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 \(d \geq b+f \).

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 \(d \geq b+f \).

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TopNewest\(\dfrac{a}{b} < \dfrac{c}{d} < \dfrac{e}{f} ;~ af - be = -1 \\ \dfrac{a}{b} < \dfrac{c}{d} \\ ad < bc \\ \boxed{ad + x = bc} \\ \dfrac{c}{d} < \dfrac{e}{f} \\ cf < de \\ \boxed{cf + y = de\\ cf = de - y} \\ \boxed{af - be = -1 \\ af = be-1}\\ ad + x = bc \\ adf + xf = bcf \\ d(af) + xf = b(cf) \\ d(be-1) + xf = b(de-y) \\ dbe - d + xf = dbe - by \\ xf - d = -by \\ d = xf + by \\ \boxed{d \ge b + f} \) – Vicky Vignesh · 6 months ago

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– Calvin Lin Staff · 6 months ago

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

– Vicky Vignesh · 6 months ago

Yeah, restricts \(x, y \in R\)Log in to reply

@Vicky Vignesh – Md Zuhair · 6 months ago

You are going to top TN Region RMO This timeLog in to reply

What have you tried? What do you know? – Calvin Lin Staff · 6 months, 1 week ago

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– Md Zuhair · 6 months ago

I did'nt understood your question?Log in to reply

What observations have you made?

For example, do you know if \( d > b \) or \( d > f \)? – Calvin Lin Staff · 6 months ago

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– Md Zuhair · 6 months ago

I could'nt do it.Log in to reply

@Vicky Vignesh .. Could you? – Md Zuhair · 6 months ago

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If you've tried nothing, then go try something. What approaches can you think of? What comparisons can we make? – Calvin Lin Staff · 6 months ago

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– Kushagra Sahni · 6 months ago

I started by assuming d<b+f. Then I did lot of manipulations and after a while a wrong result came up which means this is a contradiction. So d must be >=b+f. Is this correct?Log in to reply

– Calvin Lin Staff · 6 months ago

Assuming that the steps are correct, then that sounds like a possible proof approach. You should write up the solution and others can comment on it.Log in to reply

– Vicky Vignesh · 6 months ago

You must use the given equality with the inequality. Try adding any variable to rest of inequality with a'equal to' symbolLog in to reply

check out the third problem-http://www.isibang.ac.in/~statmath/olympiad/3sol.pdf – Sathvik Acharya · 6 days, 22 hours ago

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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. – Rajdeep Brahma · 4 weeks ago

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– Calvin Lin Staff · 4 weeks ago

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

– Rajdeep Brahma · 4 weeks ago

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. – Calvin Lin Staff · 4 weeks ago

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– Md Zuhair · 4 weeks ago

Well, I couldnt do this :PLog in to reply