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Difference between revisions of "2020 AIME I Problems/Problem 14"
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Let <math>P(x)</math> be a quadratic polynomial with complex coefficients whose <math>x^2</math> coefficient is <math>1.</math> Suppose the equation <math>P(P(x))=0</math> has four distinct solutions, <math>x=3,4,a,b.</math> Find the sum of all possible values of <math>(a+b)^2.</math>
 
Let <math>P(x)</math> be a quadratic polynomial with complex coefficients whose <math>x^2</math> coefficient is <math>1.</math> Suppose the equation <math>P(P(x))=0</math> has four distinct solutions, <math>x=3,4,a,b.</math> Find the sum of all possible values of <math>(a+b)^2.</math>
  
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== Solution ==
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== Solution 1 ==
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Either <math>P(3) = P(4)</math> or not. We first see that if <math>P(3) = P(4)</math> it's easy to obtain by Vieta's that <math>(a+b)^2 = 49</math>. Now, take <math>P(3) \neq P(4)</math> and WLOG <math>P(3) = P(a), P(4) = P(b)</math>. Now, consider the parabola formed by the graph of <math>P</math>. It has vertex <math>\frac{3+a}{2}</math>. Now, say that <math>P(x) = x^2 - (3+a)x + c</math>. We note <math>P(3)P(4) = c = P(3)(4 - 4a + \frac{8a - 1}{2}) \implies a = \frac{7P(3) + 1}{8}</math>. Now, we note <math>P(4) = \frac{7}{2}</math> by plugging in again. Now, it's easy to find that <math>a = -2.5, b = -3.5</math>, yielding a value of <math>36</math>. Finally, we add <math>49 + 36 = \boxed{085}</math>. ~awang11, charmander3333
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Either <math>P(3) = P(4)</math> or not. We first see that if <math>P(3) = P(4)</math> it's easy to obtain by Vieta's that <math>(a+b)^2 = 49</math>. Now, take <math>P(3) \neq P(4)</math> and WLOG <math>P(3) = P(a), P(4) = P(b)</math>. Now, consider the parabola formed by the graph of <math>P</math>. It has vertex <math>\frac{3+a}{2}</math>. Now, say that <math>P(x) = x^2 - (3+a)x + c</math>. We note <math>P(3)P(4) = c = P(3)\left(4 - 4a + \frac{8a - 1}{2}\right) \implies a = \frac{7P(3) + 1}{8}</math>. Now, we note <math>P(4) = \frac{7}{2}</math> by plugging in again. Now, it's easy to find that <math>a = -2.5, b = -3.5</math>, yielding a value of <math>36</math>. Finally, we add <math>49 + 36 = \boxed{085}</math>. ~awang11, charmander3333
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<b>Remark</b>: We know that <math>c=\frac{8a-1}{2}</math> from <math>P(3)+P(4)=3+a</math>.
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== Solution 2 ==
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Let the roots of <math>P(x)</math> be <math>m</math> and <math>n</math>, then we can write <math>P(x)=x^2-(m+n)x+mn</math>. The fact that <math>P(P(x))=0</math> has solutions <math>x=3,4,a,b</math> implies that some combination of <math>2</math> of these are the solution to <math>P(x)=m</math>, and the other <math>2</math> are the solution to <math>P(x)=n</math>. It's fairly easy to see there are only <math>2</math> possible such groupings: <math>P(3)=P(4)=m</math> and <math>P(a)=P(b)=n</math>, or <math>P(3)=P(a)=m</math> and <math>P(4)=P(b)=n</math> (Note that <math>a,b</math> are interchangeable, and so are <math>m</math> and <math>n</math>). We now casework:
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If <math>P(3)=P(4)=m</math>, then
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<cmath>9-3(m+n)+mn=16-4(m+n)+mn=m \implies m+n=7</cmath>
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<cmath>a^2-a(m+n)+mn=b^2-b(m+n)+mn=n \implies a+b=m+n=7</cmath>
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so this gives <math>(a+b)^2=7^2=49</math>.
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Next, if <math>P(3)=P(a)=m</math>, then
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<cmath>9-3(m+n)+mn=a^2-a(m+n)+mn=m \implies a+3=m+n</cmath>
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<cmath>16-4(m+n)+mn=b^2-b(m+n)+mn=n \implies b+4=m+n</cmath>
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Subtracting the first part of the first equation from the first part of the second equation gives
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<cmath>7-(m+n)=n-m \implies 2n=7 \implies n=\frac{7}{2} \implies m=-3</cmath>
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Hence, <math>a+b=2(m+n)-7=2\cdot \frac{1}{2}-7=-6</math>, and so <math>(a+b)^2=(-6)^2=36</math>.
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Therefore, the solution is <math>49+36=\boxed{085}</math> ~ktong
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== Solution 3 ==
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Write <math>P(x) = x^2+wx+z</math>. Split the problem into two cases: <math>P(3)\ne P(4)</math> and <math>P(3) = P(4)</math>.
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Case 1: We have <math>P(3) \ne P(4)</math>. We must have
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<cmath>w=-P(3)-P(4) = -(9+3w+z)-(16+4w+z) = -25-7w-2z.</cmath>
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Rearrange and divide through by <math>8</math> to obtain
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<cmath>w = \frac{-25-2z}{8}.</cmath>
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Now, note that
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<cmath>z = P(3)P(4) = (9+3w+z)(16+4w+z) = \left(9 + 3\cdot \frac{-25-2z}{8} + z\right)\left(16 + 4 \cdot \frac{-25-2z}{8} + z\right) =</cmath>
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<cmath>\left(-\frac{3}{8} + \frac{z}{4}\right)\left(\frac{7}{2}\right) = -\frac{21}{16} + \frac{7z}{8}.</cmath>
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Now, rearrange to get
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<cmath>\frac{z}{8} = -\frac{21}{16}</cmath>
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and thus
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<cmath>z = -\frac{21}{2}.</cmath>
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Substituting this into our equation for <math>w</math> yields <math>w = -\frac{1}{2}</math>. Then, it is clear that <math>P</math> does not have a double root at <math>P(3)</math>, so we must have <math>P(a) = P(3)</math> and <math>P(b) = P(4)</math> or vice versa. This gives <math>3+a = \frac{1}{2}</math> and <math>4+b = \frac{1}{2}</math> or vice versa, implying that <math>a+b = 1-3-4 = -6</math> and <math>(a+b)^2 = 6</math>.
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Case 2: We have <math>P(3) = P(4)</math>. Then, we must have <math>w = -7</math>. It is clear that <math>P(a) = P(b)</math> (we would otherwise get <math>P(a)=P(3)=P(4)</math> implying <math>a \in \{3,4\}</math> or vice versa), so <math>a+b=-w=7</math> and <math>(a+b)^2 = 49</math>.
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Thus, our final answer is <math>49+36=\boxed{085}</math>. ~GeronimoStilton
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==Solution 4==
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Let <math>P(x)=(x-r)(x-s)</math>. There are two cases: in the first case, <math>(3-r)(3-s)=(4-r)(4-s)</math> equals <math>r</math> (without loss of generality), and thus <math>(a-r)(a-s)=(b-r)(b-s)=s</math>. By Vieta's formulas <math>a+b=r+s=3+4=7</math>.
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In the second case, say without loss of generality <math>(3-r)(3-s)=r</math> and <math>(4-r)(4-s)=s</math>. Subtracting gives <math>-7+r+s=r-s</math>, so <math>s=7/2</math>. From this, we have <math>r=-3</math>.
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Note <math>r+s=1/2</math>, so by Vieta's, we have <math>\{a,b\}=\{1/2-3,1/2-4\}=\{-5/2,-7/2\}</math>. In this case, <math>a+b=-6</math>.
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The requested sum is <math>36+49=85</math>.~TheUltimate123
  
