AMC 1834 - If $a, b $ and $c$ are nonzero numbers such that (a+b-c)/c=(a-b+c)/b=(-a+b+c)/a and...












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I am getting stuck on this question:
If a, b, and c, are nonzero numbers such that $frac{a+b-c}{c}$=$frac{a-b+c}{b}$=$frac{-a+b+c}{a}$ and x=$frac{(a+b)(b+c)(c+a)}{abc}$ and x<0, then x=?



What I have done so far: Consider, if $frac{a}{b}$=$frac{c}{d}$, and (b+d) is non-zero, then $frac{a+c}{b+d}$=$frac{a}{b}$=$frac{c}{d}$. So with that fact, I did... $frac{a+b-c}{c}$=$frac{a-b+c}{b}$ $implies$
$frac{2a}{c+b}$ and $frac{2a}{c+b}$=$frac{-a+b+c}{a}$$implies$ $frac{(a+b+c)}{a+b+c}$ = 1. Then I took the original equations and did $frac{a+b-c}{c}$=1, so $a+b=2c$ , and similarly, $(a+c)=2b$ and $b+c=2a$. So plugging that into x=$frac{(a+b)(b+c)(c+a)}{abc}$ we have $x=frac{(2a)(2b)(2c)}{abc}$, which is $x=frac{8(abc)}{abc}=8$. This is wrong though, because it says in the problem x<0. So how do I do this?






algebra-precalculus fractions







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    $begingroup$


    I am getting stuck on this question:
    If a, b, and c, are nonzero numbers such that $frac{a+b-c}{c}$=$frac{a-b+c}{b}$=$frac{-a+b+c}{a}$ and x=$frac{(a+b)(b+c)(c+a)}{abc}$ and x<0, then x=?



    What I have done so far: Consider, if $frac{a}{b}$=$frac{c}{d}$, and (b+d) is non-zero, then $frac{a+c}{b+d}$=$frac{a}{b}$=$frac{c}{d}$. So with that fact, I did... $frac{a+b-c}{c}$=$frac{a-b+c}{b}$ $implies$
    $frac{2a}{c+b}$ and $frac{2a}{c+b}$=$frac{-a+b+c}{a}$$implies$ $frac{(a+b+c)}{a+b+c}$ = 1. Then I took the original equations and did $frac{a+b-c}{c}$=1, so $a+b=2c$ , and similarly, $(a+c)=2b$ and $b+c=2a$. So plugging that into x=$frac{(a+b)(b+c)(c+a)}{abc}$ we have $x=frac{(2a)(2b)(2c)}{abc}$, which is $x=frac{8(abc)}{abc}=8$. This is wrong though, because it says in the problem x<0. So how do I do this?






    algebra-precalculus fractions







    share|cite|improve this question











    $endgroup$















      1












      1








      1


      1



      $begingroup$


      I am getting stuck on this question:
      If a, b, and c, are nonzero numbers such that $frac{a+b-c}{c}$=$frac{a-b+c}{b}$=$frac{-a+b+c}{a}$ and x=$frac{(a+b)(b+c)(c+a)}{abc}$ and x<0, then x=?



      What I have done so far: Consider, if $frac{a}{b}$=$frac{c}{d}$, and (b+d) is non-zero, then $frac{a+c}{b+d}$=$frac{a}{b}$=$frac{c}{d}$. So with that fact, I did... $frac{a+b-c}{c}$=$frac{a-b+c}{b}$ $implies$
      $frac{2a}{c+b}$ and $frac{2a}{c+b}$=$frac{-a+b+c}{a}$$implies$ $frac{(a+b+c)}{a+b+c}$ = 1. Then I took the original equations and did $frac{a+b-c}{c}$=1, so $a+b=2c$ , and similarly, $(a+c)=2b$ and $b+c=2a$. So plugging that into x=$frac{(a+b)(b+c)(c+a)}{abc}$ we have $x=frac{(2a)(2b)(2c)}{abc}$, which is $x=frac{8(abc)}{abc}=8$. This is wrong though, because it says in the problem x<0. So how do I do this?






      algebra-precalculus fractions







      share|cite|improve this question











      $endgroup$




      I am getting stuck on this question:
      If a, b, and c, are nonzero numbers such that $frac{a+b-c}{c}$=$frac{a-b+c}{b}$=$frac{-a+b+c}{a}$ and x=$frac{(a+b)(b+c)(c+a)}{abc}$ and x<0, then x=?



