Difference between revisions of "1996 AIME Problems/Problem 5"
Mathgeek2006 (talk | contribs) m (→Solution) |
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Alternatively, we can expand the expression to get | Alternatively, we can expand the expression to get | ||
− | < | + | <cmath>\begin{align*} |
t &= -(-3-a)(-3-b)(-3-c)\\ | t &= -(-3-a)(-3-b)(-3-c)\\ | ||
&= (a+3)(b+3)(c+3)\\ | &= (a+3)(b+3)(c+3)\\ | ||
&= abc + 3[ab + bc + ca] + 9[a + b + c] + 27\\ | &= abc + 3[ab + bc + ca] + 9[a + b + c] + 27\\ | ||
− | t &= 11 + 3(4) + 9(-3) + 27 = 23\end{align*}</ | + | t &= 11 + 3(4) + 9(-3) + 27 = 23\end{align*}</cmath> |
Revision as of 11:45, 13 March 2015
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
This is just the definition for .
Alternatively, we can expand the expression to get
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.