2021 AIME II Problems/Problem 7

Problem

Let $a, b, c,$ and $d$ be real numbers that satisfy the system of equations \begin{align*} a + b &= -3, \\ ab + bc + ca &= -4, \\ abc + bcd + cda + dab &= 14, \\ abcd &= 30. \end{align*} 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 1

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 $c=-2$. Since 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

Solution 2

Note that $ab + bc + ca = -4$ can be rewritten as $ab + c(a+b) = -4$. Hence, $ab = 3c - 4$.

Rewriting $abc+bcd+cda+dab = 14$, we get $ab(c+d) + cd(a+b) = 14$. Substitute $ab = 3c - 4$ and solving, we get \[3c^{2} - 4c - 4d - 14 = 0.\] We refer to this as Equation 1.

Note that $abcd = 30$ gives $(3c-4)cd = 30$. So, $3c^{2}d - 4cd = 30$, which implies $d(3c^{2} - 4c) = 30$ or \[3c^{2} - 4c = \frac{30}{d}.\] We refer to this as Equation 2.

Substituting Equation 2 into Equation 1 gives, $\frac{30}{d} - 4d - 14 = 0$.

Solving this quadratic yields that $d \in \left\{-5, \frac{3}{2}\right\}$.

Now we just try these two cases:

For $d = \frac{3}{2}$ substituting in Equation 1 gives a quadratic in $c$ which has roots $c \in \left\{\frac{10}{3}, -2\right\}$.

Again trying cases, by letting $c = -2$, we get $ab = 3c-4$, Hence $ab = -10$. We know that $a + b = -3$, Solving these we get $a = -5, b = 2$ or $a= 2, b = -5$ (doesn't matter due to symmetry in $a,b$).

So, this case yields solutions $(a,b,c,d) = \left(-5, 2 , -2, \frac{3}{2}\right)$.

Similarly trying other three cases, we get no more solutions, Hence this is the solution for $(a,b,c,d)$.

Finally, $a^{2} + b^{2} + c^{2} + d^{2} = 25 + 4 + 4 + \frac{9}{4} = \frac{141}{4} = \frac{m}{n}$.

Therefore, $m + n = 141 + 4 = \boxed{145}$.

~Arnav Nigam

Solution 3

For simplicity purposes, we number the given equations $(1),(2),(3),$ and $(4),$ in that order.

Rearranging $(2)$ and solving for $c,$ we have \begin{align*} ab+(a+b)c&=-4 \\ ab-3c&=-4 \\ c&=\frac{ab+4}{3}. \hspace{14mm} (5) \end{align*} Substituting $(5)$ into $(4)$ and solving for $d,$ we get \begin{align*} ab\left(\frac{ab+4}{3}\right)d&=30 \\ d&=\frac{90}{ab(ab+4)}. \hspace{5mm} (6)  \end{align*} Substituting $(5)$ and $(6)$ into $(3)$ and simplifying, we rewrite the left side of $(3)$ in terms of $a$ and $b$ only: \begin{align*} ab\left[\frac{ab+4}{3}\right] + b\left[\frac{ab+4}{3}\right]\left[\frac{90}{ab(ab+4)}\right] + \left[\frac{ab+4}{3}\right]\left[\frac{90}{ab(ab+4)}\right]a + \left[\frac{90}{ab(ab+4)}\right]ab &= 14 \\ ab\left[\frac{ab+4}{3}\right] + \underbrace{\frac{30}{a} + \frac{30}{b}}_{\text{Group them.}} + \frac{90}{ab+4} &= 14 \\ ab\left[\frac{ab+4}{3}\right] + \frac{30(a+b)}{ab} + \frac{90}{ab+4} &= 14 \\ ab\left[\frac{ab+4}{3}\right] + \underbrace{\frac{-90}{ab} + \frac{90}{ab+4}}_{\text{Group them.}} &= 14 \\ ab\left[\frac{ab+4}{3}\right] - \frac{360}{ab(ab+4)}&=14. \end{align*} Let $t=ab(ab+4),$ from which \[\frac{t}{3}-\frac{360}{t}=14.\] Multiplying both sides by $3t,$ rearranging, and factoring give $(t+18)(t-60)=0.$ Substituting back and completing the squares produce \begin{align*} \left[ab(ab+4)+18\right]\left[ab(ab+4)-60\right]&=0 \\ \left[(ab)^2+4ab+18\right]\left[(ab)^2+4ab-60\right]&=0 \\ \underbrace{\left[(ab+2)^2+14\right]}_{ab+2=\pm\sqrt{14}i\implies ab\not\in\mathbb R}\underbrace{\left[(ab+2)^2-64\right]}_{ab+2=\pm8}&=0 \\ ab&=6,-10. \end{align*} If $ab=6,$ then combining this with $(1),$ we know that $a$ and $b$ are the solutions of the quadratic $x^2+3x+6=0.$ Since the discriminant is negative, neither $a$ nor $b$ is a real number.

If $ab=-10,$ then combining this with $(1),$ we know that $a$ and $b$ are the solutions of the quadratic $x^2+3x-10=0,$ or $(x+5)(x-2)=0,$ from which $\{a,b\}=\{-5,2\}.$ Substituting $ab=-10$ into $(5)$ and $(6),$ we obtain $c=-2$ and $d=\frac32,$ respectively. Together, we have \[a^2+b^2+c^2+d^2=\frac{141}{4},\] so the answer is $141+4=\boxed{145}.$

~MRENTHUSIASM

Video Solution

https://www.youtube.com/watch?v=2rrX1G7iZqg

See Also

2021 AIME II (ProblemsAnswer KeyResources)
Preceded by
Problem 6
Followed by
Problem 8
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

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