 Posers and Puzzles

1. 06 Mar '08 00:08
Find, 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.
2. 06 Mar '08 00:381 edit
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}, {-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? 😛
3. 06 Mar '08 22:43
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? 😛
Oh, you are really clever. Exactly what I was looking for. Nice work. 😀