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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?

Jan 17, 2018

#1
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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.

Jan 17, 2018
#2
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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

Jan 18, 2018