Difference between revisions of "2023 AMC 12A Problems/Problem 6"
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thus | thus | ||
− | <math> | + | <math>x_1+x_2=12</math> |
− | <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(x1)+log_2(12-x1)=log2(16)</math> |
− | <math> | + | |
+ | <math>log_2((12x1-x1^2/16))=0</math> | ||
==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 23:20, 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 taking 2 to the power of both sides (what's the word for this?) 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
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 |
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