Difference between revisions of "2023 AMC 12A Problems/Problem 6"
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==Solution 1== | ==Solution 1== | ||
− | Let <math>A(6+m,2+n)</math> and <math>B(6-m,2-n)</math>, since <math>(6,2)</math> is their midpoint. Thus, we must find <math>2m</math>. We find two equations due to <math>A,B</math> both lying on the function <math>y=\log_{2}x</math>. The two equations are then <math>\log_{2}(6+m)=2+n</math> and <math>\log_{2}(6-m)=2-n</math>. Now add these two equations to obtain <math>\log_{2}(6+m)+\log_{2}(6-m)=4</math>. By logarithm rules, we get <math>\log_{2}((6+m)(6-m))=4</math>. By raising 2 to the power of both sides, we obtain <math>(6+m)(6-m)=16</math>. We then get <cmath>36-m^2=16 \rightarrow m^2=20 \rightarrow m=2\sqrt{5}</cmath>. Since we're looking for <math>2m</math>, we obtain <math>2 | + | Let <math>A(6+m,2+n)</math> and <math>B(6-m,2-n)</math>, since <math>(6,2)</math> is their midpoint. Thus, we must find <math>2m</math>. We find two equations due to <math>A,B</math> both lying on the function <math>y=\log_{2}x</math>. The two equations are then <math>\log_{2}(6+m)=2+n</math> and <math>\log_{2}(6-m)=2-n</math>. Now add these two equations to obtain <math>\log_{2}(6+m)+\log_{2}(6-m)=4</math>. By logarithm rules, we get <math>\log_{2}((6+m)(6-m))=4</math>. By raising 2 to the power of both sides, we obtain <math>(6+m)(6-m)=16</math>. We then get <cmath>36-m^2=16 \rightarrow m^2=20 \rightarrow m=2\sqrt{5}</cmath>. Since we're looking for <math>2m</math>, we obtain <math>(2)(2\sqrt{5})=\boxed{\textbf{(D) }4\sqrt{5}}</math> |
~amcrunner (yay, my first AMC solution) | ~amcrunner (yay, my first AMC solution) | ||
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so, | so, | ||
− | <math> | + | <math>12x_1-x_1^2=16</math> |
− | <math> | + | <math>12x_1-x_1^2-16=0</math> |
for simplicity lets say <math>x_1 = x</math> | for simplicity lets say <math>x_1 = x</math> | ||
Line 66: | Line 66: | ||
Similar to above, but solve for <math>x = 2^y</math> in terms of <math>y</math>: | Similar to above, but solve for <math>x = 2^y</math> in terms of <math>y</math>: | ||
− | <math> | + | <math>(2^{y}+2^{2+(2-y)})/2= 6 </math> |
− | + | ||
− | + | <math>2^y + 2^{4-y} = 12 </math> | |
− | + | ||
− | + | <math> (2^y)^2 + 2^4 = 12(2^y) </math> | |
− | + | ||
− | + | <math> x^2 -12x + 16 = 0 </math> | |
− | </ | ||
− | Distance between roots | + | Distance between roots of the quadratic is the discriminant: <math>\sqrt{{12}^2 - 4(1)(16)} = \sqrt{80} = \boxed{\textbf{(D) }4\sqrt{5}}</math> |
~oinava | ~oinava | ||
+ | |||
+ | ==Video Solution (easy to understand) by Power Solve== | ||
+ | https://youtu.be/YXIH3UbLqK8?si=HZSSwpFx7AisyTVm&t=434 | ||
==Video Solution 1== | ==Video Solution 1== |
Latest revision as of 20:33, 10 June 2024
Contents
Problem
Points and
lie on the graph of
. The midpoint of
is
. What is the positive difference between the
-coordinates of
and
?
Solution 1
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
We have and
. The first equation becomes
and the second becomes
so
Then
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Solution 3
Basically, we can use the midpoint formula
assume that the points are and
assume that the points are (,
) and (
,
)
midpoint formula is (,
)
thus
and
since
so,
for simplicity lets say
. We rearrange to get
.
put this into quadratic formula and you should get
Therefore,
which equals
Solution 4
Similar to above, but solve for in terms of
:
Distance between roots of the quadratic is the discriminant:
~oinava
Video Solution (easy to understand) by Power Solve
https://youtu.be/YXIH3UbLqK8?si=HZSSwpFx7AisyTVm&t=434
Video Solution 1
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Video Solution 2 (🚀 Under 3 min 🚀)
~Education, the Study of Everything
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.