rawb

That may be true, but that wouldn't be what the quote implies.

and for the triangles to still actually geometrically work

it can be true for some of them right

Because the quote says any two sides, so that could be long + long or long + short

if the two sides you choose are the equal ones

I have concluded its not possible

The Pythagorean theorem - as a statement, not an equation - states that the sum of the square roots of the legs of a right triangle equal the square root of the hypotenuse, right?

Holy shit, I know how to spell hypotenuse, but typed it wrong five times before getting it right.

not root

square

so a^2 + b^2 = c^2

That's my mistake in typing.

the problem you stated reduces to finding all a such that this triangle is valid, but it can never be valid

I had the damn Scarecrow quote going in my head. X(

since this triangle violates the Pyhtagorean theorem

since the hypotenuses needs to be the largest size

is it not?

no

in this case its a

whereas one of the other sides is 2a

which is strictly always larger

for all a

I never tried to find out if it worked whatsoever with an isosceles triangle, like I never drew up an isosceles triangle, gave sides proper length for the angles, and saw if you could get c^2 to equal a^2+b^2 in even one case, because I dismissed the statement immediately because it said any two sides

oh

okay I did not do that

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

ABORT

ABORT

but then the 2a is sqrt(2)/2

uh

You're gonna blow your brain up

I actually have to check this case 1 second

And I wasn't even trying to argue math XD

i still dont see how this triangle violates Pythagorean theorem

the one I posted has 3 sides

the hypotenuse is length a

how is the hypotenuse a

arent the a's the legs

if you flip the triangle

because for our original isosceles triangle, I chose the two equal sides to be equal to a

and the bottom to be b

so b = 2a

I then solved sqrt(a) + sqrt(a) = sqrt(b)

I let Google do the work for me.

so b = 4a

right so b is the hypoteneuse

no

let me draw this out

a |_\c b

???

Mirror it, c' makes it iscosceles, right?

I can't read the writing.

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

...

Why are we on this, btw?

the first part of the problem is trivial. b = 4a

mans using an equilateral triangle in his diagram

its not

i know

its an isosceles

"its not drawn to scale" bullshit

fuck SAT questions like that

the drawing literally does not matter

i know

this is the new problem I am trying to solve

That's what I was trying to say with the ASCII above a |\ c b right triangle mirror it c' /_\ c 2b isosceles, correct?

because it is less immediately clear if there is no answer

where c' = c

This one might be possible

for all a

Let's say a = 5, b = 7