Difference between revisions of "2023 AMC 12A Problems/Problem 6"
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<math>\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</math> | <math>\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</math> | ||
− | ==Solution== | + | ==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 | + | 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>. 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> | |
+ | Then | ||
+ | <cmath> | ||
+ | \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*} | ||
+ | </cmath> | ||
− | + | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | |
− | assume that the points are (<math>x_1</math>,<math>\log_{2}(x_1)</math>) and ( | + | ==Solution 3== |
+ | |||
+ | 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</math>,<math>\log_{2}(x_1)</math>) and (<math>x_2</math>,<math>\log_{2}(x_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>) | ||
+ | |||
+ | |||
+ | thus | ||
+ | <math>x_1+x_2=12</math> | ||
+ | <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}(12-x_1)=\log_{2}(16)</math> | ||
+ | |||
+ | <math>\log_{2}((12x_1-x_1^2)/16)=0</math> | ||
+ | |||
+ | since | ||
+ | <math>2^0=1</math> | ||
+ | so, | ||
+ | |||
+ | <math>12x_1-x_1^2=16</math> | ||
+ | |||
+ | <math>12x_1-x_1^2-16=0</math> | ||
+ | for simplicity lets say <math>x_1 = x</math> | ||
+ | |||
+ | <math>12x-x^2=16</math>. We rearrange to get <math>x^2-12x+16=0</math>. | ||
+ | |||
+ | put this into quadratic formula and you should get | ||
+ | |||
+ | <math>x_1=6+2\sqrt{5}</math> | ||
+ | Therefore, | ||
+ | <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 | ||
+ | |||
+ | ~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
Contents
Problem
Points and lie on the graph of . The midpoint of is . What is the positive difference between the -coordinates of and ?
Solution 1
Let and , since is their midpoint. Thus, we must find . We find two equations due to both lying on the function . The two equations are then and . Now add these two equations to obtain . By logarithm rules, we get . By raising 2 to the power of both sides, we obtain . We then get . Since we're looking for , we obtain
~amcrunner (yay, my first AMC solution)
Solution 2
We have and . The first equation becomes and the second becomes so Then
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Solution 3
Basically, we can use the midpoint formula
assume that the points are and
assume that the points are (,) and (,)
midpoint formula is (,)
thus
and
since so,
for simplicity lets say
. We rearrange to get .
put this into quadratic formula and you should get
Therefore,
which equals
Solution 4
Similar to above, but solve for in terms of :
Distance between roots of the quadratic is the discriminant:
~oinava
Video Solution (easy to understand) by Power Solve
https://youtu.be/YXIH3UbLqK8?si=HZSSwpFx7AisyTVm&t=434
Video Solution 1
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Video Solution 2 (🚀 Under 3 min 🚀)
~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.