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Difference between revisions of "2021 Fall AMC 12A Problems/Problem 23"
(→‎Solution 10 (solution 8 but simpler))
 
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Let <math>p(x)=(x-h)^2+k</math> for some real constants <math>h</math> and <math>k.</math> Suppose that <math>p(x)</math> has real roots <math>r</math> and <math>s.</math>
 
Let <math>p(x)=(x-h)^2+k</math> for some real constants <math>h</math> and <math>k.</math> Suppose that <math>p(x)</math> has real roots <math>r</math> and <math>s.</math>
  
−
Since <math>p(p(x))=0,</math> we conclude that <math>p(x)=r</math> or <math>p(x)=s.</math> Without the loss of generality, we assume that <math>p(x)=r</math> has two real solutions and <math>p(x)=s</math> has one real solution. Therefore, we have <math>k=s,</math> from which <math>p(x)=(x-h)^2+s.</math>
+
Since <math>p(p(x))=0,</math> we conclude that <math>p(x)=r</math> or <math>p(x)=s.</math> Without loss of generality, we assume that <math>p(x)=r</math> has two real solutions and <math>p(x)=s</math> has one real solution. Therefore, we have <math>k=s,</math> from which <math>p(x)=(x-h)^2+s.</math>
  
 
As <math>p(s)=0,</math> we expand the left side to obtain <math>(s-h)^2+s=0,</math> or <cmath>s^2-(2h-1)s+h^2=0. \hspace{15mm}(\bigstar)</cmath>
 
As <math>p(s)=0,</math> we expand the left side to obtain <math>(s-h)^2+s=0,</math> or <cmath>s^2-(2h-1)s+h^2=0. \hspace{15mm}(\bigstar)</cmath>
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~MRENTHUSIASM
 
~MRENTHUSIASM
 +
 +
== Solution 2.5 (Vertex Form but Done Differently) ==
 +
Continuing from the previous solution's reasoning, we know that the <math>y</math>-coordinate of the vertex must also be a root of the polynomial. Thus, <math>\tilde{p}(x) = \left(x - \frac{s + r}{2}\right)^2 + s</math>, where <math>r</math> and <math>s</math> are roots of the polynomial.
 +
 +
We also know that in order for <math>\tilde{p}(x) = r</math> to have two solutions, <math>y = r</math> must intersect <math>\tilde{p}(x)</math> above its minimum/vertex. Thus, <math>r > s</math>. Let <math>r = s + a</math>. Then, <math>\tilde{p}(x) = \left(x - \frac {2s + a}{2}\right)^2 + s</math>.
 +
 +
Since <math>s</math> is a root, <math>\tilde{p}(s) = 0</math> so <math>\left(s - \frac{2s + a}{2}\right)^2 +s = 0 \implies \frac{a^2}{4} + s = 0 \implies s = - \frac{1}{4} a^2</math>. We wish to maximize <math>s + s + a = 2s + a = - \frac{1}{2} a^2 + a</math>.
 +
 +
This quadratic is maximized at <math>a = 1</math> (when converting the quadratic to vertex form). Then, since <math>\tilde{p}(s + a) = 0 \implies \tilde{p}(s+1) = 0</math>, <math>\left(s + 1 - s - \frac12\right)^2 + s = 0 \implies \frac{1}{4} + s = 0 \implies s = -\frac{1}{4}</math>.
 +
 +
Thus, <math>\tilde{p}(x) = \left(x - \frac{1}{4}\right)^2 - \frac{1}{4}</math>. So, <math>\tilde{p}(1) = \boxed{\textbf{(A) } \frac{5}{16}}</math>.
 +
 +
~[https://artofproblemsolving.com/wiki/index.php/User:CrazyVideoGamez CrazyVideoGamez]
  
 
== Solution 3 (Symmetry) ==
 
== Solution 3 (Symmetry) ==
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==Solution 9 (Lagrange Multipliers)==
 
==Solution 9 (Lagrange Multipliers)==
−
First note that <math>P(x)</math> must have two distinct roots, else <math>p(p(x))</math> will have less than <math>3</math> roots. Let these be <math>a,b.</math> Since <math>P(x)</math> has <math>3</math> real roots, and complex roots come in pairs, there must be a double root. So there must be <math>a_1, a_2</math> s.t <math>P(a_1) = P(a_2) = a</math> and <math>b_1, b_2</math> s.t <math>P(b_1) = P(b_2) = b</math> but two of <math>a_1, a_2, b_1, b_2</math> must be the same. WLOG either <math>a_1 = a_2</math> or <math>a_1 = b_1</math>. In the latter, case, then,  
+
First note that <math>p(x)</math> must have two distinct roots, else <math>p(p(x))</math> will have fewer than <math>3</math> roots. Let these be <math>a,b.</math> Since <math>p(x)</math> has <math>3</math> real roots, and complex roots come in pairs, there must be a double root. So there must be <math>a_1, a_2</math> s.t <math>p(a_1) = p(a_2) = a</math> and <math>b_1, b_2</math> s.t <math>p(b_1) = p(b_2) = b</math> but two of <math>a_1, a_2, b_1, b_2</math> must be the same. WLOG either <math>a_1 = a_2</math> or <math>a_1 = b_1</math>. In the latter, case, then,  
 
