2015 AIME II Problems/Problem 6

Problem

Steve says to Jon, "I am thinking of a polynomial whose roots are all positive integers. The polynomial has the form $P(x) = 2x^3-2ax^2+(a^2-81)x-c$ for some positive integers $a$ and $c$ . Can you tell me the values of $a$ and $c$ ?"

After some calculations, Jon says, "There is more than one such polynomial."

Steve says, "You're right. Here is the value of $a$ ." He writes down a positive integer and asks, "Can you tell me the value of $c$ ?"

Jon says, "There are still two possible values of $c$ ."

Find the sum of the two possible values of $c$ .

Solution 1 (Algebra)

We call the three roots (some may be equal to one another) $x_1$ , $x_2$ , and $x_3$ . Using Vieta's formulas, we get $x_1+x_2+x_3 = a$ , $x_1 \cdot x_2+x_1 \cdot x_3+x_2 \cdot x_3 = \frac{a^2-81}{2}$ , and $x_1 \cdot x_2 \cdot x_3 = \frac{c}{2}$ .

Squaring our first equation we get $x_1^2+x_2^2+x_3^2+2 \cdot x_1 \cdot x_2+2 \cdot x_1 \cdot x_3+2 \cdot x_2 \cdot x_3 = a^2$ .

We can then subtract twice our second equation to get $x_1^2+x_2^2+x_3^2 = a^2-2 \cdot \frac{a^2-81}{2}$ .

Simplifying the right side:

$a^2-2 \cdot \frac{a^2-81}{2}$

$a^2-a^2+81$

$81$

So, we know $x_1^2+x_2^2+x_3^2 = 81$ .

We can then list out all the triples of positive integers whose squares sum to $81$ :

We get $(1, 4, 8)$ , $(3, 6, 6)$ , and $(4, 4, 7)$ .

These triples give $a$ values of $13$ , $15$ , and $15$ , respectively, and $c$ values of $64$ , $216$ , and $224$ , respectively.

We know that Jon still found two possible values of $c$ when Steve told him the $a$ value, so the $a$ value must be $15$ . Thus, the two $c$ values are $216$ and $224$ , which sum to $\boxed{\text{440}}$ .

~BealsConjecture~

Solution 2 (Algebra+ Brute Force)

First things first. Vietas gives us the following:

$x_1+x_2+x_3 = a$ (1)

$x_1 \cdot x_2+x_1 \cdot x_3+x_2 \cdot x_3 = \frac{a^2-81}{2}$ (2)

$x_1 \cdot x_2 \cdot x_3 = \frac{c}{2}$ (3)

From (2), $a$ must have odd parity, meaning $a^2-81$ must be a multiple of 4, which implies that both sides of (2) are even. Then, from (1), we see that an odd number of $x_1$ , $x_2$ , and $x_3$ must be odd, because we have already deduced that $a$ is odd. In order for both sides of (2) to be even, there must only be one odd number and 2 even numbers.

Now, the theoretical maximum value of the left side of (2) is $3 \cdot \frac{a}{3}^2=\frac{a^2}{3}$ . That means that the maximum bound of $a$ is where

$\frac{a^2}{3}> \frac{a^2-81}{2}$ (4)

which simplifies to

$\sqrt{243}>a$

meaning

$16>a$ (5)

So now we have that $9<a$ from (2), $a<16$ from (5), and $a$ is odd from (2). This means that $a$ could equal $11$ , $13$ , or $15$ . At this point, we have simplified the problem to the point where we can casework+ brute force. As said above, we arrive at our solutions of $(1, 4, 8)$ , $(3, 6, 6)$ , and $(4, 4, 7)$ , of which the last two return equal $a$ values. Then, $2(3 \cdot 6 \cdot 6+4 \cdot 4 \cdot 7)=\boxed{440}$ AWD.

2015 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 5	Followed by Problem 7
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

Page

Toolbox

Search

2015 AIME II Problems/Problem 6

Contents

Problem

Solution 1 (Algebra)

Solution 2 (Algebra+ Brute Force)

See also