Problem:
The point P(a,b) in the xy-plane is first rotated counterclockwise by 90β around the point (1,5) and then reflected about the line y=βx. The image of P after these two transformations is at (β6,3). What is bβa ?
Answer Choices:
A. 1
B. 3
C. 5
D. 7
E. 9
Solution:
The final image of P is (β6,3). We know the reflection rule for reflecting over y=βx is (x,y)β(βy,βx). So before the reflection and after rotation the point is (β3,6).
By definition of rotation, the slope between (β3,6) and (1,5) must be perpendicular to the slope between (a,b) and (1,5). The first slope is 1β(β3)5β6β=4β1β. This means the slope of P and (1,5) is 4 .
Rotations also preserve distance to the center of rotation, and since we only "travelled" up and down by the slope once to get from (3,β6) to (1,5) it follows we shall only use the slope once to travel from (1,5) to P.
Therefore point P is located at (1+1,5+4)=(2,9). The answer is 9β2=7=(D) 7β.
OR
Let us reconstruct that coordinate plane as the complex plane. Then, the point P(a,b) becomes a+bβ
i. A 90β rotation around the point (1,5) can be done by translating the point (1,5) to the origin, rotating around the origin by 90β, and then translating the origin back to the point (1,5).
a+bβ
iβΉ(aβ1)+(bβ5)β
iβΉ((aβ1)+(bβ5)β
i)β
i=5βb+(aβ1)iβΉ5+1βb+(aβ1+5)i=6βb+(a+4)i.
By basis reflection rules, the reflection of (β6,3) about the line y=βx is (β3,6). Hence, we have
6βb+(a+4)i=β3+6iβΉb=9,a=2
from which bβa=9β2=(D) 7β .
The problems on this page are the property of the MAA's American Mathematics Competitions