Difference between revisions of "2023 AMC 12A Problems/Problem 6"
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− | midpoint formula is (<math> | + | midpoint formula is (<math>\frac{x_1+x_2}{2}</math>,<math>\frac{\log_{2}(x_1)+\log_{2}(x_2)}{2}</math>) |
Line 39: | Line 39: | ||
<math>x_2=12-x_1</math> | <math>x_2=12-x_1</math> | ||
and | and | ||
− | <math> | + | <math>\log_{2}(x_1)+\log_{2}(x_2)=4</math> |
− | <math> | + | <math>\log_{2}(x_1)+\log_{2}(12-x_1)=\log_{2}(16)</math> |
− | <math> | + | <math>\log_{2}((12x_1-x_1^2)/16)=0</math> |
− | + | ||
+ | since | ||
<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> | ||
Line 57: | Line 57: | ||
put this into quadratic formula and you should get | put this into quadratic formula and you should get | ||
− | <math>x_1=6+2\sqrt | + | <math>x_1=6+2\sqrt{5}</math> |
Therefore, | Therefore, | ||
− | <math>x_1=6+2\sqrt | + | <math>x_1=6+2\sqrt{5}-(6-2\sqrt{5})</math> |
− | which equals <math>6-6+4\sqrt(5 | + | which equals <math>6-6+4\sqrt{5}=\boxed{\textbf{(D) }4\sqrt{5}}</math> |
==Video Solution 1== | ==Video Solution 1== |
Revision as of 08:34, 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)
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
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.