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2020 AIME I Problems/Problem 14

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 $(a+b)^2.$

==Video solution by

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 $(a+b)^2 = 36$.

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 $P(x) = (x - r)(x - s)$ where $r$ and $s$ are its roots.

Note that $P(r) = P(s) = P(P(3)) = P(P(4)) = P(P(a)) = P(P(b)) = 0.$ Since $P(x)$ is a quadratic, it can have at most two distinct roots (in the case of this problem it can only have exactly 2), so there must be two pairs of $P(n)$ out of the four $P(3), P(4), P(a),$ and $P(b)$ which are equal to each other. (They have to be pairs because $P(x)$ can only have 2 roots, if we had something like $P(a) = P(3) = P(4) = r$ and $P(b) = s$ then $P(x)$ would have 3 roots).


Case 1: $P(3) = P(4)$ and $P(a) = P(b).$ Then, WLOG, let $(3 - r)(3 - s)=(4 - r)(4 - s) = r$ and $(a - r)(a - s)=(b - r)(b - s)=s.$ Solving for $a + b$ gives us $a + b = r + s = 3 + 4 = 7 \implies (a + b)^2 = 49.$


Case 2: WLOG, let $P(3) = P(a) = r$ and $P(4) = P(b) = s.$ So $(3 - r)(3 - s) = r = (a - r)(a - s)$ and $(4 - r)(4 - s) = s = (b - r)(b - s).$ Subtracting $(4 - r)(4 - s) = s$ from $(3 - r)(3 - s) = r$ gives $-7 + r + s = r -s$, so $s = \frac{7}{2}.$ Plugging this back into $(4-r)(4-s) = s$ gives $r = -3.$


The coefficient of the $x^3$ term in $P(P(x))$ is $-2(r+s)$, so by Vieta's Formulas the sum of the roots of $P(P(x))$ is $2$ times that of $P(x).$ We know $r + s = \frac{1}{2}$, so $3 + 4 + a + b = \frac{1}{2} \cdot 2 = 1.$ Then $a + b = -6 \implies (a + b)^2 = 36.$


The requested sum is $36 + 49 =\boxed{85}.$


~grogg007, TheUltimate123


Remark: On the last step in case 2, it's important to use Vieta's instead of just solving for $a$ and $b$ in $(a - r)(a - s) = (3 - r)(3 -s)$ and $(b - r)(b - s) = (4 - r)(4 - s)$ because not all four pairs of solutions for $a$ and $b$ are valid.

Solution 5 (Official MAA)

Note that because $P\big(P(3)\big)=P\big(P(4)\big)= 0$, $P(3)$ and $P(4)$ are roots of $P(x)$. There are two cases. CASE 1: $P(3) = P(4)$. Then $P(x)$ is symmetric about $x=\tfrac72$; that is to say, $P(r) = P(7-r)$ for all $r$. Thus the remaining two roots must sum to $7$. Indeed, the polynomials $P(x) = \left(x-\frac72\right)^2 + \frac{11}4 \pm i\sqrt3$ satisfy the conditions. CASE 2: $P(3)\neq P(4)$. Then $P(3)$ and $P(4)$ are the two distinct roots of $P(x)$, so\[P(x) = \big(x-P(3)\big)\big(x-P(4)\big)\]for all $x$. Note that any solution to $P\big(P(x)\big) = 0$ must satisfy either $P(x) = P(3)$ or $P(x) = P(4)$. Because $P(x)$ is quadratic, the polynomials $P(x) - P(3)$ and $P(x) - P(4)$ each have the same sum of roots as the polynomial $P(x)$, which is $P(3) + P(4)$. Thus the answer in this case is $2\big(P(3) + P(4)\big)-7$, and so it suffices to compute the value of $P(3)+P(4)$.

Let $P(3)=u$ and $P(4) = v$. Substituting $x=3$ and $x=4$ into the above quadratic polynomial yields the system of equations \begin{align*} u &= (3-u)(3-v) = 9 - 3u - 3v + uv\\ v &= (4-u)(4-v) = 16 - 4u - 4v + uv. \end{align*}Subtracting the first equation from the second gives $v - u = 7 - u - v$, yielding $v = \frac72.$ Substituting this value into the second equation gives\[\dfrac72 = \left(4 - u\right)\left(4 - \dfrac72\right),\]yielding $u = -3.$ The sum of the two solutions is $2\left(\tfrac72-3\right)-7 = -6$. In this case, $P(x)= (x+3)\left(x-\frac72\right)$.

