Difference between revisions of "2023 AMC 12A Problems/Problem 6"
m (→Solution) |
m (→Solution) |
||
Line 9: | Line 9: | ||
~amcrunner (yay, my first AMC solution) | ~amcrunner (yay, my first AMC solution) | ||
− | ==Solution== | + | ==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>. |
Revision as of 08:26, 4 December 2023
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
.
Therefore,
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Video Solution (🚀 Under 3 min 🚀)
~Education, the Study of Everything
Solution 2
Bascailly, we can use the midpoint formula
assume that the points are and
assume that the points are (,
) and (
,
)
midpoint formula is (,(
thus
and
thus
so,
for simplicity lets say
. We rearrange to get
.
put this into quadratic formula and you should get
.
Therefore,
which equals
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.