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Difference between revisions of "2021 Fall AMC 10A Problems/Problem 25"
(→‎Solution 2 (Do Not Do On The Test - Rotated Conics))
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~ Leo.Euler
 
~ Leo.Euler
  
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==Solution 2 (Do Not Do On The Test - Rotated Conics)==
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==Solution 2 (Factored form)==
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The disrespectful function <math>p(x)</math> has leading coefficient 1, so it can be written in factored form as <math>(x-r)(x-s)</math>. Now the problem states that all <math>p(x)</math> must satisfy <math>p(p(x)) = 0</math>. Plugging our form in, we get: <cmath> ((x-r)(x-s)-r)((x-r)(x-s)-s) = 0 </cmath>.
 +
The roots of this equation are <math>(x-r)(x-s) = r, (x-r)(x-s) = s</math>. 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 <math>(x-r)(x-s) = r</math> be the equation that produces the double root. Expanding gives <math>x^2-(r+s)x+rs-r = 0</math>. We know that if there is a double root to this equation, the discriminant must be equal to zero, so <math>(r+s)^2-4(rs-r) = 0 \implies r^2+2rs+s^2-4rs+4r = 0 \implies r^2-2rs+s^2+4r = 0</math>.
  
 
Solution in progress
 
Solution in progress
  
 
~KingRavi
 
~KingRavi

Revision as of 20:43, 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, $\tilde{p}(x)=x^2-(r_1+r_2)x+r_1r_2$. 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 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, $r_1-r_2=- 2\sqrt{-r_1}$. It follows that the sum of the roots of $\tilde{p}(x)$ is $2r_1 + 2\sqrt{-r_1}$, and its maximum value occurs when $r_1 = - \frac{1}{4}$. Therefore, $\tilde{p}(x)=x^2 - \frac{1}{2} x - \frac{3}{16}$, so $\tilde{p}(1)= \boxed{\textbf{(A) } \frac{5}{16}}$.

~ 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 $(r+s)^2-4(rs-r) = 0 \implies r^2+2rs+s^2-4rs+4r = 0 \implies r^2-2rs+s^2+4r = 0$.

Solution in progress

~KingRavi

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