<cmath>\begin{align*}
 
<cmath>\begin{align*}
−
P(x) &= a + (x - a_1)(x - a_2) \\
+
p(x) &= a + (x - a_1)(x - a_2) \\
−
P(x) &= b + (x - b_1)(x-b_2) \\
+
p(x) &= b + (x - b_1)(x-b_2) \\
 
&= b + (x - a_1)(x - b_2) \\
 
&= b + (x - a_1)(x - b_2) \\
 
\implies -(a_1 + a_2)x &= -(a_1 + b_2)x \\
 
\implies -(a_1 + a_2)x &= -(a_1 + b_2)x \\
 
\implies a_2 &= b_2
 
\implies a_2 &= b_2
 
\end{align*} </cmath>
 
\end{align*} </cmath>
−
which implies <math>a = b</math> which is not possible. Hence assume WLOG <math>a_1 = a_2.</math> Then note,
+
which implies <math>a = b</math> which is not possible. Hence assume WLOG <math>a_1 = a_2.</math> Then note that
 
<cmath>\begin{align*}
 
<cmath>\begin{align*}
−
P(x) &= (x-a)(x - b) \\
+
p(x) &= (x-a)(x - b) \\
 
  &= a + (x- a_1)^2 \\
 
  &= a + (x- a_1)^2 \\
−
-(a + b)x &= -2a_1x
+
-(a + b)x &= -2a_1x,
 
\end{align*} </cmath>
 
\end{align*} </cmath>
−
so,
+
so
 
<cmath>\begin{align*}
 
<cmath>\begin{align*}
−
P(x) &= a + (x - \frac{a + b}{2})^2 \\
+
p(x) &= a + \left(x - \frac{a + b}{2}\right)^2 \\
−
x^2 - (a + b)x + ab &= x^2 - (a + b)x + (a + \frac{1}{4}(a^2 + 2ab + b^2)) \\
+
x^2 - (a + b)x + ab &= x^2 - (a + b)x + \left(a + \frac{1}{4}(a^2 + 2ab + b^2)\right) \\
 
0 &= a^2 - 2ab + b^2 + 4a.
 
0 &= a^2 - 2ab + b^2 + 4a.
 
\end{align*} </cmath>
 
\end{align*} </cmath>
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0 &= 1 + 4a \\
 
0 &= 1 + 4a \\
 
a &= \frac{-1}{4}, b = \frac{3}{4} \\
 
a &= \frac{-1}{4}, b = \frac{3}{4} \\
−
\tilde{p(1)} &= (1 + \frac{1}{4})(1 - \frac{3}{4}) = \boxed{\frac{5}{16}}.
+
\tilde{p}(1) &= \left(1 + \frac{1}{4}\right)\left(1 - \frac{3}{4}\right) = \boxed{\textbf{(A) } \dfrac{5}{16}}.
 
\end{align*} </cmath>
 
\end{align*} </cmath>
 +
~Aaryabhatta1
  
−
~Aaryabhatta1
+
==Solution 10 (Solution 8 but Simpler)==
 +
 
 +
Let <math>p(x) = (x-r)(x-s)</math>, having roots <math>r, s</math>. WLOG, set <math>r<s</math>. The graph of <math>p(x)</math> will have a vertex at <math>(\frac{r+s}{2}, p(\frac{r+s}{2}))</math>. Since <math>p(p(x)) = 0 \iff p(x) = r </math> or <math>p(x) = s</math> has exactly 3 distinct solutions, we know that <math>r = p(\frac{r+s}{2})</math> and <math>s > p(\frac{r+s}{2})</math> providing 1 and 2 solutions respectively for a total of 3 (if both <math>r,s > p(\frac{r+s}{2})</math> there will be 4 solutions and if either <math>r,s<p(\frac{r+s}{2})</math> there will be at most 2 solutions). Then,
 +
<cmath>r = p(\frac{r+s}{2}) = \left(\frac{s-r}{2}\right)\left(\frac{r-s}{2}\right) = -\frac{(s-r)^2}{4}</cmath>
 +
<cmath>(s-r)^2 = -4r</cmath>
 +
<cmath>s+r = 2r\pm\sqrt{-4r}</cmath>
 +
The negative case clearly doesn't work (giving max <math>s+r=0</math> where <math>r,s=0</math>), leaving us with <math>s+r = 2r+\sqrt{-4r}</math>. Finding the derivative we see that <math>s+r</math> is maximized when <math>r = -\frac{1}{4}</math>, which gives <math>s = \frac{3}{4}</math>. Plugging these back into <math>p(x)</math> we see that <math>\tilde{p}(x)=(x+\frac{1}{4})(x-\frac{3}{4})</math>. Then <math>\tilde{p}(1) = \boxed{\textbf{(A) } \dfrac{5}{16}}</math>.
 +
 
 +
~Chupdogs
  
 
==Video Solution==
 
==Video Solution==

Latest revision as of 09:23, 16 July 2024

The following problem is from both the 2021 Fall AMC 10A #25 and 2021 Fall AMC 12A #23, so both problems redirect to this page.

Problem

A quadratic polynomial with real coefficients and leading coefficient $1$ is called $\emph{disrespectful}$ if the equation $p(p(x))=0$ is satisfied by exactly three real numbers. Among all the disrespectful quadratic polynomials, there is a unique such polynomial $\tilde{p}(x)$ for which the sum of the roots is maximized. What is $\tilde{p}(1)$?

