Difference between revisions of "2020 AIME I Problems/Problem 14"

Latest revision as of 18:34, 17 September 2023

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

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 $P(x)=(x-r)(x-s)$ . 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 $7^2+(-6)^2 = \boxed{085}$ .

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

@@ Line 4: / Line 4: @@
 == 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)\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
+<b>Remark</b>: We know that <math>c=\frac{8a-1}{2}</math> from <math>P(3)+P(4)=3+a</math>.
 == Solution 2 ==
-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:
+Let the roots of <math>P(x)</math> be <math>m</math> and <math>n</math>, then we can write <cmath>P(x)=x^2-(m+n)x+mn</cmath> 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:
 If <math>P(3)=P(4)=m</math>, then
 <cmath>9-3(m+n)+mn=16-4(m+n)+mn=m \implies m+n=7</cmath>
@@ Line 34: / Line 36: @@
 and thus
 <cmath>z = -\frac{21}{2}.</cmath>
-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>.
+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 = 36</math>.
 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>.
-Thus, our final answer is <math>49+36=\fbox{085)</math>. ~GeronimoStilton
+Thus, our final answer is <math>49+36=\boxed{085}</math>. ~GeronimoStilton
+==Solution 4==
+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>.
+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>.
+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>.
+The requested sum is <math>36+49=85</math>.~TheUltimate123
+==Solution 5 (Official MAA)==
+Note that because <math>P\big(P(3)\big)=P\big(P(4)\big)= 0</math>, <math>P(3)</math> and <math>P(4)</math> are roots of <math>P(x)</math>. There are two cases.
+CASE 1: <math>P(3) = P(4)</math>. Then <math>P(x)</math> is symmetric about <math>x=\tfrac72</math>; that is to say, <math>P(r) = P(7-r)</math> for all <math>r</math>. Thus the remaining two roots must sum to <math>7</math>. Indeed, the polynomials <math>P(x) = \left(x-\frac72\right)^2 + \frac{11}4 \pm i\sqrt3</math> satisfy the conditions.
+CASE 2: <math>P(3)\neq P(4)</math>. Then <math>P(3)</math> and <math>P(4)</math> are the two distinct roots of <math>P(x)</math>, so<cmath>P(x) = \big(x-P(3)\big)\big(x-P(4)\big)</cmath>for all <math>x</math>. Note that any solution to <math>P\big(P(x)\big) = 0</math> must satisfy either <math>P(x) = P(3)</math> or <math>P(x) = P(4)</math>. Because <math>P(x)</math> is quadratic, the polynomials <math>P(x) - P(3)</math> and <math>P(x) - P(4)</math> each have the same sum of roots as the polynomial <math>P(x)</math>, which is <math>P(3) + P(4)</math>. Thus the answer in this case is <math>2\big(P(3) + P(4)\big)-7</math>, and so it suffices to compute the value of <math>P(3)+P(4)</math>.
+Let <math>P(3)=u</math> and <math>P(4) = v</math>. Substituting <math>x=3</math> and <math>x=4</math> into the above quadratic polynomial yields the system of equations
+<cmath>\begin{align*}
+u &= (3-u)(3-v) = 9 - 3u - 3v + uv\\
+v &= (4-u)(4-v) = 16 - 4u - 4v + uv.
+\end{align*}</cmath>Subtracting the first equation from the second gives <math>v - u = 7 - u - v</math>, yielding <math>v = \frac72.</math> Substituting this value into the second equation gives<cmath>\dfrac72 = \left(4 - u\right)\left(4 - \dfrac72\right),</cmath>yielding <math>u = -3.</math> The sum of the two solutions is <math>2\left(\tfrac72-3\right)-7 = -6</math>. In this case, <math>P(x)= (x+3)\left(x-\frac72\right)</math>.
+The requested sum of squares is <math>7^2+(-6)^2 = \boxed{085}</math>.
+==Solution 6==
+Let <math>P(x) = (x-c)^2 - d</math> for some <math>c</math>, <math>d</math>.
+Then, we can write <math>P(P(x)) = ((x-c)^2 - d - c)^2 - d</math>. Setting the expression equal to <math>0</math> and solving for <math>x</math> gives:
+<math>x = \pm \sqrt{ \pm \sqrt{d} + d + c} + c</math>
+Therefore, we have that <math>x</math> takes on the four values <math>\sqrt{\sqrt{d} + d + c} + c</math>, <math>-\sqrt{\sqrt{d} + d + c} + c</math>, <math>\sqrt{-\sqrt{d} + d + c} + c</math>, and <math>-\sqrt{-\sqrt{d} + d + c} + c</math>. Two of these values are <math>3</math> and <math>4</math>, and the other two are <math>a</math> and <math>b</math>.
+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.
+<math>\textbf{Case 1}</math>: Both the 3 and 4 values are from the same group.
+In this case, the <math>a</math> and <math>b</math> values are both from the other group. The sum of this is just <math>2c</math> because the radical cancels out. Because of this, we can see that <math>c</math> is just the average of <math>3</math> and <math>4</math>, so we have <math>2c = 3 + 4 = 7</math>, so <math>(a+b)^2 = 7^2 = 49</math>.
+<math>\textbf{Case 2}</math>: 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 <math>(a+b)^2</math>. Without loss of generality, assume that <math>\sqrt{\sqrt{d} + d + c} + c = 4</math> and <math>\sqrt{-\sqrt{d} + d + c} + c = 3</math>. Note that we can't switch the values of these two expressions since the first one is guaranteed to be larger.
+We can write <math>\sqrt{\sqrt{d} + d + c} + c = 1 + \sqrt{-\sqrt{d} + d + c} + c</math>.
+Moving most terms to the left side and simplifying gives <math>\sqrt{\sqrt{d} + d + c} - \sqrt{-\sqrt{d} + d + c} = 1</math>.
+We can square both sides and simplify:
+<math>\sqrt{d} + d + c - \sqrt{d} + d + c - 2\sqrt{(d + c + \sqrt{d})(d + c - \sqrt{d})} = 1</math>
+<math>2d + 2c - 2\sqrt{(d + c + \sqrt{d})(d + c - \sqrt{d})} = 1</math>
+<math>\sqrt{(d + c + \sqrt{d})(d + c - \sqrt{d})} = (d+c) - \frac{1}{2}</math>
+<math>\sqrt{(d+c)^2 - (\sqrt{d})^2} = (d+c) - \frac{1}{2}</math>
+<math>\sqrt{d^2 + 2dc + c^2 - d} = (d+c) - \frac{1}{2}</math>
+Squaring both sides again gives the following:
+<math>d^2 + 2dc + c^2 - d = d^2 + 2dc + c^2 - d - c + \frac{1}{4}</math>
+Nearly all terms cancel out, yielding <math>c = \frac{1}{4}</math>.
+By substituting this back in, we obtain <math>\sqrt{\sqrt{d} + d + c} = \frac{15}{4}</math> and <math>\sqrt{-\sqrt{d} + d + c} = \frac{11}{4}</math>.
+The sum of <math>a</math> and <math>b</math> is equal to <math>-\sqrt{\sqrt{d} + d + c} - \sqrt{-\sqrt{d} + d + c} + 2c = -\frac{15}{4} - \frac{11}{4} + \frac{1}{2} = -6</math>, so <math>(a+b)^2 = 36</math>.
+Adding up both values gives <math>49 + 36 = \boxed{085}</math> as our final answer.
+==Video Solution==
+https://youtu.be/_Iji1DW7QaY?si=t6Qbn2XYAfknnIxr
+~MathProblemSolvingSkills.com
 ==See Also==

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