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>
  
== Solution ==
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== Solution 1 ==
 
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
 
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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== 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 to 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
  
 
==See Also==
 
==See Also==

Revision as of 23:51, 12 March 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 $P(3)P(4) = c = P(3)(4 - 4a + \frac{8a - 1}{2}) \implies a = \frac{7P(3) + 1}{8}$. 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

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 to 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

See Also

2020 AIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 13
Followed by
Problem 15
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

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