# 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
5 years, 10 months ago

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

- 5 years, 10 months ago

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.

- 5 years, 10 months ago

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

- 5 years, 10 months ago

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

- 5 years, 10 months ago

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

- 5 years, 10 months ago

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

- 5 years, 10 months ago

How many of the problems did you get?

- 5 years, 10 months ago

Tried the first 4 got 2.

- 5 years, 10 months ago

I got all 6.

- 5 years, 10 months ago

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

- 5 years, 10 months ago

Congrats! Nihar Mahajan.You did best!

- 5 years, 10 months ago

- 5 years, 10 months ago

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

- 5 years, 10 months ago

Congratulations! @Nihar Mahajan

- 5 years, 10 months ago

Congratulations Nihar!!

- 5 years, 10 months ago

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

- 5 years, 10 months ago

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

- 5 years, 10 months ago

- 5 years, 10 months ago

Sure.  - 5 years, 10 months ago

Can you post solution to 5,6 problem

- 5 years, 10 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{aligned} \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{aligned}

Also,

\begin{aligned} \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{aligned}

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 $\Delta ABC$.

- 5 years, 10 months ago

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

- 5 years, 10 months ago

What!Really?!My solution is a page long!

- 5 years, 10 months ago Q6, Q3

- 5 years, 10 months ago

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

- 5 years, 10 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.

- 5 years, 10 months ago

Woah!That is a big solution!Could you write the motivation please?

- 5 years, 10 months ago

Congratulations @Nihar Mahajan !

- 5 years, 10 months ago

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

- 5 years, 10 months ago

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

- 5 years, 10 months ago

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

- 5 years, 10 months ago

Thanks! to Surya Prakash for this opportunity.

- 5 years, 10 months ago

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

- 5 years, 10 months ago

When was it held?

- 5 years, 9 months ago

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

- 5 years, 7 months ago