that is the problem
rawb
Friday, April 1, 2022
also a =/= b
the first part of the problem is trivial. b = 4a
its not
its an isosceles
the drawing literally does not matter
this is the new problem I am trying to solve
because it is less immediately clear if there is no answer
This one might be possible
for all a
I don't think it violates the pythagorean theorem
since it just means you have an isosceles triangle with a height equal to sqrt(2)a/2
so yeah, the set of triangles is S = {(a,b) in R: b=sqrt(2)a}
Replying to Phijkchu_Pikachu #klombak since it just means you have an isosceles triangle with a height equal to sqrt(2)a/2 (edited)
The triangle I made by taking a right triangle and mirroring it, no, it doesn't, but that's not the definition of an isosceles triangle, and the Scarecrow's statement said any two sides of an isosceles triangle
Yeah but realistically, there is only one logical way you can do that
since the other way means one side is equal to 0
which is meaningless
since if you have a^2 + b^2 = a^2 you just have b=0
so it has to be 2a^2 = b^2
I said lets not argue math as a joke because you misunderstood my arrow notation
lol
Replying to sudo TPK #sylfar So the isosceles triangle would be c = 13, c' = 13, 2b = 24
Also this one does not work for either 2a^2 = b^2 or 2sqrt(a) = sqrt(b)
it is a valid isosceles triangle, but not one that would work for either theorem
anyways, the tl;dr of this all is, you cannot create an isosceles triangle such that if the sides are (a,a,b), 2sqrt(a) = sqrt(b)
but you can for 2a^2 = b^2