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.
- 1 Problem
- 2 Solution 1 (Standard Form)
- 3 Solution 2 (Vertex Form)
- 4 Solution 3 (Symmetry)
- 5 Solution 4 (Discriminant)
- 6 Solution 5 (Factored Form)
- 7 Solution 6 (Factored Form)
- 8 Solution 7 (Vieta’s Formulas)
- 9 Solution 8 (Calculus)
- 10 Solution 9 (Lagrange Multipliers)
- 11 Video Solution
- 12 Video Solution by SpreadTheMathLove
- 13 See Also
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 (Standard Form)
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 .
- For to have two solutions, the discriminant must be positive. From here, we get that , so . Hence, is negative, so .
- Set . Now , for which the maximum occurs when .
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
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 .
Solution 4 (Discriminant)
Equation is equivalent to the following system of equations:
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, the above equation 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 (1), 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 we have 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 , so the answer is .
~Benedict T (countmath1)
Solution 7 (Vieta’s Formulas)
Since is a disrespectful polynomial, it must satisfy . Let have roots . So, the equation becomes two separate equations: Let be the roots of . Since there are real roots, then either one of them is complex, or two of them are the same. It is known that if complex number is a root to a certain polynomial, then so is . So, if one root to is complex, then another root should also be complex — we can't have that. So, two roots are equal; WLOG, assume it is . That means the roots to are .
Now, if the roots of are and and the roots of are and , then we have giving by Vieta's Formula. That can't happen either. So, must have roots and must have a double root . Again by Vieta's, we have That implies . So, let and . Also, from , we have . By Vieta's again, we have Subtracting yields . Adding that with brings us to Thus, . Now, by Vieta's. Plugging in gives Simplifying yields by Trivial Inequality. Thus, . Since have the maximum value to the sum of the roots, which is , we want to minimize . That means that for is . That means and , and thus giving us .
Solution 8 (Calculus)
Let . Then As noted in other solutions, in order for to have three (distinct) real roots, one of the roots must be repeated twice.
For a general polynomial with an -times repeated root , write , where is another polynomial. Taking the derivative, This shows that a repeated root / linear factor will appear in the derivative as well.
Returning to the problem, consider the derivative of , which is By the observation above, the repeated root of must be a root of this derivative. Therefore, a search through its roots will eventually give information to resolve and . Setting the derivative equal to zero yields the candidate equations Case 1: Starting with the first equation, if is a solution, then rearrange to obtain Then This shows that . However, this gives and , which only has at most two distinct real roots. This contradicts the three-root condition given by the problem.
Case 2: The failure of Case 1 shows that the second equation must contain the repeated root. The solution for the second equation is readily . Then, Solving for yields four candidate equations: Since it is desired to maximise the sum of roots, add to both sides to consider For the first family of solutions, the "plus" version is immediately larger, so maximise Taking the derivative and setting equal to zero, yielding and , which can be verified to be a maximum. Since , . A similar calculation on the second family of solutions yields the complementary version and .
In summary, this calculation shows that . Finally, .
Solution 9 (Lagrange Multipliers)
First note that must have two distinct roots, else will have fewer than roots. Let these be Since has real roots, and complex roots come in pairs, there must be a double root. So there must be s.t and s.t but two of must be the same. WLOG either or . In the latter, case, then, which implies which is not possible. Hence assume WLOG Then note that so The problem wants us to maximize the sum of the roots, Thus define By the method of Lagrange Multipliers, we need to solve, In the first equation, note, Substituting into the second equation, ~Aaryabhatta1
Video Solution by SpreadTheMathLove
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