Math trivia problem of the day

[GoldenBear]

I got this cool book with lots of fun math problems in it, so let's see how you guys stack up with totally useless problems..

(a + b - c)/c = (a - b + c)/b = (-a + b + c)/a

Prove that a, b, and c are equal but not through way of number plugging.

 

[PassatVR6]

Im not THAT bored

 

[GoldenBear]

Uh okay, thanks for pointing that out..

Are you going to do the same for every thread you're not interested in now?

 

[Viper GTS]

I'd have to agree with that... There are lots of things I'd like to do when I'm bored (sex, sleep, etc.) but math is NOT one of them. I did one step & said "Eff it, I'm not doing this." Maybe Napalm will do it. He's crazy like that.

Viper GTS

 

[GoldenBear]

<< There are lots of things I'd like to do when I'm bored (sex, >>

So abstinence only comes as a result of finding better things to do? Riight..

 

[stndn]

i'm working on that and go around in circles
can i cheat?

 

[DannyLove]

<< I'd have to agree with that... There are lots of things I'd like to do when I'm bored (sex, sleep, etc.) but math is NOT one of them. I did one step &amp; said &quot;Eff it, I'm not doing this.&quot; Maybe Napalm will do it. He's crazy like that.

Viper GTS >>

sex??? i thought you were a virgin?

danny~!

 

[Viper GTS]

Hey, I said they were things I would like to do, I never said they all happened. Suffice it to say I do a lot more sleeping than I do screwing.



Viper GTS

 

[GoldenBear]

<< i'm working on that and go around in circles
can i cheat? >>

Define cheating.

<< sex??? i thought you were a virgin? >>

I think the way he does it, it's not considered losing your virginity.

 

[DannyLove]

<< Hey, I said they were things I would like to do, I never said they all happened. Suffice it to say I do a lot more sleeping than I do screwing.



Viper GTS >>

yea, after i re-read it i figured it out, i didnt want to edit it, i'm rather lazy today :frown:

danny~!

 

[AmazonRasta]

I don't know if I'm on the right track... but I got this far... Hope it's right. Please don't laugh if I'm dumb .

(a + b + c)
--------------------- = 0
(-a - b + c)

 

[dopcombo]

lemme try

whats the name of the book btw?

 

[thEnEuRoMancER]

There are two possible solutions for this problem:

either a=b=c or a+b+c=0

 

[Napalm381]

You can rearrange
(a + b - c)/c = (a - b + c)/b
=>
(a + b - c)/(a - b + c) = (c / b).

Therefore,
a + b - c = c, or a + b = 2c.

Similarly, you can show a + c = 2b, and b + c = 2a.

Add those three and rearrange, gets you 0 = 0.

Eh?

 

[dopcombo]

napalm

i dun think u can equate numerators and denominators like that

4/2 = 8/4
does not mean that 4 = 8

or am i reading u wrongly?

damn, this is a hard one.

 

[Napalm381]

Hmmm, good point. Oh well, that's what you get on three hours sleep .

Multiply the first two terms to get ab + b^2 - cb = ac -bc +c^2.
Same thing for second and third, and first and third.

Add, rearrange, 0 = 0.

 

[dopcombo]

ok, i think i got it.

let a, b, c be the 3 sides of a triangle.
a+b-c is the difference between the sum of sides a+b and side c.
by triangle inequality this has to be a positive value.

(a+b-c)/c is therefore the percentage difference between the other sides and side c.
similarly for the other 2 expressions as well.

therefore, the only way the % difference of the sides is the same is if the triangle is an equilateral triangle.

And therefore a=b=c

 

[thEnEuRoMancER]

dopcombo: why do you assume the triangle inequality (a+b>c etc.) holds for these three numbers?

First rearrange the equations like this (it's not too hard):

(a+b)/c = (a+c)/b = (b+c)/a

let's rearrange first equality some more:

ab + b^2 = ac + c^2 and

a(b-c) = -(b-c)*(c+b)

This is true if a=-(c+b) or if (b-c)=0, this means if a+b+c=0 or if b=c

Rearrange the other two equalities and get a=c and a=b

So either a=b=c or a+b+c=0

 

[BilltheAvenger]

Ok, this is an extremely stupid and elementary problem. If we have to think to solve this one, how about just go on and prove that square root of 2 is irrational?

 

[BilltheAvenger]

Prove that if given six points (no three of them are in a straight line) all interconnected with one another, and if these connecting lines (edges) are either blue or red, then there has to be at least one triangle that has all three sides of the same color(either blue or red). (Diagrams are helpful)

 

[thEnEuRoMancER]

That's already been done here

edit: I mean the proof that sqrt(2) is irrational

 

[daddyo]

Napalm was on the right track.

You now have 3 equations and 3 unknowns.

1)a+b=2c
2)a+c=2b
3)b+c=2a

Solve (1) for a, substitute into (2), yields c=b
Solve (1) for b, substitute into (3), yields a=c
Solve (3) for c, substitute into (2), yields a=b

There you have it, a=b=c, *proof*