USAMO 1972 #4 July 26, 2009
Posted by lumixedia in Problem-solving.
Tags: algebra, contest math, olympiad math, USAMO, USAMO 1972
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USAMO 1972 #4. Let 
denote a non-negative rational number. Determine a fixed set of integers 
, 
, 
, 
, 
, 
such that, for every choice of ,
![\displaystyle |\frac{aR^2+bR+c}{dR^2+eR+f}-\sqrt[3]{2}|<|R-\sqrt[3]{2}|.](http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%7C%5Cfrac%7BaR%5E2%2BbR%2Bc%7D%7BdR%5E2%2BeR%2Bf%7D-%5Csqrt%5B3%5D%7B2%7D%7C%3C%7CR-%5Csqrt%5B3%5D%7B2%7D%7C.&bg=ffffff&fg=000000&s=0 "\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}|](http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+0%5Cle%5Clim_%7BR%5Crightarrow%5Csqrt%5B3%5D%7B2%7D%7D%7C%5Cfrac%7BaR%5E2%2BbR%2Bc%7D%7BdR%5E2%2BeR%2Bf%7D-%5Csqrt%5B3%5D%7B2%7D%7C&bg=ffffff&fg=000000&s=0 "\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](http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cle%5Clim_%7BR%5Crightarrow%5Csqrt%5B3%5D%7B2%7D%7D%7C%7CR-%5Csqrt%5B3%5D%7B2%7D%7C%3D0&bg=ffffff&fg=000000&s=0 "\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](http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Clim_%7BR%5Crightarrow%5Csqrt%5B3%5D%7B2%7D%7D%7C%5Cfrac%7BaR%5E2%2BbR%2Bc%7D%7BdR%5E2%2BeR%2Bf%7D-%5Csqrt%5B3%5D%7B2%7D%7C%3D0&bg=ffffff&fg=000000&s=0 "\displaystyle \lim_{R\rightarrow\sqrt[3]{2}}|\frac{aR^2+bR+c}{dR^2+eR+f}-\sqrt[3]{2}|=0")
Since ![{\sqrt[3]{2}}](http://s0.wp.com/latex.php?latex=%7B%5Csqrt%5B3%5D%7B2%7D%7D&bg=ffffff&fg=000000&s=0 "{\sqrt[3]{2}}")
is not the root of any quadratic polynomial with integer coefficients, 
is continuous at 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](http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cfrac%7Ba%28%5Csqrt%5B3%5D%7B2%7D%29%5E2%2Bb%5Csqrt%5B3%5D%7B2%7D%2Bc%7D%7Bd%28%5Csqrt%5B3%5D%7B2%7D%29%5E2%2Be%5Csqrt%5B3%5D%7B2%7D%2Bf%7D-%5Csqrt%5B3%5D%7B2%7D%3D0&bg=ffffff&fg=000000&s=0 "\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")
For this equation to be satisfied, we must have 
, 
, 
. So the LHS of the inequality becomes the absolute value of
![\displaystyle \frac{aR^2+bR+2d}{dR^2+aR+b}-\sqrt[3]{2}](http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cfrac%7BaR%5E2%2BbR%2B2d%7D%7BdR%5E2%2BaR%2Bb%7D-%5Csqrt%5B3%5D%7B2%7D&bg=ffffff&fg=000000&s=0 "\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}](http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%3D%5Cfrac%7BaR%5E2%2BbR%2B2d-dR%5E2%5Csqrt%5B3%5D%7B2%7D-aR%5Csqrt%5B3%5D%7B2%7D-b%5Csqrt%5B3%5D%7B2%7D%7D%7BdR%5E2%2BaR%2Bb%7D&bg=ffffff&fg=000000&s=0 "\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}](http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%3D%5Cfrac%7BaR%28R-%5Csqrt%5B3%5D%7B2%7D%29%2Bb%28R-%5Csqrt%5B3%5D%7B2%7D%29-d%5Csqrt%5B3%5D%7B2%7D%28R%2B%5Csqrt%5B3%5D%7B2%7D%29%28R-%5Csqrt%5B3%5D%7B2%7D%29%7D%7BdR%5E2%2BaR%2Bb%7D&bg=ffffff&fg=000000&s=0 "\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}](http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%3D%28R-%5Csqrt%5B3%5D%7B2%7D%29%5Cfrac%7BaR%2Bb-d%5Csqrt%5B3%5D%7B2%7D%28R%2B%5Csqrt%5B3%5D%7B2%7D%29%7D%7BdR%5E2%2BaR%2Bb%7D&bg=ffffff&fg=000000&s=0 "\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}|}](http://s0.wp.com/latex.php?latex=%7B%7CR-%5Csqrt%5B3%5D%7B2%7D%7C%7D&bg=ffffff&fg=000000&s=0 "{|R-\sqrt[3]{2}|}")
and multiplying by 
, we find that it is equivalent to
![\displaystyle |aR+b-d\sqrt[3]{2}(R+\sqrt[3]{2})|<|dR^2+aR+b|](http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%7CaR%2Bb-d%5Csqrt%5B3%5D%7B2%7D%28R%2B%5Csqrt%5B3%5D%7B2%7D%29%7C%3C%7CdR%5E2%2BaR%2Bb%7C&bg=ffffff&fg=000000&s=0 "\displaystyle |aR+b-d\sqrt[3]{2}(R+\sqrt[3]{2})|<|dR^2+aR+b|")
This is certainly satisfied if we choose 
so both ![{aR+b-d\sqrt[3]{2}(R+\sqrt[3]{2})}](http://s0.wp.com/latex.php?latex=%7BaR%2Bb-d%5Csqrt%5B3%5D%7B2%7D%28R%2B%5Csqrt%5B3%5D%7B2%7D%29%7D&bg=ffffff&fg=000000&s=0 "{aR+b-d\sqrt[3]{2}(R+\sqrt[3]{2})}")
and 
are always positive for nonnegative . Any with ,,
, ![{a>d\sqrt[3]{2}}](http://s0.wp.com/latex.php?latex=%7Ba%3Ed%5Csqrt%5B3%5D%7B2%7D%7D&bg=ffffff&fg=000000&s=0 "{a>d\sqrt[3]{2}}")
, and 
will work. One example is 
.
It might possibly be interesting to find conditions on which are necessary and sufficient for the inequality to hold. It might also not be very interesting at all.
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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}](http://s0.wp.com/latex.php?latex=%5Csqrt%5B3%5D%7B2%7D&bg=ffffff&fg=666666&s=0 "\sqrt[3]{2}")
to convergents. It’s probable no choice with this property exists since the continued fraction of non-square roots are aperiodic.
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.
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.