Difference between revisions of "2023 AMC 12A Problems/Problem 6"
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put this into quadratic formula and you should get | put this into quadratic formula and you should get | ||
− | </math>x_1=6+ | + | </math>x_1=6+2\sqrt(5)<math> |
then | then | ||
− | </math>x_1=6+ | + | </math>x_1=6+2\sqrt(5)-(6-2\sqrt(5)<math> |
− | which equals </math>6-6+ | + | which equals </math>6-6+4\sqrt(5)<math> |
==Solution== | ==Solution== |
Revision as of 18:28, 10 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 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
(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)$==Solution==
We have$ (Error compiling LaTeX. Unknown error_msg)\frac{x_A + x_B}{2} = 6\frac{\log_2 x_A + \log_2 x_B}{2} = 2
$\begin{align*}
\left| x_A - x_B \right|
& = \sqrt{\left( x_A + x_B \right)^2 - 4 x_A x_B} \\
& = \boxed{\textbf{(D)$ (Error compiling LaTeX. Unknown error_msg)4 \sqrt{5}$}}.
\end{align*}$ (Error compiling LaTeX. Unknown error_msg)$
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
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 |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.