## USAMO 1972 #4 July 26, 2009

Posted by lumixedia in Problem-solving.
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USAMO 1972 #4. Let ${R}$ denote a non-negative rational number. Determine a fixed set of integers ${a}$, ${b}$, ${c}$, ${d}$, ${e}$, ${f}$ such that, for every choice of ${R}$, $\displaystyle |\frac{aR^2+bR+c}{dR^2+eR+f}-\sqrt{2}|<|R-\sqrt{2}|.$

Solution. From the desired inequality, we can conclude that $\displaystyle 0\le\lim_{R\rightarrow\sqrt{2}}|\frac{aR^2+bR+c}{dR^2+eR+f}-\sqrt{2}|$ $\displaystyle \le\lim_{R\rightarrow\sqrt{2}}||R-\sqrt{2}|=0$

So $\displaystyle \lim_{R\rightarrow\sqrt{2}}|\frac{aR^2+bR+c}{dR^2+eR+f}-\sqrt{2}|=0$

Since ${\sqrt{2}}$ is not the root of any quadratic polynomial with integer coefficients, ${\frac{aR^2+bR+c}{dR^2+eR+f}}$ is continuous at ${\sqrt{2}}$ and therefore $\displaystyle \frac{a(\sqrt{2})^2+b\sqrt{2}+c}{d(\sqrt{2})^2+e\sqrt{2}+f}-\sqrt{2}=0$ $\displaystyle a2^{2/3}+b2^{1/3}+c=e2^{2/3}+f2^{1/3}+2d$

For this equation to be satisfied, we must have ${a=e}$, ${b=f}$, ${c=2d}$. So the LHS of the inequality becomes the absolute value of $\displaystyle \frac{aR^2+bR+2d}{dR^2+aR+b}-\sqrt{2}$ $\displaystyle =\frac{aR^2+bR+2d-dR^2\sqrt{2}-aR\sqrt{2}-b\sqrt{2}}{dR^2+aR+b}$ $\displaystyle =\frac{aR(R-\sqrt{2})+b(R-\sqrt{2})-d\sqrt{2}(R+\sqrt{2})(R-\sqrt{2})}{dR^2+aR+b}$ $\displaystyle =(R-\sqrt{2})\frac{aR+b-d\sqrt{2}(R+\sqrt{2})}{dR^2+aR+b}$

Dividing both sides of the inequality by ${|R-\sqrt{2}|}$ and multiplying by ${|dR^2+aR+b|}$, we find that it is equivalent to $\displaystyle |aR+b-d\sqrt{2}(R+\sqrt{2})|<|dR^2+aR+b|$

This is certainly satisfied if we choose ${(a,b,d)}$ so both ${aR+b-d\sqrt{2}(R+\sqrt{2})}$ and ${dR^2+aR+b}$ are always positive for nonnegative ${R}$. Any ${(a,b,d)}$ with ${a}$, ${b}$, ${d>0}$, ${a>d\sqrt{2}}$, and ${b>d2^{2/3}}$ will work. One example is ${(a,b,c,d,e,f)=(2,2,2,1,2,2)}$.

It might possibly be interesting to find conditions on ${(a,b,d)}$ which are necessary and sufficient for the inequality to hold. It might also not be very interesting at all. 1. Qiaochu Yuan - July 26, 2009
I suggested in the AoPS thread on the problem here that an interesting follow-up would be to figure out which choices take convergents of the continued fraction of $\sqrt{2}$ to convergents. It’s probable no choice with this property exists since the continued fraction of non-square roots are aperiodic. 2. lumixedia - July 27, 2009 3. Qiaochu Yuan - July 28, 2009