another math question for you peeps, try and solve this bad boy!?

Question

There are exactly three right triangles with integral sides having the property that its area is equal to twice its perimeter. Find the side lengths all three triangles.

please show all work!!!

Answer 1

(9,40,41) : A = 180 ; P = 90
(12,16,20) : A = 96 ; P = 48
(10,24,26) : A = 120 ; P = 60

Proof:
Let the sides be a, b, and √(a^2+b^2)
Area = ab/2
perimeter = a+b+ √(a^2+b^2)
so ab/2 = 2(a+b+ √(a^2+b^2))

ab = 4(a+b+ √(a^2+b^2))
ab - 4a - 4b = 4√(a^2+b^2)
(ab - 4a - 4b)^2 = 16(a^2+b^2)
(ab)^2-8a^2b-8ab^2 +16a^2+32ab+16b^2 = 16a^2 + 16b^2
(ab)^2 + 32ab = 8ab(a+b)
ab + 32 = 8(a+b)
ab - 8a = 8b - 32
a(b-8) = 8(b-4)
a = 8(b-4)/(b-8)

Let c = b-8
a = 8(c+4)/c = 8(1 + 4/c) = 8 + 32/c
The only values of c that make 32/c (and thus a) an integer are 1, 2, 4, 8, 16 and 32
[note: to be rigorous, it should be pointed out that the negatives of these values would also produce an integer, but that these would imply a non-positive a or b]
and since b = c+8 ,
then b = 9, 10, 12, 16, 24, or 40
while a = 40, 24, 16, 12, 10 or 9 respectively.

These lead to the three solutions given, (as well as the three equivalent solutions where a and b are switched.)

Answer 2

Hi,

9,40,41 because area = 1/2*9*40 = 180
and perimeter = 9 + 40 + 41 = 90, half the area

12,16,20 because area = 1/2*12*16 = 96
and perimeter = 12 + 16 + 20 = 48, half the area

10,24,26 because area = 1/2*10*24 = 120
and perimeter = 10 + 24 + 26 = 60, half the area

I hope that does it !!!! :-)

Answer 3

It's true that
9, 40, 41
12,16, 20
and
10, 24, 26
are 3 such triangles.
BUT
How do we know there are no more??
I'll return to prove this soon!

Answer 4

i+d+o+n+t+k+n+o+w= i dont know.