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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.

Note by Surya Prakash
1 year, 7 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. · 1 year, 7 months ago

It would be better if you even mention proof to Archimedes broken chord. · 1 year, 7 months ago

All I did was prove congruences and did a bit of angle chasing. My solution is longer but simpler. · 1 year, 7 months ago

Names of the ebooks prizes ? ( I guess you won't mind to reveal :) ). · 1 year, 7 months ago

Thank you everyone :) Also thanks @Surya Prakash for holding this contest. Hope to see more challenging upcoming contests :) · 1 year, 7 months ago

Congrats! Nihar Mahajan.You did best! · 1 year, 7 months ago

Can you post the solutions to the first and the second questions?@Surya Prakash · 1 year, 7 months ago

Yes I too want to see the solution for the second one. · 1 year, 7 months ago

How many of the problems did you get? · 1 year, 7 months ago

I got all 6. · 1 year, 7 months ago

Tried the first 4 got 2. · 1 year, 7 months ago

@Surya Prakash Could u please conduct an INMO mock test similar to this ? · 1 year, 4 months ago

When was it held? · 1 year, 7 months ago

Hey guys. The ebooks i.e. prizes are sent to the participants. Please check your mails. · 1 year, 7 months ago

Thanks! to Surya Prakash for this opportunity. · 1 year, 7 months ago

I also want to participate insuch contests . Can any one suggest me the pathway and dates for future such contests · 1 year, 7 months ago

@Surya Prakash Thank you very much for holding the content. When will the second one start? · 1 year, 7 months ago

I can't say when because I am somewhat busy with preparation for NSEP. · 1 year, 7 months ago

Congratulations @Nihar Mahajan ! · 1 year, 7 months ago

Congrats nihar but want to ask you did you solve the paper in 3 hrs because I did . · 1 year, 7 months ago

I did the first 5 questions in 2 hours. The 6th question required time though... · 1 year, 7 months ago

Can you please post your solution for the second question? · 1 year, 7 months ago

Sure.

· 1 year, 7 months ago

Can you post solution to 5,6 problem · 1 year, 7 months ago

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 $$\Delta ABC$$. · 1 year, 7 months ago

I would not recommend my solution for Q6 since it is (a) 4 pages long and (b) Proves a massive generalisation. · 1 year, 7 months ago

What!Really?!My solution is a page long! · 1 year, 7 months ago

Q6

,

Q3

· 1 year, 7 months ago

@Shivam Jadhavcould you post the solution to the first one? · 1 year, 7 months ago

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

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. · 1 year, 7 months ago

Woah!That is a big solution!Could you write the motivation please? · 1 year, 7 months ago

Congratulations Nihar!! · 1 year, 7 months ago

Congratulations! @Nihar Mahajan · 1 year, 7 months ago

Comment deleted Oct 12, 2015

Nope, the fourth question's answer $$9 \choose 4$$. · 1 year, 7 months ago

@Nihar Mahajan took 10 mins more than u to score 2 marks more than u😛 · 1 year, 7 months ago

No, it took him 4 more hours. · 1 year, 7 months ago

OK whatever is it. He scored more than u 😛😛😛. He scored more than "THE SHARKY KESA". · 1 year, 7 months ago

:( Don't make me cry! Stop bullying me! :P · 1 year, 7 months ago

That hurt u??😞 · 1 year, 7 months ago

Note the :P. · 1 year, 7 months ago

I noted the" :(" · 1 year, 7 months ago

It was part of the :P. · 1 year, 7 months ago

U should've enclosed that in brackets :P. It was just like the difference between 1+23 and (1+2)3. · 1 year, 7 months ago

Not really. We both did perfect 5 questions. And please don't taunt Sharky by this. · 1 year, 7 months ago

That was not a taunt man. I was just joking. · 1 year, 7 months ago

Thanks. · 1 year, 7 months ago