Two Rectangles

[BMAD]

Find two rectangles, with integral sides, such that the area of the first is three times the area of the second, and the perimeter of the second is three times the perimeter of the first.

[Pickett]

There are MANY(at least 51) that meet these criteria...here are all 51 with side lengths under 1000...

10x87 1x290

11x48 1x176

12x35 1x140

15x22 1x110

20x174 2x580

21x122 2x427

22x96 2x352

24x70 2x280

26x57 2x247

30x44 2x220

30x261 3x870

31x42 2x217

33x144 3x528

36x105 3x420

40x81 3x360

42x244 4x854

44x192 4x704

45x66 3x330

48x140 4x560

49x132 4x539

52x114 4x494

53x54 3x318

55x240 5x880

58x195 5x754

60x88 4x440

60x175 5x700

62x84 4x434

68x75 4x425

70x123 5x574

71x120 5x568

72x210 6x840

75x110 5x550

78x171 6x741

80x162 6x720

81x158 6x711

84x95 5x532

84x245 7x980

89x210 7x890

90x132 6x660

93x126 6x651

102x161 7x782

104x228 8x988

105x154 7x770

106x108 6x636

111x200 8x925

112x141 7x752

120x176 8x880

124x168 8x868

135x198 9x990

136x150 8x850

159x162 9x954

...

[bonanova]

There are MANY(at least 51) that meet these criteria...here are all 51 with side lengths under 1000...10x87 1x29011x48 1x17612x35 1x14015x22 1x11020x174 2x58021x122 2x42722x96 2x35224x70 2x28026x57 2x24730x44 2x22030x261 3x87031x42 2x21733x144 3x52836x105 3x42040x81 3x36042x244 4x85444x192 4x70445x66 3x33048x140 4x56049x132 4x53952x114 4x49453x54 3x31855x240 5x88058x195 5x75460x88 4x44060x175 5x70062x84 4x43468x75 4x42570x123 5x57471x120 5x56872x210 6x84075x110 5x55078x171 6x74180x162 6x72081x158 6x71184x95 5x53284x245 7x98089x210 7x89090x132 6x66093x126 6x651102x161 7x782104x228 8x988105x154 7x770106x108 6x636111x200 8x925112x141 7x752120x176 8x880124x168 8x868135x198 9x990136x150 8x850159x162 9x954...

Nice.

Curious about your algorithm.

Mine ran forever with n=100, (and came up empty.)

I tried direct solutions also, restricting the first to be square and the second to be 1xm.

Came up empty again.

[Pickett]

There are MANY(at least 51) that meet these criteria...here are all 51 with side lengths under 1000...10x87 1x29011x48 1x17612x35 1x14015x22 1x11020x174 2x58021x122 2x42722x96 2x35224x70 2x28026x57 2x24730x44 2x22030x261 3x87031x42 2x21733x144 3x52836x105 3x42040x81 3x36042x244 4x85444x192 4x70445x66 3x33048x140 4x56049x132 4x53952x114 4x49453x54 3x31855x240 5x88058x195 5x75460x88 4x44060x175 5x70062x84 4x43468x75 4x42570x123 5x57471x120 5x56872x210 6x84075x110 5x55078x171 6x74180x162 6x72081x158 6x71184x95 5x53284x245 7x98089x210 7x89090x132 6x66093x126 6x651102x161 7x782104x228 8x988105x154 7x770106x108 6x636111x200 8x925112x141 7x752120x176 8x880124x168 8x868135x198 9x990136x150 8x850159x162 9x954...

Nice.

Curious about your algorithm.

Mine ran forever with n=100, (and came up empty.)

I tried direct solutions also, restricting the first to be square and the second to be 1xm.

Came up empty again.

It's unbelievably inefficient and very brute force, but it worked like a charm:

public class Dummy {
   public static void main(String... args) {
      for (int a = 1; a < 1000; a++) {
         for (int b = a; b < 1000; b++) {
            for (int c = 1; c < 1000; c++) {
               for (int d = c; d < 1000; d++) {
                  if ((a * b) == 3*(c * d) && 3 * (2 * a + 2 * b) == (2 * c + 2 * d)) {
                     System.out.println(a + "x" + b + "   " + c + "x" + d);
                  }
               }
            }
         }
      }
   }
}