A rectangular sheet of paper is folded so that two diagonally opposite corners come together. The crease thus formed is as long as the longer side of the rectangle.
What is the ratio of the longer side of the rectangle to the shorter?
(In reply to
Puzzle Answer by K Sengupta)
We construct the rectangle PQRS.
Assume PQ > QR with PQ=y and QR =1
We will label the endpoints of the resultant crease TV with T on PQ and V on RS.
We now observe that U is the midpoint of PR and also the midpoint of TV.
Obviously, PR and TV are perpendicular to each other.
The triangles PTU and PQR are simila Accordingly, we must have:
TU/QR = PU/PQ = PT/PR
Hence, we must obtain:
TU = y/2
PU^2 = (y^2+1)/2
PT^2 = (2y^2+1)/2
PR^2 = y^2+1
Now, we have:
TU/QR = PU/PQ
V(y^2+1)
=> y/2 = ---------------
2y
=> y^4=y^2+1
=> y^4-y^2-1=0
Substituting m=y^2, we obtain:
m^2-m-1 =0
=> m = (1 +/- V5)/2
=> y^2 = (1 + V5)/2, since V5>1, so 1-V5 <0, and accordingly inadmissible.
=> y = V{(1+V5)/2} ~= 1.272
Consequently, the required ratio of the longer side to the shorter side is approximately 1.272.
Edited on July 15, 2022, 2:47 am
Explanation to Puzzle Answer
