Difference between revisions of "2023 AMC 12A Problems/Problem 6"
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<math></math> | <math></math> | ||
\begin{align*} | \begin{align*} | ||
− | &(<math>2^{y}+2^{2+(2-y)})/2= 6 | + | &(<math>2^{y}+2^{2+(2-y)})/2= 6 \ |
− | & 2^y + 2^{4-y} = 12 | + | & 2^y + 2^{4-y} = 12 \ |
− | & (2^y)^2 + 2^4 = 12{2^y} | + | & (2^y)^2 + 2^4 = 12{2^y} \ |
− | & (2^y)^2 -12(2^y) + 16 = 0 | + | & (2^y)^2 -12(2^y) + 16 = 0 \ |
\end{align*} | \end{align*} | ||
</math><math> | </math><math> |
Revision as of 08:28, 22 February 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
:
$$ (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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.