1996 AIME Problems/Problem 5
Problem
Suppose that the roots of 
are 
, 
, and 
, and that the roots of 
are 
, 
, and 
. Find 
.
Video Solution
https://youtu.be/3dfbWzOfJAI?t=2785
~ pi_is_3.14
Solution 1
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
![\begin{align*} 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*}](http://latex.artofproblemsolving.com/d/4/1/d41a23c58ab7b5d1fe76c063b04b35cfc89d9fc2.png)
Solution 2
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:
Solution 3
We have that for roots 
In the second cubic function 
the roots are 
By Vieta's formulae, we see that 
As we know that the sum of the roots of the first polynomial, 
is 
by applying Vieta's again.
Using this fact, we can rewrite as 
Seeing this, we can find the value of the product of the roots by applying this to the first equation.
This can be done by setting 
so, from this, we see that we should plug in 
for 
in .
After simplifying, we get that the polynomial is 
Given that the product of the roots of this equation is equivalent to our desired value for , we can apply Vieta's formulae for a third time to find that 
~MathWhiz35
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. 
