Difference between revisions of "2021 Fall AMC 10A Problems/Problem 25"
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− | Now that we know the roots of <math>\tilde{p}(1)</math>, we can plug in our equation: <math>(x-r)(x-s) = (1-(-\frac{1}{4}))(1-\frac{3}{4}) = \frac{5}{4} \cdot \frac{1}{4} = \frac{5}{16} = \boxed{ | + | Now that we know the roots of <math>\tilde{p}(1)</math>, we can plug in our equation: <math>(x-r)(x-s) = (1-(-\frac{1}{4}))(1-\frac{3}{4}) = \frac{5}{4} \cdot \frac{1}{4} = \frac{5}{16} = \boxed{A}</math> |
~KingRavi | ~KingRavi |
Revision as of 23:53, 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 (Factored Form)
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
, 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
.
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 1. 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
and
are
and
, respectively.
Solution 2 (rejoined)
Now that we know the roots of , we can plug in our equation:
~KingRavi
Solution 3 (Vertex Form)
~MRENTHUSIASM
See Also
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.