Difference between revisions of "2021 Fall AMC 12A Problems/Problem 23"

Revision as of 10:59, 16 April 2022

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 (Vieta’s Formulas)

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 $r_1 = - \frac{1}{4}$ . Solving for $r_2$ yields $r_2 = \frac{3}{4}$ . Therefore, $\tilde{p}(x)=x^2 - \frac{1}{2} x - \frac{3}{16}$ , so $\tilde{p}(1)= \boxed{\textbf{(A) } \frac{5}{16}}$ .

Remark

$*$ 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 $r_1-r_2=-2\sqrt{-r_1}$ .

~ 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 the 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 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 $p^2 - 4 \left( q - r_1 \right) = 0 . \hspace{1cm} (1)$

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

Hence, Equation (1) can be written as $p^2 - 4 \left( q - \frac{- p - \sqrt{p^2 - 4 q}}{2} \right) = 0 .$

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

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 (2), 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.

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, $m-r = s \implies \boldsymbol{s = \frac{3}{4}}$ .

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 $\boldsymbol{s=\frac{3}{4}}$ .

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, but different)

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 $c=q^2,$ the value of $p$ is

$\frac{2q-4\pm\sqrt{(-(2q-4))^2-4(1(q^2))}}{2(1)} \rightarrow$ $p=\frac{2q-4\pm4\sqrt{1-q}}{2} \rightarrow$ $p=q-2+2\sqrt{1-q}.$

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

$p+q=2(q-1+\sqrt{1-q}).$

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 $(x-\frac{3}{4})(x+\frac{1}{4}).$ Letting $1$ be $x$ , $\tilde{p}(1)=\frac{5}{16}$ , which is $\textbf{(A)}$ . ~Benedict T (countmath1)

@@ Line 209: / Line 209: @@
 Because the roots of this quadratic are <math>\frac{3}{4}</math> and <math>\frac{-1}{4},</math>, our quadratic is <math>(x-\frac{3}{4})(x+\frac{1}{4}).</math> Letting <math>1</math> be <math>x</math>, <math>\tilde{p}(1)=\frac{5}{16}</math>, which is <math>\textbf{(A)}</math>.
-~Benedict (countmath1)
+~Benedict T (countmath1)
 ==See Also==

2021 Fall AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 22	Followed by Problem 24
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

2021 Fall AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 24	Followed by Last Problem
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

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