      What I have done so far: Consider, if $frac{a}{b}$=$frac{c}{d}$, and (b+d) is non-zero, then $frac{a+c}{b+d}$=$frac{a}{b}$=$frac{c}{d}$. So with that fact, I did... $frac{a+b-c}{c}$=$frac{a-b+c}{b}$ $implies$
      $frac{2a}{c+b}$ and $frac{2a}{c+b}$=$frac{-a+b+c}{a}$$implies$ $frac{(a+b+c)}{a+b+c}$ = 1. Then I took the original equations and did $frac{a+b-c}{c}$=1, so $a+b=2c$ , and similarly, $(a+c)=2b$ and $b+c=2a$. So plugging that into x=$frac{(a+b)(b+c)(c+a)}{abc}$ we have $x=frac{(2a)(2b)(2c)}{abc}$, which is $x=frac{8(abc)}{abc}=8$. This is wrong though, because it says in the problem x<0. So how do I do this?






      algebra-precalculus fractions




      algebra-precalculus fractions






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      edited Dec 20 '18 at 17:49









      Ankit Kumar

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      1,494221










      asked Dec 20 '18 at 17:26









      Kenneth DangKenneth Dang

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          3 Answers
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          $begingroup$

          $$frac{a+b-c}{c}=frac{a-b+c}{b}$$
          $$ab+b^2-bc=ac-bc+c^2implies a(b-c)=(b+c)(c-b)$$
          $$implies b=c quad ORquad a+b+c=0$$
          If we take $b=c$, $$frac{a}{c}=frac{-a+2c}{a}quad using (1)and (3).$$
          $$implies ({frac{a}{c}})^2=1quad ORquad -2$$
          Obviosuly $-2$ isn't possible, $a=c$ will give $x=1$ and $a=-c$ will give $x=0$.



          So, $a+b+c=0implies a=-(b+c), b=-(a+c), c=-(a+b)implies x=-1.$






          share|cite|improve this answer









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          • 1




            $begingroup$
            Dear Mr. Kumar, how do you know from a(b-c)=(b+c)(c-b) that b=c or a+b+c=0?
            $endgroup$
            – Kenneth Dang
            Dec 20 '18 at 18:44






          • 2




            $begingroup$
            $a(b-c)=(b+c)(c-b)$ is same as $a(b-c)+(b-c)(b+c)=0$ which is same as $(b-c)(a+b+c)=0$. Btw thank you, no one addressed me like that before ;)
            $endgroup$
            – Ankit Kumar
            Dec 20 '18 at 18:49





















          1












          $begingroup$

          Hint: $$a=-frac{b}{2},b=b,c=-frac{b}{2}$$ solves your problem and $$x=-1$$






          share|cite|improve this answer









          $endgroup$





















            0












            $begingroup$

            You correctly derived:
            $$frac{2a}{c+b}=frac{-a+b+c}{a}$$
            Your mistake is here:




            Then I took the original equations and did $frac{a+b-c}{c}=1$




            Instead, denote: $frac{b+c}{a}=t$, then the above equation becomes:
            $$frac2{t}=-1+t Rightarrow t^2-t-2=0 Rightarrow t_1=-1 text{and} t_2=2.$$



            Also note that:
            $$frac{a+b-c}{c}=frac{a-b+c}{b}=frac{-a+b+c}{a}iff \
            frac{a+b}{c}-1=frac{a+c}{b}-1=frac{b+c}{a}-1 iff \
            frac{a+b}{c}=frac{a+c}{b}=frac{b+c}{a}=t.$$
            Hence:
            $$x=frac{(a+b)(b+c)(c+a)}{abc}=t^3=(-1)^3=-1;\
            x=frac{(a+b)(b+c)(c+a)}{abc}=t^3=2^3not< 0.$$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Wow, I never thought of it like that. Do you an assertion as to when I should think similarly to this when dealing with problem solving?
              $endgroup$
              – Kenneth Dang
              Dec 20 '18 at 20:59











            Your Answer





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            3 Answers
            3






            active

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            3 Answers
            3






            active

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            active

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            active

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            1












            $begingroup$

            $$frac{a+b-c}{c}=frac{a-b+c}{b}$$
            $$ab+b^2-bc=ac-bc+c^2implies a(b-c)=(b+c)(c-b)$$
            $$implies b=c quad ORquad a+b+c=0$$
            If we take $b=c$, $$frac{a}{c}=frac{-a+2c}{a}quad using (1)and (3).$$
            $$implies ({frac{a}{c}})^2=1quad ORquad -2$$
            Obviosuly $-2$ isn't possible, $a=c$ will give $x=1$ and $a=-c$ will give $x=0$.



