Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2023 AMC 10A Problems/Problem 19"
Line 6: Line 6:
 
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).  
 
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:  
 
Thus we will square our equations to yield:  
−
(1-r)^2+(2-s)^2=(3-r)^2+(1-s)^2, and (3-r)^2+(3-s)^2=(4-r)^2+(3-s)^2.
+
<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>.
−
Cancelling (3-s)^2 from the second equation makes it clear that r equals 3.5.
+
Cancelling <math>(3-s)^2</math> from the second equation makes it clear that r equals 3.5.
−
Now substituting will yield (2.5)^2+(2-s)^2=(-0.5)^2+(1-s)^2.
+
Now substituting will yield <math>(2.5)^2+(2-s)^2=(-0.5)^2+(1-s)^2</math>.
−
6.25+4-4s+s^2=0.25+1-2s+s^2
+
<math>6.25+4-4s+s^2=0.25+1-2s+s^2</math>
−
2s = 9, s = 4.5.
+
<math>2s = 9</math>, <math>s = 4.5</math>.
−
Now |r-s| = |3.5-4.5| = 1.
+
Now <math>|r-s| = |3.5-4.5| = 1</math>.
  
 
== Video Solution 1 by OmegaLearn ==
 
== Video Solution 1 by OmegaLearn ==
 
https://youtu.be/88F18qth0xI
 
https://youtu.be/88F18qth0xI

Revision as of 15:56, 9 November 2023

The line segment formed by $A(1, 2)$ and $B(3, 3)$ is rotated to the line segment formed by $A'(3, 1)$ and $B'(4, 3)$ about the point $P(r, s)$. What is $|r-s|$?

$\text{A) } \frac{1}{4} \qquad \text{B) } \frac{1}{2} \qquad \text{C) } \frac{3}{4} \qquad \text{D) } \frac{2}{3} \qquad \text{E) } 1$

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: $(1-r)^2+(2-s)^2=(3-r)^2+(1-s)^2$, and $(3-r)^2+(3-s)^2=(4-r)^2+(3-s)^2$. Cancelling $(3-s)^2$ from the second equation makes it clear that r equals 3.5. Now substituting will yield $(2.5)^2+(2-s)^2=(-0.5)^2+(1-s)^2$. $6.25+4-4s+s^2=0.25+1-2s+s^2$ $2s = 9$, $s = 4.5$. Now $|r-s| = |3.5-4.5| = 1$.

Video Solution 1 by OmegaLearn

https://youtu.be/88F18qth0xI

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2023_AMC_10A_Problems/Problem_19&oldid=201073"