Question

For fun (:



The drawing messed up and the link doesn't work. The question was:

\(\Large \frac{a}{b+c} + \frac{b}{a+c} = \frac{c}{a+b}=1\)

Find

\(\Large \frac{a^2}{b+c} + \frac{b^2}{a+c} + \frac{c^2}{a+b}=??\)

Answer 1

Link in case the drawing goes crazy in the future: https://i.imgsafe.org/7ec1a7e7ca.png

Answer 2

@mayankdevnani let's do this :)

Answer 3

yw :D

Answer 4

i got it !

Answer 5

mayankdevnani: hint:- multiply a number/variable with given expression

Answer 6

Another Hint :- Carefully observe both expressions, i mean compare their numerator and denominator part

Answer 7

mhmmm, I'm just going to do something to get started

let's call

b + c = x
a + c = y
a + b = z

so we have

a/x + b/y + c/z = 1

and we're trying to find the value of a^2/x + b^2/y + c^2/z

mhmmm, still stuck ;p

Answer 8

mhmm, a different way...multiply it by a

\(\Large a\left(\frac{a}{b+c} + \frac{b}{a+c} +\frac{c}{a+b}\right) = a(1)\)

\(\Large \frac{a^2}{b+c} + \frac{ab}{a+c} +\frac{ac}{a+b} = a\)

ummm... multiply it by b

\(\Large \frac{ab}{b+c} + \frac{b^2}{a+c} +\frac{bc}{a+b} = b\)

and multiply it by c

\(\Large \frac{ac}{b+c} + \frac{bc}{a+c} +\frac{c^2}{a+b} = c\)

now let's see... ummm... mhmmm

Answer 9

oh, I meant multiply the original equation by a, and then original expression by b, and then the original expression by c, btw

now we have three expressions, mhmmm.... ummmm

Answer 10

well going

Answer 11

why dun you add them ?

Answer 12

I tried doing it mentally but there are the 3 fractions we want the a^2 + b^2 + c^2 and then the others have a different denominators so I got confused :P

Answer 13

add them
show your work

Answer 14

I found the interesting fact that

a^3+b^3+c^3 = -abc

but I don't know how useful it is

Answer 15

seems unimportant

Answer 16

That's interesting Jim! :)

Adding them:
\[\frac{a^2}{b+c} + \frac{b^2}{a+c} +\frac{c^2}{a+b}+ \frac{ab}{a+c} +\frac{ab}{b+c} +\frac{ac}{a+b} +\frac{bc}{a+b}+\\ \frac{bc}{a+c}+ \frac{ac}{b+c} = a+b+c\]

Answer 17

oh nvm, that's a much easier way that I didn't see

Answer 18

good tso,now take common factors out

Answer 19

from the numerator or the denominator? :b

Answer 20

denominator

Answer 21

i get 0 , which makes the first condition seem impossible.

Answer 22

first condition seems impossible
how ?

Answer 23

if 2nd expression is 0, this is only satisfied if a=b=c = 0 . Then you get 0/0 terms in 1st condition.

Answer 24

\[\frac{a^2}{b+c} + \frac{b^2}{a+c} +\frac{c^2}{a+b} +\frac{ac}{a+b} +\frac{bc}{a+b}+\\ \frac{bc}{a+c}+ \frac{ab}{a+c} + \frac{ac}{b+c}+\frac{ab}{b+c} = a+b+c\]

\[\frac{a^2}{b+c} + \frac{b^2}{a+c} +\frac{c^2}{a+b} +\frac{ac+bc}{a+b} + \frac{bc+ab}{a+c} + \frac{ac+ab}{b+c} = a+b+c\]

OH I SEE IT NOW

\[\frac{a^2}{b+c} + \frac{b^2}{a+c} +\frac{c^2}{a+b} +\frac{c\cancel{(a+b)}}{\cancel{a+b}} + \frac{b\cancel{(a+c)}}{\cancel{a+c}} + \frac{a\cancel{(b+c)}}{\cancel{b+c}} = a+b+c\]

\[\frac{a^2}{b+c} + \frac{b^2}{a+c} +\frac{c^2}{a+b} +c+ b + a = a+b+c\]

\[\frac{a^2}{b+c} + \frac{b^2}{a+c} +\frac{c^2}{a+b} =0\]

The answer is 0!

Answer 25

Well, they are adding up to 0 not multiplying to 0, mhmmm

Answer 26

oh nevermind, was ignoring denominators. a,b,c can still be negative

Answer 27

:D

Answer 28

Thanks for the interesting question, Mayank. :D

I might use it to test a few people in real life maybe xD

Answer 29

TSO that is not fun. That is a question for 8th grade.

Answer 30

such easy

Answer 31

Nice!