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

(Created page with "==Problem== ==Solution== ==See Also== {{AMC12 box|year=2015|ab=B|num-a=13|num-b=11}} {{MAA Notice}}")
 
(Solution 1)
 
(6 intermediate revisions by 4 users not shown)
Line 1: Line 1:
 
==Problem==
 
==Problem==
 +
Let <math>a</math>, <math>b</math>, and <math>c</math> be three distinct one-digit numbers. What is the maximum value of the sum of the roots of the equation <math>(x-a)(x-b)+(x-b)(x-c)=0</math> ?
  
 +
<math>\textbf{(A)}\; 15 \qquad\textbf{(B)}\; 15.5 \qquad\textbf{(C)}\; 16 \qquad\textbf{(D)} 16.5 \qquad\textbf{(E)}\; 17</math>
  
 +
== Video Solution ==
 +
https://youtu.be/ba6w1OhXqOQ?t=423
  
==Solution==
+
~ pi_is_3.14
  
 +
==Solution 1==
 +
The left-hand side of the equation can be factored as <math>(x-b)(x-a+x-c) = (x-b)(2x-(a+c))</math>, from which it follows that the roots of the equation are <math>x=b</math>, and <math>x=\tfrac{a+c}{2}</math>. The sum of the roots is therefore <math>b + \tfrac{a+c}{2}</math>, and the maximum is achieved by choosing <math>b=9</math>, and <math>\{a,c\}=\{7,8\}</math>. Therefore the answer is <math>9 + \tfrac{7+8}{2} = 9 + 7.5 = \boxed{\textbf{(D)}\; 16.5}.</math>
 +
 +
==Solution 2==
 +
Expand the polynomial. We get <math>(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).</math>
 +
 +
Now, consider a general quadratic equation <math>ax^2+bx+c=0.</math> The two solutions to this are <cmath>\dfrac{-b}{2a}+\dfrac{\sqrt{b^2-4ac}}{2a}, \dfrac{-b}{2a}-\dfrac{\sqrt{b^2-4ac}}{2a}.</cmath> The sum of these roots is <cmath>\dfrac{-b}{a}.</cmath>
 +
 +
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.
  
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2015|ab=B|num-a=13|num-b=11}}
 
{{AMC12 box|year=2015|ab=B|num-a=13|num-b=11}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 04:07, 15 January 2021

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.

See Also

2015 AMC 12B (ProblemsAnswer KeyResources)
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