Difference between revisions of "2023 AMC 12A Problems/Problem 6"
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~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ||
− | == | + | ==Solution 3== |
− | + | Basically, we can use the midpoint formula | |
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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,y_1)</math> and <math>(x_2,y_2)</math> | ||
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which equals <math>6-6+4\sqrt(5)</math> | which equals <math>6-6+4\sqrt(5)</math> | ||
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+ | ==Video Solution 1== | ||
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+ | https://youtu.be/R_OdhW85yUc | ||
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+ | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ||
+ | |||
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+ | ==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}} |
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)
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
thus
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.