4
$\begingroup$

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?

$\endgroup$

1 Answer 1

5
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.