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| − | == Problem ==
| + | #REDIRECT [[2021_Fall_AMC_12A_Problems/Problem_23]] |
| − | A quadratic polynomial with real coefficients and leading coefficient <math>1</math> is called <math>\emph{disrespectful}</math> if the equation <math>p(p(x))=0</math> is satisfied by exactly three real numbers. Among all the disrespectful quadratic polynomials, there is a unique such polynomial <math>\tilde{p}(x)</math> for which the sum of the roots is maximized. What is <math>\tilde{p}(1)</math>?
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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>
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| − | == Solution 1 (Factored Form)==
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| − | 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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| − | <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 \rightarrow 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>.
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| − | ~ Leo.Euler
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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>
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| − | 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>.
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| − | From here two solutions can progress.
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| − | ===Solution 2.1 (Fastest)===
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| − | 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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| − | We can rewrite our equation in terms of m as <math>(2r-m)^2 + 4r = 0 \implies m- 4rm + 4r^2+4r = 0 </math>.
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| − | This is a quadratic in m, so we can use the quadratic formula:
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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>
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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>\boxed{r = -\frac{1}{4}}</math>.
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| − | <math>m = -2q(q-1) \implies m = \frac{1}{2}. m-r = s \implies \boxed{s = \frac{3}{4}}</math>.
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| − | ===Solution 2.2 (Derivation-Rotated Conics)===
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| − | 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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| − | 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 <math>(a,b)</math> and <math>y = mx+c</math>. Then we can try to find the general form of a rotated parabola in terms of <math>a, b, m, </math>and<math>c</math>.
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| − | The distance between two points <math>(x,y)</math> and <math>(a,b)</math> is <math>\sqrt{(x-a)^2+(y-b)^2}</math>. Therefore this is the distance from any point on the parabola to the focus.
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| − | The distance from a point <math>(x,y)</math> to a line <math>y = mx+c \implies mx-y+c = 0</math> is <math>\frac{|mx-y+c|}{\sqrt{m^2+1}}</math>.
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| − | We can set these two equal to each other and we get: <cmath>\sqrt{(x-a)^2+(y-b)^2} = \frac{|mx-y+c|}{\sqrt{m^2+1}}</cmath>
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| − | Squaring both sides of the equation, we get <cmath>(x-a)^2+(y-b)^2 = \frac{(mx-y+c)^2}{m^2+1}</cmath>.
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| − | Expanding both sides of the equation gives <cmath>x^2-2ax+a^2+y^2-2by+b^2 = \frac{m^2x^2+y^2+c^2+2mxc-2yc-2mxy}{m^2+1}</cmath>
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| − | Multiplying both sides of the equation by <math>m^2+1</math> and rearranging gives <cmath>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</cmath>
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| − | Solution in progress
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| − | ~KingRavi
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| − | == Solution 3 (Vertex Form) ==
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| − | ~MRENTHUSIASM
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