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

(→‎Solution 2)
(→‎Solution 2)
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Rewriting <math>abc+bcd+cda+dab = 14</math>, we get <math>ab(c+d) + cd(a+b) = 14</math>.  
 
Rewriting <math>abc+bcd+cda+dab = 14</math>, we get <math>ab(c+d) + cd(a+b) = 14</math>.  
 
Substitute <math>ab = 3c - 4</math> and solving, we get,
 
Substitute <math>ab = 3c - 4</math> and solving, we get,
−
<math>3c^{2} - 4c - 4d - 14 = 0</math>
+
<math>3c^{2} - 4c - 4d - 14 = 0</math> call this Equation 1
  
 
<math>abcd = 30</math> gives <math>(3c-4)cd = 30</math>.
 
<math>abcd = 30</math> gives <math>(3c-4)cd = 30</math>.
−
So, <math>3c^{2}d - 4cd = 30</math>, which implies <math>d(3c^{2} - 4c) = 30</math> or <math>3c^{2} - 4c = \frac{30}{d}</math>
+
So, <math>3c^{2}d - 4cd = 30</math>, which implies <math>d(3c^{2} - 4c) = 30</math> or <math>3c^{2} - 4c = \frac{30}{d}</math> call this equation 2.
 +
 
 +
Substituting Eq 2 in Eq 1 gives, <math>\frac{30}{d} - 4d - 14 = 0</math>
 +
 
 +
Solving this quadratic yields that <math>d \in {-5, \frac{3}{2}}</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 00:59, 23 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, $3c^{2} - 4c - 4d - 14 = 0$ 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 $d \in {-5, \frac{3}{2}}$

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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