Question

Find all right triangles having sides with integral lengths that form an arithmetic sequence.

... and that is all it says ... :??

Answer 1

what grade / course is this for?

Answer 2

Pre-calculus, sorry, this answer is a little late. We just finished a unit on vectors, if that's helpful...

Answer 3

Let A, B, C be the sides of the triangle


Furthermore, let


A = x

B = x+d

C = x+2d


By the pythagorean theorem


A^2 + B^2 = C^2


Plug in the given values and solve for x


x^2 + (x+d)^2 = (x+2d)^2

x^2 + x^2 + 2xd + d^2 = x^2 + 4xd + 4d^2

2x^2 + 2xd + d^2 = x^2 + 4xd + 4d^2

2x^2 + 2xd + d^2 - x^2 - 4xd - 4d^2 = 0

x^2 - 2xd - 3d^2 = 0

(x + d)(x - 3d) = 0

x+d = 0 or x-3d = 0

x = -d or x = 3d


Now if the side lengths are integers, then d must be an integer


Assuming that x is the smallest side length and all side lengths are positive (negative side lengths do not make sense), this means that x = -d is not possible.


So the only solution is x = 3d


So this means that the smallest side length is x = 3(1) = 3


So the smallest triangle possible is a 3,4,5 triangle


Any other triangle that fits this description will be multiples of 3,4,5

Answer 4

Thank you!

Answer 5

you're welcome