 
==See Also==
 
==See Also==

Revision as of 14:07, 3 June 2020

Problem

Let $P(x)$ be a quadratic polynomial with complex coefficients whose $x^2$ coefficient is $1.$ Suppose the equation $P(P(x))=0$ has four distinct solutions, $x=3,4,a,b.$ Find the sum of all possible values of $(a+b)^2.$

Solution 1

Either $P(3) = P(4)$ or not. We first see that if $P(3) = P(4)$ it's easy to obtain by Vieta's that $(a+b)^2 = 49$. Now, take $P(3) \neq P(4)$ and WLOG $P(3) = P(a), P(4) = P(b)$. Now, consider the parabola formed by the graph of $P$. It has vertex $\frac{3+a}{2}$. Now, say that $P(x) = x^2 - (3+a)x + c$. We note $P(3)P(4) = c = P(3)\left(4 - 4a + \frac{8a - 1}{2}\right) \implies a = \frac{7P(3) + 1}{8}$. Now, we note $P(4) = \frac{7}{2}$ by plugging in again. Now, it's easy to find that $a = -2.5, b = -3.5$, yielding a value of $36$. Finally, we add $49 + 36 = \boxed{085}$. ~awang11, charmander3333

Remark: We know that $c=\frac{8a-1}{2}$ from $P(3)+P(4)=3+a$.

