So, RMO online mock test is completed.

Only 6 members had sent their solutions. So, here are the results.

So, Congratulations Nihar. You stood first.

Your prizes(ebooks) will be sent to your mail soon.

So, RMO online mock test is completed.

Only 6 members had sent their solutions. So, here are the results.

So, Congratulations Nihar. You stood first.

Your prizes(ebooks) will be sent to your mail soon.

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TopNewestHey Guys I will just give you an easy solution to problem 2.

It can be easily observed that \(E\) is the midpoint of arc \(AC\) containing \(B\). Let us drop a perpendicular \(N\) from \(E\) to on to \(BC\). According to Archimedes Broken Chord Theorem, \(CN = BN + AB = BN + BF = FN\). So, \(N\) is midpoint of \(CF\). It implies that \(EN\) is perpendicular bisector of \(CF\). Now, it is easy to prove that \(DE\) is perpendicular bisector of \(AC\). As perpendicular bisectors of \(CF\) and \(AC\) i.e. \(EN\) and \(ED\) respectively meet at \(E\). So, \(E\) is the circumcenter of \(\Delta AFC\).

Hence Proved. – Surya Prakash · 1 year, 3 months ago

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– Surya Prakash · 1 year, 3 months ago

It would be better if you even mention proof to Archimedes broken chord.Log in to reply

– Sharky Kesa · 1 year, 3 months ago

All I did was prove congruences and did a bit of angle chasing. My solution is longer but simpler.Log in to reply

Names of the ebooks prizes ? ( I guess you won't mind to reveal :) ). – Karthik Venkata · 1 year, 3 months ago

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Thank you everyone :) Also thanks @Surya Prakash for holding this contest. Hope to see more challenging upcoming contests :) – Nihar Mahajan · 1 year, 3 months ago

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– Siddharth Singh · 1 year, 3 months ago

Congrats! Nihar Mahajan.You did best!Log in to reply

Can you post the solutions to the first and the second questions?@Surya Prakash – Adarsh Kumar · 1 year, 3 months ago

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– Svatejas Shivakumar · 1 year, 3 months ago

Yes I too want to see the solution for the second one.Log in to reply

– Adarsh Kumar · 1 year, 3 months ago

How many of the problems did you get?Log in to reply

– Sharky Kesa · 1 year, 3 months ago

I got all 6.Log in to reply

– Svatejas Shivakumar · 1 year, 3 months ago

Tried the first 4 got 2.Log in to reply

@Surya Prakash Could u please conduct an INMO mock test similar to this ? – Shrihari B · 1 year ago

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When was it held? – Anandhu Raj · 1 year, 2 months ago

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Hey guys. The ebooks i.e. prizes are sent to the participants. Please check your mails. – Surya Prakash · 1 year, 3 months ago

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Thanks! to Surya Prakash for this opportunity. – Siddharth Singh · 1 year, 3 months ago

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I also want to participate insuch contests . Can any one suggest me the pathway and dates for future such contests – Aakash Khandelwal · 1 year, 3 months ago

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@Surya Prakash Thank you very much for holding the content. When will the second one start? – Svatejas Shivakumar · 1 year, 3 months ago

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– Surya Prakash · 1 year, 3 months ago

I can't say when because I am somewhat busy with preparation for NSEP.Log in to reply

Congratulations @Nihar Mahajan ! – Adarsh Kumar · 1 year, 3 months ago

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Congrats nihar but want to ask you did you solve the paper in 3 hrs because I did . – Shivam Jadhav · 1 year, 3 months ago

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– Nihar Mahajan · 1 year, 3 months ago

I did the first 5 questions in 2 hours. The 6th question required time though...Log in to reply

– Svatejas Shivakumar · 1 year, 3 months ago

Can you please post your solution for the second question?Log in to reply

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– Shivam Jadhav · 1 year, 3 months ago

Can you post solution to 5,6 problemLog in to reply

Extend AI to meet the circumcircle of \(\Delta ABC\) at \(X\), \(BI\) at \(Y\) and \(CI\) at \(Z\). Note that \(\angle BAX=\angle BCX\) and \(\angle CAX = \angle CBX\) by cyclic bowties (Angles subtended by the same chord). Also, \(\angle BAX=\angle CAX\) because \(AX\) is the angle bisector of \(\angle BAC\). Therefore, \(\angle CBX=\angle BCX\), which implies \(\Delta BXC\) is an isosceles triangle, which further implies \(BX=CX\). Now, we have to prove \(BX=IX\).

