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

(→‎Solution 2.2 (Derivation-Rotated Conics))
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<math>\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}</math>
 
<math>\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}</math>
  
−
== Solution 1==
+
== Solution 1 (Factored Form)==
 
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 <math>x^2-(r_1+r_2)x+(r_1r_2-r_1)=0</math> and <math>x^2-(r_1+r_2)x+(r_1r_2-r_2)=0</math>. 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 <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>.
 
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 <math>x^2-(r_1+r_2)x+(r_1r_2-r_1)=0</math> and <math>x^2-(r_1+r_2)x+(r_1r_2-r_2)=0</math>. 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 <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>.
  
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~ Leo.Euler
 
~ Leo.Euler
  
−
==Solution 2 (Factored form)==
+
==Solution 2 (Factored Form)==
  
 
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 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>  
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From here two solutions can progress.
 
From here two solutions can progress.
  
−
==Solution 2.1 (Fastest)==
+
===Solution 2.1 (Fastest)===
  
 
We can rewrite <math>r^2-2rs+s^2+4r = 0</math> as <math>(r-s)^2+4r = 0</math>. Let's keep our eyes on the ball; we want to find the disrespectful quadratic that maximizes the sum of the roots, which is <math>r+s</math>. Let this be equal to a new variable, <math>m</math>, so that our problem is reduced to maximizing this variable.  
 
We can rewrite <math>r^2-2rs+s^2+4r = 0</math> as <math>(r-s)^2+4r = 0</math>. Let's keep our eyes on the ball; we want to find the disrespectful quadratic that maximizes the sum of the roots, which is <math>r+s</math>. Let this be equal to a new variable, <math>m</math>, so that our problem is reduced to maximizing this variable.  
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<math>m = -2q(q-1) \implies m = \frac{1}{2}. m-r = s \implies \boxed{s = \frac{3}{4}}</math>.
 
<math>m = -2q(q-1) \implies m = \frac{1}{2}. m-r = s \implies \boxed{s = \frac{3}{4}}</math>.
  
−
==Solution 2.2 (Derivation-Rotated Conics)==
+
===Solution 2.2 (Derivation-Rotated Conics)===
  
 
We see that the equation <math>r^2-2rs+s^2+4r = 0</math> is in the form of the general equation of a rotated conic - <math>Ax^2+Bxy+Cy^2+Dx+Ey+F = 0</math>. Because <math>B^2 -4AC = (-2)^2 - 4(1)(1) = 0</math>, this rotated conic is a parabola.  
 
We see that the equation <math>r^2-2rs+s^2+4r = 0</math> is in the form of the general equation of a rotated conic - <math>Ax^2+Bxy+Cy^2+Dx+Ey+F = 0</math>. Because <math>B^2 -4AC = (-2)^2 - 4(1)(1) = 0</math>, this rotated conic is a parabola.  
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~KingRavi
 
~KingRavi
 +
 +
== Solution 3 (Vertex Form) ==
 +
 +
~MRENTHUSIASM

Revision as of 22:30, 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 (Factored 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 $x^2-(r_1+r_2)x+(r_1r_2-r_1)=0$ 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 $r = -q^2$, $\boxed{r = -\frac{1}{4}}$. $m = -2q(q-1) \implies m = \frac{1}{2}. m-r = s \implies \boxed{s = \frac{3}{4}}$.

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 \[x^2+2mx+m^2y^2-2((m^2+1)*a + mc)x-2((m^2+1)*b - c)y +(m^2+1)(a^2+b^2)-c^2\]


Solution in progress

~KingRavi

Solution 3 (Vertex Form)

~MRENTHUSIASM

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