If a/(b+c) = b/(c+a) = c/(a+b) = r,then r cannot any value except?

If a/(b+c) = b/(c+a) = c/(a+b) = r,then r cannot any value except


(a) 1/2


(b) -1


(c) 1/2 or -1


(d) -1/2 or -1


The answer is c but how

If a/(b+c) = b/(c+a) = c/(a+b) = r,then r cannot any value except?
a/(b+c)=b/(c+a) =%26gt; a(c+a)=b(c+a)


ac+a^2=bc+b^2


a^2-b^2=-ac+bc


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


1) a-b=0


a=b


or 2) (a-b)(a+b)=-c(a-b) /:(a-b)


=%26gt; a+b=-c


a+b+c=0


=%26gt; a/(b+c)=a/(-a)=r=-1





In the same way with b/(c+a)=c/(a+b) =%26gt;b=c or a+b+c=0


from b=c and a=c =%26gt;a/(b+c)=a/2a=1/2


that's all! if you have some questions mail me...
Reply:I a/(b+c)=b/(a+c) =%26gt; a2+ac=b2+bc


II a/(b+c)=c/(a+b) =%26gt; a2+ab=c2+bc


III b/(a+c)=c/(a+b) =%26gt; b2+ab=c2+ac





I,II =%26gt; a(b-c)=c2-b2 =%26gt; a(b-c)=(-b-c)(b-c)=%26gt; b=c or a+b+c=0


I, III=%26gt;b(c-a)=a2-c2 =%26gt; b(c-a)=(-a-c)(c-a)=%26gt; c=a or a+b+c=0


II,III=%26gt;c(b-a)=a2-b2 =%26gt; c(b-a)=(-b-a)(b-a)=%26gt; b=a or a+b+c=0





so you have to way


1- a=b=c =%26gt; r=a/(b+c)=a/(a+a)=1/2=0.5


2- a+b+c=0 =%26gt; r =a/(b+c)=a/(-a)=-1
Reply:Ok, all these answers would burst a Math novice's head blood vessels, so I come with a simple logic.


The answer is C, as everyone knows, and until u do a whole lot of manipulations, u might not know how they come up with the answer. But you can easily see that the answer is correct.


The C options show 1/2 and -1. Now, 1/2 could be written as 1/ (1+1), and 1/2 could be re-written as 1/ (1/2 + 1/2). In each instance, let each variable have one of the numbers, and try plugging them in. It'll work for both options. now, if u decide to try any other combo of numbers, it won't work for all of the variables.


Try it, it works!
Reply:Equate the last 2 parts of the statement by the 1st statement to make 2 equations


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


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





Multiply each by their LCD


a² + ac = b² + bc


a² + ab = c² + bc





Subtract the 1st equation from the 2nd equation


c² - b² = ab - ac





Transpose to the left


c² - b² + ac - ab = 0





Factor monomially and by special factoring


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





Factor the groups


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





There are 2 cases


c - b = 0 or c + b + a = 0


Thus,


b = c (Let it be case A) or a + b + c = 0 (Let it be case B)








---case A---


Since b = c and


a² + ac = b² + bc,





Substitute b for all c


a² + ab = b² + b²





Add and transpose to the left


a² + ab - 2b² = 0





factor


(a - b)(a + 2b) = 0





There are 2 cases for case A


a = b or a = -2b





For a = b = c, The value for r is r = a/(b + c) = a/(a + a) = a/2a = 1/2


For b = c,a = -2b, The value for r is r = a/(b + c) = (-2b)/(b + b) = -2b/2b = -1








---case B---


a + b + c = 0


Thus,


b + c = -a


The value for r is r = a/(b + c) = a/(-a) = -1





Thus, the only values for r that we got is 1/2 (from case A) and -1 (from both cases). Therefore, the answer is (c) 1/2 or -1.





(Note that there are some undefined values for r, specifically when b + c = 0, a + c = 0, and/or a + b = 0, but this does not affect our answer.)








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Reply:yes


the answer is C
Reply:a/(b+c) = b/(c+a) -%26gt; a^2 + ac = b^2 + bc (1)


b/(c+a) = c/(a+b) -%26gt; c^2 + ac = b^2 + ab (2)





(1) - (2) -%26gt; a^2 - c^2 = bc - ab


-%26gt; (a-c)*(a+c) = b*(c-a)


-%26gt; a = c (both sides are 0) or a + c = -b (cancel the random factor a-c)





If a = c then similarly we have b = c -%26gt; r = a/(b+c) = c/(c+c) = 1/2


If a+c = -b -%26gt; r = b/(c+a) = b/-b = -1
Reply:To explain r=1/2 is direct, when a=b=c, then sum of any 2 is the twice of any of them.


To explain r=-1, need to use complex numbers.


a=b=c=e^(ix), x=0, 2pi/3, 4pi/3


Visualize these 3 points in the complex plane, you will see that the addition of any two at the 180 degree rotation of the remaining one, and this rotation corresponds to a muliply by -1.


thus a = -1 x (b+c), b = -1 x (c+a) and c = -1 x (a+b).

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