Difference between revisions of "2021 Fall AMC 10A Problems/Problem 25"
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<cmath>m=2(-q^2+q) \implies m = 2\cdot-q(q-1) \implies m = -2q(q-1)</cmath> | <cmath>m=2(-q^2+q) \implies m = 2\cdot-q(q-1) \implies m = -2q(q-1)</cmath> | ||
− | 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>r = -\frac{1}{4}</math>. | + | 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>. |
− | <math>m = -2q(q-1) \implies m = \frac{1}{2}. m-r = s \implies 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)== |
Revision as of 22:47, 22 November 2021
Contents
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 (a double root) 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 must be the case that
*. It follows that the sum of the roots of
is
, whose maximum value occurs when
. Solving for
yields
. Therefore,
, so
.
For
to have two solutions, the discriminant
must be positive. From here, we get that
, so
. Hence,
is negative, 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
.
From here two solutions can progress.
Solution 2.1 (Fastest)
We can rewrite as
. Let's keep our eyes on the ball; we want to find the disrespectful quadratic that maximizes the sum of the roots, which is
. Let this be equal to a new variable,
, so that our problem is reduced to maximizing this variable.
We can rewrite our equation in terms of m as
.
This is a quadratic in m, so we can use the quadratic formula:
It'll be easier to think without the square root, so let : then we can rewrite the equation as
. We want to maximize m, so we take the plus value of the right-hand-side of the equation. Then;
To maximize m, we find the vertex of the right-hand side of the equation. The vertex of is the average of the roots of the equation which is
. This means that since
,
.
.
Solution 2.2 (Derivation-Rotated Conics)
We see that the equation is in the form of the general equation of a rotated conic -
. Because $B^2
Solution in progress
~KingRavi