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 Solution (using infinite descent):

This solution is similar to the above solution. We can invert a number with $\tan \sin^{-1} \cos \tan^{-1} \sqrt{n/m} = \sqrt{m/n}$ and we can get $\tan \sin^{-1} \cos \tan^{-1}\cos\tan^{-1}\sqrt{x} = \sqrt{x + 1}.$ Then we do $\sqrt{x}$ to $\sqrt{x + 1}$ in reverse, trying to get 1. Subtracting one inside the radical for a fraction $\sqrt{m/n}$ will eventually result in a fraction less than or equal to 1. Taking the reciprocal of this will result in a fraction greater than or equal to 1, and we can keep subtracting one inside the radical.

We do this repeatedly, and this should eventually result in 1 in a finite number of steps. This is because each time we do this, m+n is smaller. This is true even if $\sqrt{m/n}$ can be simplified, as simplifying will only result in a smaller value. The smallest possible value of the fraction is 1/1, so this will be achieved after finite steps.

Finally, we know that 1 is achievable as cos(0) = 1.

~Unum

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

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

Page

Toolbox

Search

1995 USAMO Problems/Problem 2

Contents

Problem

Solution

Alternate Solution (using infinite descent):

See Also