I found this questions from past year maths competition in my country, I've tried any possible way to find it, but it is just way too hard.

If $a, b, c$ are distinct numbers such that $a^2 - bc = 2014$, $b^2 + ac = 2014$, $c^2 + ab = 2014$. Then compute $a^2 + b^2 + c^2$

(A)$4030$ (B)$4028$ (C)$4026$ (D)$4000$ (E)$2014$

Adding these three equations together

$$a^2 + b^2 + c^2 + ab + ac - bc = 3\times2014 \quad(1)$$

And also found that

\begin{align} (a-b-c)^2 &= a^2 + b^2 + c^2 - 2ab - 2ac + 2bc\\ (a-b-c)^2 &= a^2 + b^2 + c^2 - 2(ab + ac - bc)\\ \end{align}

I don't know how to continue to reduce $ ab + ac - bc$, or am I using the wrong way to reducing it?

I'm very appreciate for those who have helped me to hint/explain me on how to do all these questions (I'm currently 10th grade (in US grade system), so I don't understand these much, all of these are outside my syllabus)

  • 3
    $\begingroup$ It looks more promising to try subtracting the equations. For example if we subtract the second equation from the first, we get $a^2 - b^2 - bc - ac = 0$, and a common factor of $a+b$ can be found in the left hand side. $\endgroup$
    – hardmath
    Jul 26 '15 at 13:43
  • $\begingroup$ This problem becomes symmetric if you replace $b'=-b$ and $c'=-c$ then $a^2-b'c'=2014$, $(b')^2-ac'=2014$ and $(c')^2-ab'=2014$. $\endgroup$ Jul 26 '15 at 13:46
  • $\begingroup$ Are $a,b,c$ positive? Integer? Otherwise a silly solution works: set $c=0$, then $a^2=b^2=ab=2014$, so $a=b=\sqrt{2014}$ and $a^2+b^2+c^2=4028$ :) $\endgroup$
    – A.Γ.
    Jul 26 '15 at 13:47
  • $\begingroup$ @A.G. If they are positive and distinct (see question) then $a^2=b^2=ab$ is not possible. $\endgroup$ Jul 26 '15 at 13:50
  • 1
    $\begingroup$ @wuiyang I've learned this trick from my students - there is always a very short wrong solution that gives the right answer :) $\endgroup$
    – A.Γ.
    Jul 26 '15 at 14:02

Note that $$b^2+ac=c^2+ab\iff b^2-c^2+ac-ab=0\iff (b-c)(b+c)-a(b-c)=0$$ $$\iff (b-c)(b+c-a)=0\iff a=b+c.$$

Now $$a^2+b^2+c^2=(b+c)^2+b^2+c^2=2(b^2+c^2+bc)$$ $$=2((b+c)^2-bc)=2(a^2-bc)=2\times 2014$$

  • 1
    $\begingroup$ If $b=c$ then $a^2-b^2=2014$ and $b^2+ab=2014$. So $a^2+ab=2014$. This implies $a=b=c$. But then $a^2-b^2=0$, a contradiction. $\endgroup$
    – ajotatxe
    Jul 26 '15 at 13:52
  • $\begingroup$ @ajotatxe Distinct! $\endgroup$
    – A.Γ.
    Jul 26 '15 at 13:53
  • $\begingroup$ Yes, but this proves that the conditions can be relaxed a bit. $\endgroup$
    – ajotatxe
    Jul 26 '15 at 13:54

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