Waste less time on Facebook — follow Brilliant.
×

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 \(d \geq b+f \).

Note by Md Zuhair
9 months, 4 weeks ago

No vote yet
1 vote

Comments

Sort by:

Top Newest

\(\dfrac{a}{b} < \dfrac{c}{d} < \dfrac{e}{f} ;~ af - be = -1 \\ \dfrac{a}{b} < \dfrac{c}{d} \\ ad < bc \\ \boxed{ad + x = bc} \\ \dfrac{c}{d} < \dfrac{e}{f} \\ cf < de \\ \boxed{cf + y = de\\ cf = de - y} \\ \boxed{af - be = -1 \\ af = be-1}\\ ad + x = bc \\ adf + xf = bcf \\ d(af) + xf = b(cf) \\ d(be-1) + xf = b(de-y) \\ dbe - d + xf = dbe - by \\ xf - d = -by \\ d = xf + by \\ \boxed{d \ge b + f} \) Vicky Vignesh · 9 months, 3 weeks ago

Log in to reply

@Vicky Vignesh Great! The interesting fact is that \( d = xf + by \), which is extremely restrictive. Calvin Lin Staff · 9 months, 3 weeks ago

Log in to reply

@Calvin Lin Yeah, restricts \(x, y \in R\) Vicky Vignesh · 9 months, 3 weeks ago

Log in to reply

@Vicky Vignesh You are going to top TN Region RMO This time @Vicky Vignesh Md Zuhair · 9 months, 3 weeks ago

Log in to reply

What have you tried? What do you know? Calvin Lin Staff · 9 months, 4 weeks ago

Log in to reply

@Calvin Lin I did'nt understood your question? Md Zuhair · 9 months, 4 weeks ago

Log in to reply

@Md Zuhair 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 · 9 months, 4 weeks ago

Log in to reply

@Calvin Lin I could'nt do it. Md Zuhair · 9 months, 4 weeks ago

Log in to reply

@Md Zuhair @Vicky Vignesh .. Could you? Md Zuhair · 9 months, 4 weeks ago

Log in to reply

@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 · 9 months, 4 weeks ago

Log in to reply

@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 · 9 months, 3 weeks ago

Log in to reply

@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 · 9 months, 3 weeks ago

Log in to reply

@Md Zuhair You must use the given equality with the inequality. Try adding any variable to rest of inequality with a'equal to' symbol Vicky Vignesh · 9 months, 4 weeks ago

Log in to reply

check out the third problem-http://www.isibang.ac.in/~statmath/olympiad/3sol.pdf Sathvik Acharya · 3 months, 4 weeks ago

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. Rajdeep Brahma · 4 months, 2 weeks ago

Log in to reply

@Rajdeep Brahma 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 months, 2 weeks ago

Log in to reply

@Calvin Lin 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 months, 2 weeks ago

Log in to reply

@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 months, 2 weeks ago

Log in to reply

@Rajdeep Brahma Well, I couldnt do this :P Md Zuhair · 4 months, 2 weeks ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...