Difference between revisions of "2023 AMC 12A Problems/Problem 6"
(→Solution 2) |
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assume that the points are (x1,y1) and (x2,y2) | assume that the points are (x1,y1) and (x2,y2) | ||
− | assume that the points are (<math>x_1</math>,<math>\log_{2}( | + | assume that the points are (<math>x_1</math>,<math>\log_{2}(x_1)</math>) and (x2,<math>\log_{2}(x2)</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 23:13, 9 November 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
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 (x1,y1) and (x2,y2)
assume that the points are (,) and (x2,)
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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