$\textbf{(A) } \dfrac{5}{16} \qquad\textbf{(B) } \dfrac{1}{2} \qquad\textbf{(C) } \dfrac{5}{8} \qquad\textbf{(D) } 1 \qquad\textbf{(E) } \dfrac{9}{8}$

Solution 1 (Standard Form)

Let $r_1$ and $r_2$ be the roots of $\tilde{p}(x)$. Then, $\tilde{p}(x)=(x-r_1)(x-r_2)=x^2-(r_1+r_2)x+r_1r_2$. The solutions to $\tilde{p}(\tilde{p}(x))=0$ is the union of the solutions to \[\tilde{p}(x)-r_1=x^2-(r_1+r_2)x+(r_1r_2-r_1)=0\] and \[\tilde{p}(x)-r_2=x^2-(r_1+r_2)x+(r_1r_2-r_2)=0.\] Note that one of these two quadratics has one solution (a double root) and the other has two as there are exactly three solutions. WLOG, assume that the quadratic with one root is $x^2-(r_1+r_2)x+(r_1r_2-r_1)=0$. Then, the discriminant is $0$, so $(r_1+r_2)^2 = 4r_1r_2 - 4r_1$. Thus, $r_1-r_2=\pm 2\sqrt{-r_1}$, but for $x^2-(r_1+r_2)x+(r_1r_2-r_2)=0$ to have two solutions, it must be the case that $r_1-r_2=- 2\sqrt{-r_1} (*)$. It follows that the sum of the roots of $\tilde{p}(x)$ is $2r_1 + 2\sqrt{-r_1}$, whose maximum value occurs when $r_1 = - \frac{1}{4} (\star)$. Solving for $r_2$ yields $r_2 = \frac{3}{4}$. Therefore, $\tilde{p}(x)=x^2 - \frac{1}{2} x - \frac{3}{16}$, so $\tilde{p}(1)= \boxed{\textbf{(A) } \frac{5}{16}}$.

Remarks

  • For $x^2-(r_1+r_2)x+(r_1r_2-r_2)=0$ to have two solutions, the discriminant $(r_1+r_2)^2-4r_1r_2+4r_2$ must be positive. From here, we get that $(r_1-r_2)^2>-4r_2$, so $-4r_1>-4r_2 \implies r_1<r_2$. Hence, $r_1-r_2$ is negative, so $r_1-r_2=-2\sqrt{-r_1}$.
  • Set $\sqrt{-r_1}=x$. Now $r_1+\sqrt{-r_1}=-x^2+x$, for which the maximum occurs when $x=\frac{1}{2} \rightarrow r_1=-\frac{1}{4}$.

~ Leo.Euler

Solution 2 (Vertex Form)

Let $p(x)=(x-h)^2+k$ for some real constants $h$ and $k.$ Suppose that $p(x)$ has real roots $r$ and $s.$

Since $p(p(x))=0,$ we conclude that $p(x)=r$ or $p(x)=s.$ Without loss of generality, we assume that $p(x)=r$ has two real solutions and $p(x)=s$ has one real solution. Therefore, we have $k=s,$ from which $p(x)=(x-h)^2+s.$

As $p(s)=0,$ we expand the left side to obtain $(s-h)^2+s=0,$ or \[s^2-(2h-1)s+h^2=0. \hspace{15mm}(\bigstar)\] Since $(\bigstar)$ has real solutions for $s,$ the discriminant is nonnegative: $(2h-1)^2-4h^2\geq0.$ We solve this inequality to get $h\leq \frac14.$

Either by the axis of symmetry or Vieta's Formulas, note that $r+s=2h.$ As we wish to maximize $2h,$ we maximize $h.$ Substituting $h=\frac14$ into $(\bigstar),$ we obtain $s^2+\frac12s+\frac{1}{16}=0.$ We factor the left side to get $\left(s+\frac14\right)^2=0,$ or $s=-\frac14.$

Finally, the unique such polynomial is \[\tilde{p}(x)=\left(x-\frac14\right)^2-\frac14,\] from which $\tilde{p}(1)=\boxed{\textbf{(A) } \frac{5}{16}}.$

~MRENTHUSIASM

Solution 2.5 (Vertex Form but Done Differently)

Continuing from the previous solution's reasoning, we know that the $y$-coordinate of the vertex must also be a root of the polynomial. Thus, $\tilde{p}(x) = \left(x - \frac{s + r}{2}\right)^2 + s$, where $r$ and $s$ are roots of the polynomial.

We also know that in order for $\tilde{p}(x) = r$ to have two solutions, $y = r$ must intersect $\tilde{p}(x)$ above its minimum/vertex. Thus, $r > s$. Let $r = s + a$. Then, $\tilde{p}(x) = \left(x - \frac {2s + a}{2}\right)^2 + s$.

Since $s$ is a root, $\tilde{p}(s) = 0$ so $\left(s - \frac{2s + a}{2}\right)^2 +s = 0 \implies \frac{a^2}{4} + s = 0 \implies s = - \frac{1}{4} a^2$. We wish to maximize $s + s + a = 2s + a = - \frac{1}{2} a^2 + a$.

This quadratic is maximized at $a = 1$ (when converting the quadratic to vertex form). Then, since $\tilde{p}(s + a) = 0 \implies \tilde{p}(s+1) = 0$, $\left(s + 1 - s - \frac12\right)^2 + s = 0 \implies \frac{1}{4} + s = 0 \implies s = -\frac{1}{4}$.

