Difference between revisions of "2023 AMC 12A Problems/Problem 6"
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We have <math>\frac{x_A + x_B}{2} = 6</math> and <math>\frac{\log_2 x_A + \log_2 x_B}{2} = 2</math>. | We have <math>\frac{x_A + x_B}{2} = 6</math> and <math>\frac{\log_2 x_A + \log_2 x_B}{2} = 2</math>. | ||
Therefore, | Therefore, | ||
+ | <cmath> | ||
+ | \begin{align*} | ||
+ | u & = \pm d \cos 2 \alpha , \ | ||
+ | v & = \pm d \cos 2 \alpha \frac{1 - \cos 2 \alpha}{\sin 2 \alpha} \ | ||
+ | & = \pm d \cos 2 \alpha \tan \alpha . | ||
+ | \end{align*} | ||
+ | </cmath> | ||
+ | |||
+ | |||
<math></math> | <math></math> | ||
\begin{align*} | \begin{align*} |
Revision as of 17:31, 10 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
We have and . Therefore,
$$ (Error compiling LaTeX. Unknown error_msg)
\begin{align*}
\left| x_A - x_B \right|
& = \sqrt{\left( x_A + x_B \right)^2 - 4 x_A x_B} \
& = \boxed{\textbf{(D) }}.
\end{align*}
$$ (Error compiling LaTeX. Unknown error_msg)
~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
2^0=1(12x_1)-(x_1^2)=16(12x_1)-(x_1^2)-16=0$for simplicty lets say x1=x
12x-x^2=16 x^2-12x+16
put this into quadratic formula and you should get$ (Error compiling LaTeX. Unknown error_msg)x_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.