            So, $a+b+c=0implies a=-(b+c), b=-(a+c), c=-(a+b)implies x=-1.$






            share|cite|improve this answer









            $endgroup$









            • 1




              $begingroup$
              Dear Mr. Kumar, how do you know from a(b-c)=(b+c)(c-b) that b=c or a+b+c=0?
              $endgroup$
              – Kenneth Dang
              Dec 20 '18 at 18:44






            • 2




              $begingroup$
              $a(b-c)=(b+c)(c-b)$ is same as $a(b-c)+(b-c)(b+c)=0$ which is same as $(b-c)(a+b+c)=0$. Btw thank you, no one addressed me like that before ;)
              $endgroup$
              – Ankit Kumar
              Dec 20 '18 at 18:49


















            1












            $begingroup$

            $$frac{a+b-c}{c}=frac{a-b+c}{b}$$
            $$ab+b^2-bc=ac-bc+c^2implies a(b-c)=(b+c)(c-b)$$
            $$implies b=c quad ORquad a+b+c=0$$
            If we take $b=c$, $$frac{a}{c}=frac{-a+2c}{a}quad using (1)and (3).$$
            $$implies ({frac{a}{c}})^2=1quad ORquad -2$$
            Obviosuly $-2$ isn't possible, $a=c$ will give $x=1$ and $a=-c$ will give $x=0$.



            So, $a+b+c=0implies a=-(b+c), b=-(a+c), c=-(a+b)implies x=-1.$






            share|cite|improve this answer









            $endgroup$









            • 1




              $begingroup$
              Dear Mr. Kumar, how do you know from a(b-c)=(b+c)(c-b) that b=c or a+b+c=0?
              $endgroup$
              – Kenneth Dang
              Dec 20 '18 at 18:44






            • 2




              $begingroup$
              $a(b-c)=(b+c)(c-b)$ is same as $a(b-c)+(b-c)(b+c)=0$ which is same as $(b-c)(a+b+c)=0$. Btw thank you, no one addressed me like that before ;)
              $endgroup$
              – Ankit Kumar
              Dec 20 '18 at 18:49
















            1












            1








            1





            $begingroup$

            $$frac{a+b-c}{c}=frac{a-b+c}{b}$$
            $$ab+b^2-bc=ac-bc+c^2implies a(b-c)=(b+c)(c-b)$$
            $$implies b=c quad ORquad a+b+c=0$$
            If we take $b=c$, $$frac{a}{c}=frac{-a+2c}{a}quad using (1)and (3).$$
            $$implies ({frac{a}{c}})^2=1quad ORquad -2$$
            Obviosuly $-2$ isn't possible, $a=c$ will give $x=1$ and $a=-c$ will give $x=0$.



            So, $a+b+c=0implies a=-(b+c), b=-(a+c), c=-(a+b)implies x=-1.$






            share|cite|improve this answer









            $endgroup$



            $$frac{a+b-c}{c}=frac{a-b+c}{b}$$
            $$ab+b^2-bc=ac-bc+c^2implies a(b-c)=(b+c)(c-b)$$
            $$implies b=c quad ORquad a+b+c=0$$
            If we take $b=c$, $$frac{a}{c}=frac{-a+2c}{a}quad using (1)and (3).$$
            $$implies ({frac{a}{c}})^2=1quad ORquad -2$$
            Obviosuly $-2$ isn't possible, $a=c$ will give $x=1$ and $a=-c$ will give $x=0$.



