Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

2023 AMC 12A Problems/Problem 6

Revision as of 15:43, 11 November 2023 by Thestudyofeverything (talk | contribs) (→‎Video Solution)

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

We have $\frac{x_A + x_B}{2} = 6$ and $\frac{\log_2 x_A + \log_2 x_B}{2} = 2$. Therefore, \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)

Video Solution

https://youtu.be/R_OdhW85yUc

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


Video Solution (Quick and Easy!)

https://youtu.be/DOXmoQlMS7Y

~Education, the Study of Everything

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,

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


$(12x_1)-(x_1^2)-16=0$ 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 $6-6+4\sqrt(5)$

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2023_AMC_12A_Problems/Problem_6&oldid=202358"