band for the other) only work if you’re prepared to do a lot of tedious arithmetic. There are ugly solutions involving binomial expansions. There are red herrings galore ($e^{\frac{6}{19}}$, for instance, lies between the solutions. This is of no use). The solution I came across, I came across by accident, fiddling with the powers and following my nose. First, let me present the solution, and then talk through some of the thought process.

Obviously, spoilers below the line.

**Lemma 1**: $\sqrt{\frac{3}{2}} > \frac{60}{49}$

**Proof**: $\left(\frac{60}{49}\right)^2 = \frac{3600}{2401} < \frac{3}{2}$

**Lemma 2** $\ln\left(1 + \frac{1}{x}\right) = 2\left(\frac{1}{2x+1} + \frac{1}{3}\cdot \frac{1}{(2x+1)^3} + \dots + \frac{1}{2n+1} \cdot \frac{1}{(2x+1)^{2n+1}} + \dots \right)$

**Proof**

- $\left(1 + \frac{1}{2x+1}\right) \div \left(1 - \frac{1}{2x+1}\right) = \frac{2x+2}{2x}$
- $\dots = 1 + \frac{1}{x}$
- So $\ln\left(1 + \frac{1}{x}\right) = \ln\left(1 + \frac{1}{2x+1}\right) - \ln\left(1 - \frac{1}{2x+1}\right)$ (*)
- Let $D = \frac{1}{2x+1}$ and expand (*):
- $\ln\left(1 + \frac{1}{x}\right) = \left(D - \frac{1}{2}D^2 + \frac{1}{3}D^3 \dots - \frac{1}{N}(-D)^N - \dots \right) - \left(-D - \frac{1}{2}D^2 - \frac{1}{3}D^3 -\dots - \frac{1}{N}D^N - \dots \right)$
- The even-powered terms vanish, leaving
- $\dots = 2D + \frac{2}{3}D^3 + \frac{2}{5}D^5 + \dots + \frac{2}{2n+1} D^{2n+1} + \dots$
- $\dots = 2\left(\frac{1}{2x+1} + \frac{1}{3}\cdot \frac{1}{(2x+1)^3} + \dots + \frac{1}{2n+1} \cdot \frac{1}{(2x+1)^{2n+1}} + \dots \right)$ as required.

**Proposition**: $\left(\frac{6}{5}\right)^{\sqrt{3}} > \left( \frac{5}{4}\right)^{\sqrt{2}}$

**Proof**:

- $11 \ln\left( \frac{6}{5} \right) = 2\left(1 + \frac{1}{3} \cdot \frac{1}{11^3} + \dots\right)$
- $49 \ln\left( \frac{25}{24}\right) = 2\left(1 + \frac{1}{3} \cdot \frac{1}{49^3} + \dots\right)$
- Comparing terms, $\left(\frac{6}{5}\right)^{11} > \left(\frac{25}{24}\right)^{49}$

Therefore:

- $\left(\frac{6}{5}\right)^{11} > \left(\frac{6}{5}\right)^{-49} \cdot \left(\frac{5}{4}\right)^{49}$
- $\left(\frac{6}{5}\right)^{60} > \left(\frac{5}{4}\right)^{49}$
- $\left(\frac{6}{5}\right)^{\frac{60}{49}} > \frac{5}{4}$
- By lemma 1, $\left(\frac{6}{5}\right)^{\sqrt{\frac{3}{2}}} >\left(\frac{6}{5}\right)^{\frac{60}{49}}$
- So $\left(\frac{6}{5}\right)^{\sqrt{\frac{3}{2}}} > \frac{5}{4}$
- And $\left(\frac{6}{5}\right)^{\sqrt{3}} > \left(\frac{5}{4}\right)^{\sqrt{2}}$ $\blacksquare$

Lemma 2, the log trick, is one of my favorites, and one that I’d hoped to use from the off. Trouble is, it gives bounds on the result that aren’t really tight enough to be useful if you do it naively.

However, it was *there and on my mind*. For all that Feynman was a… problematic role model, he did have some good techniques for problem-solving. One was to always have several interesting problems testing in your mind, and to apply any novel approaches you come across to them.

Bulgaria 3 certainly qualifies as an interesting problem.

The breakthrough came when I came up with $\frac{60}{49}$ as a reasonably tight lower bound on $\sqrt{\frac{3}{2}}$. (This solution was calculator-assisted, of which more later.) I had the expression $\left(\frac{6}{5}\right)^{60}$ vs $\left(\frac{5}{4}\right)^{49}$ on the board, and noticed that the difference in powers was 11, just like the sum of 6 and 5. I wondered if I could show that $\left(\frac{6}{5}\right)^{11}$ was greater than $e$, and whatever was left over was smaller. That isn’t the case - but it was a step in the right sort of direction.

Dividing across by $\left(\frac{6}{5}\right)^{49}$ was where I started to get excited: the second number became $\left( \frac{25}{24} \right)^{49}$ and I now had a similar form to both numbers. And from there, applying the log trick makes the whole thing fairly obvious - it’s clear from comparing coefficients that $11 \ln \left(\frac{6}{5}\right) > 49 \ln \left( \frac{25}{24}\right)$, and then it’s just a case of working backwards.

This is a solution that, in principle, could be done without a calculator in exam conditions in half an hour, In practice, it relies on finding precisely the correct approximation for the square root, and precisely the correct way of splitting the number, and precisely the correct expansion for the logarithm. That’s a lot to spot.

Of course, it is the Olympiad and it’s supposed to be difficult. I still feel it’s a trick question and my initial feelings of revulsion were perfectly justified.

Do you have a neater solution? I’d love to see it if so!