            So, $a+b+c=0implies a=-(b+c), b=-(a+c), c=-(a+b)implies x=-1.$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 20 '18 at 17:43









            Ankit KumarAnkit Kumar

            1,494221




            1,494221








            • 1




              $begingroup$
              Dear Mr. Kumar, how do you know from a(b-c)=(b+c)(c-b) that b=c or a+b+c=0?
              $endgroup$
              – Kenneth Dang
              Dec 20 '18 at 18:44






            • 2




              $begingroup$
              $a(b-c)=(b+c)(c-b)$ is same as $a(b-c)+(b-c)(b+c)=0$ which is same as $(b-c)(a+b+c)=0$. Btw thank you, no one addressed me like that before ;)
              $endgroup$
              – Ankit Kumar
              Dec 20 '18 at 18:49
















            • 1




              $begingroup$
              Dear Mr. Kumar, how do you know from a(b-c)=(b+c)(c-b) that b=c or a+b+c=0?
              $endgroup$
              – Kenneth Dang
              Dec 20 '18 at 18:44






            • 2




              $begingroup$
              $a(b-c)=(b+c)(c-b)$ is same as $a(b-c)+(b-c)(b+c)=0$ which is same as $(b-c)(a+b+c)=0$. Btw thank you, no one addressed me like that before ;)
              $endgroup$
              – Ankit Kumar
              Dec 20 '18 at 18:49










            1




            1




            $begingroup$
            Dear Mr. Kumar, how do you know from a(b-c)=(b+c)(c-b) that b=c or a+b+c=0?
            $endgroup$
            – Kenneth Dang
            Dec 20 '18 at 18:44




            $begingroup$
            Dear Mr. Kumar, how do you know from a(b-c)=(b+c)(c-b) that b=c or a+b+c=0?
            $endgroup$
            – Kenneth Dang
            Dec 20 '18 at 18:44




            2




            2




            $begingroup$
            $a(b-c)=(b+c)(c-b)$ is same as $a(b-c)+(b-c)(b+c)=0$ which is same as $(b-c)(a+b+c)=0$. Btw thank you, no one addressed me like that before ;)
            $endgroup$
            – Ankit Kumar
            Dec 20 '18 at 18:49






            $begingroup$
            $a(b-c)=(b+c)(c-b)$ is same as $a(b-c)+(b-c)(b+c)=0$ which is same as $(b-c)(a+b+c)=0$. Btw thank you, no one addressed me like that before ;)
            $endgroup$
            – Ankit Kumar
            Dec 20 '18 at 18:49













            1












            $begingroup$

            Hint: $$a=-frac{b}{2},b=b,c=-frac{b}{2}$$ solves your problem and $$x=-1$$






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              Hint: $$a=-frac{b}{2},b=b,c=-frac{b}{2}$$ solves your problem and $$x=-1$$






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                Hint: $$a=-frac{b}{2},b=b,c=-frac{b}{2}$$ solves your problem and $$x=-1$$






                share|cite|improve this answer









                $endgroup$



                Hint: $$a=-frac{b}{2},b=b,c=-frac{b}{2}$$ solves your problem and $$x=-1$$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 20 '18 at 17:44









                Dr. Sonnhard GraubnerDr. Sonnhard Graubner

                76.5k42866




                76.5k42866























                    0












                    $begingroup$

                    You correctly derived:
                    $$frac{2a}{c+b}=frac{-a+b+c}{a}$$
                    Your mistake is here:




                    Then I took the original equations and did $frac{a+b-c}{c}=1$




                    Instead, denote: $frac{b+c}{a}=t$, then the above equation becomes:
                    $$frac2{t}=-1+t Rightarrow t^2-t-2=0 Rightarrow t_1=-1 text{and} t_2=2.$$



                    Also note that:
                    $$frac{a+b-c}{c}=frac{a-b+c}{b}=frac{-a+b+c}{a}iff \
                    frac{a+b}{c}-1=frac{a+c}{b}-1=frac{b+c}{a}-1 iff \
                    frac{a+b}{c}=frac{a+c}{b}=frac{b+c}{a}=t.$$
                    Hence:
                    $$x=frac{(a+b)(b+c)(c+a)}{abc}=t^3=(-1)^3=-1;\
                    x=frac{(a+b)(b+c)(c+a)}{abc}=t^3=2^3not< 0.$$






                    share|cite|improve this answer









                    $endgroup$













                    • $begingroup$
                      Wow, I never thought of it like that. Do you an assertion as to when I should think similarly to this when dealing with problem solving?
                      $endgroup$
                      – Kenneth Dang
                      Dec 20 '18 at 20:59
















                    0












                    $begingroup$

                    You correctly derived:
                    $$frac{2a}{c+b}=frac{-a+b+c}{a}$$
                    Your mistake is here:




