Waste less time on Facebook — follow Brilliant.
×

Olympiad Corner

1) Determine all pairs of real numbers \((x,y)\) such that

\(x^6 = y^4 + 18\)

\(y^6 = x^4 + 18\)

2) Let \(a,b,c\) be integers such that \(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}=3\). Prove that \(abc\) is a perfect cube.

Note by Dev Sharma
10 months, 3 weeks ago

No vote yet
1 vote

Comments

Sort by:

Top Newest

First of all replace \(({x}^2, {y}^2)\) by a and b. Then it is easy to see that if \(a \geq b\) then \( {a}^3 \geq {b}^3\) but that would also mean that \( {a}^2 \leq {b}^2\) comparing the rhs of the two equations . But we have \(a \geq b\) so we also have \({a}^2 \geq {b}^2\) . Hence a = b now we write the equation as \( {a}^3 - {a}^2 - 18 = 0 \) it is easy to see that a = 3 satisfies the equation and so we get \( x = y = \pm \sqrt {3}\) we then get another quadratic factor whose roots are obtained and then we get two more values of x as ;- \( \sqrt {-1 \pm \sqrt{-5}}\) so we have only two real solutions x = y = \( \pm \sqrt{3}\) Akash Deep · 10 months, 2 weeks ago

Log in to reply

@Akash Deep Nice way @akash deep Dev Sharma · 10 months, 2 weeks ago

Log in to reply

For question no.2

By AM-GM we see that:-

\[\Large{\frac { a }{ b } +\frac { b }{ c } +\frac { c }{ a } \ge 3\sqrt [ 3 ]{ \frac { a }{ b } \times \frac { b }{ c } \times \frac { c }{ a } } =3}\]

So we see the minimum value of expression and its value are same therefore \(a=1,b=1,c=1\) for the equality.

We see \(abc=1\) which is a perfect cube! Lakshya Sinha · 10 months, 2 weeks ago

Log in to reply

@Lakshya Sinha I too did it the same way but the question is not restricted to natural numbers it is for integers which can be negative also. So we can't use AM -GM . Akash Deep · 10 months, 2 weeks ago

Log in to reply

@Akash Deep Yes that's a problem Lakshya Sinha · 10 months, 2 weeks ago

Log in to reply

Comment deleted 10 months ago

Log in to reply

@Dev Sharma Are you going celebrated Diwali today with crackers Lakshya Sinha · 10 months, 2 weeks ago

Log in to reply

@Lakshya Sinha NO. Today, I would celebrate it without crackers... I would not fire crackers because it causes pollution and I dont find any good reason to fire them Dev Sharma · 10 months, 2 weeks ago

Log in to reply

@Lakshya Sinha @Lakshya Sinha you?? Dev Sharma · 10 months, 2 weeks ago

Log in to reply

@Dev Sharma Nope it's waste of time Lakshya Sinha · 10 months, 2 weeks ago

Log in to reply

@Lakshya Sinha Come on google + hangout Dev Sharma · 10 months, 2 weeks ago

Log in to reply

@Lakshya Sinha AGREEEE Dev Sharma · 10 months, 2 weeks ago

Log in to reply

@Dev Sharma Well I have uninstalled Hangouts, give me your email id I will add you on circles in Google+ Lakshya Sinha · 10 months, 2 weeks ago

Log in to reply

@Lakshya Sinha sdev3377@gmail.com Dev Sharma · 10 months, 2 weeks ago

Log in to reply

@Dev Sharma I think you added me in your circles so did I Lakshya Sinha · 10 months, 2 weeks ago

Log in to reply

@Lakshya Sinha @Harsh Shrivastava , please check it. @Dev Sharma Lakshya Sinha · 10 months, 2 weeks ago

Log in to reply

@Pi Han Goh @Nihar Mahajan @Surya Prakash Adarsh Kumar Dev Sharma · 10 months, 2 weeks ago

Log in to reply

@Dev Sharma @Swapnil Das @Kushagra Sahni Dev Sharma · 10 months, 2 weeks ago

Log in to reply

@Dev Sharma I have posted a question related to the first one with the name; "inspired by dev sharma's note" u can check that out. Akash Deep · 10 months, 2 weeks ago

Log in to reply

Hey can u plz post the solution for the second question for the cases in which a,b,c may be negative. I have done it upto natural numbers. Please post the solution. Akash Deep · 10 months, 2 weeks ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...