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RMO Online Mock Test - Results

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.

Note by Surya Prakash
1 year, 5 months ago

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Hey 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, 5 months ago

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@Surya Prakash It would be better if you even mention proof to Archimedes broken chord. Surya Prakash · 1 year, 5 months ago

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@Surya Prakash All I did was prove congruences and did a bit of angle chasing. My solution is longer but simpler. Sharky Kesa · 1 year, 5 months ago

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Names of the ebooks prizes ? ( I guess you won't mind to reveal :) ). Karthik Venkata · 1 year, 5 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, 5 months ago

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@Nihar Mahajan Congrats! Nihar Mahajan.You did best! Siddharth Singh · 1 year, 5 months ago

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Can you post the solutions to the first and the second questions?@Surya Prakash Adarsh Kumar · 1 year, 5 months ago

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@Adarsh Kumar Yes I too want to see the solution for the second one. Svatejas Shivakumar · 1 year, 5 months ago

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@Svatejas Shivakumar How many of the problems did you get? Adarsh Kumar · 1 year, 5 months ago

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@Adarsh Kumar I got all 6. Sharky Kesa · 1 year, 5 months ago

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@Adarsh Kumar Tried the first 4 got 2. Svatejas Shivakumar · 1 year, 5 months ago

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@Surya Prakash Could u please conduct an INMO mock test similar to this ? Shrihari B · 1 year, 2 months ago

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When was it held? Anandhu Raj · 1 year, 5 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, 5 months ago

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Thanks! to Surya Prakash for this opportunity. Siddharth Singh · 1 year, 5 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, 5 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, 5 months ago

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@Svatejas Shivakumar I can't say when because I am somewhat busy with preparation for NSEP. Surya Prakash · 1 year, 5 months ago

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Congratulations @Nihar Mahajan ! Adarsh Kumar · 1 year, 5 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, 5 months ago

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@Shivam Jadhav I did the first 5 questions in 2 hours. The 6th question required time though... Nihar Mahajan · 1 year, 5 months ago

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@Nihar Mahajan Can you please post your solution for the second question? Svatejas Shivakumar · 1 year, 5 months ago

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@Svatejas Shivakumar Sure.

Nihar Mahajan · 1 year, 5 months ago

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@Nihar Mahajan Can you post solution to 5,6 problem Shivam Jadhav · 1 year, 5 months ago

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@Shivam Jadhav Q5)

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, 5 months ago

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@Sharky Kesa I would not recommend my solution for Q6 since it is (a) 4 pages long and (b) Proves a massive generalisation. Sharky Kesa · 1 year, 5 months ago

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@Sharky Kesa What!Really?!My solution is a page long! Adarsh Kumar · 1 year, 5 months ago

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@Adarsh Kumar

Q6

Q6

,
Q3

Q3

Adarsh Kumar · 1 year, 5 months ago

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@Adarsh Kumar @Shivam Jadhavcould you post the solution to the first one? Adarsh Kumar · 1 year, 5 months ago

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@Adarsh Kumar \(ab(a+b)+ac(a+c)+bc(b+c)= \dfrac{ab(a+b)+ac(a+c)+bc(b+c)+ab(a+b)+ac(a+c)+bc(b+c)}{2}=\dfrac{a^2(b+c)+b^2(a+c)+c^2(a+b)+(a^2b+b^2c+a^2c)+(ab^2+bc^2+ac^2)}{2}\)

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, 5 months ago

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@Svatejas Shivakumar Woah!That is a big solution!Could you write the motivation please? Adarsh Kumar · 1 year, 5 months ago

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Congratulations Nihar!! Svatejas Shivakumar · 1 year, 5 months ago

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Congratulations! @Nihar Mahajan Swapnil Das · 1 year, 5 months ago

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Comment deleted Oct 12, 2015

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@Svatejas Shivakumar Nope, the fourth question's answer \(9 \choose 4\). Sharky Kesa · 1 year, 5 months ago

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@Sharky Kesa @Nihar Mahajan took 10 mins more than u to score 2 marks more than u😛 Aditya Kumar · 1 year, 5 months ago

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@Aditya Kumar No, it took him 4 more hours. Sharky Kesa · 1 year, 5 months ago

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@Sharky Kesa OK whatever is it. He scored more than u 😛😛😛. He scored more than "THE SHARKY KESA". Aditya Kumar · 1 year, 5 months ago

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@Aditya Kumar :( Don't make me cry! Stop bullying me! :P Sharky Kesa · 1 year, 5 months ago

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@Sharky Kesa That hurt u??😞 Aditya Kumar · 1 year, 5 months ago

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@Aditya Kumar Note the :P. Sharky Kesa · 1 year, 5 months ago

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@Sharky Kesa I noted the" :(" Aditya Kumar · 1 year, 5 months ago

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@Aditya Kumar It was part of the :P. Sharky Kesa · 1 year, 5 months ago

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@Sharky Kesa U should've enclosed that in brackets :P. It was just like the difference between 1+23 and (1+2)3. Aditya Kumar · 1 year, 5 months ago

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@Aditya Kumar Not really. We both did perfect 5 questions. And please don't taunt Sharky by this. Nihar Mahajan · 1 year, 5 months ago

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@Nihar Mahajan That was not a taunt man. I was just joking. Aditya Kumar · 1 year, 5 months ago

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@Nihar Mahajan Thanks. Sharky Kesa · 1 year, 5 months ago

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@Surya Prakash Can we get our individual question marks and best solutions for each problem? Sharky Kesa · 1 year, 5 months ago

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