                    Then I took the original equations and did $frac{a+b-c}{c}=1$




                    Instead, denote: $frac{b+c}{a}=t$, then the above equation becomes:
                    $$frac2{t}=-1+t Rightarrow t^2-t-2=0 Rightarrow t_1=-1 text{and} t_2=2.$$



                    Also note that:
                    $$frac{a+b-c}{c}=frac{a-b+c}{b}=frac{-a+b+c}{a}iff \
                    frac{a+b}{c}-1=frac{a+c}{b}-1=frac{b+c}{a}-1 iff \
                    frac{a+b}{c}=frac{a+c}{b}=frac{b+c}{a}=t.$$
                    Hence:
                    $$x=frac{(a+b)(b+c)(c+a)}{abc}=t^3=(-1)^3=-1;\
                    x=frac{(a+b)(b+c)(c+a)}{abc}=t^3=2^3not< 0.$$






                    share|cite|improve this answer









                    $endgroup$













                    • $begingroup$
                      Wow, I never thought of it like that. Do you an assertion as to when I should think similarly to this when dealing with problem solving?
                      $endgroup$
                      – Kenneth Dang
                      Dec 20 '18 at 20:59














                    0












                    0








                    0





                    $begingroup$

                    You correctly derived:
                    $$frac{2a}{c+b}=frac{-a+b+c}{a}$$
                    Your mistake is here:




                    Then I took the original equations and did $frac{a+b-c}{c}=1$




                    Instead, denote: $frac{b+c}{a}=t$, then the above equation becomes:
                    $$frac2{t}=-1+t Rightarrow t^2-t-2=0 Rightarrow t_1=-1 text{and} t_2=2.$$



                    Also note that:
                    $$frac{a+b-c}{c}=frac{a-b+c}{b}=frac{-a+b+c}{a}iff \
                    frac{a+b}{c}-1=frac{a+c}{b}-1=frac{b+c}{a}-1 iff \
                    frac{a+b}{c}=frac{a+c}{b}=frac{b+c}{a}=t.$$
                    Hence:
                    $$x=frac{(a+b)(b+c)(c+a)}{abc}=t^3=(-1)^3=-1;\
                    x=frac{(a+b)(b+c)(c+a)}{abc}=t^3=2^3not< 0.$$






                    share|cite|improve this answer









                    $endgroup$



                    You correctly derived:
                    $$frac{2a}{c+b}=frac{-a+b+c}{a}$$
                    Your mistake is here:




                    Then I took the original equations and did $frac{a+b-c}{c}=1$




                    Instead, denote: $frac{b+c}{a}=t$, then the above equation becomes:
                    $$frac2{t}=-1+t Rightarrow t^2-t-2=0 Rightarrow t_1=-1 text{and} t_2=2.$$



                    Also note that:
                    $$frac{a+b-c}{c}=frac{a-b+c}{b}=frac{-a+b+c}{a}iff \
                    frac{a+b}{c}-1=frac{a+c}{b}-1=frac{b+c}{a}-1 iff \
                    frac{a+b}{c}=frac{a+c}{b}=frac{b+c}{a}=t.$$
                    Hence:
                    $$x=frac{(a+b)(b+c)(c+a)}{abc}=t^3=(-1)^3=-1;\
                    x=frac{(a+b)(b+c)(c+a)}{abc}=t^3=2^3not< 0.$$







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Dec 20 '18 at 19:40









                    farruhotafarruhota

                    20.5k2739




                    20.5k2739












                    • $begingroup$
                      Wow, I never thought of it like that. Do you an assertion as to when I should think similarly to this when dealing with problem solving?
                      $endgroup$
                      – Kenneth Dang
                      Dec 20 '18 at 20:59


















                    • $begingroup$
                      Wow, I never thought of it like that. Do you an assertion as to when I should think similarly to this when dealing with problem solving?
                      $endgroup$
                      – Kenneth Dang
                      Dec 20 '18 at 20:59
















                    $begingroup$
                    Wow, I never thought of it like that. Do you an assertion as to when I should think similarly to this when dealing with problem solving?
                    $endgroup$
                    – Kenneth Dang
                    Dec 20 '18 at 20:59




                    $begingroup$
                    Wow, I never thought of it like that. Do you an assertion as to when I should think similarly to this when dealing with problem solving?
                    $endgroup$
                    – Kenneth Dang
                    Dec 20 '18 at 20:59


















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