If \(a,b,c,d,e,f\) are positive integers and \(\dfrac{a}{b}<\dfrac{c}{d}<\dfrac{e}{f}\).

Suppose \(af-be=-1\) ,then prove that \(d \geq b+f \).

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

## Comments

Sort by:

TopNewest\(\dfrac{a}{b} < \dfrac{c}{d} < \dfrac{e}{f} ;~ af - be = -1 \\ \dfrac{a}{b} < \dfrac{c}{d} \\ ad < bc \\ \boxed{ad + x = bc} \\ \dfrac{c}{d} < \dfrac{e}{f} \\ cf < de \\ \boxed{cf + y = de\\ cf = de - y} \\ \boxed{af - be = -1 \\ af = be-1}\\ ad + x = bc \\ adf + xf = bcf \\ d(af) + xf = b(cf) \\ d(be-1) + xf = b(de-y) \\ dbe - d + xf = dbe - by \\ xf - d = -by \\ d = xf + by \\ \boxed{d \ge b + f} \)

Log in to reply

Great! The interesting fact is that \( d = xf + by \), which is extremely restrictive.

Log in to reply

Yeah, restricts \(x, y \in R\)

Log in to reply

You are going to top TN Region RMO This time @Vicky Vignesh

Log in to reply

What have you tried? What do you know?

Log in to reply

I did'nt understood your question?

Log in to reply

How have you tried to solve this problem?

What observations have you made?

For example, do you know if \( d > b \) or \( d > f \)?

Log in to reply

Log in to reply

@Vicky Vignesh .. Could you?

Log in to reply

If you've tried nothing, then go try something. What approaches can you think of? What comparisons can we make?

Log in to reply

Log in to reply

Log in to reply

Log in to reply

check out the third problem-http://www.isibang.ac.in/~statmath/olympiad/3sol.pdf

Log in to reply

Note that bc-ad>=0.We have bc-ad>=1.Similarly de-fc>=1.So d=d(be-af)=dbe-daf=dbe-bfc+bfc-adf=b(de-fc)+f(bc-ad)>=b+f This might be a shorter proof.

Log in to reply

Very nicely done. What is the motivation / intuition behind how you arrived at this solution? What led you to think about it?

Log in to reply

It is easy to get bc-ad>0.b,c,a,d,e,f are integers.So I was wondering how can I use it in the problem?Then I thought that bc-ad>=1 might help as after all I have to show something >= something.So using bc-ad>=1 is better than using bc-ad>0.Hence I used it.Now I broke (dbe - bfc) in the following way so as to use the fact bc-ad>=1 & de-fc>=1 & finally arrived the solution.

Log in to reply

You've now added "Here is my reasoning for why I considered these steps, and how I was wanting to use the conditions of the problem". This will help @Md Zuhair (and everyone else) learn from this and apply it to other scenarios.

Log in to reply

Well, I couldnt do this :P

Log in to reply