Difference between revisions of "2023 AMC 10A Problems/Problem 19"
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Thus we will square our equations to yield: | Thus we will square our equations to yield: | ||
<math>(1-r)^2+(2-s)^2=(3-r)^2+(1-s)^2</math>, and <math>(3-r)^2+(3-s)^2=(4-r)^2+(3-s)^2</math>. | <math>(1-r)^2+(2-s)^2=(3-r)^2+(1-s)^2</math>, and <math>(3-r)^2+(3-s)^2=(4-r)^2+(3-s)^2</math>. | ||
− | + | Canceling <math>(3-s)^2</math> from the second equation makes it clear that <math>r</math> equals <math>3.5</math>. | |
− | + | ||
− | + | Substituting will yield | |
− | + | ||
+ | <cmath>\begin{align*}(2.5)^2+(2-s)^2 &= (-0.5)^2+(1-s)^2 \\ | ||
+ | 6.25+4-4s+s^2 &= 0.25+1-2s+s^2 \\ | ||
+ | 2s &= 9 \\ | ||
+ | s &=4.5 \\ | ||
+ | \end{align*} | ||
+ | </cmath>. | ||
+ | |||
Now <math>|r-s| = |3.5-4.5| = \boxed{\textbf{(E) } 1}</math>. | Now <math>|r-s| = |3.5-4.5| = \boxed{\textbf{(E) } 1}</math>. | ||
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When we solve for the perpendicular bisector of <math>y = -\frac{1}{2}x + \frac{5}{2}</math>, we determine that it has a slope of 2, and it runs through <math>(2, 1.5)</math>. Plugging in <math>1.5 = 2(2)-n</math>, we get than <math>n = \frac{5}{2}</math>. Therefore our perpendicular bisector is <math>y=2x-\frac{5}{2}</math>. Next, we solve for the perpendicular of <math>y = 3</math>. We know that it has an undefined slope, and it runs through <math>(3.5, 3)</math>. We can determine that our second perpendicular bisector is <math>x = 3.5</math>. | When we solve for the perpendicular bisector of <math>y = -\frac{1}{2}x + \frac{5}{2}</math>, we determine that it has a slope of 2, and it runs through <math>(2, 1.5)</math>. Plugging in <math>1.5 = 2(2)-n</math>, we get than <math>n = \frac{5}{2}</math>. Therefore our perpendicular bisector is <math>y=2x-\frac{5}{2}</math>. Next, we solve for the perpendicular of <math>y = 3</math>. We know that it has an undefined slope, and it runs through <math>(3.5, 3)</math>. We can determine that our second perpendicular bisector is <math>x = 3.5</math>. | ||
− | Setting the equations equal to each other, we get <math> | + | Setting the equations equal to each other, we get <math>y = \frac{9}{2}</math>. Therefore, <math>|r - s| = |3.5 - 4.5| = \boxed{\textbf{(E) } 1}</math>. |
<asy> | <asy> | ||
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~aydenlee & wuwang2002 | ~aydenlee & wuwang2002 | ||
− | ==Solution== | + | ==Solution 4== |
We use the complex numbers approach to solve this problem. | We use the complex numbers approach to solve this problem. | ||
Line 77: | Line 84: | ||
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ||
− | + | == Video Solution == | |
+ | https://youtu.be/aYHwBpdWkAk | ||
+ | |||
+ | == Video Solution by == | ||
+ | |||
+ | https://www.youtube.com/watch?v=fIzCR4x4x-M | ||
+ | |||
== Video Solution 1 by OmegaLearn == | == Video Solution 1 by OmegaLearn == | ||
https://youtu.be/88F18qth0xI | https://youtu.be/88F18qth0xI | ||
+ | ==Video Solution by SpreadTheMathLove== | ||
+ | https://www.youtube.com/watch?v=oHMJwiEwOS0 | ||
+ | |||
+ | ==Video Solution== | ||
+ | |||
+ | https://youtu.be/va3ZCFeKfzg | ||
+ | |||
+ | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ||
+ | == Video Solution by TheBeautyofMath== | ||
+ | https://youtu.be/Q_4uMxMbQlI | ||
+ | |||
+ | ~IceMatrix | ||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2023|ab=A|num-b=18|num-a=20}} | {{AMC10 box|year=2023|ab=A|num-b=18|num-a=20}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 22:21, 6 August 2024
Contents
Problem
The line segment formed by and
is rotated to the line segment formed by
and
about the point
. What is
?
Solution 1
Due to rotations preserving an equal distance, we can bash the answer with the distance formula. , and
.
Thus we will square our equations to yield:
, and
.
Canceling
from the second equation makes it clear that
equals
.
Substituting will yield
.
Now .
-Antifreeze5420
Solution 2
Due to rotations preserving distance, we have that , as well as
. From here, we can see that P must be on the perpendicular bisector of
due to the property of perpendicular bisectors keeping the distance to two points constant.
From here, we proceed to find the perpendicular bisector of . We can see that this is just a horizontal line segment with midpoint at
. This means that the equation of the perpendicular bisector is
.
Similarly, we find the perpendicular bisector of . We find the slope to be
, so our new slope will be
. The midpoint of
and
is
, which we can use with our slope to get another equation of
.
Now, point P has to lie on both of these perpendicular bisectors, meaning that it has to satisfy both equations. Plugging in the value of we found earlier, we find that
. This means that
.
-DEVSAXENA
Solution 3 (Coordinate Geometry)
To find the center of rotation, we find the intersection point of the perpendicular bisectors of and
.
We can find that the equation of the line is
, and that the equation of the line
is
.
When we solve for the perpendicular bisector of , we determine that it has a slope of 2, and it runs through
. Plugging in
, we get than
. Therefore our perpendicular bisector is
. Next, we solve for the perpendicular of
. We know that it has an undefined slope, and it runs through
. We can determine that our second perpendicular bisector is
.
Setting the equations equal to each other, we get . Therefore,
.
~aydenlee & wuwang2002
Solution 4
We use the complex numbers approach to solve this problem.
Denote by the angle that
rotates about
in the counterclockwise direction.
Thus, and
.
Taking ratio of these two equations, we get
By solving this equation, we get .
Therefore,
.
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Video Solution
Video Solution by
https://www.youtube.com/watch?v=fIzCR4x4x-M
Video Solution 1 by OmegaLearn
Video Solution by SpreadTheMathLove
https://www.youtube.com/watch?v=oHMJwiEwOS0
Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Video Solution by TheBeautyofMath
~IceMatrix
See Also
2023 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 18 |
Followed by Problem 20 | |
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 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.