2021 Fall AMC 12A Problems/Problem 23
- The following problem is from both the 2021 Fall AMC 10A #25 and 2021 Fall AMC 12A #23, so both problems redirect to this page.
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 (Vieta’s Formulas)
Let and
be the roots of
. Then,
. The solutions to
is the union of the solutions to
and
Note that one of these two quadratics has one solution (a double root) and the other has two as there are exactly three solutions. 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
.
Remark
For
to have two solutions, the discriminant
must be positive. From here, we get that
, so
. Hence,
is negative, so
.
~ Leo.Euler
Solution 2 (Vertex Form)
Let for some real constants
and
Suppose that
has real roots
and
Since we conclude that
or
Without the loss of generality, we assume that
has two real solutions and
has one real solution. Therefore, we have
from which
As we expand the left side to obtain
or
Since
has real solutions for
the discriminant is nonnegative:
We solve this inequality to get
Either by the axis of symmetry or Vieta's Formulas, note that As we wish to maximize
we maximize
Substituting
into
we obtain
We factor the left side to get
or
Finally, the unique such polynomial is from which
~MRENTHUSIASM
Solution 3 (Symmetry)
Let . We seek to maximize the sum of the roots
, so we maximize
.
Let . Note that
is symmetric about
.
has
real solutions, Due to the complex conjugate theorem,
) must have
real roots. Therefore,
must have exactly
double root.
This root cannot be to the left or to the right of , as the symmetry of the function would mean that there would be another double root reflected across the
. It follows that the double root could only be situated at
.
. Expanding and writing this out in terms of k,
.
In order for this to have a solution, the discriminant has to be non-negative. In other words, .
This simplifies to , or
.
As we seek to maximize , we set
and see that
.
Therefore, , and
.
~ConcaveTriangle
Solution 4 (Discriminant)
Equation is equivalent to the following system of equations:
Denote .
Denote by and
two roots of
.
Because has three real roots, we must have the properties that
and
are real with
.
Without loss of generality, we assume
.
We notice that all roots of
are the collection of all roots of
and all roots of
.
Because each of these two equations is quadratic, it has two roots (may be identical). To get a total number of three roots, one equation must have two identical roots.
Because , equation
has two identical roots.
Hence, the discriminant of this equation satisfies
Because ,
.
Hence, Equation (1) can be written as
This can be reorganized as
Define .
Hence, the value of
should ensure that equation
has at least one real nonnegative root.
This condition can be satisfied if the discriminant of this equation is nonnegative: .
Hence,
.
Now, we are ready to find .
Following from Vieta's formula, .
Hence, to get
maximized, we need to find smallest
.
Because , the smallest
is
.
Plugging this value into Equation (2), we get
.
Hence,
.
Thus, .
Therefore,
.
~Steven Chen (www.professorchenedu.com)
Solution 5 (Factored Form)
The disrespectful function has leading coefficient
, 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
and
. By the fundamental theorem of algebra, each root must have two roots for a total of four possible values of
yet the problem states that this equation is satisfied by three values of
. 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 5.1 (Quadratic Formula)
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
as
.
This is a quadratic in , so we can use the quadratic formula:
It will be easier to think without the square root, so let
. We can rewrite the equation as
. We want to maximize
, so we take the plus value of the right-hand-side of the equation. Then,
To maximize
, 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
and
. Therefore,
.
Solution 5.2 (Derivation-Rotated Conics)
We see that the equation is in the form of the general equation of a rotated conic
Because
, 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 and
. Then we can try to find the general form of a rotated parabola in terms of
and
.
The distance between two points and
is
. Therefore this is the distance from any point on the parabola to the focus.
The distance from a point to a line
is
.
We can set these two equal to each other and we get
Squaring both sides of the equation, we get
Expanding both sides of the equation gives
Multiplying both sides of the equation by
and rearranging gives
Now we can compare to our rotated parabola,
. From this,
or
. From here we have a system of three equations:
Plugging in
we get
Solving for the first equation,
.
Subtracting the first two equations, .
Plugging into the third equation, .
Substituting in, we get
.
Now and
.
This means that the focus of the parabola is and the directrix is
. The maximum value of
would lie at the vertex of the parabola, which is the midpoint of the focus and the foot of the focus at the directrix. The line that the vertex and focus lie on is perpendicular to the directrix, so it has slope
. It can be written as
and must go through
so
. This perpendicular line intersects the directrix, so to find the point at which this foot occurs, we set the equation of the lines equal to each other:
Adding, we get
or
and
. The vertex of the parabola is now at the midpoint of
and
which is
.
Therefore, we have and
.
Solutions 5.1 and 5.2 Rejoined
Now that we know the roots of , we can plug in our equation:
~KingRavi
Solution 6 (Factored Form)
Let Then,
, which means that either
or
. Both of these equations are quadratics, so
has four roots, unless there's a double root.
Without loss of generality, let be the expression that produces the double root, so its discriminant is zero. When expanded,
The value of is irrelevant to the discriminant, which is
Setting this equal to zero and simplifing, this equation becomes
which is a quadratic in
.
Now, we seek the value of . The previous quadratic is equivalent to
. Using the quadratic formula by having
, and
the value of
is
Our focus is on maximizing , so we need the maximum values of
and
respectively (by taking the positive square root). Adding
we see that
Let Then,
so
The graph of is a parabola that opens up downwards, and has its maximum value at
-value of the vertex. The
coordinate of the vertex is the average of the two roots, which in this case, are
and
, so the average of these two is
. This means that the maximum value of
occurs at
Substituting we see that
Since we can plug
in, to see that
Because the roots of this quadratic are and
, our quadratic is
Letting
be
,
, which is
.
~Benedict T (countmath1)
See Also
2021 Fall AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 22 |
Followed by Problem 24 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
2021 Fall AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 24 |
Followed by Last Problem | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
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