USAMO 1973 #4 August 19, 2009Posted by lumixedia in algebra, Problem-solving.
Tags: algebra, contest math, olympiad math, USAMO, USAMO 1973
A fairly straightforward algebra problem. Could appear on a modern AMC-12, though the decoy answers would have to be carefully written.
USAMO 1973 #4. Determine all the roots, real or complex, of the system of simultaneous equations
Solution. First note that
Let , so that are the roots of the polynomial . Then we have
Summing these three equalities together gives
from which . So are the roots of the polynomial , and the only solution is .
An interesting note is that the scan of the original test shows the last inequality as . While the official solution makes clear that was intended, the problem is only slightly trickier as originally printed. Here’s how to do the typoed-up version. Let . The calculation of stands. We also have, just like before,
so we can write . By multiplying one of the equalities we used before by , we get
as well as the two analogous equations in and , and summing these up gives
Similarly, since we know
as well as the two analogous equations, we can sum them up to get
which gives as before, so we conclude that as before.
You can see from the intermediate step that giving for the last equation would also have worked and given the same answer. Using makes things a little more complicated. The analogous solution gives the quadratic , so in addition to as before, we can also take , making the polynomial or , so we have a second solution , , where is an imaginary cube root of unity. For still higher powers things probably continue to get worse.