Is this geometry problem solvable? EDIT: Answer is no

[homercles337]

Originally posted by: DrPizza

Originally posted by: fanerman91
Yeah, I'm pretty sure the problem isn't solvable analytically. Thanks for all the help guys.

I might try to write some script that tries to find the unknown incrementally, checking to see whether the numbers given actually form a legit triangle or not.

Did you even read my second post? I spelled it out for you: c> (a-b)/2. And, the conditions on a and b are that a must be greater than b. *sigh*

Oh, i see what youre saying with the constraint that the lengths c = c'. I was thinking algorithmically, not analytically. My bad. Disregard the PM.

 

[FleshLight]

I think you can do it if the angle between A and B is known.

 

[BrownTown]

too bad it isn't...

 

[mjrpes3]

Originally posted by: BrownTown

Originally posted by: mjrpes3
what are you saying? based on that diagram, a has to be greater than b.

no

no what? no, that's not my point? or no, a can be less than b?

 

[BrownTown]

no, your wrong a and b can be whatever they want to be. To be physically relevant we could say rational numbers greater than 0 if you want.

 

[mjrpes3]

I don't know how to geometrically prove it, but I'm pretty sure you're wrong about that, given that c and b are on the same line, and c = c. You can, though, picture the triangle in your head and play around with the angles to understand more fully why. As the angle AC approaches 180 degrees (causing AB and CC to approach 0), a approaches the length of b, but will never be less than b. Just picture it in your head. Vice versa, as the angle of AC gets smaller and CC and AB get larger, a gets longer compared to b.

 

[crt1530]

Originally posted by: FleshLight
I think you can only do it if the angle between A and B is known.

Fixed.

It can't be found. The real question is whether or not you have to prove it.

 

[pennylane]

mjrpes3 is right.

One of the properties of a triangle is that the sum of two of the sides must be greater than the remaining side.

So, (a) + (c) > (b + c)
which simplifies to a > b

but then what do I know.

 

[crt1530]

Bah...I wouldn't have bothered replying if I saw DrPizza had already posted.

 

[smack Down]

Originally posted by: fanerman91
mjrpes3 is right.

One of the properties of a triangle is that the sum of two of the sides must be greater than the remaining side.

So, (a) + (c) > (b + c)
which simplifies to a > b

but then what do I know.

The triangle inequality doesn't say which side must be longer. We can have
a + (b+c) > c Which is just
a + b > 0

 

[potoba]

lol, this is not too hard; I think it's solvable. Law of cosine to find the side between a and b, what kind of triangle you will have? You know the 3 angles and one side of that triangle already!!!

 

[sao123]

Originally posted by: smack Down

Originally posted by: fanerman91
mjrpes3 is right.

One of the properties of a triangle is that the sum of two of the sides must be greater than the remaining side.

So, (a) + (c) > (b + c)
which simplifies to a > b

but then what do I know.

The triangle inequality doesn't say which side must be longer. We can have
a + (b+c) > c Which is just
a + b > 0

Acutally The triangle inequality says this must work for all sides.
3 sides...
a
c
(b + c)

a+c > b+c
a > b

a + b + c > c
a + b > 0

c + b + c > a
2c > a - b
c > (a-b) / 2

All three constraints apply... not just 1 chosen arbitrarily.

 

[Bryophyte]

Without reading the whole thread to see if someone else has posted it, I'll pass on what Mr Bryo just sent me regarding this problem, by IM:

The answer is c = (a^2 - b^2)/(2*(b+a*cos x))

 

[DrPizza]

Originally posted by: Alone
Here.

smartass
I loled pretty hard.

 

[BrownTown]

TBH if you are doing any math on this problem then you are either way overcomplicating it or have something wrong.

 

[Alone]

Originally posted by: DrPizza

Originally posted by: Alone
Here.

smartass
I loled pretty hard.

I try.

 

[xtknight]

I'm not sure it could be solved even if you knew the triangle on the bottom was a right triangle.

 

[chuckywang]

Let a=5 and b=1.

Then if c=5, you would have a 5, 5, 6 triangle, which is perfectly constructable.

If c=4, you would have a 5, 4, 5 triangle, again perfectly constructable.

So no, c cannot be uniquely determined by the above counterexample.

 

[chuckywang]

Originally posted by: Bryophyte
Without reading the whole thread to see if someone else has posted it, I'll pass on what Mr Bryo just sent me regarding this problem, by IM:

The answer is c = (a^2 - b^2)/(2*(b+a*cos x))

That would be right except that x is not an angle, it's just a label to a vertex. If the angle of x is known, then yes, c can be uniquely determined.

 

[sdifox]

Unless you know the angle between A and B, then there are infinite answers.

P.S. Yes, I am a little slow... I just had to think through everything and think if there is some way... counldn't find one.

Even if B is 0, we still can't determine C by knowing A.