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

m
(LaTeX edit)
Line 32: Line 32:
  
  
midpoint formula is (<math>(x_1+x_2)/2</math>,(<math>(\log_{2}(x_1)+\log_{2}(x_2))/2</math>
+
midpoint formula is (<math>\frac{x_1+x_2}{2}</math>,<math>\frac{\log_{2}(x_1)+\log_{2}(x_2)}{2}</math>)
  
  
Line 39: Line 39:
 
<math>x_2=12-x_1</math>
 
<math>x_2=12-x_1</math>
 
and  
 
and  
<math>log_2(x_1)+log_2(x_2)=4</math>
+
<math>\log_{2}(x_1)+\log_{2}(x_2)=4</math>
<math>log_2(x_1)+log_2(12-x_1)=log_2(16)</math>
+
<math>\log_{2}(x_1)+\log_{2}(12-x_1)=\log_{2}(16)</math>
  
<math>log_2((12x_1-x_1^2/16))=0</math>
+
<math>\log_{2}((12x_1-x_1^2)/16)=0</math>
thus
+
 
 +
since
 
<math>2^0=1</math>
 
<math>2^0=1</math>
 
so,
 
so,
  
 
<math>(12x_1)-(x_1^2)=16</math>
 
<math>(12x_1)-(x_1^2)=16</math>
 
  
 
<math>(12x_1)-(x_1^2)-16=0</math>
 
<math>(12x_1)-(x_1^2)-16=0</math>
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put this into quadratic formula and you should get
 
put this into quadratic formula and you should get
  
<math>x_1=6+2\sqrt(5)</math>.
+
<math>x_1=6+2\sqrt{5}</math>
 
Therefore,
 
Therefore,
<math>x_1=6+2\sqrt(5)-(6-2\sqrt(5)</math>
+
<math>x_1=6+2\sqrt{5}-(6-2\sqrt{5})</math>
  
which equals <math>6-6+4\sqrt(5)</math>
+
which equals <math>6-6+4\sqrt{5}=\boxed{\textbf{(D) }4\sqrt{5}}</math>
  
 
==Video Solution 1==
 
==Video Solution 1==

Revision as of 07:34, 4 December 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 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 $\frac{\log_2 x_A + \log_2 x_B}{2} = 2$. 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)

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

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

See Also

2023 AMC 12A (ProblemsAnswer KeyResources)
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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