A right triangle with integer leg lengths is called cool if the number of square units in its area is equal to five times the number of units in the sum of the lengths of its legs. What is the sum of all the different possible areas of cool right triangles?

Answer 1

Let the legs of the right triangle be a and b. The area of the triangle is (1/2)*a*b and the sum of the lengths of the legs is a+b.

According to the given condition, the area is equal to 5 times the sum of the lengths of the legs:
(1/2)*a*b = 5*(a+b)
a*b = 10*(a+b)

Expanding the right side:
a*b = 10a + 10b

Rearranging terms:
a*b - 10a - 10b = 0
a*b - 10a - 10b + 100 = 100
(a-10)(b-10) = 100

Now we need to find all the pairs of integers whose difference is 10 and product is 100:
(1,100), (2,50), (4,25), (5,20), (10,10)

Therefore, the possible values of a and b are (11,110), (12,60), (14,35), (15,30), (20,20).

The possible areas of the cool right triangles are:
(1/2)*11*110 = 605
(1/2)*12*60 = 360
(1/2)*14*35 = 245
(1/2)*15*30 = 225
(1/2)*20*20 = 200

The sum of all possible areas is 605 + 360 + 245 + 225 + 200 = 1635.

Therefore, the sum of all the different possible areas of cool right triangles is 1635.