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Difference between revisions of "2015 AMC 12B Problems/Problem 12"

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Therefore, reconsidering the polynomial of the problem, the sum of the roots is <cmath>\dfrac{a+2b+c}{2}.</cmath> Now, to maximize this, it is clear that <math>b=9.</math> Also, we must have <math>a=8, c=7</math> (or vice versa). The reason <math>a,c</math> have to equal these values instead of larger values is because each of <math>a,b,c</math> is distinct.
 
Therefore, reconsidering the polynomial of the problem, the sum of the roots is <cmath>\dfrac{a+2b+c}{2}.</cmath> Now, to maximize this, it is clear that <math>b=9.</math> Also, we must have <math>a=8, c=7</math> (or vice versa). The reason <math>a,c</math> have to equal these values instead of larger values is because each of <math>a,b,c</math> is distinct.
 +
 +
==Solution 3 (Vieta's Formula)==
 +
Expanding the formula gives: <math>2x^2-(a+2b+c)x+(ab+bc).</math>
 +
 +
Dividing by the coefficient of the polynomial, 2, and applying Vieta's formulas we know that the sum of the roots of the polynomial is equal to <math>\frac{a + 2b + c}{2}</math>. Obviously, the largest value should occur when b = 9, and due to symmetrically a = 8, and c = 7, or a = 7 and c = 8. Both of these results give the same final answer of (D).
 +
 +
More on Vieta's Formulas: https://artofproblemsolving.com/wiki/index.php/Vieta%27s_formulas
 +
 +
~PeterDoesPhysics
  
 
==Solution 3 (Vieta's Formula)==
 
==Solution 3 (Vieta's Formula)==

Revision as of 00:02, 25 August 2024

Problem

Let $a$, $b$, and $c$ be three distinct one-digit numbers. What is the maximum value of the sum of the roots of the equation $(x-a)(x-b)+(x-b)(x-c)=0$ ?

$\textbf{(A)}\; 15 \qquad\textbf{(B)}\; 15.5 \qquad\textbf{(C)}\; 16 \qquad\textbf{(D)} 16.5 \qquad\textbf{(E)}\; 17$

Video Solution

https://youtu.be/ba6w1OhXqOQ?t=423

~ pi_is_3.14

Solution 1

The left-hand side of the equation can be factored as $(x-b)(x-a+x-c) = (x-b)(2x-(a+c))$, from which it follows that the roots of the equation are $x=b$, and $x=\tfrac{a+c}{2}$. The sum of the roots is therefore $b + \tfrac{a+c}{2}$, and the maximum is achieved by choosing $b=9$, and $\{a,c\}=\{7,8\}$. Therefore the answer is $9 + \tfrac{7+8}{2} = 9 + 7.5 = \boxed{\textbf{(D)}\; 16.5}.$

Solution 2

Expand the polynomial. We get $(x-a)(x-b)+(x-b)(x-c)=x^2-(a+b)x+ab+x^2-(b+c)x+bc=2x^2-(a+2b+c)x+(ab+bc).$

Now, consider a general quadratic equation $ax^2+bx+c=0.$ The two solutions to this are \[\dfrac{-b}{2a}+\dfrac{\sqrt{b^2-4ac}}{2a}, \dfrac{-b}{2a}-\dfrac{\sqrt{b^2-4ac}}{2a}.\] The sum of these roots is \[\dfrac{-b}{a}.\]

Therefore, reconsidering the polynomial of the problem, the sum of the roots is \[\dfrac{a+2b+c}{2}.\] Now, to maximize this, it is clear that $b=9.$ Also, we must have $a=8, c=7$ (or vice versa). The reason $a,c$ have to equal these values instead of larger values is because each of $a,b,c$ is distinct.

Solution 3 (Vieta's Formula)

Expanding the formula gives: $2x^2-(a+2b+c)x+(ab+bc).$

Dividing by the coefficient of the polynomial, 2, and applying Vieta's formulas we know that the sum of the roots of the polynomial is equal to $\frac{a + 2b + c}{2}$. Obviously, the largest value should occur when b = 9, and due to symmetrically a = 8, and c = 7, or a = 7 and c = 8. Both of these results give the same final answer of (D).

More on Vieta's Formulas: https://artofproblemsolving.com/wiki/index.php/Vieta%27s_formulas

~PeterDoesPhysics

Solution 3 (Vieta's Formula)

Expanding the formula gives: $2x^2-(a+2b+c)x+(ab+bc).$

Dividing by the coefficient of the polynomial, 2, and applying Vieta's formulas we know that the sum of the roots of the polynomial is equal to $\frac{a + 2b + c}{2}$. Obviously, the largest value should occur when b = 9, and due to symmetrically a = 8, and c = 7, or a = 7 and c = 8. Both of these results give the same final answer of (D).

More on Vieta's Formulas: https://artofproblemsolving.com/wiki/index.php/Vieta%27s_formulas

~PeterDoesPhysics

See Also

2015 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 11 		Followed by
Problem 13
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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