Difference between revisions of "2023 AMC 10A Problems/Problem 19"
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==Solution 2== | ==Solution 2== | ||
− | Due to rotations preserving distance, we have that <math>BP = B^\ | + | Due to rotations preserving distance, we have that <math>BP = B^\prime P</math>, as well as <math>AP = A^\prime P</math>. From here, we can see that P must be on the perpendicular bisector of <math>\overline{BB^\prime}</math> due to the property of perpendicular bisectors keeping the distance to two points constant. |
From here, we proceed to find the perpendicular bisector of <math>\overline{BB^\prime}</math>. We can see that this is just a horizontal line segment with midpoint at <math>(3.5, 3)</math>. This means that the equation of the perpendicular bisector is <math>x = 3.5</math>. | From here, we proceed to find the perpendicular bisector of <math>\overline{BB^\prime}</math>. We can see that this is just a horizontal line segment with midpoint at <math>(3.5, 3)</math>. This means that the equation of the perpendicular bisector is <math>x = 3.5</math>. | ||
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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 <math>x</math> we found earlier, we find that <math>y=4.5</math>. This means that <math>|r - s| = |3.5 - 4.5| = \boxed{1}</math>. | 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 <math>x</math> we found earlier, we find that <math>y=4.5</math>. This means that <math>|r - s| = |3.5 - 4.5| = \boxed{1}</math>. | ||
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+ | -DEVSAXENA | ||
== Video Solution 1 by OmegaLearn == | == Video Solution 1 by OmegaLearn == | ||
https://youtu.be/88F18qth0xI | https://youtu.be/88F18qth0xI |
Revision as of 19:34, 9 November 2023
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 distance, we can bash the answer with the distance formula. D(A, P) = D(A', P), and D(B, P) = D(B',P). Thus we will square our equations to yield: , and . Cancelling from the second equation makes it clear that r equals 3.5. Now 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