Difference between revisions of "2021 AIME II Problems/Problem 7"

Revision as of 00:56, 28 March 2021

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 (Easy Algebra)

We can factor $d$ out of the last two equations. Therefore, it becomes $abc + d(bc + ac + ab) = 14$ . Notice this is just $abc -4d$ , since $bc + ac + ab = -4$ . We now have $abc -4d = 14$ and $abcd = 30$ . We then find $d$ in terms of $abc$ , so $abc = \frac{30}{d}-4d=14$ . We solve for $d$ and find that it is either $\dfrac32$ or $-5$ . We can now try for these two values, and plug the rest into the equation. Thus, we have $33 + \dfrac94 = \dfrac{33 \cdot 4 + 9}{4} = \dfrac{132+9}{4} = \dfrac{141}{4}$ . We have $141 + 4 = \boxed{145}$ and we're done.

~Arcticturn

Solution 3 (Easy Algebra)

$ab + bc + ca = -4$ 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, $3c^{2} - 4c - 4d - 14 = 0$ call this Equation 1

$abcd = 30$ gives $(3c-4)cd = 30$ . So, $3c^{2}d - 4cd = 30$ , which implies $d(3c^{2} - 4c) = 30$ or $3c^{2} - 4c = \frac{30}{d}$ call this equation 2.

Substituting Eq 2 in Eq 1 gives, $\frac{30}{d} - 4d - 14 = 0$

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

Now we just try these 2 cases.

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

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 $(a,b,c,d) = (-5, 2 , -2, \frac{3}{2})$

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

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

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

- Arnav Nigam

@@ Line 1: / Line 1: @@
 ==Problem==
 Let <math>a, b, c,</math> and <math>d</math> be real numbers that satisfy the system of equations
-<cmath>a + b = -3</cmath><cmath>ab + bc + ca = -4</cmath><cmath>abc + bcd + cda + dab = 14</cmath><cmath>abcd = 30.</cmath>There exist relatively prime positive integers <math>m</math> and <math>n</math> such that
+<cmath>\begin{align*}
-<cmath>a^2 + b^2 + c^2 + d^2 = \frac{m}{n}.</cmath>Find <math>m + n</math>.
+a + b &= -3, \\
+ab + bc + ca &= -4, \\
+abc + bcd + cda + dab &= 14, \\
+abcd &= 30.
+\end{align*}</cmath>
+There exist relatively prime positive integers <math>m</math> and <math>n</math> such that
+<cmath>a^2 + b^2 + c^2 + d^2 = \frac{m}{n}. </cmath>Find <math>m + n</math>.
 ==Solution 1==

2021 AIME II (Problems • Answer Key • Resources)
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