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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[3]{2}|<|R-\sqrt[3]{2}|.

Solution. From the desired inequality, we can conclude that

\displaystyle 0\le\lim_{R\rightarrow\sqrt[3]{2}}|\frac{aR^2+bR+c}{dR^2+eR+f}-\sqrt[3]{2}|

\displaystyle \le\lim_{R\rightarrow\sqrt[3]{2}}||R-\sqrt[3]{2}|=0

So

\displaystyle \lim_{R\rightarrow\sqrt[3]{2}}|\frac{aR^2+bR+c}{dR^2+eR+f}-\sqrt[3]{2}|=0

Since {\sqrt[3]{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[3]{2}} and therefore

\displaystyle \frac{a(\sqrt[3]{2})^2+b\sqrt[3]{2}+c}{d(\sqrt[3]{2})^2+e\sqrt[3]{2}+f}-\sqrt[3]{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[3]{2}

\displaystyle =\frac{aR^2+bR+2d-dR^2\sqrt[3]{2}-aR\sqrt[3]{2}-b\sqrt[3]{2}}{dR^2+aR+b}

\displaystyle =\frac{aR(R-\sqrt[3]{2})+b(R-\sqrt[3]{2})-d\sqrt[3]{2}(R+\sqrt[3]{2})(R-\sqrt[3]{2})}{dR^2+aR+b}

\displaystyle =(R-\sqrt[3]{2})\frac{aR+b-d\sqrt[3]{2}(R+\sqrt[3]{2})}{dR^2+aR+b}

Dividing both sides of the inequality by {|R-\sqrt[3]{2}|} and multiplying by {|dR^2+aR+b|}, we find that it is equivalent to

\displaystyle |aR+b-d\sqrt[3]{2}(R+\sqrt[3]{2})|<|dR^2+aR+b|

This is certainly satisfied if we choose {(a,b,d)} so both {aR+b-d\sqrt[3]{2}(R+\sqrt[3]{2})} and {dR^2+aR+b} are always positive for nonnegative {R}. Any {(a,b,d)} with {a},{b},{d>0}, {a>d\sqrt[3]{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.

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Comments»

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[3]{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

Thanks for the link and the suggestion. Do you have any ideas as to how to tackle a problem like that? I know very little about continued fractions.

3. Qiaochu Yuan - July 28, 2009

No, not really; it’s not easy to identify the convergents of an aperiodic continued fraction. Maybe a more reasonable question would be to identify choices that take best rational approximations to best rational approximations.


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