Difference between revisions of "2023 AMC 12A Problems/Problem 6"

Revision as of 07:27, 22 February 2024

Problem

Points $A$ and $B$ lie on the graph of $y=\log_{2}x$ . The midpoint of $\overline{AB}$ is $(6, 2)$ . What is the positive difference between the $x$ -coordinates of $A$ and $B$ ?

$\textbf{(A)}~2\sqrt{11}\qquad\textbf{(B)}~4\sqrt{3}\qquad\textbf{(C)}~8\qquad\textbf{(D)}~4\sqrt{5}\qquad\textbf{(E)}~9$

Solution 1

Let $A(6+m,2+n)$ and $B(6-m,2-n)$ , since $(6,2)$ is their midpoint. Thus, we must find $2m$ . We find two equations due to $A,B$ both lying on the function $y=\log_{2}x$ . The two equations are then $\log_{2}(6+m)=2+n$ and $\log_{2}(6-m)=2-n$ . Now add these two equations to obtain $\log_{2}(6+m)+\log_{2}(6-m)=4$ . By logarithm rules, we get $\log_{2}((6+m)(6-m))=4$ . By raising 2 to the power of both sides, we obtain $(6+m)(6-m)=16$ . We then get $36-m^2=16 \rightarrow m^2=20 \rightarrow m=2\sqrt{5}$ . Since we're looking for $2m$ , we obtain $2*2\sqrt{5}=\boxed{\textbf{(D) }4\sqrt{5}}$

~amcrunner (yay, my first AMC solution)

Solution 2

We have $\frac{x_A + x_B}{2} = 6$ and $\frac{\log_2 x_A + \log_2 x_B}{2} = 2$ . The first equation becomes $x_A + x_B = 12,$ and the second becomes $\log_2(x_A x_B) = 4,$ so $x_A x_B = 16.$ Then $\begin{align*} \left| x_A - x_B \right| & = \sqrt{\left( x_A + x_B \right)^2 - 4 x_A x_B} \\ & = \boxed{\textbf{(D) } 4 \sqrt{5}}. \end{align*}$

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

Solution 3

Basically, we can use the midpoint formula

assume that the points are $(x_1,y_1)$ and $(x_2,y_2)$

assume that the points are ( $x_1$ , $\log_{2}(x_1)$ ) and ( $x_2$ , $\log_{2}(x_2)$ )

midpoint formula is ( $\frac{x_1+x_2}{2}$ , $\frac{\log_{2}(x_1)+\log_{2}(x_2)}{2}$ )

thus $x_1+x_2=12$ $x_2=12-x_1$ and $\log_{2}(x_1)+\log_{2}(x_2)=4$ $\log_{2}(x_1)+\log_{2}(12-x_1)=\log_{2}(16)$

$\log_{2}((12x_1-x_1^2)/16)=0$

since $2^0=1$ so,

$(12x_1)-(x_1^2)=16$

$(12x_1)-(x_1^2)-16=0$ for simplicity lets say $x_1 = x$

$12x-x^2=16$ . We rearrange to get $x^2-12x+16=0$ .

put this into quadratic formula and you should get

$x_1=6+2\sqrt{5}$ Therefore, $x_1=6+2\sqrt{5}-(6-2\sqrt{5})$

which equals $6-6+4\sqrt{5}=\boxed{\textbf{(D) }4\sqrt{5}}$

Solution 4

Similar to above, but solve for $x = 2^y$ in terms of $y$ :

$$ (Error compiling LaTeX. Unknown error_msg) \begin{align*} &($2^{y}+2^{2+(2-y)})/2= 6 & 2^y + 2^{4-y} = 12 & (2^y)^2 + 2^4 = 12{2^y} & (2^y)^2 -12(2^y) + 16 = 0 \end{align*}$ (Error compiling LaTeX. Unknown error_msg) $Distance between roots ($ 2^y}) of the quadratic is the discriminant: $\sqrt{{12}^2 - 4(1)(16)} = \sqrt{80} = boxed{\textbf{(D) }4\sqrt{5}}$

~oinava

Video Solution 1

https://youtu.be/R_OdhW85yUc

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

Video Solution 2 (🚀 Under 3 min 🚀)

https://youtu.be/DOXmoQlMS7Y

~Education, the Study of Everything

@@ Line 62: / Line 62: @@
 which equals <math>6-6+4\sqrt{5}=\boxed{\textbf{(D) }4\sqrt{5}}</math>
+==Solution 4==
+Similar to above, but solve for <math>x = 2^y</math> in terms of <math>y</math>:
+<math></math>
+\begin{align*}
+&(<math>2^{y}+2^{2+(2-y)})/2= 6
+& 2^y + 2^{4-y} = 12
+& (2^y)^2 + 2^4 = 12{2^y}
+& (2^y)^2 -12(2^y) + 16 = 0
+\end{align*}
+</math><math>
+Distance between roots (</math>2^y}) of the quadratic is the discriminant: <math>\sqrt{{12}^2 - 4(1)(16)} = \sqrt{80} = boxed{\textbf{(D) }4\sqrt{5}}</math>
+~oinava
 ==Video Solution 1==

2023 AMC 12A (Problems • Answer Key • Resources)
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