AIME 2015 II · 第 6 题

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:

$$\begin{aligned} a^2-2 \cdot \frac{a^2-81}{2} &= a^2-a^2+81\\ &= 81.\\ \end{aligned}$$ 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:

$$\begin{aligned} x_1+x_2+x_3 = a\\ x_1 \cdot x_2+x_1 \cdot x_3+x_2 \cdot x_3 = \frac{a^2-81}{2}\\ x_1 \cdot x_2 \cdot x_3 = \frac{c}{2} \end{aligned}$$ 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 two even numbers.

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

$$\frac{a^2}{3} > \frac{a^2-81}{2},$$ which simplifies to $\sqrt{243} > a$, meaning

$$16 > a.$$ So now we have that $9 from$(2)$,$a<16$, and$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.

Solution 3 (Basic calculus)

Since each of the roots is positive, the local maximum of the function must occur at a positive value of $x$. Taking $\frac{d}{dx}$ of the polynomial yields $6x^2-4ax+a^2-81$, which is equal to $0$ at the local maximum. Since this is a quadratic in $a$, we can find an expression for $a$ in terms of $x$. The quadratic formula gives $a=\frac{4x\pm\sqrt{324-8x^2}}{2}$, which simplifies to $a=2x\pm\sqrt{81-2x^2}$. We know that $a$ is a positive integer, and testing small positive integer values of $x$ yields $a=15$ or $a=1$ when $x=4$, and $a=15$ or $a=9$ when $x=6$. Because the value of $a$ alone does not determine the polynomial, $a$, $a$ must equal $15$.

Now our polynomial equals $2x^3-30x^2+144x-c$. Because one root is less than (or equal to) the $x$ value at the local maximum (picture the graph of a cubic equation), it suffices to synthetically divide by small integer values of $x$ to see if the resulting quadratic also has positive integer roots. Dividing by $x=3$ leaves a quotient of $2x^2-24x+72=2(x-6)^2$, and dividing by $x=4$ leaves a quotient of $2x^2-22x+56=2(x-4)(x-7)$. Thus, $c=2\cdot 3\cdot 6\cdot 6=216$, or $c=2\cdot 4\cdot 4\cdot 7=224$. Our answer is $216+224=\boxed{440}$ ~bad_at_mathcounts

Video Solution

https://www.youtube.com/watch?v=9re2qLzOKWk&t=427s

~MathProblemSolvingSkills.com