AIME 2011 I · 第 15 题

Solution 1

From Vieta's formulas, we know that $a+b+c = 0$, and $ab+bc+ac = -2011$. Thus $a = -(b+c)$. All three of $a$, $b$, and $c$ are non-zero: say, if $a=0$, then $b=-c=\pm\sqrt{2011}$ (which is not an integer). $\textsc{wlog}$, let $|a| \ge |b| \ge |c|$. If $a > 0$, then $b,c < 0$ and if $a < 0$, then $b,c > 0,$ from the fact that $a+b+c=0$. We have

$$-2011=ab+bc+ac = a(b+c)+bc = -a^2+bc$$ Thus $a^2 = 2011 + bc$. We know that $b$, $c$ have the same sign, so product $bc$ is always positive. So $|a| \ge 45 = \lceil \sqrt{2011} \rceil$.

Also, if we fix $a$, $b+c$ is fixed, so $bc$ is maximized when $b = c$ . Hence,

$$2011 = a^2 - bc > \tfrac{3}{4}a^2 \qquad \Longrightarrow \qquad a ^2 < \tfrac{4}{3}\cdot 2011 = 2681+\tfrac{1}{3}$$ So $|a| \le 51$. Thus we have bounded $a$ as $45\le |a| \le 51$, i.e. $45\le |b+c| \le 51$ since $a=-(b+c)$. Let's analyze $bc=(b+c)^2-2011$. Here is a table:

$|a|$ $bc=a^2-2011$ $45$ $14$ $46$ $105$ $47$ $198$ $48$ $293$ $49$ $390$

We can tell we don't need to bother with $45$,

$105 = (3)(5)(7)$, So $46$ won't work. $198/47 > 4$,

$198$ is not divisible by $5$, $198/6 = 33$, which is too small to get $47$.

$293/48 > 6$, $293$ is not divisible by $7$ or $8$ or $9$, we can clearly tell that $10$ is too much.

Hence, $|a| = 49$, $a^2 -2011 = 390$. $b = 39$, $c = 10$.

Answer: $\boxed{098}$

Solution 2

Starting off like the previous solution, we know that $a + b + c = 0$, and $ab + bc + ac = -2011$.

Therefore, $c = -b-a$.

Substituting, $ab + b(-b-a) + a(-b-a) = ab-b^2-ab-ab-a^2 = -2011$.

Factoring the perfect square, we get: $ab-(b+a)^2=-2011$ or $(b+a)^2-ab=2011$.

Therefore, a sum ($a+b$) squared minus a product ($ab$) gives $2011$..

We can guess and check different $a+b$’s starting with $45$ since $44^2 < 2011$.

$45^2 = 2025$ therefore $ab = 2025-2011 = 14$.

Since no factors of $14$ can sum to $45$ ($1+14$ being the largest sum), a + b cannot equal $45$.

$46^2 = 2116$ making $ab = 105 = 3 * 5 * 7$.

$5 * 7 + 3 < 46$ and $3 * 5 * 7 > 46$ so $46$ cannot work either.

We can continue to do this until we reach $49$.

$49^2 = 2401$ making $ab = 390 = 2 * 3 * 5* 13$.

$3 * 13 + 2* 5 = 49$, so one root is $10$ and another is $39$. The roots sum to zero, so the last root must be $-49$.

$|-49|+10+39 = \boxed{098}$.

Solution 3

Let us first note the obvious that is derived from Vieta's formulas: $a+b+c=0, ab+bc+ac=-2011$. Now, due to the first equation, let us say that $a+b=-c$, meaning that $a,b>0$ and $c<0$. Now, since both $a$ and $b$ are greater than 0, their absolute values are both equal to $a$ and $b$, respectively. Since $c$ is less than 0, it equals $-a-b$. Therefore, $|c|=|-a-b|=a+b$, meaning $|a|+|b|+|c|=2(a+b)$. We now apply Newton's sums to get that $a^2+b^2+ab=2011$,or $(a+b)^2-ab=2011$. Solving, we find that $49^2-390$ satisfies this, meaning $a+b=49$, so $2(a+b)=\boxed{098}$.

