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

Revision as of 15:07, 24 November 2021

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 $\tilde{p}(1)$ ?

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

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 (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 $(x-r)(x-s) = r, (x-r)(x-s) = s$ . 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 2.1 (Fastest)

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:

$m = \frac{4r \pm \sqrt{16r^2-4(4r^2+4r)}}{2} = 2r \pm \sqrt{-4r} = 2(r \pm \sqrt{-r})$

It'll be easier to think without the square root, so let $q = \sqrt{-r}$ : then 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;

$m=2(-q^2+q) \implies m = 2\cdot-q(q-1) \implies m = -2q(q-1)$

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 since $r = -q^2$ , $\boxed{r = -\frac{1}{4}}$ . $m = -2q(q-1) \implies m = \frac{1}{2}. m-r = s \implies \boxed{s = \frac{3}{4}}$ .

Solution 2.2 (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:

$-2((m^2+1)a + mc) = 4$ $-2((m^2+1)b - c) = 0$ $(m^2+1)(a^2+b^2)-c^2 = 0$

Plugging in $m=-1$ we get:

$-2(2a-c) = 4$ $-2(2b-c) = 0$ $2(a^2+b^2) - c^2 = 0$

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 $(-\frac{1}{2}, \frac{1}{2})$ 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 $(-\frac{1}{2}, \frac{1}{2})$ 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; $y = x+1$ $y = -x + 1$

Adding, we get $2y = 2$ or $y = 1$ and $x = 0$ . The vertex of the parabola is now at the midpoint of $(-\frac{1}{2}, \frac{1}{2})$ and $(0, 1)$ which is $(-\frac{1}{4}, \frac{3}{4})$ . Therefore $r$ and $s$ are $-\frac{1}{4}$ and $\frac{3}{4}$ , respectively.

Solutions 2.1 and 2.2 Rejoined

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

~KingRavi

Solution 3 (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, or $(2h-1)^2-4h^2\geq0.$ 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, we have $\tilde{p}(x)=\left(x-\frac14\right)^2-\frac14,$ from which $\tilde{p}(1)=\boxed{\textbf{(A) } \frac{5}{16}}.$

~MRENTHUSIASM

@@ Line 8: / Line 8: @@
 == Solution 1 (Vieta’s Formulas)==
 Let <math>r_1</math> and <math>r_2</math> be the roots of <math>\tilde{p}(x)</math>. Then, <math>\tilde{p}(x)=(x-r_1)(x-r_2)=x^2-(r_1+r_2)x+r_1r_2</math>. The solutions to <math>\tilde{p}(\tilde{p}(x))=0</math> is the union of the solutions to <cmath> \tilde{p}(x)-r_1=x^2-(r_1+r_2)x+(r_1r_2-r_1)=0 </cmath> and <cmath> \tilde{p}(x)-r_2=x^2-(r_1+r_2)x+(r_1r_2-r_2)=0. </cmath> 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 <math>x^2-(r_1+r_2)x+(r_1r_2-r_1)=0</math>. Then, the discriminant is <math>0</math>, so <math>(r_1+r_2)^2 = 4r_1r_2 - 4r_1</math>. Thus, <math>r_1-r_2=\pm 2\sqrt{-r_1}</math>, but for <math>x^2-(r_1+r_2)x+(r_1r_2-r_2)=0</math> to have two solutions, it must be the case that <math>r_1-r_2=- 2\sqrt{-r_1}</math> *. It follows that the sum of the roots of <math>\tilde{p}(x)</math> is <math>2r_1 + 2\sqrt{-r_1}</math>, whose maximum value occurs when <math>r_1 = - \frac{1}{4}</math>. Solving for <math>r_2</math> yields <math>r_2 = \frac{3}{4}</math>. Therefore, <math>\tilde{p}(x)=x^2 - \frac{1}{2} x - \frac{3}{16}</math>, so <math>\tilde{p}(1)= \boxed{\textbf{(A) } \frac{5}{16}}</math>.
+<u><b>Remark</b></u>
 <math>*</math> For <math>x^2-(r_1+r_2)x+(r_1r_2-r_2)=0</math> to have two solutions, the discriminant <math>(r_1+r_2)^2-4r_1r_2+4r_2</math> must be positive. From here, we get that <math>(r_1-r_2)^2>-4r_2</math>, so <math>-4r_1>-4r_2 \implies r_1<r_2</math>. Hence, <math>r_1-r_2</math> is negative, so <math>r_1-r_2=-2\sqrt{-r_1}</math>.

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

Page

Toolbox

Search