Difference between revisions of "2021 Fall AMC 10A Problems/Problem 25"
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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^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 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{- | + | 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^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 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, <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>, and its maximum value occurs when <math>r_1 = - \frac{1}{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>. |
~ Leo.Euler | ~ Leo.Euler |
Revision as of 19:36, 22 November 2021
Problem
A quadratic polynomial with real coefficients and leading coefficient is called
if the equation
is satisfied by exactly three real numbers. Among all the disrespectful quadratic polynomials, there is a unique such polynomial
for which the sum of the roots is maximized. What is
?
Solution 1
Let and
be the roots of
. Then,
. The solutions to
is the union of the solutions to
and
. 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
. Then, the discriminant is
, so
. Thus,
, but for
to have two solutions,
. It follows that the sum of the roots of
is
, and its maximum value occurs when
. Therefore,
, so
.
~ Leo.Euler