- 06 Mar '08 00:08Find, and prove that you have found every right triangle that satisfies the following conditions:

-Its side lengths are integers

-Its area is numerically equivalent to its perimeter

You can use Euclid's formula for calculating Pythagorean triples for this problem, but if you're really clever, you won't need it. I just want to see what people come up with. - 06 Mar '08 00:38 / 1 editLets see:

Area = ab/2

Perimeter = a+b+sqrt(a^2+b^2)

ab/2 = a+b+sqrt(a^2+b^2)

ab/2-a-b = sqrt(a^2+b^2)

ab-2a-2b=2sqrt(a^2+b^2)

squaring:

a^2*b^2+4a^2+4b^2-4a^2*b-4ab^2+8ab = 4a^2+4b^2

a^2*b^2-4a^2*b-4ab^2+8ab = 0

Divide by ab:

ab-4a-4b+8=0

ab-4a-4b+16=8

(a-4)(b-4)=8

a-4 and b-4 are integers, so the possibilities are

{a-4,b-4} = {-1,-8}, {-2,-4}, {1,8}, {2,4}

{a,b} = {3,-4}, {2,0}, {5,12}, {6,8}

Of course only the last two are OK so the solutions are:

The triangles' sides are 5,12,13 or 6,8,10.

Am I really clever? - 06 Mar '08 22:43

Oh, you are really clever. Exactly what I was looking for. Nice work.*Originally posted by David113***Lets see:**

Area = ab/2

Perimeter = a+b+sqrt(a^2+b^2)

ab/2 = a+b+sqrt(a^2+b^2)

ab/2-a-b = sqrt(a^2+b^2)

ab-2a-2b=2sqrt(a^2+b^2)

squaring:

a^2*b^2+4a^2+4b^2-4a^2*b-4ab^2+8ab = 4a^2+4b^2

a^2*b^2-4a^2*b-4ab^2+8ab = 0

Divide by ab:

ab-4a-4b+8=0

ab-4a-4b+16=8

(a-4)(b-4)=8

a-4 and b-4 are integers, so the possibilities are

{a-4,b-4} = {-1,-8}, { ...[text shortened]... so the solutions are:

The triangles' sides are 5,12,13 or 6,8,10.

Am I really clever?