Difference between revisions of "2023 AMC 12A Problems/Problem 6"
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<math>log_2(x_1)+log_2(12-x_1)=log_2(16)</math> | <math>log_2(x_1)+log_2(12-x_1)=log_2(16)</math> | ||
− | <math>log_2((12x_1-x_1^2/16))=0 | + | <math>log_2((12x_1-x_1^2/16))=0</math> |
thus | thus | ||
− | < | + | <math>2^0=1</math> |
so, | so, | ||
− | < | + | <math>(12x_1)-(x_1^2)=16</math> |
e | e | ||
− | < | + | <math>(12x_1)-(x_1^2)-16=0</math> |
− | for simplicty lets say | + | for simplicty lets say <math>x_1 = x</math> |
− | 12x-x^2=16 | + | <math>12x-x^2=16 \$ |
− | x^2-12x+16 | + | </math>x^2-12x+16<math> |
put this into quadratic formula and you should get | put this into quadratic formula and you should get |
Revision as of 14:28, 11 November 2023
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
We have and . Therefore,
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
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
thus so,
e for simplicty lets say
$12x-x^2=16 $ (Error compiling LaTeX. Unknown error_msg)x^2-12x+16x_1=6+2\sqrt(5)x_1=6+2\sqrt(5)-(6-2\sqrt(5)6-6+4\sqrt(5)$
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.