Thus, $\tilde{p}(x) = \left(x - \frac{1}{4}\right)^2 - \frac{1}{4}$. So, $\tilde{p}(1) = \boxed{\textbf{(A) } \frac{5}{16}}$.

~CrazyVideoGamez

Solution 3 (Symmetry)

Let $\tilde p(x)=(x-h)^2+k$. We seek to maximize the sum of the roots $2h$, so we maximize $h$.

Let $P(x)=\tilde p(\tilde p(x))=((x-h)^2+k-h)^2+k$. Note that $P(x)$ is symmetric about $x=h$.

$P(x)=0$ has $3$ real solutions, Due to the complex conjugate theorem, $P(x$) must have $4$ real roots. Therefore, $P(x)$ must have exactly $1$ double root.

This root cannot be to the left or to the right of $x=h$, as the symmetry of the function would mean that there would be another double root reflected across the $x=h$. It follows that the double root could only be situated at $x=h$.

$0=P(h)=((h-h)^2+k-h)^2+k=(k-h)^2+k$. Expanding and writing this out in terms of k, $k^2+(1-2h)k+h^2=0$.

In order for this to have a solution, the discriminant has to be non-negative. In other words, $(1-2h)^2-4h^2\geq0$.

This simplifies to $1-4h\geq0$, or $h\leq\frac{1}{4}$.

As we seek to maximize $h$, we set $h=\frac{1}{4}$ and see that $k=-\frac{1}{4}$.

Therefore, $\tilde p(x)=\left(x-\frac{1}{4}\right)^2-\frac{1}{4}$, and $\tilde p(1)=\left(1-\frac{1}{4}\right)^2-\frac{1}{4}=\frac{9}{16}-\frac{1}{4}=\boxed{\textbf{(A) } \frac{5}{16}}$.

~ConcaveTriangle

Solution 4 (Discriminant)

Equation $p \left( p \left( x \right) \right) = 0$ is equivalent to the following system of equations: \begin{align*} p \left( u \right) &= 0, \\ p \left( x \right) - u &= 0. \end{align*}

Denote $p \left( x \right) = x^2 + p x + q$.

Denote by $r_1$ and $r_2$ two roots of $p \left( x \right) = 0$.

Because $p \left( p \left( x \right) \right) = 0$ has three real roots, we must have the properties that $r_1$ and $_2$ are real with $r_1 \neq r_2$. Without loss of generality, we assume $r_1 < r_2$.

We notice that all roots of $p \left( p \left( x \right) \right) = 0$ are the collection of all roots of $p \left( x \right) - r_1 = 0$ and all roots of $p \left( x \right) - r_2 = 0$.

Because each of these two equations is quadratic, it has two roots (may be identical). To get a total number of three roots, one equation must have two identical roots.

Because $r_1 < r_2$, equation $p \left( x \right) - r_1 = 0$ has two identical roots. Hence, the discriminant of this equation satisfies \[ p^2 - 4 \left( q - r_1 \right) = 0 . \hspace{1cm} \]

Because $r_1 < r_2$, $r_1 = \frac{- p - \sqrt{p^2 - 4 q}}{2}$.

Hence, the above equation can be written as \[ p^2 - 4 \left( q - \frac{- p - \sqrt{p^2 - 4 q}}{2} \right) = 0 . \]

This can be reorganized as \[ \left( p^2 - 4 q \right) - 2 \sqrt{p^2 - 4 q} - 2 p = 0 . \hspace{1cm} (1) \]

Define $x = \sqrt{p^2 - 4 q}$. Hence, the value of $p$ should ensure that equation $x^2 - 2 x - 2 p = 0$ has at least one real nonnegative root.

This condition can be satisfied if the discriminant of this equation is nonnegative: $\left( -2 \right)^2 - 4 \cdot 1 \cdot \left( - 2 p \right) \geq 0$. Hence, $p \geq - \frac{1}{2}$.

Now, we are ready to find $\tilde p \left( x \right)$.

Following from Vieta's formula, $r_1 + r_2 = - p$. Hence, to get $r_1 + r_2$ maximized, we need to find smallest $p$.

Because $p \geq - \frac{1}{2}$, the smallest $p$ is $-\frac{1}{2}$. Plugging this value into Equation (1), we get $\sqrt{p^2 - 4 q} = 1$. Hence, $q = - \frac{3}{16}$.

Thus, $\tilde p \left( x \right) = x^2 - \frac{1}{2} x - \frac{3}{16}$. Therefore, $\tilde p \left( 1 \right) = \boxed{\textbf{(A) } \frac{5}{16}}$.

~Steven Chen (www.professorchenedu.com)

Solution 5 (Factored Form)

The disrespectful function $p(x)$ has leading coefficient $1$, so it can be written in factored form as $(x-r)(x-s)$. Now the problem states that all $p(x)$ must satisfy $p(p(x)) = 0$. Plugging our form in, we get \[((x-r)(x-s)-r)((x-r)(x-s)-s) = 0.\] The roots of this equation are $(x-r)(x-s) = r$ and $(x-r)(x-s) = s$. By the fundamental theorem of algebra, each root must have two roots for a total of four possible values of $x$ yet the problem states that this equation is satisfied by three values of $x$. Therefore one equation must give a double root. Without loss of generality, let the equation $(x-r)(x-s) = r$ be the equation that produces the double root. Expanding gives $x^2-(r+s)x+rs-r = 0$. We know that if there is a double root to this equation, the discriminant must be equal to zero, so $(r+s)^2-4(rs-r) = 0 \implies r^2+2rs+s^2-4rs+4r = 0 \implies r^2-2rs+s^2+4r = 0$.

