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

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==Solution 3 (Vieta's Formula)==
 
==Solution 3 (Vieta's Formula)==
−
Expanding the formula gives: <math>2x^2-(a+2b+c)x+(ab+bc).</math>
+
Expanding 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).  
 
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).  

Latest revision as of 23:02, 24 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 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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