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!

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TopNewestI took the Junior paper though, and it was easier than I had thought previously, to be honest. I solved all 5 questions :) – Yuxuan Seah · 2 years, 3 months ago

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– Yuxuan Seah · 2 years, 3 months ago

Good luck to all who took the SMO! Wish you all the best! :DLog in to reply

– Joshua Ong · 2 years, 3 months ago

5 was pretty obvious though... For 2 I used another method.Log in to reply

– Victor Loh · 2 years, 3 months ago

How does Question 4 pop out without much effort at allLog in to reply

– Victor Loh · 2 years, 3 months ago

Its the hardest questionLog in to reply

– Joshua Ong · 2 years, 3 months ago

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

– Yuxuan Seah · 2 years, 3 months ago

Okay okay fine. I'm removing that with all this debate going on... D:Log in to reply

– Victor Loh · 2 years, 3 months ago

How about your first commentLog in to reply

– Fengyu Seah · 2 years, 3 months ago

Q3 just spam lah hahahLog in to reply

Answer to second is \((0,0,00\) and \((4,4,4)\) – Dinesh Chavan · 2 years, 3 months ago

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– Mietantei Conan · 2 years, 3 months ago

Question asked for only positive reals.Log in to reply

– Dinesh Chavan · 2 years, 3 months ago

Sorry, Then only \((4,4,4)\)Log in to reply

– Happy Melodies · 2 years, 3 months ago

Whats the proof?Log in to reply

– Fengyu Seah · 2 years, 3 months ago

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.Log in to reply

@Victor Loh did you do that to? – Fengyu Seah · 2 years, 3 months ago

I think that's itLog in to reply

– Daniel Remo · 2 years, 3 months ago

There is also a solution with AM-GM i thinkLog in to reply

– Dinesh Chavan · 2 years, 3 months ago

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 resultLog in to reply

– Mathh Mathh · 2 years, 3 months ago

right... and how did you prove that the huge polynomial has no positive real roots?Log in to reply

– Joshua Ong · 2 years, 3 months ago

Pure evil.Log in to reply

If anyone requires, the SMO Junior Round 2 2014 Qns can be found here – Ryan Wong Jun Hui · 2 years, 3 months ago

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Still weak at solving symmetric systems... Number 2.... – John Ashley Capellan · 2 years, 3 months ago

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