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.

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

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

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Names of the ebooks prizes ? ( I guess you won't mind to reveal :) ).

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Thank you everyone :) Also thanks @Surya Prakash for holding this contest. Hope to see more challenging upcoming contests :)

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Congrats! Nihar Mahajan.You did best!

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Can you post the solutions to the first and the second questions?@Surya Prakash

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

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How many of the problems did you get?

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@Surya Prakash Could u please conduct an INMO mock test similar to this ?

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When was it held?

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Hey guys. The ebooks i.e. prizes are sent to the participants. Please check your mails.

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Thanks! to Surya Prakash for this opportunity.

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I also want to participate insuch contests . Can any one suggest me the pathway and dates for future such contests

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@Surya Prakash Thank you very much for holding the content. When will the second one start?

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I can't say when because I am somewhat busy with preparation for NSEP.

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Congratulations @Nihar Mahajan !

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Congrats nihar but want to ask you did you solve the paper in 3 hrs because I did .

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

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Can you please post your solution for the second question?

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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 \(\Delta ABC\).

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Q6

Q3

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@Shivam Jadhavcould you post the solution to the first one?

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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.

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Congratulations Nihar!!

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Congratulations! @Nihar Mahajan

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@Nihar Mahajan @Sharky Kesa @Shivam Jadhav @Siddharth Singh @Svatejas Shivakumar @Sarthak Behera

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Comment deleted Oct 12, 2015

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Nope, the fourth question's answer \(9 \choose 4\).

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@Nihar Mahajan took 10 mins more than u to score 2 marks more than u😛

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3 and (1+2)3.Log in to reply

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Can we get our individual question marks and best solutions for each problem?

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