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Difference between revisions of "2021 Fall AMC 10A Problems/Problem 25"

(→‎Solution 2.1 (Fastest))
(→‎Solution 2.1 (Fastest))
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<math>m = \frac{4r \pm \sqrt{16r^2-4(4r^2+4r)}}{2} = 2r \pm \sqrt{-4r} = 2(r \pm \sqrt{-r})</math>
 
<math>m = \frac{4r \pm \sqrt{16r^2-4(4r^2+4r)}}{2} = 2r \pm \sqrt{-4r} = 2(r \pm \sqrt{-r})</math>
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 +
It'll be easier to think without the square root, so let <math>q = sqrt{-r}</math>: then we can rewrite the equation as <math>m = 2(-q^2 \pm q)</math>. We want to maximize m, so we take the plus value of the right-hand-side of the equation. Then;
 +
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<cmath>m=2(-q^2+q) \implies m = 2\cdot-q(q-1) \implies m = -2q(q-1)</cmath>
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To maximize m, we find the vertex of the right-hand side of the equation. The vertex of <math>-2q(q-1)</math> is the average of the roots of the equation which is <math>\frac{0+1}{2} = \frac{1}{2}</math>. This means that since <math>r = -q^2</math>, <math>r = -1/4</math>.
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<math>m = -2q(q-1) \implies m = 1/2. m-r = s \implies s = 3/4</math>.
  
 
==Solution 2.2 (Derivation-Rotated Conics)==
 
==Solution 2.2 (Derivation-Rotated Conics)==

Revision as of 21:45, 22 November 2021

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

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 $x^2-(r_1+r_2)x+(r_1r_2-r_1)=0$ and $x^2-(r_1+r_2)x+(r_1r_2-r_2)=0$. It follows that one of these two quadratics has one solution (a double root) and the other has two. 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 $\tilde{p}(1)= \boxed{\textbf{(A) } \frac{5}{16}}$.

$*$ 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 $-4r_1>-4r_2 \rightarrow r_1<r_2$. 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 $(2r-m)^2 + 4r = 0 \implies m- 4rm + 4r^2+4r = 0$.

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 $r = -q^2$, $r = -1/4$. $m = -2q(q-1) \implies m = 1/2. m-r = s \implies s = 3/4$.

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


Solution in progress

~KingRavi

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