Olympiad Corner

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

x6=y4+18x^6 = y^4 + 18

y6=x4+18y^6 = x^4 + 18

2) Let a,b,ca,b,c be integers such that ab+bc+ca=3\frac{a}{b}+\frac{b}{c}+\frac{c}{a}=3. Prove that abcabc is a perfect cube.

Note by Dev Sharma
3 years, 11 months ago

No vote yet
1 vote

  Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

  • Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
  • Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
  • Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
  • Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 2×3 2 \times 3
2^{34} 234 2^{34}
a_{i-1} ai1 a_{i-1}
\frac{2}{3} 23 \frac{2}{3}
\sqrt{2} 2 \sqrt{2}
\sum_{i=1}^3 i=13 \sum_{i=1}^3
\sin \theta sinθ \sin \theta
\boxed{123} 123 \boxed{123}

Comments

Sort by:

Top Newest

First of all replace (x2,y2)({x}^2, {y}^2) by a and b. Then it is easy to see that if aba \geq b then a3b3 {a}^3 \geq {b}^3 but that would also mean that a2b2 {a}^2 \leq {b}^2 comparing the rhs of the two equations . But we have aba \geq b so we also have a2b2{a}^2 \geq {b}^2 . Hence a = b now we write the equation as a3a218=0 {a}^3 - {a}^2 - 18 = 0 it is easy to see that a = 3 satisfies the equation and so we get x=y=±3 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 ;- 1±5 \sqrt {-1 \pm \sqrt{-5}} so we have only two real solutions x = y = ±3 \pm \sqrt{3}

akash deep - 3 years, 11 months ago

Log in to reply

Nice way @akash deep

Dev Sharma - 3 years, 11 months 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 - 3 years, 11 months ago

Log in to reply

@Pi Han Goh @Nihar Mahajan @Surya Prakash Adarsh Kumar

Dev Sharma - 3 years, 11 months ago

Log in to reply

@Swapnil Das @Kushagra Sahni

Dev Sharma - 3 years, 11 months ago

Log in to reply

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 - 3 years, 11 months ago

Log in to reply

For question no.2

By AM-GM we see that:-

ab+bc+ca3ab×bc×ca3=3\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=1a=1,b=1,c=1 for the equality.

We see abc=1abc=1 which is a perfect cube!

Department 8 - 3 years, 11 months ago

Log in to reply

@Harsh Shrivastava , please check it. @Dev Sharma

Department 8 - 3 years, 11 months ago

Log in to reply

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 - 3 years, 11 months ago

Log in to reply

Yes that's a problem

Department 8 - 3 years, 11 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...