1995 USAMO Problems/Problem 2


A calculator is broken so that the only keys that still work are the $\, \sin, \; \cos,$ $\tan, \; \sin^{-1}, \; \cos^{-1}, \,$ and $\, \tan^{-1} \,$ buttons. The display initially shows 0. Given any positive rational number $\, q, \,$ show that pressing some finite sequence of buttons will yield $\, q$. Assume that the calculator does real number calculations with infinite precision. All functions are in terms of radians.


We will prove the following, stronger statement : If $m$ and $n$ are relatively prime nonnegative integers such that $n>0$, then the some finite sequence of buttons will yield $\sqrt{m/n}$.

To prove this statement, we induct strongly on $m+n$. For our base case, $m+n=1$, we have $n=1$ and $m=0$, and $\sqrt{m/n} = 0$, which is initially shown on the screen. For the inductive step, we consider separately the cases $m=0$, $0<m\le n$, and $n<m$.

If $m=0$, then $n=1$, and we have the base case.

If $0< m \le n$, then by inductive hypothesis, $\sqrt{(n-m)/m}$ can be obtained in finitely many steps; then so can \[\cos \tan^{-1} \sqrt{(n-m)/m} = \sqrt{m/n} .\]

If $n<m$, then by the previous case, $\sqrt{n/m}$ can be obtained in finitely many steps. Since $\cos \tan^{-1} \sqrt{n/m} = \sin \tan^{-1} \sqrt{m/n}$, it follows that \[\tan \sin^{-1} \cos \tan^{-1} \sqrt{n/m} = \sqrt{m/n}\] can be obtained in finitely many steps. Thus the induction is complete. $\blacksquare$

Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

See Also

1995 USAMO (ProblemsResources)
Preceded by
Problem 1
Followed by
Problem 3
1 2 3 4 5
All USAMO Problems and Solutions

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