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

Latest revision as of 23:20, 27 April 2024

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 ~minor edit by Mathkiddie

Solution 2

Note that $ab + bc + ca = -4$ 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 $3c^{2} - 4c - 4d - 14 = 0.$ We refer to this as Equation 1.

Note that $abcd = 30$ gives $(3c-4)cd = 30$ . So, $3c^{2}d - 4cd = 30$ , which implies $d(3c^{2} - 4c) = 30$ or $3c^{2} - 4c = \frac{30}{d}.$ We refer to this as Equation 2.

Substituting Equation 2 into Equation 1 gives, $\frac{30}{d} - 4d - 14 = 0$ .

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

Now we just try these two cases:

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

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

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

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

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

~Arnav Nigam

Solution 3

For simplicity purposes, we number the given equations $(1),(2),(3),$ and $(4),$ in that order.

Rearranging $(2)$ and solving for $c,$ we have $\begin{align*} ab+(a+b)c&=-4 \\ ab-3c&=-4 \\ c&=\frac{ab+4}{3}. \hspace{14mm} (5) \end{align*}$ Substituting $(5)$ into $(4)$ and solving for $d,$ we get $\begin{align*} ab\left(\frac{ab+4}{3}\right)d&=30 \\ d&=\frac{90}{ab(ab+4)}. \hspace{5mm} (6) \end{align*}$ Substituting $(5)$ and $(6)$ into $(3)$ and simplifying, we rewrite the left side of $(3)$ in terms of $a$ and $b$ only: $\begin{align*} ab\left[\frac{ab+4}{3}\right] + b\left[\frac{ab+4}{3}\right]\left[\frac{90}{ab(ab+4)}\right] + \left[\frac{ab+4}{3}\right]\left[\frac{90}{ab(ab+4)}\right]a + \left[\frac{90}{ab(ab+4)}\right]ab &= 14 \\ ab\left[\frac{ab+4}{3}\right] + \underbrace{\frac{30}{a} + \frac{30}{b}}_{\text{Group them.}} + \frac{90}{ab+4} &= 14 \\ ab\left[\frac{ab+4}{3}\right] + \frac{30(a+b)}{ab} + \frac{90}{ab+4} &= 14 \\ ab\left[\frac{ab+4}{3}\right] + \underbrace{\frac{-90}{ab} + \frac{90}{ab+4}}_{\text{Group them.}} &= 14 \\ ab\left[\frac{ab+4}{3}\right] - \frac{360}{ab(ab+4)}&=14. \end{align*}$ Let $t=ab(ab+4),$ from which $\frac{t}{3}-\frac{360}{t}=14.$ Multiplying both sides by $3t,$ rearranging, and factoring give $(t+18)(t-60)=0.$ Substituting back and completing the squares produce $\begin{align*} \left[ab(ab+4)+18\right]\left[ab(ab+4)-60\right]&=0 \\ \left[(ab)^2+4ab+18\right]\left[(ab)^2+4ab-60\right]&=0 \\ \underbrace{\left[(ab+2)^2+14\right]}_{ab+2=\pm\sqrt{14}i\implies ab\not\in\mathbb R}\underbrace{\left[(ab+2)^2-64\right]}_{ab+2=\pm8}&=0 \\ ab&=6,-10. \end{align*}$ If $ab=6,$ then combining this with $(1),$ we know that $a$ and $b$ are the solutions of the quadratic $x^2+3x+6=0.$ Since the discriminant is negative, neither $a$ nor $b$ is a real number.

If $ab=-10,$ then combining this with $(1),$ we know that $a$ and $b$ are the solutions of the quadratic $x^2+3x-10=0,$ or $(x+5)(x-2)=0,$ from which $\{a,b\}=\{-5,2\}.$ Substituting $ab=-10$ into $(5)$ and $(6),$ we obtain $c=-2$ and $d=\frac32,$ respectively. Together, we have $a^2+b^2+c^2+d^2=\frac{141}{4},$ so the answer is $141+4=\boxed{145}.$

~MRENTHUSIASM

Solution 4 (way too long)

