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

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assume that the points are (x1,y1) and (x2,y2)
 
assume that the points are (x1,y1) and (x2,y2)
  
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assume that the points are (x1,<math>/log_{2}(x1)</math>) and (x2,<math>/log_{2}(x2)</math>)
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assume that the points are (x1,<math>\log_{2}(x1)</math>) and (x2,<math>\log_{2}(x2)</math>)
  
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2023|ab=A|num-b=5|num-a=7}}
 
{{AMC12 box|year=2023|ab=A|num-b=5|num-a=7}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 23:12, 9 November 2023

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 taking 2 to the power of both sides (what's the word for this?) 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 (x1,y1) and (x2,y2)

assume that the points are (x1,$\log_{2}(x1)$) and (x2,$\log_{2}(x2)$)

See Also

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

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