## Rectangle Perimeters and Areas

1. Find the dimensions of all rectangles that have area equal numerically to M times the perimeter, where both the length and the width are whole numbers. (M=1,2,3,4,5)
2. Repeat for perimeter equal to M times the area.
Source: Extended from a puzzle by Joe (cschen@xnet.com) on rec.puzzles January, 1996.
Solutions were received from Stan Morrice, Alan Zimmermann, Henry Bottomley, Richard Winterstein, Ravi Subramanian, Kirk Bresniker, Robert Mattson, Joseph DeVincentis, Michael Moyer, Philippe Fondanaiche, and Sandy Thompson. Two representative solutions are included here.

From Al Zimmermann:

```M=1: (4,4), (3,6)
M=2: (8,8), (6,12), (5,20)
M=3: (12,12), (10,15), (9,18), (8,24), (7,42)
M=4: (16,16), (12,24), (10,40), (9,72)
M=5: (20,20), (15,30), (14,35), (12,60), (11,110)

Some of these rectangles are just "multiples" of smaller rectangles.
Specifically, if M = Area(x,y) / Perimeter(x,y) then kM = Area(kx,ky) /
Perimeter(kx,ky).  I think such rectangles are inherently less
interesting.
The abbreviated list of rectangles which are not multiples of smaller
ones
(that is, where M, x, and y have no common divisor other than 1) is:

M=1: (4,4), (3,6)
M=2: (5,20)
M=3: (10,15), (8,24), (7,42)
M=4: (9,72)
M=5: (14,35), (12,60), (11,110)

Extending this list of "primitive" rectangles to M=10 yields:

M=6: (21,28), (13,156)
M=7: (18,63), (16,112), (15,210)
M=8: (17,272)
M=9: (22,99), (20,180), (19,342)
M=10: (36,45), (21,420)

The largest rectangle for a given M appears to always be
(2M+1,4M^2+2M).  [KD: Stan Morrice stated this as "the dimensions
are X=2*M+1, and Y=2*M*X."  I thought that was a nice compact statement.]

(2M+2,2M^2+2M) and (2M+4,M^2+2M) also work, although neither is
primitive if M is even.

> Repeat for perimeter equal to M times the area.

M=1: (4,4), (3,6)
M=2: (2,2)
M=3: (1,2)
M=4: (1,1)
M=5:
```
From Ravi Subramanian:
```Let the dimensions of the rectangle be a, b.
Area = a * b
Perimeter = 2 * (a + b)

First part : Area = M * Perimeter
a * b = M * 2 * (a + b)
1 / a + 1 / b = 1 / (2 * M)
let (2 * M) = N
1 / a + 1 / b = 1 / N
Let a <= b.
Hence N < a <= 2 * N.
Let a = (N + d) with 0 < d <= N
So
1 / (N + d) + 1 / b = 1 / N
so b = N * (N + d) / d => d is a factor N^2 => d is a product of 2
factors of N.

So considering various M's

M    N    d    a    b    P    A
1    2    1    3    6   18   18
2    4    4   16   16
2    4    1    5   20   50  100
2    6   12   36   72
4    8    8   32   64
3    6    1    7   42   98  294
2    8   24   64  192
3    9   18   54  162
4   10   15   50  150
6   12   12   48  144
4    8    1    9   72  162  648
2   10   40  100  400
4   12   24   72  288
8   16   16   64  256
5   10    1   11  110  242 1210
2   12   60  144  720
4   14   35   98  490
5   15   30   90  450
10   20   20   80  400

Second part : Perimeter = M * Area
2 * (a + b) = M * a * b
1 / a + 1 / b = M / 2
Let a <= b.
So
1 / a >= M / 4
a <= 4 / M
and a > 2 / M

No solution for M >= 5!

M    a    b    A    P
1    3    6   18   18
4    4   16   16
2    2    2    4    8
3    1    2    2    6
4    1    1    1    4
```

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