2015 AIME II Problems/Problem 6

Revision as of 15:51, 28 May 2020 by Dawangk (talk | contribs) (Solution 2 (Algebra+ Brute Force))


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:

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



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 $\boxed{\text{440}}$.


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



$16>a$ (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.

See also

2015 AIME II (ProblemsAnswer KeyResources)
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

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