Difference between revisions of "2021 Fall AMC 12A Problems/Problem 23"
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Let <math>\tilde p(x)=(x-h)^2+k</math>. We seek to maximize <math>h</math>. | Let <math>\tilde p(x)=(x-h)^2+k</math>. We seek to maximize <math>h</math>. | ||
Revision as of 15:25, 24 November 2021
- 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
[hide]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 (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 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 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 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 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 and are and , respectively.
Solutions 2.1 and 2.2 Rejoined
Now that we know the roots of , we can plug in our equation:
~KingRavi
Solution 3 (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, or 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, we have from which ~MRENTHUSIASM
3.1 (Symmetry)
Let . We seek to maximize .
Let . Note that is symmetric about .
has 3 real solutions, Due to the complex conjugate theorem, ) must have 4 real roots. Therefore, must have exactly 1 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 h, we set and see that .
Therefore, , and - ConcaveTriangle
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.