2021 AIME II Problem 7

Question

Let

a,

b,

c,

and

d

be real numbers that satisfy the system of equations

\begin{aligned} a + b &= -3, \\ ab + bc + ca &= -4, \end{aligned}

\begin{aligned} abc + bcd + cda + dab &= 14, \\ abcd &= 30. \end{aligned}

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.

Answer

Answer: 145

Concepts:system of equations symmetry (algebra)factoring

Difficulty rating: 2650

Small hint:

The second equation is

ab + c(a + b) = -4,

giving

ab = 3c - 4;

group the third as

ab(c + d) + cd(a + b) = 14

Big hint:

Eliminating

d

and

cd

leaves a quartic in

c

whose real roots are

-2

and

\frac{10}{3};

one of them makes

a

and

b

non-real

Solution:

Since

a + b = -3,

the second equation reads

ab + c(a + b) = ab - 3c = -4,

so

ab = 3c - 4.

Grouping the third equation as

ab(c + d) + cd(a + b) = 14

gives

(3c - 4)(c + d) - 3cd = 14,

which simplifies to

3c^2 - 4c - 4d = 14,

so

d = \frac{3c^2 - 4c - 14}{4}.

The fourth equation becomes

(3c - 4)\,cd = 30.

Substituting for

d

yields

c(3c - 4)(3c^2 - 4c - 14) = 120,

i.e.

\begin{aligned} &9c^4 - 24c^3 - 26c^2 \\ &\quad {}+ 56c - 120 \\ &= (c + 2)(3c - 10) \\ &\quad {}\cdot (3c^2 - 4c + 6) \\ &= 0. \end{aligned}

The quadratic factor has negative discriminant, so

c = -2

or

c = \frac{10}{3}.

If

c = \frac{10}{3},

then

ab = 6

with

a + b = -3,

impossible for real

a, b

since

9 - 24 \lt 0.

So

c = -2,