1995 USAMO Problems/Problem 2

Problem

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.

Solution

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.

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1995 USAMO (Problems • Resources)
Preceded by Problem 1	Followed by Problem 3
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1995 USAMO Problems/Problem 2

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