Solution 2

Let the roots of $P(x)$ be $m$ and $n$, then we can write $P(x)=x^2-(m+n)x+mn$. The fact that $P(P(x))=0$ has solutions $x=3,4,a,b$ implies that some combination of $2$ of these are the solution to $P(x)=m$, and the other $2$ are the solution to $P(x)=n$. It's fairly easy to see there are only $2$ possible such groupings: $P(3)=P(4)=m$ and $P(a)=P(b)=n$, or $P(3)=P(a)=m$ and $P(4)=P(b)=n$ (Note that $a,b$ are interchangeable, and so are $m$ and $n$). We now casework: If $P(3)=P(4)=m$, then \[9-3(m+n)+mn=16-4(m+n)+mn=m \implies m+n=7\] \[a^2-a(m+n)+mn=b^2-b(m+n)+mn=n \implies a+b=m+n=7\] so this gives $(a+b)^2=7^2=49$. Next, if $P(3)=P(a)=m$, then \[9-3(m+n)+mn=a^2-a(m+n)+mn=m \implies a+3=m+n\] \[16-4(m+n)+mn=b^2-b(m+n)+mn=n \implies b+4=m+n\] Subtracting the first part of the first equation from the first part of the second equation gives \[7-(m+n)=n-m \implies 2n=7 \implies n=\frac{7}{2} \implies m=-3\] Hence, $a+b=2(m+n)-7=2\cdot \frac{1}{2}-7=-6$, and so $(a+b)^2=(-6)^2=36$. Therefore, the solution is $49+36=\boxed{085}$ ~ktong

Solution 3

Write $P(x) = x^2+wx+z$. Split the problem into two cases: $P(3)\ne P(4)$ and $P(3) = P(4)$.

Case 1: We have $P(3) \ne P(4)$. We must have \[w=-P(3)-P(4) = -(9+3w+z)-(16+4w+z) = -25-7w-2z.\] Rearrange and divide through by $8$ to obtain \[w = \frac{-25-2z}{8}.\] Now, note that \[z = P(3)P(4) = (9+3w+z)(16+4w+z) = \left(9 + 3\cdot \frac{-25-2z}{8} + z\right)\left(16 + 4 \cdot \frac{-25-2z}{8} + z\right) =\] \[\left(-\frac{3}{8} + \frac{z}{4}\right)\left(\frac{7}{2}\right) = -\frac{21}{16} + \frac{7z}{8}.\] Now, rearrange to get \[\frac{z}{8} = -\frac{21}{16}\] and thus \[z = -\frac{21}{2}.\] Substituting this into our equation for $w$ yields $w = -\frac{1}{2}$. Then, it is clear that $P$ does not have a double root at $P(3)$, so we must have $P(a) = P(3)$ and $P(b) = P(4)$ or vice versa. This gives $3+a = \frac{1}{2}$ and $4+b = \frac{1}{2}$ or vice versa, implying that $a+b = 1-3-4 = -6$ and $(a+b)^2 = 6$.

Case 2: We have $P(3) = P(4)$. Then, we must have $w = -7$. It is clear that $P(a) = P(b)$ (we would otherwise get $P(a)=P(3)=P(4)$ implying $a \in \{3,4\}$ or vice versa), so $a+b=-w=7$ and $(a+b)^2 = 49$.

Thus, our final answer is $49+36=\boxed{085}$. ~GeronimoStilton

Solution 4

Let $P(x)=(x-r)(x-s)$. There are two cases: in the first case, $(3-r)(3-s)=(4-r)(4-s)$ equals $r$ (without loss of generality), and thus $(a-r)(a-s)=(b-r)(b-s)=s$. By Vieta's formulas $a+b=r+s=3+4=7$.

In the second case, say without loss of generality $(3-r)(3-s)=r$ and $(4-r)(4-s)=s$. Subtracting gives $-7+r+s=r-s$, so $s=7/2$. From this, we have $r=-3$.

Note $r+s=1/2$, so by Vieta's, we have $\{a,b\}=\{1/2-3,1/2-4\}=\{-5/2,-7/2\}$. In this case, $a+b=-6$.

The requested sum is $36+49=85$.~TheUltimate123

See Also

2020 AIME I (Problems • Answer Key • Resources)
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