Let the four equations from top to bottom be listed 1 through 4 respectively. We factor equation 3 like so: $abc+d(ab+bc+ca)=14$ Then we plug in equation 2 to receive $abc-4d=14$ . By equation 4 we get $abc=\frac{30}{d}$ . Plugging in, we get $\frac{30}{d}-4d=14$ . Multiply by $d$ on both sides to get the quadratic equation $4d^2+14d-30=0$ . Solving using the quadratic equation, we receive $d=\frac{3}{2},d=-5$ . So, we have to test which one is correct. We repeat a similar process as we did above for equations 1 and 2. We factor equation 2 to get $ab+c(a+b)=-4$ After plugging in equation 1, we get $ab-3c=-4$ . Now we convert it into a quadratic to receive $3c^2-4c-abc=0$ . The value of $abc$ will depend on $d$ . So we obtain the discriminant $16+12abc$ . Let d = -5. Then $abc = \frac{30}{-5}$ , so $abc=-6$ , discriminant is $16-72$ , which makes this a dead end. Thus $d=\frac{3}{2}$ For $d=\frac{3}{2}$ , making $abc=20$ . This means the discriminant is just $256$ , so we obtain two values for $c$ as well. We get either $c=\frac{10}{3}$ or $c=-2$ . So, we must AGAIN test which one is correct. We know $ab=3c-4$ , and $a+b=-3$ , so we use these values for testing. Let $c=\frac{10}{3}$ . Then $ab=6$ , so $a=\frac{6}{b}$ . We thus get $\frac{6}{b}+b=-3$ , which leads to the quadratic $b^2+3b+6$ . The discriminant for this is $9-24$ . That means this value of $c$ is wrong, so $c=-2$ . Thus we get polynomial $b^2+3b-10$ . The discriminant this time is $49$ , so we get two values for $b$ . Through simple inspection, you may see they are interchangeable, as if you take the value $b=2$ , you get $a=-5$ . If you take the value $b=-5$ , you get $a=2$ . So it doesn't matter. That means the sum of all their squares is $\frac{9}{4}+4+4+25=\frac{141}{4}$ so the answer is $141+4=\boxed{145}.$ ~amcrunner

Solution 5

Let the four equations from top to bottom be listed $(1)$ through $(4)$ respectively. Multiplying both sides of $(3)$ by $d$ and factoring some terms gives us $abcd + d^2(ab + ac + bc) = 14d$ . Substituting using equations $(4)$ and $(2)$ gives us $30 -4 d^2 = 14d$ , and solving gives us $d = -5$ or $d = \frac{3}{2}$ . Plugging this back into $(3)$ gives us $abc + d(ab + ac + bc) = abc + (-5)(-4) = abc + 20 = 14$ , or using the other solution for $d$ gives us $abc - 6 = 14$ . Solving both of these equations gives us $abc = -6$ when $d = -5$ and $abc = 14$ when $d = \frac{3}{2}$ .

Multiplying both sides of $(2)$ by $c$ and factoring some terms gives us $abc + c^2 (a + b) = abc -3c^2 = -4c$ . Testing $abc = -6$ will give us an imaginary solution for $c$ , so therefore $abc = 14$ and $d = \frac{3}{2}$ . This gets us $14 - 3c^2 = -4c$ . Solving for $c$ gives us $c = \frac{3}{10}$ or $c = -2$ . With a bit of testing, we can see that the correct value of $c$ is $c=-2$ . Now we know $a+b = -3$ and $ab + bc + ca = ab + c(a+b) = ab + 6 = -4$ , $ab = -10$ , and it is obvious that $a = -5$ and $b = 2$ or the other way around, and therefore, $a^2 + b^2 + c^2 + d^2 = 25 + 4 + 4 + \frac{9}{4} = \frac{141}{4}$ , giving us the answer $141 + 4 = \boxed{145}$ .

