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Discussion of SMO Senior Round 2 Qns 2014

The link to the Round 2 Paper for this year's Singapore Mathematical Olympiad is here. The Round 2 competition takes place just yesterday (28 June) so it's pretty new.

Here's the place to discuss the answers to these qns. :) So for those interested to try them or discuss them, feel free to do so!

Note by Happy Melodies
3 years, 3 months ago

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I took the Junior paper though, and it was easier than I had thought previously, to be honest. I solved all 5 questions :)

Yuxuan Seah - 3 years, 3 months ago

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Good luck to all who took the SMO! Wish you all the best! :D

Yuxuan Seah - 3 years, 3 months ago

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5 was pretty obvious though... For 2 I used another method.

Joshua Ong - 3 years, 3 months ago

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How does Question 4 pop out without much effort at all

Victor Loh - 3 years, 3 months ago

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@Victor Loh Its the hardest question

Victor Loh - 3 years, 3 months ago

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Wut! For 4 you have to prove that a,b and c HAVE to be equal man.

Joshua Ong - 3 years, 3 months ago

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@Joshua Ong Okay okay fine. I'm removing that with all this debate going on... D:

Yuxuan Seah - 3 years, 3 months ago

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@Yuxuan Seah How about your first comment

Victor Loh - 3 years, 3 months ago

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Q3 just spam lah hahah

Fengyu Seah - 3 years, 3 months ago

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Answer to second is \((0,0,00\) and \((4,4,4)\)

Dinesh Chavan - 3 years, 3 months ago

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Question asked for only positive reals.

Mietantei Conan - 3 years, 3 months ago

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Sorry, Then only \((4,4,4)\)

Dinesh Chavan - 3 years, 3 months ago

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@Dinesh Chavan Whats the proof?

Happy Melodies - 3 years, 3 months ago

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@Happy Melodies Something like WLOG \(a \geq b \geq c > 0\). Then \(a+b \geq a+c\), yet \(a\sqrt{b} \geq b\sqrt{c}\), which implies \(a+c \geq a+b\) instead, so \(b = c\). Similarly, \(a = b = c\). And we are done.

Fengyu Seah - 3 years, 3 months ago

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@Fengyu Seah I think that's it @Victor Loh did you do that to?

Fengyu Seah - 3 years, 3 months ago

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@Happy Melodies There is also a solution with AM-GM i think

Daniel Remo - 3 years, 3 months ago

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@Happy Melodies An easy way is to substitute \(x=\sqrt{a},y=\sqrt{b},z=\sqrt{c}\) . The you get 3 equations. Try to find ab expression involving just one variable. So, we get \(x\left( x-2\right)\left(x^{12}-3x^{11}+6x^{10}-14x^{9}+22x^{8}-28x^{7}+37x^{6}-35x^{5}+26x^{4}-21x^{3}+14x^{2}-12x+8\right) =0\). . Now this gives \(x=2,x=0\). So, we get \(a=4,0\). So, leaving \(0\), we get the desired result

Dinesh Chavan - 3 years, 3 months ago

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@Dinesh Chavan right... and how did you prove that the huge polynomial has no positive real roots?

Mathh Mathh - 3 years, 3 months ago

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@Mathh Mathh Pure evil.

Joshua Ong - 3 years, 3 months ago

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If anyone requires, the SMO Junior Round 2 2014 Qns can be found here

Ryan Wong Jun Hui - 3 years, 3 months ago

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Still weak at solving symmetric systems... Number 2....

John Ashley Capellan - 3 years, 3 months ago

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