2021 AIME II Problems/Problem 7

Problem

Let $a, b, c,$ and $d$ be real numbers that satisfy the system of equations $a + b = -3$ $ab + bc + ca = -4$ $abc + bcd + cda + dab = 14$ $abcd = 30.$ There exist relatively prime positive integers $m$ and $n$ such that $a^2 + b^2 + c^2 + d^2 = \frac{m}{n}.$ Find $m + n$ .

Solution

From the fourth equation we get $d=\frac{30}{abc}.$ substitute this into the third equation and you get $abc + \frac{30(ab + bc + ca)}{abc} = abc - \frac{120}{abc} = 14$ . Hence $(abc)^2 - 14(abc)-120 = 0$ . Solving we get $abc = -6$ or $abc = 20$ . From the first and second equation we get $ab + bc + ca = ab-3c = -4 \Longrightarrow ab = 3c-4$ , if $abc=-6$ , substituting we get $c(3c-4)=-6$ . If you try solving this you see that this does not have real solutions in $c$ , so $abc$ must be $20$ . So $d=\frac{3}{2}$ . Since $c(3c-4)=20$ , $c=-2$ or $c=\frac{10}{3}$ . If $c=\frac{10}{3}$ , then the system $a+b=-3$ and $ab = 6$ does not give you real solutions. So $c=-2$ . From here you already know $d=\frac{3}{2}$ and $c=-2$ , so you can solve for $a$ and $b$ pretty easily and see that $a^{2}+b^{2}+c^{2}+d^{2}=\frac{141}{4}$ . So the answer is $\boxed{145}$ .

~ math31415926535

2021 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 6	Followed by Problem 8
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2021 AIME II Problems/Problem 7

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