Olympiad Proof's

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 $d \geq b+f$ .

#Algebra

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`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
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`\frac{2}{3}`	$\frac{2}{3}$
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`\sum_{i=1}^3`	$\sum_{i=1}^3$
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Comments

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

Viki Zeta - 4 years, 7 months ago

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

Md Zuhair - 4 years, 7 months ago

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

Calvin Lin Staff - 4 years, 7 months ago

Yeah, restricts $x, y \in R$

Viki Zeta - 4 years, 7 months ago

What have you tried? What do you know?

Calvin Lin Staff - 4 years, 7 months ago

I did'nt understood your question?

Md Zuhair - 4 years, 7 months ago

How have you tried to solve this problem?

What observations have you made?

For example, do you know if $d > b$ or $d > f$ ?

Calvin Lin Staff - 4 years, 7 months ago

@Calvin Lin – I could'nt do it.

Md Zuhair - 4 years, 7 months ago

@Md Zuhair – @Vicky Vignesh .. Could you?

Md Zuhair - 4 years, 7 months ago

@Md Zuhair – What have you tried?

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

Calvin Lin Staff - 4 years, 7 months ago

@Calvin Lin – 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?

Kushagra Sahni - 4 years, 7 months ago

@Kushagra Sahni – 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.

Calvin Lin Staff - 4 years, 7 months ago

@Md Zuhair – You must use the given equality with the inequality. Try adding any variable to rest of inequality with a'equal to' symbol

Viki Zeta - 4 years, 7 months ago

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 years, 2 months ago

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

Calvin Lin Staff - 4 years, 2 months 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.

rajdeep brahma - 4 years, 2 months ago

@Rajdeep Brahma – Wonderful! Thanks for explaining. This helps to demystify what you did. Previously when stated as 1 sentence, it seems like "I magically created this equation and got the answer".

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 years, 2 months ago

Well, I couldnt do this :P

Md Zuhair - 4 years, 2 months ago

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

Sathvik Acharya - 4 years, 1 month ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$