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

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==Solution 1==
 
==Solution 1==
−
Let <math>A(6+m,2+n)</math> and <math>B(6-m,2-n)</math>, since <math>(6,2)</math> is their midpoint. Thus, we must find <math>2m</math>. We find two equations due to <math>A,B</math> both lying on the function <math>y=\log_{2}x</math>. The two equations are then <math>\log_{2}(6+m)=2+n</math> and <math>\log_{2}(6-m)=2-n</math>. Now add these two equations to obtain <math>\log_{2}(6+m)+\log_{2}(6-m)=4</math>. By logarithm rules, we get <math>\log_{2}((6+m)(6-m))=4</math>. By raising 2 to the power of  both sides, we obtain <math>(6+m)(6-m)=16</math>. We then get <cmath>36-m^2=16 \rightarrow m^2=20 \rightarrow m=2\sqrt{5}</cmath>. Since we're looking for <math>2m</math>, we obtain <math>2*2\sqrt{5}=\boxed{\textbf{(D) }4\sqrt{5}}</math>
+
Let <math>A(6+m,2+n)</math> and <math>B(6-m,2-n)</math>, since <math>(6,2)</math> is their midpoint. Thus, we must find <math>2m</math>. We find two equations due to <math>A,B</math> both lying on the function <math>y=\log_{2}x</math>. The two equations are then <math>\log_{2}(6+m)=2+n</math> and <math>\log_{2}(6-m)=2-n</math>. Now add these two equations to obtain <math>\log_{2}(6+m)+\log_{2}(6-m)=4</math>. By logarithm rules, we get <math>\log_{2}((6+m)(6-m))=4</math>. By raising 2 to the power of  both sides, we obtain <math>(6+m)(6-m)=16</math>. We then get <cmath>36-m^2=16 \rightarrow m^2=20 \rightarrow m=2\sqrt{5}</cmath>. Since we're looking for <math>2m</math>, we obtain <math>(2)(2\sqrt{5})=\boxed{\textbf{(D) }4\sqrt{5}}</math>
  
 
~amcrunner (yay, my first AMC solution)
 
~amcrunner (yay, my first AMC solution)
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==Solution 2==
 
==Solution 2==
  
−
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>.
+
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>. The first equation becomes <math>x_A + x_B = 12,</math> and the second becomes <math>\log_2(x_A x_B) = 4,</math> so <math>x_A x_B = 16.</math>
−
Therefore,
+
Then
 
<cmath>
 
<cmath>
 
\begin{align*}
 
\begin{align*}
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~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
 
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
  
−
==Video Solution==
+
==Solution 3==
−
 
 
−
https://youtu.be/R_OdhW85yUc
 
−
 
 
−
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
 
−
 
 
−
 
 
−
==Video Solution (🚀 Under 3 min 🚀)==
 
−
https://youtu.be/DOXmoQlMS7Y
 
−
 
 
−
~Education, the Study of Everything
 
−
 
 
−
==Solution 2==
 
  
−
Bascailly, we can use the midpoint formula  
+
Basically, we can use the midpoint formula  
  
 
assume that the points are <math>(x_1,y_1)</math> and <math>(x_2,y_2)</math>
 
assume that the points are <math>(x_1,y_1)</math> and <math>(x_2,y_2)</math>
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−
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>)
  
  
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<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>
+
since
−
thus
 
 
<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>
 
for simplicity lets say <math>x_1 = x</math>
 
for simplicity lets say <math>x_1 = x</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}=\boxed{\textbf{(D) }4\sqrt{5}}</math>
 +
 
 +
==Solution 4==
 +
Similar to above, but solve for <math>x = 2^y</math> in terms of <math>y</math>:
 +
 
 +
<math>(2^{y}+2^{2+(2-y)})/2= 6 </math>
 +
 
 +
<math>2^y + 2^{4-y} = 12 </math>
 +
 
 +
<math> (2^y)^2 + 2^4 = 12(2^y) </math>
 +
 
 +
<math> x^2 -12x + 16 = 0 </math>
 +
 
 +
Distance between roots of the quadratic is the discriminant: <math>\sqrt{{12}^2 - 4(1)(16)} = \sqrt{80} = \boxed{\textbf{(D) }4\sqrt{5}}</math>
 +
 
 +
~oinava
 +
 
 +
==Video Solution (easy to understand) by Power Solve==
 +
https://youtu.be/YXIH3UbLqK8?si=HZSSwpFx7AisyTVm&t=434
 +
 
 +
==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
  
−
which equals <math>6-6+4\sqrt(5)</math>
+
~Education, the Study of Everything
  
 
==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}}

Latest revision as of 19:33, 10 June 2024

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 $\frac{\log_2 x_A + \log_2 x_B}{2} = 2$. The first equation becomes $x_A + x_B = 12,$ and the second becomes $\log_2(x_A x_B) = 4,$ so $x_A x_B = 16.$ Then \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 $6-6+4\sqrt{5}=\boxed{\textbf{(D) }4\sqrt{5}}$

Solution 4

Similar to above, but solve for $x = 2^y$ in terms of $y$:

$(2^{y}+2^{2+(2-y)})/2= 6$

$2^y + 2^{4-y} = 12$

$(2^y)^2 + 2^4 = 12(2^y)$

$x^2 -12x + 16 = 0$

Distance between roots of the quadratic is the discriminant: $\sqrt{{12}^2 - 4(1)(16)} = \sqrt{80} = \boxed{\textbf{(D) }4\sqrt{5}}$

~oinava

Video Solution (easy to understand) by Power Solve

https://youtu.be/YXIH3UbLqK8?si=HZSSwpFx7AisyTVm&t=434

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 (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

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