~hihitherethere

Video Solution

https://www.youtube.com/watch?v=2rrX1G7iZqg

Video Solution by Interstigation

https://youtu.be/fGgbCgIHRHM

~Interstigation

@@ Line 1: / Line 1: @@
-These problems will not be available until the <math>2021</math> AIME II is released on Thursday, March <math>25, 2021</math>.
+==Problem==
+Let <math>a, b, c,</math> and <math>d</math> be real numbers that satisfy the system of equations
+<cmath>\begin{align*}
+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==
+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>. Since 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>.
+~math31415926535 ~minor edit by [[Mathkiddie]]
+==Solution 2==
+Note that <math>ab + bc + ca = -4</math> can be rewritten as <math>ab + c(a+b) = -4</math>. Hence, <math>ab = 3c - 4</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 <cmath>3c^{2} - 4c - 4d - 14 = 0.</cmath> We refer to this as Equation 1.
+Note that <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 <cmath>3c^{2} - 4c = \frac{30}{d}.</cmath> We refer to this as Equation 2.
+Substituting Equation 2 into Equation 1 gives, <math>\frac{30}{d} - 4d - 14 = 0</math>.
+Solving this quadratic yields that <math>d \in \left\{-5, \frac{3}{2}\right\}</math>.
+Now we just try these two cases:
+For <math>d = \frac{3}{2}</math> substituting in Equation 1 gives a quadratic in <math>c</math> which has roots <math>c \in \left\{\frac{10}{3}, -2\right\}</math>.
+Again trying cases, by letting <math>c = -2</math>, we get <math>ab = 3c-4</math>, Hence <math>ab = -10</math>.
+We know that <math>a + b = -3</math>, Solving these we get <math>a = -5, b = 2</math> or <math>a= 2, b = -5</math> (doesn't matter due to symmetry in <math>a,b</math>).
+So, this case yields solutions <math>(a,b,c,d) = \left(-5, 2 , -2, \frac{3}{2}\right)</math>.
+Similarly trying other three cases, we get no more solutions, Hence this is the solution for <math>(a,b,c,d)</math>.
+Finally, <math>a^{2} + b^{2} + c^{2} + d^{2} = 25 + 4 + 4 + \frac{9}{4} = \frac{141}{4} = \frac{m}{n}</math>.
+Therefore, <math>m + n = 141 + 4 = \boxed{145}</math>.
+~Arnav Nigam
+==Solution 3==
+For simplicity purposes, we number the given equations <math>(1),(2),(3),</math> and <math>(4),</math> in that order.
+Rearranging <math>(2)</math> and solving for <math>c,</math> we have
+<cmath>\begin{align*}
+ab+(a+b)c&=-4 \\
+ab-3c&=-4 \\
+c&=\frac{ab+4}{3}. \hspace{14mm} (5)
+\end{align*}</cmath>
+Substituting <math>(5)</math> into <math>(4)</math> and solving for <math>d,</math> we get
+<cmath>\begin{align*}
+ab\left(\frac{ab+4}{3}\right)d&=30 \\
+d&=\frac{90}{ab(ab+4)}. \hspace{5mm} (6)
+\end{align*}</cmath>
+Substituting <math>(5)</math> and <math>(6)</math> into <math>(3)</math> and simplifying, we rewrite the left side of <math>(3)</math> in terms of <math>a</math> and <math>b</math> only:
+<cmath>\begin{align*}
+ab\left[\frac{ab+4}{3}\right] + b\left[\frac{ab+4}{3}\right]\left[\frac{90}{ab(ab+4)}\right] + \left[\frac{ab+4}{3}\right]\left[\frac{90}{ab(ab+4)}\right]a + \left[\frac{90}{ab(ab+4)}\right]ab &= 14 \\
+ab\left[\frac{ab+4}{3}\right] + \underbrace{\frac{30}{a} + \frac{30}{b}}_{\text{Group them.}} + \frac{90}{ab+4} &= 14 \\
+ab\left[\frac{ab+4}{3}\right] + \frac{30(a+b)}{ab} + \frac{90}{ab+4} &= 14 \\
+ab\left[\frac{ab+4}{3}\right] + \underbrace{\frac{-90}{ab} + \frac{90}{ab+4}}_{\text{Group them.}} &= 14 \\
+ab\left[\frac{ab+4}{3}\right] - \frac{360}{ab(ab+4)}&=14.
+\end{align*}</cmath>
+Let <math>t=ab(ab+4),</math> from which <cmath>\frac{t}{3}-\frac{360}{t}=14.</cmath> Multiplying both sides by <math>3t,</math> rearranging, and factoring give <math>(t+18)(t-60)=0.</math> Substituting back and completing the squares produce
+<cmath>\begin{align*}
+\left[ab(ab+4)+18\right]\left[ab(ab+4)-60\right]&=0 \\
+\left[(ab)^2+4ab+18\right]\left[(ab)^2+4ab-60\right]&=0 \\
+\underbrace{\left[(ab+2)^2+14\right]}_{ab+2=\pm\sqrt{14}i\implies ab\not\in\mathbb R}\underbrace{\left[(ab+2)^2-64\right]}_{ab+2=\pm8}&=0 \\
+ab&=6,-10.
+\end{align*}</cmath>
+If <math>ab=6,</math> then combining this with <math>(1),</math> we know that <math>a</math> and <math>b</math> are the solutions of the quadratic <math>x^2+3x+6=0.</math> Since the discriminant is negative, neither <math>a</math> nor <math>b</math> is a real number.
