1996 AIME Problems/Problem 5
Revision as of 19:31, 4 July 2013 by Nathan wailes (talk | contribs)
Problem
Suppose that the roots of are
,
, and
, and that the roots of
are
,
, and
. Find
.
Solution
By Vieta's formulas on the polynomial , we have
,
, and
. Then
![$t = -(a+b)(b+c)(c+a) = -(s-a)(s-b)(s-c) = -(-3-a)(-3-b)(-3-c)$](http://latex.artofproblemsolving.com/c/1/9/c19770805040b313fafffdedbfb2ee4726cf0414.png)
This is just the definition for .
Alternatively, we can expand the expression to get
t &= -(-3-a)(-3-b)(-3-c)\\
&= (a+3)(b+3)(c+3)\\ &= abc + 3[ab + bc + ca] + 9[a + b + c] + 27\\t &= 11 + 3(4) + 9(-3) + 27 = 23\end{align*}$ (Error compiling LaTeX. Unknown error_msg)
A third solution arises if it is seen that each term in the expansion of has a total degree of 3. Another way to get terms with degree 3 is to multiply out
. Expanding both of these expressions and comparing them shows that:
See also
1996 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.