From here two solutions can progress.

~KingRavi

Solution 5.1 (Quadratic Formula)

We can rewrite $r^2-2rs+s^2+4r = 0$ as $(r-s)^2+4r = 0$. Let's keep our eyes on the ball; we want to find the disrespectful quadratic that maximizes the sum of the roots, which is $r+s$. Let this be equal to a new variable, $m$, so that our problem is reduced to maximizing this variable. We can rewrite our equation in terms of $m$ as $(2r-m)^2 + 4r = 0 \implies m^2- 4rm + 4r^2+4r = 0$.

This is a quadratic in $m$, so we can use the quadratic formula: \[m = \frac{4r \pm \sqrt{16r^2-4(4r^2+4r)}}{2} = 2r \pm \sqrt{-4r} = 2\left(r \pm \sqrt{-r}\right).\] It will be easier to think without the square root, so let $q = \sqrt{-r}$. We can rewrite the equation as $m = 2(-q^2 \pm q)$. We want to maximize $m$, so we take the plus value of the right-hand-side of the equation. Then, \[m=2(-q^2+q) \implies m = -2q(q-1).\] To maximize $m$, we find the vertex of the right-hand side of the equation. The vertex of $-2q(q-1)$ is the average of the roots of the equation which is $\frac{0+1}{2} = \frac{1}{2}$. This means that $r = -q^2 \implies \boldsymbol{r = -\frac{1}{4}}$ and $m = -2q(q-1) \implies m = \frac{1}{2}$. Therefore, $m-r = s \implies \boldsymbol{s = \frac{3}{4}}$.

~KingRavi

Solution 5.2 (Derivation-Rotated Conics)

We see that the equation $r^2-2rs+s^2+4r = 0$ is in the form of the general equation of a rotated conic \[Ax^2+Bxy+Cy^2+Dx+Ey+F = 0.\] Because $B^2 -4AC = (-2)^2 - 4(1)(1) = 0$, this rotated conic is a parabola.

The definition of a parabola is the locus of all points that are equidistant from a point (focus) and line (directrix). Let the focus and directrix of this particular parabola be $(a,b)$ and $y = mx+c$. Then we can try to find the general form of a rotated parabola in terms of $a, b, m,$and$c$.

The distance between two points $(x,y)$ and $(a,b)$ is $\sqrt{(x-a)^2+(y-b)^2}$. Therefore this is the distance from any point on the parabola to the focus.

The distance from a point $(x,y)$ to a line $y = mx+c \implies mx-y+c = 0$ is $\frac{|mx-y+c|}{\sqrt{m^2+1}}$.

We can set these two equal to each other and we get \[\sqrt{(x-a)^2+(y-b)^2} = \frac{|mx-y+c|}{\sqrt{m^2+1}}.\] Squaring both sides of the equation, we get \[(x-a)^2+(y-b)^2 = \frac{(mx-y+c)^2}{m^2+1}.\] Expanding both sides of the equation gives \[x^2-2ax+a^2+y^2-2by+b^2 = \frac{m^2x^2+y^2+c^2+2mxc-2yc-2mxy}{m^2+1}.\] Multiplying both sides of the equation by $m^2+1$ and rearranging gives \[x^2+2mx+m^2y^2-2((m^2+1)a + mc)x-2((m^2+1)b - c)y +(m^2+1)(a^2+b^2)-c^2.\] Now we can compare to our rotated parabola, $r^2-2rs+s^2+4r = 0$. From this, $-2 = 2m$ or $m = -1$. From here we have a system of three equations: \begin{align*} -2((m^2+1)a + mc) &= 4, \\ -2((m^2+1)b - c) &= 0, \\ (m^2+1)(a^2+b^2)-c^2 &= 0. \end{align*} Plugging in $m=-1$ we get \begin{align*} -2(2a-c) &= 4, \\ -2(2b-c) &= 0, \\ 2(a^2+b^2) - c^2 &= 0. \end{align*} Solving for the first equation, $c = 2+2a$.

Subtracting the first two equations, $-4a+4b = 4 \implies b = a+1$.

Plugging into the third equation, $2a^2+2a^2+4a+2 = c^2 \implies 4a^2+4a+2 = c^2$.

Substituting $c$ in, we get $4a^2+4a+2 = 4a^2+8a+4 \implies 4a+2 = 0 \implies a = -\frac{1}{2}$.

Now $b = a+1 = \frac{1}{2}$ and $c = 2+2a = 1$.

