Difference between revisions of "2021 Fall AMC 10A Problems/Problem 25"
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~ Leo.Euler | ~ Leo.Euler | ||
− | ==Solution 2 ( | + | ==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 21:43, 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
Solution 2 (Factored form)
The disrespectful function has leading coefficient 1, so it can be written in factored form as
. Now the problem states that all
must satisfy
. Plugging our form in, we get:
.
The roots of this equation are
. 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
be the equation that produces the double root. Expanding gives
. We know that if there is a double root to this equation, the discriminant must be equal to zero, so
.
Solution in progress
~KingRavi