Difference between revisions of "2023 AMC 12A Problems/Problem 6"
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</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- | + | </math>(12x_1)-(x_1^2)=16<math> |
+ | </math>(12x_1)-(x_1^2)-16=0$ | ||
==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 23:23, 9 November 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
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
Bascailly, we can use the midpoint formula
assume that the points are and
assume that the points are (,) and (,)
midpoint formula is (,(
thus
and
2^0=1(12x_1)-(x_1^2)=16$$ (Error compiling LaTeX. Unknown error_msg)(12x_1)-(x_1^2)-16=0$
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.