Difference between revisions of "1996 AIME Problems/Problem 5"
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&= 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*}</math></center> | t &= 11 + 3(4) + 9(-3) + 27 = 23\end{align*}</math></center> | ||
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| + | A third solution arises if it is seen that each term in the expansion of <math>(a+b)(b+c)(c+a)</math> has a total degree of 3. Another way to get terms with degree 3 is to multiply out <math>(a+b+c)(ab+bc+ca)</math>. Expanding both of these expressions and comparing them shows that: | ||
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| + | <math>(a+b)(b+c)(c+a) = (ab+bc+ca)(a+b+c)-abc</math> | ||
| + | <math>t = -(a+b)(b+c)(c+a) = abc-(ab+bc+ca)(a+b+c) = 11-(4)(-3) = 23</math> | ||
== See also == | == See also == | ||
Revision as of 02:04, 17 December 2008
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
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 | ||
