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 $(a+b)^2.$

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

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 $7^2+(-6)^2 = {85}$ .

2020 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
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