This means that the focus of the parabola is $\left(-\frac{1}{2}, \frac{1}{2}\right)$ and the directrix is $y = -x+1$. The maximum value of $r+s$ would lie at the vertex of the parabola, which is the midpoint of the focus and the foot of the focus at the directrix. The line that the vertex and focus lie on is perpendicular to the directrix, so it has slope $1$. It can be written as $y = x+d$ and must go through $\left(-\frac{1}{2}, \frac{1}{2}\right)$ so $d = 1$. This perpendicular line intersects the directrix, so to find the point at which this foot occurs, we set the equation of the lines equal to each other: \begin{align*} y &= x+1, \\ y &= -x+1. \end{align*} Adding, we get $2y = 2$ or $y = 1$ and $x = 0$. The vertex of the parabola is now at the midpoint of $\left(-\frac{1}{2}, \frac{1}{2}\right)$ and $(0, 1)$ which is $\left(-\frac{1}{4}, \frac{3}{4}\right)$.

Therefore, we have $\boldsymbol{r=-\frac{1}{4}}$ and $\boldsymbol{s=\frac{3}{4}}$.

~KingRavi

Solutions 5.1 and 5.2 Rejoined

Now that we know the roots of $\tilde{p}(1)$, we can plug in our equation: \[(x-r)(x-s) = \left(1-\left(-\frac{1}{4}\right)\right)\left(1-\frac{3}{4}\right) = \frac{5}{4} \cdot \frac{1}{4} = \frac{5}{16} = \boxed{\textbf{(A) } \frac{5}{16}}.\] ~KingRavi

Solution 6 (Factored Form)

Let $p(x)=(x-p)(x-q).$ Then, \[p(p(x))=((x-p)(x-q)-p)((x-p)(x-q)-q)=0,\] which means that either $(x-p)(x-q)-p=0$ or $(x-p)(x-q)-q=0$. Both of these equations are quadratics, so $p(p(x))$ has four roots, unless there's a double root.

Without loss of generality, let $(x-p)(x-q)-p$ be the expression that produces the double root, so its discriminant is zero. When expanded, \[(x-p)(x-q)-p=x^2-(p+q)x+pq-p=0.\] The value of $x$ is irrelevant to the discriminant, which is $(-(p+q))^2-4(pq-p).$ Setting this equal to zero and simplifing, this equation becomes $p^2-2pq+q^2+4p=0,$ which is a quadratic in $p$.


Now, we seek the value of $p$. The previous quadratic is equivalent to $p^2-(2q-4)p+q^2=0$. Using the quadratic formula by having $a=1, b=-(2q-4)$, and $c=q^2,$ we have \[p = \frac{2q-4\pm\sqrt{(-(2q-4))^2-4(1(q^2))}}{2(1)} = \frac{2q-4\pm4\sqrt{1-q}}{2}=q-2+2\sqrt{1-q}.\] Our focus is on maximizing $p+q$, so we need the maximum values of $p$ and $q$ respectively (by taking the positive square root). Adding $q,$ we see that \[p+q=2\left(q-1+\sqrt{1-q}\right).\] Let $k=\sqrt{1-q}.$ Then, $-k^2=q-1,$ so $p+q=2(-k^2+k).$


The graph of $-k^2+k$ is a parabola that opens up downwards, and has its maximum value at $y$-value of the vertex. The $x$ coordinate of the vertex is the average of the two roots, which in this case, are $0$ and $1$, so the average of these two is $\frac{1}{2}$. This means that the maximum value of $-k^2+k$ occurs at $k=\frac{1}{2}.$


Substituting $k=\sqrt{1-q},$ we see that $q=\frac{3}{4}.$ Since $p=q-2+2\sqrt{1-q},$ we can plug $q=\frac{3}{4}$ in, to see that $p=-\frac{1}{4}.$


Because the roots of this quadratic are $\frac{3}{4}$ and $-\frac{1}{4},$ our quadratic is $\left(x-\frac{3}{4}\right)\left(x+\frac{1}{4}\right).$ $\tilde{p}(1)=\frac{5}{16}$, so the answer is $\boxed{\textbf{(A) } \frac{5}{16}}$.

~Benedict T (countmath1)

Solution 7 (Vieta’s Formulas)

Since $\tilde{p}(x)$ is a disrespectful polynomial, it must satisfy $\tilde{p}(\tilde{p}(x))$. Let $\tilde{p}(x)$ have roots $x_1, x_2$. So, the equation $\tilde{p}(\tilde{p}(x))$ becomes two separate equations: \begin{align*} x^2+bx+c=x_1, \\ x^2+bx+c=x_2. \end{align*} Let $r_1,r_2,r_3,r_4$ be the roots of $\tilde{p}(\tilde{p}(x))$. Since there are $3$ real roots, then either one of them is complex, or two of them are the same. It is known that if complex number $w$ is a root to a certain polynomial, then so is $\overline{w}$. So, if one root to $\tilde{p}(\tilde{p}(x))$ is complex, then another root should also be complex — we can't have that. So, two roots are equal; WLOG, assume it is $r_3=r_4$. That means the roots to $\tilde{p}(\tilde{p}(x))$ are $r_1,r_2,r_3,r_3$.