+If <math>ab=-10,</math> then combining this with <math>(1),</math> we know that <math>a</math> and <math>b</math> are the solutions of the quadratic <math>x^2+3x-10=0,</math> or <math>(x+5)(x-2)=0,</math> from which <math>\{a,b\}=\{-5,2\}.</math> Substituting <math>ab=-10</math> into <math>(5)</math> and <math>(6),</math> we obtain <math>c=-2</math> and <math>d=\frac32,</math> respectively. Together, we have <cmath>a^2+b^2+c^2+d^2=\frac{141}{4},</cmath> so the answer is <math>141+4=\boxed{145}.</math>
+~MRENTHUSIASM
+==Solution 4 (way too long)==
+Let the four equations from top to bottom be listed 1 through 4 respectively. We factor equation 3 like so: <cmath>abc+d(ab+bc+ca)=14</cmath>
+Then we plug in equation 2 to receive <math>abc-4d=14</math>. By equation 4 we get <math>abc=\frac{30}{d}</math>. Plugging in, we get <math>\frac{30}{d}-4d=14</math>. Multiply by <math>d</math> on both sides to get the quadratic equation <math>4d^2+14d-30=0</math>. Solving using the quadratic equation, we receive <math>d=\frac{3}{2},d=-5</math>. So, we have to test which one is correct. We repeat a similar process as we did above for equations 1 and 2. We factor equation 2 to get <cmath>ab+c(a+b)=-4</cmath>
+After plugging in equation 1, we get <math>ab-3c=-4</math>. Now we convert it into a quadratic to receive <math>3c^2-4c-abc=0</math>. The value of <math>abc</math> will depend on <math>d</math>. So we obtain the discriminant <math>16+12abc</math>.
+Let d = -5.
+Then <math>abc = \frac{30}{-5}</math>, so <math>abc=-6</math>, discriminant is <math>16-72</math>, which makes this a dead end. Thus <math>d=\frac{3}{2}</math>
+For <math>d=\frac{3}{2}</math>, making <math>abc=20</math>. This means the discriminant is just <math>256</math>, so we obtain two values for <math>c</math> as well. We get either <math>c=\frac{10}{3}</math> or <math>c=-2</math>. So, we must AGAIN test which one is correct.
+We know <math>ab=3c-4</math>, and <math>a+b=-3</math>, so we use these values for testing.
+Let <math>c=\frac{10}{3}</math>.
+Then <math>ab=6</math>, so <math>a=\frac{6}{b}</math>. We thus get <math>\frac{6}{b}+b=-3</math>, which leads to the quadratic <math>b^2+3b+6</math>. The discriminant for this is <math>9-24</math>. That means this value of <math>c</math> is wrong, so <math>c=-2</math>. Thus we get polynomial <math>b^2+3b-10</math>. The discriminant this time is <math>49</math>, so we get two values for <math>b</math>. Through simple inspection, you may see they are interchangeable, as if you take the value <math>b=2</math>, you get <math>a=-5</math>. If you take the value <math>b=-5</math>, you get <math>a=2</math>. So it doesn't matter. That means the sum of all their squares is
+<cmath>\frac{9}{4}+4+4+25=\frac{141}{4}</cmath>
+so the answer is <math>141+4=\boxed{145}.</math>
+~amcrunner
+==Solution 5==
+Let the four equations from top to bottom be listed <math>(1)</math> through <math>(4)</math> respectively. Multiplying both sides of <math>(3)</math> by <math>d</math> and factoring some terms gives us <math>abcd + d^2(ab + ac + bc) = 14d</math>. Substituting using equations <math>(4)</math> and <math>(2)</math> gives us <math>30 -4 d^2 = 14d</math>, and solving gives us <math>d = -5</math> or <math>d = \frac{3}{2}</math>. Plugging this back into <math>(3)</math> gives us <math>abc + d(ab + ac + bc) = abc + (-5)(-4) = abc + 20 = 14</math>, or using the other solution for <math>d</math> gives us <math>abc - 6 = 14</math>. Solving both of these equations gives us <math>abc = -6</math> when <math>d = -5</math> and <math>abc = 14</math> when <math>d = \frac{3}{2}</math>.
+Multiplying both sides of <math>(2)</math> by <math>c</math> and factoring some terms gives us <math>abc + c^2 (a + b) = abc -3c^2 = -4c</math>. Testing <math>abc = -6</math> will give us an imaginary solution for <math>c</math>, so therefore <math>abc = 14</math> and <math>d = \frac{3}{2}</math>. This gets us <math>14 - 3c^2 = -4c</math>. Solving for <math>c</math> gives us <math>c = \frac{3}{10}</math> or <math>c = -2</math>. With a bit of testing, we can see that the correct value of <math>c</math> is <math>c=-2</math>. Now we know <math>a+b = -3</math> and <math>ab + bc + ca = ab + c(a+b) = ab + 6 = -4</math>, <math>ab = -10</math>, and it is obvious that <math>a = -5</math> and <math>b = 2</math> or the other way around, and therefore, <math>a^2 + b^2 + c^2 + d^2 = 25 + 4 + 4 + \frac{9}{4} = \frac{141}{4}</math>, giving us the answer <math>141 + 4 = \boxed{145}</math>.
+~hihitherethere
+==Video Solution==
+https://www.youtube.com/watch?v=2rrX1G7iZqg
+==Video Solution by Interstigation==
+https://youtu.be/fGgbCgIHRHM
+~Interstigation
+==See Also==
+{{AIME box|year=2021|n=II|num-b=6|num-a=8}}
+{{MAA Notice}}

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