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) | ||
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− | + | 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 07:26, 4 December 2023
Contents
[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 . 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.