2023 AMC 12A Problems/Problem 6

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

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

Bascailly, 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 ( $(x_1+x_2)/2$ ,( $(\log_{2}(x_1)+\log_{2}(x_2))/2$

thus $x_1+x_2=12$ $x_2=12-x_1$ and $log_2(x_1)+log_2(x_2)=4$ $log_2(x_1)+log_2(12-x_1)=log_2(16)$

$log_2((12x_1-x_1^2/16))=0 thus$ 2^0=1 $so,$ (12x_1)-(x_1^2)=16$$ (Error compiling LaTeX. Unknown error_msg)(12x_1)-(x_1^2)16=0$

2023 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 5	Followed by Problem 7
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All AMC 12 Problems and Solutions

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2023 AMC 12A Problems/Problem 6

Contents

Problem

Solution

Solution 2

See Also