A rectangle is called ``cool'' if the number of square units in its area is equal to twice the number of units in its perimeter. A cool rectangle also must have integer side lengths. What is the sum of all the different possible areas of cool rectangles?
Area = L x W
Perimeter =2[L + W]
L x W = 4L + 4W
There are 5 pairs of solutions:
L= 5 and W=20 =5 x 20 =100
L= 6 and W =12 =6 x 12 =72
L =8 and W = 8 = 8 x 8 = 64
L = 12 and W = 6 = 12 x 6 =72
L=20 and W = 5 = 20 x 5 =100
100+72+64+72 + 100 = 408 total sums of the areas.
+128475 Let the area = L * W and the perimeter is 2(L + W )
So
Area Units = Twice Perimeter Units
(L * W) = 2 [ 2 ( L + W)]
L * W = 4L + 4W
L * W - 4L = 4W
L (W - 4) = 4W ⇒ W > 4
L = [ 4W ] / ( W - 4)
Possible Values of W and L
W L Ratio of [4W ] / ( W - 4) = L
5 20 20
6 12 12
8 8 8
12 6 6
20 5 5
Note that the last two dimensions are exactly the same as the first two...it just depends on the orientation of the rectangle
Note that as W ⇒ infinity, the ratio of [4W] / (W - 4) ⇒ 4 but is always > 4
So..... no other integer values are possible for L
So.....the possible dimensions are :
5, 20 6,12 and 8,8
So....the sum of the possible areas is
5*20 + 6*12 + 8*8 =
100 + 72 + 64 =
100 + 136 =
236
