Problem:
What is the length of the boundary of the region in the xy plane consisting of points of the form (2u−3w,v+4w) where 0≤u≤1,0≤v≤1, and 0≤w≤1?
Answer Choices:
A. 103
B. 13
C. 15
D. 18
E. 16
Solution:
Call the region R. Let f(u,v,w)=(2u−3w,v+4w). Then A=(0,0)=f(0,0,0), B=(2,0)=f(1,0,0),C=(2,1)=f(1,1,0),D=(−1,5)=f(1,1,1),E=(−3,5)=\ f(0,1,1), and F=(−3,4)=f(0,0,1) are points in R. Note that if (x1,y1)=f(u1,v1,w1) and (x2,y2)=f(u2,v2,w2), then
for all real numbers t. If 0≤t≤1, then tu1+(1−t)u2 lies between u1 and u2. Thus if also 0≤u1≤1 and 0≤u2≤1, then 0≤tu1+(1−t)u2≤1, and similarly for the vi and wi. Thus if p1=(x1,y1) and p2=(x2,y2) are points of R that arise from triples (u1,v1,w1) and (u2,v2,w2) in the unit cube [0,1]3, then the line segment tp1+(1−t)p2,0≤t≤1, joining them also lies entirely in R. Hence the edges of the hexagon ABCDEF all lie in R, and thus also the interior of this hexagon.
To see that there are no other points in R, note that because x=2u−3w, it follows that −3≤x≤2. The segments BC and EF lie on the boundary of this strip. Also, because y=v+4w, it follows that 0≤y≤5. The segments AB and DE lie on the boundary of this strip. Finally, because 4x+3y=4(2u−3w)+3(v+4w)=8u+3v, it follows that 0≤4x+3y≤11. The segments CD and AF lie on the boundary of this strip. Because all points of R must lie in the intersection of these three strips, all points of R have been identified.\
The requested length of the boundary of R is the perimeter of the hexagon:
AB+BC+CD+DE+EF+FA=2+1+32+42+2+1+32+42=(E)16.
Note: The interval [0,1] on the real line is convex, so the unit cube [0,1]3 is convex. The image of a convex set under a linear transformation is convex. Also, a convex set in the plane can be described by its supporting half-planes. This is particularly simple when the set in question is a convex polygon.
