This is a puzzle from an old Mensa book. The book listed a single solution with no explanation.
Consider two rectangles. R1 is 8 times the area of R2. R2 is 2 times the perimeter of R1.
The dimensions of both rectangles are integer values.
What are the dimensions of R1 and R2?
So obviously there are multiple solutions, but how do you solve for any one of them?
The two starting equations, where a and b are the dimensions of the large rectangle (R1) and c and d are the dimensions of the smaller rectangle (R2):
ab = 8cd
2 * (2a + 2b) = 2c + 2d
The third equation which may or may not be redundant is that the area/perimeter ratio of the large rectangle (R1) is 16 times the area/perimeter ratio for the small rectangle (R2).
ab / (2a + 2b) = 16 * cd / (2c + 2d)
I’m not sure what the next step towards “solving” is, but we know that the second rectangle is going to be short/skinny so I assigned the value of 1 to c (the short/skinny side) and then in later iterations, I assigned 2, and then 3. With c = 1, we have:
ab = 8d
2 * (2a + 2b) = 2 + 2d -> 2a + 2b = 1 + d
ab / (2a + 2b) = 16d / (2 + 2d) -> ab / (2a + 2b) = 8d / (1 + d)
I wasn’t able to do progress further with that, but an online solver generated the following solutions:
R1: 17 x 264, R2 = 1 x 561
R1: 18 x 140, R2 = 1 x 315
R1: 20 x 78, R2: 1 x 195
R1: 24 x 47, R2: 1 x 141
And for c = 2:
R1: 33 x 1024, R2: 2 x 2112
R1: 63 x 64, R2: 2 x 252
And for c = 3:
R1: 49 x 2280, R2: 3 x 4655
R1: 50 x 1164, R2: 3 x 2425
R1: 52 x 606, R2: 3 x 1313
R1: 56 x 327, R2: 3 x 763
R1: 57 x 296, R2: 3 x 703
R1: 79 x 120, R2: 3 x 395
R1: 84 x 110, R2: 3 x 385
Can someone show me the math to produce any one of these solutions or name the method of solving?
Graphically, we're looking at pairs of rectangles where the second has twice the perimeter of the first and then looking at all the different possible areas for each rectangle and trying to find an area in the first rectangle that is eight times an area in the second rectangle. The attached screenshot illustrates (or tries to) the 4th solution listed above.