okay I did not do that
rawb
Friday, April 1, 2022
I proved it is not possible for 2sqrt(a) = sqrt(b)
since you said roots originally
proof is the same regardless actually
the 4 is just replaced with a sqrt(2)
wait
but then the 2a is sqrt(2)/2
I actually have to check this case 1 second
the one I posted has 3 sides
the hypotenuse is length a
we don't know the other side, but one side is 2a
because for our original isosceles triangle, I chose the two equal sides to be equal to a
and the bottom to be b
I then solved sqrt(a) + sqrt(a) = sqrt(b)
so b = 4a
let me draw this out
want all a,b in the set of real numbers such that 2sqrt(a) = sqrt(b) and the triangle is valid
that is the problem
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