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

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

@@ 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>
 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>
 so,
 <math>(12x_1)-(x_1^2)=16</math>
 <math>(12x_1)-(x_1^2)-16=0</math>
@@ Line 57: / Line 57: @@
 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,
-<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==

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