Now, if the roots of $x^2+bx+c=x_1$ are $r_1$ and $r_3$ and the roots of $x^2+bx+c=x_2$ are $r_2$ and $r_3$, then we have $-b = r_1+r_3 = r_2+r_3$ giving $r_1=r_2$ by Vieta's Formula. That can't happen either. So, $x^2+bx+c=x_1$ must have roots $r_1, r_2$ and $x^2+bx+c=x_2$ must have a double root $r_3$. Again by Vieta's, we have \begin{align*} r_1+r_2&=-b, \\ 2r_3&=-b. \end{align*} That implies $r_2-r_3=r_3-r_1$. So, let $r_1=r_3-k$ and $r_2=r_3+k$. Also, from $2r_3=-b$, we have $r_3=-\frac{b}{2}$. By Vieta's again, we have \begin{align*} r_1\cdot r_2 &= c-x_1 = r_1\cdot r_2 = (r_3-k)(r_3-k) = r_3^2-k^2, \\ r_3^2&=c-x_2. \end{align*} Subtracting yields $x_1-x_2=k^2$. Adding that with $x_1+x_2=-b$ brings us to \begin{align*} x_1 &= \frac{k^2-b}{2}, \\ x_2 &= \frac{-k^2-b}{2}. \end{align*} Thus, $c=x_1\cdot x_2=\frac{b^2-k^4}{4}$. Now, $c-x_2 = r_3^2$ by Vieta's. Plugging in $c, x_2, r_3=-\frac{b}{2}$ gives \[\frac{b^2-k^4}{4} + \frac{k^2+b}{2} = \frac{b^2-k^4+2k^2-2b}{4} = \frac{b^2}{4}.\] Simplifying yields $2b=k^4-2k^2=(k^2-1)^2-1 \ge -1$ by Trivial Inequality. Thus, $\text{min}(b) = -\frac{1}{2}$. Since $\tilde{p}(x)$ have the maximum value to the sum of the roots, which is $-b$, we want to minimize $b$. That means that $b$ for $\tilde{p}(x)$ is $\frac{1}{2}$. That means $k^2=1$ and $c=-\frac{3}{16}$, and thus giving us $\tilde{p}(1) = 1+b+c = 1-\frac{1}{2}-\frac{3}{16} = \boxed{\textbf{(A) } \frac{5}{16}}$.

~sml1809

Solution 8 (Calculus)

Let $p(x)=(x-r_1)(x-r_2)$. Then \[p(p(x)) = ((x-r_1)(x-r_2)-r_1)((x-r_1)(x-r_2)-r_2)\] As noted in other solutions, in order for $p(p(x))$ to have three (distinct) real roots, one of the roots must be repeated twice.

For a general polynomial $q(x)$ with an $n$-times repeated root $s$, write $q(x)=(x-s)^n u(x)$, where $u(x)$ is another polynomial. Taking the derivative, \begin{align*} q'(x)&= n(x-s)^{n-1} u(x) + (x-s)^n u'(x) \\ &= (x-s)^{n-1} \left[nu(x) + (x-s)u'(x)\right] \end{align*} This shows that a repeated root / linear factor will appear in the derivative as well.

Returning to the problem, consider the derivative of $p(p(x))$, which is \[[p(p(x))]' = p'(p(x))p'(x) = \left[2(x-r_1)(x-r_2)-r_1-r_2\right] \left[2x-r_1-r_2\right]\] By the observation above, the repeated root of $p(p(x))$ must be a root of this derivative. Therefore, a search through its roots will eventually give information to resolve $r_1$ and $r_2$. Setting the derivative equal to zero yields the candidate equations \[2(x-r_1)(x-r_2)-r_1-r_2 = 0 \hspace{.5cm}\text{or}\hspace{.5cm} 2x-r_1-r_2 = 0\] Case 1: Starting with the first equation, if $x_0$ is a solution, then rearrange to obtain \[(x_0-r_1)(x_0-r_2) = \frac{r_1+r_2}{2}\] Then \begin{align*} p(p(x_0)) &= ((x_0-r_1)(x_0-r_2)-r_1)((x_0-r_1)(x_0-r_2)-r_2) \\ &= \left(\frac{r_1+r_2}{2} - r_1\right)\left(\frac{r_1+r_2}{2} - r_2\right) \\ &= \frac{1}{4}(r_1-r_2)^2 = 0 \end{align*} This shows that $r_1=r_2$. However, this gives $p(x) = (x-r_1)^2$ and $p(p(x)) = \left[(x-r_1)^2-r_1\right]^2$, which only has at most two distinct real roots. This contradicts the three-root condition given by the problem.

Case 2: The failure of Case 1 shows that the second equation must contain the repeated root. The solution for the second equation is readily $x_0 = (r_1+r_2)/2$. Then, \[p(p(x_0)) = \left[\left(\frac{r_1+r_1}{2}-r_1\right)\left(\frac{r_1+r_2}{2}-r_2\right)-r_1\right]\left[\left(\frac{r_1+r_2}{2}-r_1\right)\left(\frac{r_1+r_2}{2}-r_2\right)-r_2\right] = 0\] Solving for $r_1$ yields four candidate equations: \[r_1 = r_2\pm 2\sqrt{-r_2} \hspace{.5cm}\text{or}\hspace{.5cm} r_1 = r_2\pm 2\sqrt{1-r_2} - 2\] Since it is desired to maximise the sum of roots, add $r_2$ to both sides to consider \[r_1+r_2 = 2r_2\pm 2\sqrt{-r_2} \hspace{.5cm}\text{or}\hspace{.5cm} r_1+r_2 = 2r_2\pm 2\sqrt{1-r_2} - 2\] For the first family of solutions, the "plus" version is immediately larger, so maximise \[f(r_2) = 2r_2 + 2\sqrt{-r_2}\] Taking the derivative and setting equal to zero, \[f'(r_2) = 2 - \frac{1}{\sqrt{-r_2}} = 0\] yielding $r_2=-1/4$ and $f(-1/4)=1/2$, which can be verified to be a maximum. Since $f(r_2)=r_1+r_2$, $r_1=3/4$. A similar calculation on the second family of solutions yields the complementary version $r_1=-1/4$ and $r_2=3/4$.

