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

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==Solution 2==
 
==Solution 2==
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<math>ab + bc + ca = -4</math> can be rewritten as <math>ab + c(a+b) = -4</math>
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Hence, <math>ab = 3c - 4</math>
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Rewriting <math>abc+bcd+cda+dab = 14</math>, we get <math>ab(c+d) + cd(a+b) = 14</math>
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Substitute <math>ab = 3c - 4</math> and solving, we get
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<math>3c^{2} - 4c - 4d - 14 = 0</math>
  
 
==See also==
 
==See also==
 
{{AIME box|year=2021|n=II|num-b=6|num-a=8}}
 
{{AIME box|year=2021|n=II|num-b=6|num-a=8}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 23:54, 22 March 2021

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 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 $c=-2$. 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

Solution 2

$ab + bc + ca = -4$ 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 $3c^{2} - 4c - 4d - 14 = 0$

See also

2021 AIME II (Problems • Answer Key • Resources)
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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