8
$\begingroup$

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)

$\endgroup$
  • 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 $(c')^2-ab'=2014$. $\endgroup$ – Thomas Andrews 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. $\endgroup$ – Thomas Andrews 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
17
$\begingroup$

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

$\endgroup$
  • 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.