Waste less time on Facebook — follow Brilliant.
×

Olympiad Proof's

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

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

Note by Md Zuhair
1 month, 1 week 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 · 1 month, 1 week ago

Log in to reply

@Vicky Vignesh Great! The interesting fact is that \( d = xf + by \), which is extremely restrictive. Calvin Lin Staff · 1 month, 1 week ago

Log in to reply

@Calvin Lin Yeah, restricts \(x, y \in R\) Vicky Vignesh · 1 month, 1 week ago

Log in to reply

@Vicky Vignesh You are going to top TN Region RMO This time @Vicky Vignesh Md Zuhair · 1 month, 1 week ago

Log in to reply

What have you tried? What do you know? Calvin Lin Staff · 1 month, 1 week ago

Log in to reply

@Calvin Lin I did'nt understood your question? Md Zuhair · 1 month, 1 week 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 · 1 month, 1 week ago

Log in to reply

@Calvin Lin I could'nt do it. Md Zuhair · 1 month, 1 week ago

Log in to reply

@Md Zuhair @Vicky Vignesh .. Could you? Md Zuhair · 1 month, 1 week 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 · 1 month, 1 week 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 · 1 month, 1 week 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 · 1 month, 1 week 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 · 1 month, 1 week ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...