rawb

True or False: The sum of the square roots any two sides of an isosceles triangle is equal to the square root of the remaining side?

lost 20

so a net of -10ish

-7 i guess

is that the simpsons math thing you are referencing

I know they stated something like that

as a joke

No, it's a quote from The Wizard of Oz, when the Wizard gives the Scarecrow a degree, as a symbol of his brains, as Th.D, Doctor of Thinkology

oh lol

And it's 100% wrong, which makes it funnier.

I think the simpsons also referenced it then at some point because that is where I thought of

It's like they took the Pythagorean theorem and bastardized it as much as they could while having it be just close enough to recognize what they did.

That theorem is valid for isosceles triangles where the one side is 1/4th the size of the two equal size

although

I'd have to check to see if thats possible

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

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

we don't know the other side, but one side is 2a

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