Difference between revisions of "2023 AMC 10A Problems/Problem 19"
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<math>\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</math> | <math>\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</math> | ||
| + | ==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 == | == Video Solution 1 by OmegaLearn == | ||
https://youtu.be/88F18qth0xI | https://youtu.be/88F18qth0xI | ||
Revision as of 15:55, 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: (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.
