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

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<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>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>a^2 + b^2 + c^2 + d^2 = \frac{m}{n}.</cmath>Find <math>m + n</math>.
 
<cmath>a^2 + b^2 + c^2 + d^2 = \frac{m}{n}.</cmath>Find <math>m + n</math>.
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==Solution (Easy Algebra)==
 
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We can factor out the d out of the last 2 equations. Therefore, it becomes abc + d(bc + ac + ab) = 14. Notice this is just abc -4b since bc + ac + ab = -4. We now have abc -4d = 14 and abcd = 30.
 
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We then put d in terms of abc, so abc = <math>\frac{30}{d}-4d=14</math>. We solve fore d and it is either <math>\dfrac32</math> or -5. We can now try for these two values, and plug the rest in to the equation. Thus, we have 33 + <math>\dfrac94</math> = <math>\dfrac{33 \cdot 4 + 9}{4}</math> = <math>\dfrac{132+9}{4}</math> = <math>\dfrac{141}{4}</math>. We have 141 + 4 = <math>\boxed{145}</math> and we're done.
 
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~Arcticturn
 
  
 
==Solution 1==
 
==Solution 1==

Revision as of 11:21, 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 $c=-2$. 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)

$ab + bc + ca = -4$ 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

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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