If $m$ and $n$ are positive real numbers satisfying the equation $$m+4\sqrt{mn}-2\sqrt{m}-4\sqrt{n}+4n=3$$ find the value of $$\frac{\sqrt{m}+2\sqrt{n}+2014}{4-\sqrt{m}-2\sqrt{n}}$$

I came across this question in a Math Olympiad Competition and had no idea how to solve it. Can anyone help? Thanks.

  • 3
    $\begingroup$ The question has been edited because the formulae were unclear. Can you please confirm if this is your actual question? $\endgroup$ – Bart Michels Jun 3 '14 at 17:41
  • $\begingroup$ Yes this is my actual question. $\endgroup$ – snivysteel Jun 4 '14 at 5:01
  • $\begingroup$ Simply substitute $m=n=1$ and you will see that the equation holds. Hence the answer becomes 2017. If the IMO is for smaller grades, then I'm sure this is the method. $\endgroup$ – N.S.JOHN Apr 5 '16 at 3:21

It can be helpful to get rid of the square root symbols. If you let $m=x^2$ and $n=y^2$ with $x$ and $y$ understood to be positive, the equation becomes $x^2+4xy-2x-4y+4y^2=3$, which can be rewritten as

$$(x+2y)^2-2(x+2y)=3$$ or $$u^2-2u-3=0\qquad\text{with}\quad u=x+2y$$

The quadratic factors as $(u-3)(u+1)=0$. Recalling that $x=\sqrt m$ and $y=\sqrt n$ are supposed to be positive, this gives $u=x+2y=3$ as the only meaningful solution. We also see that the value we're after is


| cite | improve this answer | |

The trivial answer is to realize that m = 1 and n = 1 meets the first criteria, so you can use those in the second equation.

But that's not much fun.

| cite | improve this answer | |
  • $\begingroup$ this is an excellent observation $\endgroup$ – Jonathan Jun 4 '14 at 3:09
  • $\begingroup$ Or $(m,n) = (9,0)$. $\endgroup$ – user21820 Jun 4 '14 at 10:55
  • $\begingroup$ You are assuming that $\frac{\sqrt{m}+2\sqrt{n}+2014}{4-\sqrt{m}-2\sqrt{n}}$ does not depend on the values of $m$ and/or $n$, that is, you are assuming it is equal to some number instead of a function of $m$ or/and $n$. You would not get all the points for this kind of a solution in a math competition. I'm just adding this so that everyone knows - this fact wasn't emphasized in the answer. $\endgroup$ – user26486 Jun 4 '14 at 14:19

Given: $m$ and $n$ are positive real numbers satisfying the equation


Just to get a better feeling, substitute



Now your equation becomes


Combining first, second and last term of L.H.S. , we get,


Substitute: $(x+2y)=t$ to get,


$\implies t=3$ or $t=-1$

But since $t=x+2y=\sqrt{m}+2\sqrt{n} \implies t \ge 0$ (Since $\sqrt{}$ gives positive value in its domain)

$\implies \sqrt{m}+2\sqrt{n}=3$

$\implies \dfrac{\sqrt{m} +2\sqrt{n} +2014}{4-(\sqrt{m} +2\sqrt{n})}= \boxed{2017}$

| cite | improve this answer | |
  • 3
    $\begingroup$ I think my answer was maybe four seconds ahead of yours! Great minds, and all that.... $\endgroup$ – Barry Cipra Jun 3 '14 at 18:05

We have $$ \begin{align*} m+4\sqrt{mn}-2\sqrt{m}-4\sqrt{n}+4n&=3\\ m+4\sqrt{mn}+4n-2\sqrt{m}-4\sqrt{n}-3&=0\\ (\sqrt{m}+2\sqrt{n})^2-2(\sqrt{m}+2\sqrt{n})-3&=0\\ (\sqrt{m}+2\sqrt{n}-3)(\sqrt{m}+2\sqrt{n}+1)&=0 \end{align*} $$ which gives $$ \sqrt{m}+2\sqrt{n}=3\,\,\text{or}\,\,\sqrt{m}+2\sqrt{n}=-1. $$ We can disregard the second solution as $m$ and $n$ are real so $$ \frac{\sqrt{m}+2\sqrt{n}+2014}{4-\sqrt{m}-2\sqrt{n}}=\frac{3+2014}{4-3}=2017. $$

| cite | improve this answer | |

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.