In summary, this calculation shows that $\tilde{p}(x)=\left(x+\frac{1}{4}\right)\left(x-\frac{3}{4}\right)$. Finally, $\tilde{p}(1)=\left(1+\frac{1}{4}\right)\left(1-\frac{3}{4}\right)=\boxed{\textbf{(A) } \frac{5}{16}}$.

~BinaryField

Solution 9 (Lagrange Multipliers)

First note that $p(x)$ must have two distinct roots, else $p(p(x))$ will have fewer than $3$ roots. Let these be $a,b.$ Since $p(x)$ has $3$ real roots, and complex roots come in pairs, there must be a double root. So there must be $a_1, a_2$ s.t $p(a_1) = p(a_2) = a$ and $b_1, b_2$ s.t $p(b_1) = p(b_2) = b$ but two of $a_1, a_2, b_1, b_2$ must be the same. WLOG either $a_1 = a_2$ or $a_1 = b_1$. In the latter, case, then, \begin{align*} p(x) &= a + (x - a_1)(x - a_2) \\ p(x) &= b + (x - b_1)(x-b_2) \\ &= b + (x - a_1)(x - b_2) \\ \implies -(a_1 + a_2)x &= -(a_1 + b_2)x \\ \implies a_2 &= b_2 \end{align*} which implies $a = b$ which is not possible. Hence assume WLOG $a_1 = a_2.$ Then note that \begin{align*} p(x) &= (x-a)(x - b) \\ &= a + (x- a_1)^2 \\ -(a + b)x &= -2a_1x, \end{align*} so \begin{align*} p(x) &= a + \left(x - \frac{a + b}{2}\right)^2 \\ x^2 - (a + b)x + ab &= x^2 - (a + b)x + \left(a + \frac{1}{4}(a^2 + 2ab + b^2)\right) \\ 0 &= a^2 - 2ab + b^2 + 4a. \end{align*} The problem wants us to maximize the sum of the roots, $a + b.$ Thus define $f(a,b) = a + b, g(a,b) = a^2 - 2ab + b^2 + 4a.$ By the method of Lagrange Multipliers, we need to solve, \begin{align*} \vec{\nabla} f &= \lambda \vec{\nabla} g \\ g &= 0. \end{align*} In the first equation, note, \begin{align*} \vec{\nabla} f &= \lambda \vec{\nabla} g \\ \langle 1, 1 \rangle &= \lambda \langle 2a - 2b + 4, -2a + 2b \rangle \\ 2a - 2b + 4 &= -2a + 2b \\ a - b &= -1. \end{align*} Substituting into the second equation, \begin{align*} g(a,b) &= (a-b)^2 + 4a \\ 0 &= 1 + 4a \\ a &= \frac{-1}{4}, b = \frac{3}{4} \\ \tilde{p}(1) &= \left(1 + \frac{1}{4}\right)\left(1 - \frac{3}{4}\right) = \boxed{\textbf{(A) } \dfrac{5}{16}}. \end{align*} ~Aaryabhatta1

Solution 10 (Solution 8 but Simpler)

Let $p(x) = (x-r)(x-s)$, having roots $r, s$. WLOG, set $r<s$. The graph of $p(x)$ will have a vertex at $(\frac{r+s}{2}, p(\frac{r+s}{2}))$. Since $p(p(x)) = 0 \iff p(x) = r$ or $p(x) = s$ has exactly 3 distinct solutions, we know that $r = p(\frac{r+s}{2})$ and $s > p(\frac{r+s}{2})$ providing 1 and 2 solutions respectively for a total of 3 (if both $r,s > p(\frac{r+s}{2})$ there will be 4 solutions and if either $r,s<p(\frac{r+s}{2})$ there will be at most 2 solutions). Then, \[r = p(\frac{r+s}{2}) = \left(\frac{s-r}{2}\right)\left(\frac{r-s}{2}\right) = -\frac{(s-r)^2}{4}\] \[(s-r)^2 = -4r\] \[s+r = 2r\pm\sqrt{-4r}\] The negative case clearly doesn't work (giving max $s+r=0$ where $r,s=0$), leaving us with $s+r = 2r+\sqrt{-4r}$. Finding the derivative we see that $s+r$ is maximized when $r = -\frac{1}{4}$, which gives $s = \frac{3}{4}$. Plugging these back into $p(x)$ we see that $\tilde{p}(x)=(x+\frac{1}{4})(x-\frac{3}{4})$. Then $\tilde{p}(1) = \boxed{\textbf{(A) } \dfrac{5}{16}}$.

~Chupdogs

Video Solution

https://youtu.be/byo2vnCi6tk

~MathProblemSolvingSkills.com

Video Solution by SpreadTheMathLove

https://www.youtube.com/watch?v=fUcBjeP4XQ0

See Also

2021 Fall AMC 12A (Problems • Answer Key • Resources)
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All AMC 12 Problems and Solutions
2021 Fall AMC 10A (Problems • Answer Key • Resources)
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Problem 24 		Followed by
Last Problem
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