Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1995 USAMO Problems/Problem 2"
m (→‎Resources)
Line 21: Line 21:
 
{{alternate solutions}}
 
{{alternate solutions}}
  
−
== Resources ==
+
== See Also ==
  
 
{{USAMO box|year=1995|num-b=1|num-a=3}}
 
{{USAMO box|year=1995|num-b=1|num-a=3}}

Revision as of 08:05, 19 July 2016

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.

See Also

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1995_USAMO_Problems/Problem_2&oldid=79541"