2023 AMC 12A Problems/Problem 6
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[hide]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 :
$$ (Error compiling LaTeX. Unknown error_msg) \begin{align*} &($2^{y}+2^{2+(2-y)})/2= 6 & 2^y + 2^{4-y} = 12 & (2^y)^2 + 2^4 = 12{2^y} & (2^y)^2 -12(2^y) + 16 = 0 \end{align*}$ (Error compiling LaTeX. Unknown error_msg)2^y\sqrt{{12}^2 - 4(1)(16)} = \sqrt{80} = boxed{\textbf{(D) }4\sqrt{5}}$
~oinava
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 |
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