Problem:
The line segment from A(1,2) to B(3,3) can be transformed to the line segment from A′(3,1) to B′(4,3), sending A to A′ and B to B′, by a rotation centered at the point P(s,t). What is ∣s−t∣?
Answer Choices:
A. 41
B. 21
C. 32
D. 43
E. 1
Solution:
Because A and A′ are the same distance from the center of rotation, P(s,t) must lie on the perpendicular bisector of AA′; similarly it must lie on the perpendicular bisector of BB′. The line AA′ passes through the midpoint (2,23) and has slope −1−23−1=2, so its equation is
y=2(x−2)+23.
The line BB′ is vertical and passes through (27,3), so its equation is x=27. Solving this system of equations gives P(s,t)=(27,29), as shown in the figure. The requested absolute difference of coordinates is ∣∣∣27−29∣∣∣=(E)1.
Note: As a check, note that the angle of rotation must be ∠APA′ and also ∠BPB′; it is about 37∘. By the SSS Congruence Theorem, △APB≅△A′PB′, so indeed
∠APA′=∠APB+∠BPA′=∠A′PB′+∠BPA′=∠BPB′.
Thus the rotation of this amount around (27,29) is the required transformation.
The problems on this page are the property of the MAA's American Mathematics Competitions
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