Difference between revisions of "2024 AIME II Problems/Problem 11"

Revision as of 20:16, 18 August 2024

Problem

Find the number of triples of nonnegative integers $(a, b, c)$ satisfying $a + b + c = 300$ and $a^{2} b + a^{2} c + b^{2} a + b^{2} c + c^{2} a + c^{2} b = 6, 000, 000.$

Solution 1

$ab(a+b)+bc(b+c)+ac(a+c)=300(ab+bc+ac)-3abc=6000000, 100(ab+bc+ac)-abc=2000000$

Note $(100-a)(100-b)(100-c)=1000000-10000(a+b+c)+100(ab+bc+ac)-abc=0$ . Thus, $a/b/c=100$ . There are $201$ cases for each but we need to subtract $2$ for $(100,100,100)$ . The answer is $\boxed{601}$

~Bluesoul

Solution 2

$a^2(b+c)+b^2(a+c)+c^2(a+b) = 6000000$ , thus $a^2(300-a)+b^2(300-b)+c^2(300-c) = 6000000$ . Complete the cube to get $-(a-100)^3-(b-100)^3+(c-100)^3 = 9000000-30000(a+b+c)$ , which so happens to be 0. Then we have $(a-100)^3+(b-100)^3+(c-100)^3 = 0$ . We can use Fermat's last theorem here to note that one of $a, b, c$ has to be 100. We have $200+200+200+1 = 601.$

Solution 3

We have $\begin{aligned} a^{2} b + a^{2} c + b^{2} a + b^{2} c + c^{2} a + c^{2} b \\ = a b (a + b) + b c (b + c) + c a (c + a) \\ = a b (300 - c) + b c (300 - a) + c a (300 - b) \\ = 300 (a b + b c + c a) - 3 a b c \\ = - 3 ((a - 100) (b - 100) (c - 100) - 10^{4} (a + b + c) + 10^{6}) \\ = - 3 ((a - 100) (b - 100) (c - 100) - 2 \cdot 10^{6}) \\ = 6 \cdot 10^{6} . \end{aligned}$ The first and the fifth equalities follow from the condition that $a+b+c = 300$ .

Therefore, $\left( a - 100 \right) \left( b - 100 \right) \left( c - 100 \right) = 0 .$

Case 1: Exactly one out of $a - 100$ , $b - 100$ , $c - 100$ is equal to 0.

Step 1: We choose which term is equal to 0. The number ways is 3.

Step 2: For the other two terms that are not 0, we count the number of feasible solutions.

W.L.O.G, we assume we choose $a - 100 = 0$ in Step 1. In this step, we determine $b$ and $c$ .

Recall $a + b + c = 300$ . Thus, $b + c = 200$ . Because $b$ and $c$ are nonnegative integers and $b - 100 \neq 0$ and $c - 100 \neq 0$ , the number of solutions is 200.

Following from the rule of product, the number of solutions in this case is $3 \cdot 200 = 600$ .

Case 2: At least two out of $a - 100$ , $b - 100$ , $c - 100$ are equal to 0.

Because $a + b + c = 300$ , we must have $a = b = c = 100$ .

Therefore, the number of solutions in this case is 1.

Putting all cases together, the total number of solutions is $600 + 1 = \boxed{\textbf{(601) }}$ .

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

Solution 4

We will use Vieta's formulas to solve this problem. We assume $a + b + c = 300$ , $ab + bc + ca = m$ , and $abc = n$ . Thus $a$ , $b$ , $c$ are the three roots of a cubic polynomial $f(x)$ .

We note that $300m = (a + b + c)(ab + bc + ca)=\sum_{cyc} a^2b + 3abc = 6000000 + 3n$ , which simplifies to $100m - 2000000 = n$ .

Our polynomial $f(x)$ is therefore equal to $x^3 - 300x^2 + mx - (100m - 2000000)$ . Note that $f(100) = 0$ , and by polynomial division we obtain $f(x) = (x - 100)(x^2 - 200x - (m-20000))$ .

We now notice that the solutions to the quadratic equation above are $x = 100 \pm \frac{\sqrt{200^2 - 4(m - 20000)}}{2} = 100 \pm \sqrt{90000 - 4m}$ , and that by changing the value of $m$ we can let the roots of the equation be any pair of two integers which sum to $200$ . Thus any triple in the form $(100, 100 - x, 100 + x)$ where $x$ is an integer between $0$ and $100$ satisfies the conditions.

Now to count the possible solutions, we note that when $x \ne 0$ , the three roots are distinct; thus there are $3! = 6$ ways to order the three roots. As we can choose $x$ from $1$ to $100$ , there are $100 \cdot 3! = 600$ triples in this case. When $x = 0$ , all three roots are equal to $100$ , and there is only one triple in this case.

In total, there are thus $\boxed{601}$ distinct triples.

~GaloisTorrent <3

- minor edit made by MEPSPSPSOEODODODO

Solution 5

Let's define $a=100+x$ , $b=100+y$ , $c=100+z$ . Then we have $x+y+z=0$ and $6000000 = \sum a^2(b+c)$

$= \sum (100+x)^2(200-x) = \sum (10000+200x+x^2)(200-x) = \sum (20000 - 10000 x + x(40000-x^2))$

$= \sum (20000 + 30000 x -x^3) = 6000000 - \sum x^3$ , so we get $x^3 + y^3 + z^3 = 0$ . Then from $x+y+z = 0$ , we can find $0 = x^3+y^3+z^3 = x^3+y^3-(x+y)^3 = 3xyz$ , which means that one of $a$ , $b$ , $c$ must be 0. There are 201 solutions for each of $a=0$ , $b=0$ and $c=0$ , and subtract the overcounting of 2 for solution $(200, 200, 200)$ , the final result is $201 \times 3 - 2 = \boxed{601}$ .

Dan Li

dan

Video Solution

https://youtu.be/YMYe9chPLdY

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

@@ Line 14: / Line 14: @@
 ==Solution 2==
-<math>a^2(b+c)+b^2(a+c)+c^2(a+b) = 6000000</math>, thus <math>a^2(300-a)+b^2(300-b)+c^2(300-c) = 6000000</math>. Complete the cube to get <math>-(a-100)^3-(b-100)^3+(c-100)^3 = 9000000-30000(a+b+c)</math>, which so happens to be 0. Then we have <math>(a-100)^3+(b-100)^3+(c-100)^3 = 0</math>. We can use Fermat's last theorem here to note that one of a, b, c has to be 100. We have 200+200+200+1 = 601.
+<math>a^2(b+c)+b^2(a+c)+c^2(a+b) = 6000000</math>, thus <math>a^2(300-a)+b^2(300-b)+c^2(300-c) = 6000000</math>. Complete the cube to get <math>-(a-100)^3-(b-100)^3+(c-100)^3 = 9000000-30000(a+b+c)</math>, which so happens to be 0. Then we have <math>(a-100)^3+(b-100)^3+(c-100)^3 = 0</math>. We can use Fermat's last theorem here to note that one of <math>a, b, c</math> has to be 100. We have <math>200+200+200+1 = 601.</math>
 ==Solution 3==

2024 AIME II (Problems • Answer Key • Resources)
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