We have

\[\begin{align} \angle XBI&=\angle XBC+\angle CBI\\ &=\angle XAC+\angle CBI\\ &=\angle IAC+\angle CBI\\ &=\dfrac {1}{2} \angle BAC + \dfrac {1}{2} \angle ABC \end{align}\]

Also,

\[\begin{align} \angle XIB&=\angle IAB+ \angle IBA \text{(}\angle XIB \text{ is the external angle of } \Delta AIB\text{)}\\ &=\dfrac{1}{2} \angle BAC+\dfrac{1}{2} \angle ABC \end{align}\]

Thus, \(\angle XBI=\angle XIB\), which implies \(\Delta XIB\) is isosceles, which implies \(XI=XB\). We already know that \(XB=XC\), so \(X\) is the circumcentre of \(\Delta IBC\). By definition, point \(X\) is on the circumcircle of \(\Delta ABC\). Therefore, the circumcentre of \(\Delta IBC\) lies on the circumcircle of \(\Delta ABC\). Similarly, it can be proven that the circumcentre (\(Y\)) of \(\Delta ICA\) lies on the circumcircle of \(\Delta ABC\) and the circumcentre (\(Z\)) of \(\Delta IAB\) lies on the circumcircle of \(\Delta ABC\). – Sharky Kesa · 1 year, 3 months ago

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– Sharky Kesa · 1 year, 3 months ago

I would not recommend my solution for Q6 since it is (a) 4 pages long and (b) Proves a massive generalisation.Log in to reply

– Adarsh Kumar · 1 year, 3 months ago

What!Really?!My solution is a page long!Log in to reply

Q6

Q3

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@Shivam Jadhavcould you post the solution to the first one? – Adarsh Kumar · 1 year, 3 months ago

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By AM-GM inequality, \(ab^2+bc^2+ac^2 \ge 3abc\)

Therefore,

\(\dfrac{a^2(b+c)+b^2(a+c)+c^2(a+b)+(a^2b+b^2c+a^2c)+(ab^2+bc^2+ac^2)}{2} \ge \dfrac{a^2(b+c)+b^2(a+c)+c^2(a+b)+(a^2b+b^2c+a^2c)+3abc}{2}=\dfrac{a^2(b+c)+b^2(a+c)+c^2(a+b)+ab(a+c)+bc(a+b)+ac(b+c)}{2}\)

Grouping and again applying AM-GM inequality, (Applying AM-GM inequality on each term and adding all the terms)

\(\dfrac{(a^2(b+c)+ab(a+c))}{2}+\dfrac{(b^2(a+c)+bc(a+b))+}{2}+\dfrac{(c^2(a+b)+ac(b+c))}{2} \ge \sqrt{a^3b(b+c)(a+c)}+\sqrt{b^3c(a+c)(a+b)}+\sqrt{c^3a(a+b)(a+c)}=\displaystyle\sum_{cyc}ab\sqrt{\dfrac{a}{b}(b+c)(c+a)}.\) Equality occurs when \(a=b=c\)

Hence proved. – Svatejas Shivakumar · 1 year, 3 months ago

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– Adarsh Kumar · 1 year, 3 months ago

Woah!That is a big solution!Could you write the motivation please?Log in to reply

Congratulations Nihar!! – Svatejas Shivakumar · 1 year, 3 months ago

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Congratulations! @Nihar Mahajan – Swapnil Das · 1 year, 3 months ago

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@Nihar Mahajan @Sharky Kesa @Shivam Jadhav @Siddharth Singh @Svatejas Shivakumar @Sarthak Behera – Surya Prakash · 1 year, 3 months ago

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– Sharky Kesa · 1 year, 3 months ago

Nope, the fourth question's answer \(9 \choose 4\).Log in to reply

@Nihar Mahajan took 10 mins more than u to score 2 marks more than u😛 – Aditya Kumar · 1 year, 3 months ago

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– Sharky Kesa · 1 year, 3 months ago

No, it took him 4 more hours.Log in to reply

– Aditya Kumar · 1 year, 3 months ago

OK whatever is it. He scored more than u 😛😛😛. He scored more than "THE SHARKY KESA".Log in to reply

– Sharky Kesa · 1 year, 3 months ago

:( Don't make me cry! Stop bullying me! :PLog in to reply

– Aditya Kumar · 1 year, 3 months ago

That hurt u??😞Log in to reply

– Sharky Kesa · 1 year, 3 months ago

Note the :P.Log in to reply

– Aditya Kumar · 1 year, 3 months ago

I noted the" :("Log in to reply

– Sharky Kesa · 1 year, 3 months ago

It was part of the :P.Log in to reply

3 and (1+2)3. – Aditya Kumar · 1 year, 3 months agoLog in to reply

– Nihar Mahajan · 1 year, 3 months ago

Not really. We both did perfect 5 questions. And please don't taunt Sharky by this.Log in to reply

– Aditya Kumar · 1 year, 3 months ago

That was not a taunt man. I was just joking.Log in to reply

– Sharky Kesa · 1 year, 3 months ago

Thanks.Log in to reply

– Sharky Kesa · 1 year, 3 months ago

Can we get our individual question marks and best solutions for each problem?Log in to reply