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

(→‎Solution 2)
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</math>x_1=6+2sqrt(5)-(6-2sqrt(5)<math>
 
</math>x_1=6+2sqrt(5)-(6-2sqrt(5)<math>
  
−
which equals </math>6-6+4sqrt(5)$
+
which equals </math>6-6+4sqrt(5)<math>
 +
 
 +
==Solution==
 +
 
 +
We have </math>\frac{x_A + x_B}{2} = 6<math> and </math>\frac{\log_2 x_A + \log_2 x_B}{2} = 2<math>.
 +
Therefore,
 +
</math><math>
 +
\begin{align*}
 +
\left| x_A - x_B \right|
 +
& = \sqrt{\left( x_A + x_B \right)^2 - 4 x_A x_B} \\
 +
& = \boxed{\textbf{(D) </math>4 \sqrt{5}<math>}}.
 +
\end{align*}
 +
</math>$
 +
 
 +
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
  
 
==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 17:27, 10 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 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,$(12x_1)-(x_1^2)=16$e$(12x_1)-(x_1^2)-16=0$for simplicty lets say x1=x

12x-x^2=16 x^2-12x+16

put this into quadratic formula and you should get$ (Error compiling LaTeX. Unknown error_msg)x_1=6+2sqrt(5)$then$x_1=6+2sqrt(5)-(6-2sqrt(5)$which equals$6-6+4sqrt(5)$==Solution==

We have$ (Error compiling LaTeX. Unknown error_msg)\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)$ (Error compiling LaTeX. Unknown error_msg)4 \sqrt{5}$}}. \end{align*}$ (Error compiling LaTeX. Unknown error_msg)$

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

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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