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

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==Solution==
 
==Solution==
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From the fourth equation we get <cmath> d=\frac{30}{abc}. </cmath>, substitute this into the third equation and you get <cmath>abc + \frac{30(ab + bc + ca)}{abc} = abc - \frac{120}{abc} = 14</cmath>. Hence <math>(abc)^2 - 14(abc)-120 = 0</math>. Solving we get <math>abc = -6</math> or <math>abc = 20</math>. From the first and second equation we get <math>ab + bc + ca = ab-3c = -4 \Longrightarrow ab = 3c-4</math>, if <math>abc=-6</math>, substituting we get <math>c(3c-4)=-6</math>. If you try solving this you see that this does not have real solutions in <math>c</math>, so <math>abc</math> must be <math>20</math>. Since <math>abc=20</math>, <math>c=-2 or c=\frac{10}{3}</math>. If <math>c=\frac{10}{3}</math>, then the system <math>a+b=-3</math> and <math>ab = 6</math> does not give you real solutions. So <math>c=-2</math>. From here you already know <math>d=\frac{3}{2}</math> and <math>c=-2</math>, so you can solve for <math>a</math> and <math>b</math> pretty easily and see that <math>a^{2}+b^{2}+c^{2}+d^{2}=\frac{141}{4}</math>. So the answer is <math>\boxed{145}</math>.
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From the fourth equation we get <math> d=\frac{30}{abc}. </math> substitute this into the third equation and you get <math>abc + \frac{30(ab + bc + ca)}{abc} = abc - \frac{120}{abc} = 14</math>. Hence <math>(abc)^2 - 14(abc)-120 = 0</math>. Solving we get <math>abc = -6</math> or <math>abc = 20</math>. From the first and second equation we get <math>ab + bc + ca = ab-3c = -4 \Longrightarrow ab = 3c-4</math>, if <math>abc=-6</math>, substituting we get <math>c(3c-4)=-6</math>. If you try solving this you see that this does not have real solutions in <math>c</math>, so <math>abc</math> must be <math>20</math>. So <math>d=\frac{3}{2}</math>. Since <math>c(3c-4)=20</math>, <math>c=-2</math> or <math>c=\frac{10}{3}</math>. If <math>c=\frac{10}{3}</math>, then the system <math>a+b=-3</math> and <math>ab = 6</math> does not give you real solutions. So <math>c=-2</math>. From here you already know <math>d=\frac{3}{2}</math> and <math>c=-2</math>, so you can solve for <math>a</math> and <math>b</math> pretty easily and see that <math>a^{2}+b^{2}+c^{2}+d^{2}=\frac{141}{4}</math>. So the answer is <math>\boxed{145}</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 14:49, 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

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 $\boxed{145}$.

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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