Difference between revisions of "2023 AMC 12A Problems/Problem 6"
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</math>x_1=6+2sqrt(5)-(6-2sqrt(5)<math> | </math>x_1=6+2sqrt(5)-(6-2sqrt(5)<math> | ||
− | which equals </math>6-6+4sqrt(5)$ | + | which equals </math>6-6+4sqrt(5)<math> |
+ | |||
+ | ==Solution== | ||
+ | |||
+ | 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, | ||
+ | </math><math> | ||
+ | \begin{align*} | ||
+ | \left| x_A - x_B \right| | ||
+ | & = \sqrt{\left( x_A + x_B \right)^2 - 4 x_A x_B} \ | ||
+ | & = \boxed{\textbf{(D) </math>4 \sqrt{5}<math>}}. | ||
+ | \end{align*} | ||
+ | </math>$ | ||
+ | |||
+ | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ||
==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 17:27, 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 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+2sqrt(5)x_1=6+2sqrt(5)-(6-2sqrt(5)6-6+4sqrt(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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.