Solution 4 (Quadratic)

From Vieta's Formulas we have $a + b + c = 0$ and $a(b + c) + bc = -2011.$ $b + c = -a \implies -a^2 + bc = -2011.$ $a^2 = (b + c)^2 \implies bc + 2011 = (b + c)^2.$ So now we have a simple looking two-variable quadratic equation. From here, we can solve for $c$ in terms of $b$ using the quadratic formula and see if we can do something with the discriminant. $bc + 2011 = b^2 + 2bc + c^2.$ $c^2 + (b)c + (b^2 - 2011) = 0.$ So $c = \frac{-b \pm \sqrt{8044 - 3b^2}}{2}.$ So $8044 - 3b^2$ must be a perfect square. $8044$ is around $89^2$, so we can start from here and work downwards. Immediately, we see that if $8044 - 3b^2 = 88^2, b = 10.$ Will this work? If $b = 10,$ then $c$ can be $-49$ or $39$. So we have two cases:

Case 1: $c = -49$

$b + c = -39$ and $bc = -490$

$a(-39) - 490 = -2011 \implies a = -39.$ But this doesn't satisfy the equation $a + b + c = 0,$ so this case won't work.

Case 2: $c = 39$

$b + c = 49$ and $bc = 390$

$a(49) + 390 = -2011 \implies a = -49.$ Since $a + b + c = 10 + 39 - 49 = 0$ this works.

So our answer is $10 + 39 + 49 = \boxed{98}.$

~grogg007

Solution 5 (mod to help bash)

First, derive the equations $a=-b-c$ and $ab+bc+ca=-2011\implies b^2+bc+c^2=2011$. Since the product is negative, $a$ is negative, and $b$ and $c$ positive. Now, a simple mod 3 testing of all cases shows that $b\equiv \{1,2\} \pmod{3}$, and $c$ has the repective value. We can choose $b$ not congruent to 0, make sure you see why. Now, we bash on values of $b$, testing the quadratic function to see if $c$ is positive. You can also use a delta argument like solution 4, but this is simpler. We get that for $b=10$, $c=39, -49$. Choosing $c$ positive we get $a=-49$, so $|a|+|b|+|c|=10+49+39=\boxed{098}$ ~firebolt360

Solution 6

Note that $-c=b+a$, so $c^2=a^2+2ab+b^2$, or $-c^2+ab=-a^2-ab-b^2$. Also, $ab+bc+ca=-2011$, so $(a+b)c+ab=-c^2+ab=-2011$. Substituting $-c^2+ab=-a^2-ab-b^2$, we can obtain $a^2+ab+b^2=2011$, or $\frac{a^3-b^3}{a-b}=2011$. If it is not known that $2011$ is prime, it may be proved in $5$ minutes or so by checking all primes up to $43$. If $2011$ divided either of $a, b$, then in order for $a^3-b^3$ to contain an extra copy of $2011$, both $a, b$ would need to be divisible by $2021$. But then $c$ would also be divisible by $2011$, and the sum $ab+bc+ca$ would clearly be divisible by $2011^2$.

By LTE, $v_{2011}(a^3-b^3)=v_{2011}(a-b)$ if $a-b$ is divisible by $2011$ and neither $a,b$ are divisible by $2011$. Thus, the only possibility remaining is if $a-b$ did not divide $2011$. Let $a=k+b$. Then, we have $(b+k)^3-b^3=2011k$. Rearranging gives $3b(b+k)=2011-k^2$. As in the above solutions, we may eliminate certain values of $k$ by using mods. Then, we may test values until we obtain $k=29$, and $a=10$. Thus, $b=39$, $c=-49$, and our answer is $49+39+10=098$.

Solution 7

For the monic cubic $x^3 - 2011x + m$ with integer roots $a,b,c$, we have

$a+b+c=0,\quad ab+ac+bc=-2011,\quad abc=-m.$

Set $c=-(a+b).$ Then

$$ab + a(-a-b) + b(-a-b) = -2011,$$ which simplifies to

$$-a^2 - ab - b^2 = -2011 \;\;\Longrightarrow\;\; a^2+ab+b^2=2011.$$ Treat this as a quadratic in $a$. Its discriminant must be a perfect square:

$$\Delta = b^2 - 4(b^2 - 2011) = 8044 - 3b^2 = k^2 \quad \text{for some integer }k.$$ Thus $3b^2 + k^2 = 8044.$ Checking integer $b$ with $|b|\leq 51$ (since $3b^2 \leq 8044$) yields solutions only when $b=10,\;39,\;49 \quad (\text{up to sign}),$ which produce the root triples (up to permutation)

$$(39,\,10,\,-49)$$ and the sign-permuted equivalent

$(-39,\,-10,\,49).$

These satisfy $a+b+c=0 \quad \text{and} \quad ab+ac+bc=-2011.$ Therefore $|a|+|b|+|c| = 39+10+49 =\boxed{98}$.

Video Solution

https://www.youtube.com/watch?v=QNbfAu5rdJI&t=26s ~ MathEx