The requested sum of squares is $7^2+(-6)^2 = \boxed{085}$.

Solution 6

Let $P(x) = (x-c)^2 - d$ for some $c$, $d$.

Then, we can write $P(P(x)) = ((x-c)^2 - d - c)^2 - d$. Setting the expression equal to $0$ and solving for $x$ gives:

$x = \pm \sqrt{ \pm \sqrt{d} + d + c} + c$

Therefore, we have that $x$ takes on the four values $\sqrt{\sqrt{d} + d + c} + c$, $-\sqrt{\sqrt{d} + d + c} + c$, $\sqrt{-\sqrt{d} + d + c} + c$, and $-\sqrt{-\sqrt{d} + d + c} + c$. Two of these values are $3$ and $4$, and the other two are $a$ and $b$.

We can split these four values into two "groups" based on the radicand in the expression - for example, the first group consists of the first two values listed above, and the second group consists of the other two values.

$\textbf{Case 1}$: Both the 3 and 4 values are from the same group.

In this case, the $a$ and $b$ values are both from the other group. The sum of this is just $2c$ because the radical cancels out. Because of this, we can see that $c$ is just the average of $3$ and $4$, so we have $2c = 3 + 4 = 7$, so $(a+b)^2 = 7^2 = 49$.

$\textbf{Case 2}$: The 3 and 4 values come from different groups.

It is easy to see that all possibilities in this case are basically symmetric and yield the same value for $(a+b)^2$. Without loss of generality, assume that $\sqrt{\sqrt{d} + d + c} + c = 4$ and $\sqrt{-\sqrt{d} + d + c} + c = 3$. Note that we can't switch the values of these two expressions since the first one is guaranteed to be larger.

We can write $\sqrt{\sqrt{d} + d + c} + c = 1 + \sqrt{-\sqrt{d} + d + c} + c$.

Moving most terms to the left side and simplifying gives $\sqrt{\sqrt{d} + d + c} - \sqrt{-\sqrt{d} + d + c} = 1$.

We can square both sides and simplify:

$\sqrt{d} + d + c - \sqrt{d} + d + c - 2\sqrt{(d + c + \sqrt{d})(d + c - \sqrt{d})} = 1$

$2d + 2c - 2\sqrt{(d + c + \sqrt{d})(d + c - \sqrt{d})} = 1$

$\sqrt{(d + c + \sqrt{d})(d + c - \sqrt{d})} = (d+c) - \frac{1}{2}$

$\sqrt{(d+c)^2 - (\sqrt{d})^2} = (d+c) - \frac{1}{2}$

$\sqrt{d^2 + 2dc + c^2 - d} = (d+c) - \frac{1}{2}$

Squaring both sides again gives the following:

$d^2 + 2dc + c^2 - d = d^2 + 2dc + c^2 - d - c + \frac{1}{4}$

Nearly all terms cancel out, yielding $c = \frac{1}{4}$.

By substituting this back in, we obtain $\sqrt{\sqrt{d} + d + c} = \frac{15}{4}$ and $\sqrt{-\sqrt{d} + d + c} = \frac{11}{4}$.

The sum of $a$ and $b$ is equal to $-\sqrt{\sqrt{d} + d + c} - \sqrt{-\sqrt{d} + d + c} + 2c = -\frac{15}{4} - \frac{11}{4} + \frac{1}{2} = -6$, so $(a+b)^2 = 36$.

Adding up both values gives $49 + 36 = \boxed{085}$ as our final answer.


Video Solution

https://youtu.be/_Iji1DW7QaY?si=t6Qbn2XYAfknnIxr

~MathProblemSolvingSkills.com


See Also

2020 AIME I (Problems • Answer Key • Resources)
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Problem 13 		Followed by
Problem 15
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All AIME Problems and Solutions

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