Puzzling Stack Exchange is a question and answer site for those who create, solve, and study puzzles. It only takes a minute to sign up.
Sign up to join this community
a) Six different rectangles, none a square, have all integer sides chosen from a, b, c, and d. If I take any two of these rectangles with no common side (there are three ways of doing this), the possible sum of their two areas is 34, 38, or 43.
What are a, b, c, and d?
b) Can the sums of the areas of each of those pairs of rectangles with no common side be three consecutive integers rather than 34, 38, and 43? Three integers in arithmetic progression?
a)
Assuming a>b>c>d we have
(a+b)(c+d) = 38+34 = 72 (1)
(a-b)(c-d) = 38-34 = 4 (2)
(a+c)(b+d) = 43+34 = 77 (3)
(a-c)(b-d) = 43-34 = 9 (4)
(a+d)(b+c) = 43+38 = 81 (5)
(a-d)(b-c) = 43-38 = 5 (6)
We'll only need (5) and (6)
By (6)
a-d = 5 and b-c = 1
Together with (5) this implies
a=7,b=5,c=4,d=2
It is easily verified that this is indeed a solution.
b)
In this scenario the r.h.s of (2,4,6) would be 1,2,1 for the consecutive integers case and k,2k,k for the arithmetic progression case. 1,2,1 is clearly impossible.
A possible solution for k,2k,k would be
k=6,c-d=2,b-c=1,a-b=3. For example, a=7,b=4,c=3,d=1.
