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
4 years, 1 month ago

No vote yet
1 vote

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

Karthik Venkata - 4 years, 1 month ago

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

It can be easily observed that EE is the midpoint of arc ACAC containing BB. Let us drop a perpendicular NN from EE to on to BCBC. According to Archimedes Broken Chord Theorem, CN=BN+AB=BN+BF=FNCN = BN + AB = BN + BF = FN. So, NN is midpoint of CFCF. It implies that ENEN is perpendicular bisector of CFCF. Now, it is easy to prove that DEDE is perpendicular bisector of ACAC. As perpendicular bisectors of CFCF and ACAC i.e. ENEN and EDED respectively meet at EE. So, EE is the circumcenter of ΔAFC\Delta AFC.

Hence Proved.

Surya Prakash - 4 years, 1 month ago

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

Surya Prakash - 4 years, 1 month ago

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

Sharky Kesa - 4 years, 1 month ago

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

Adarsh Kumar - 4 years, 1 month ago

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

A Former Brilliant Member - 4 years, 1 month ago

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How many of the problems did you get?

Adarsh Kumar - 4 years, 1 month ago

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@Adarsh Kumar Tried the first 4 got 2.

A Former Brilliant Member - 4 years, 1 month ago

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@Adarsh Kumar I got all 6.

Sharky Kesa - 4 years, 1 month 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 - 4 years, 1 month ago

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Congrats! Nihar Mahajan.You did best!

Siddharth Singh - 4 years, 1 month ago

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

Sharky Kesa - 4 years, 1 month ago

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Congratulations! @Nihar Mahajan

Swapnil Das - 4 years, 1 month ago

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Congratulations Nihar!!

A Former Brilliant Member - 4 years, 1 month 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 - 4 years, 1 month ago

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

Nihar Mahajan - 4 years, 1 month ago

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Can you please post your solution for the second question?

A Former Brilliant Member - 4 years, 1 month ago

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@A Former Brilliant Member Sure.

Nihar Mahajan - 4 years, 1 month ago

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@Nihar Mahajan Can you post solution to 5,6 problem

Shivam Jadhav - 4 years, 1 month ago

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@Shivam Jadhav Q5)

Extend AI to meet the circumcircle of ΔABC\Delta ABC at XX, BIBI at YY and CICI at ZZ. Note that BAX=BCX\angle BAX=\angle BCX and CAX=CBX\angle CAX = \angle CBX by cyclic bowties (Angles subtended by the same chord). Also, BAX=CAX\angle BAX=\angle CAX because AXAX is the angle bisector of BAC\angle BAC. Therefore, CBX=BCX\angle CBX=\angle BCX, which implies ΔBXC\Delta BXC is an isosceles triangle, which further implies BX=CXBX=CX. Now, we have to prove BX=IXBX=IX.

We have

XBI=XBC+CBI=XAC+CBI=IAC+CBI=12BAC+12ABC\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,

XIB=IAB+IBA(XIB is the external angle of ΔAIB)=12BAC+12ABC\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, XBI=XIB\angle XBI=\angle XIB, which implies ΔXIB\Delta XIB is isosceles, which implies XI=XBXI=XB. We already know that XB=XCXB=XC, so XX is the circumcentre of ΔIBC\Delta IBC. By definition, point XX is on the circumcircle of ΔABC\Delta ABC. Therefore, the circumcentre of ΔIBC\Delta IBC lies on the circumcircle of ΔABC\Delta ABC. Similarly, it can be proven that the circumcentre (YY) of ΔICA\Delta ICA lies on the circumcircle of ΔABC\Delta ABC and the circumcentre (ZZ) of ΔIAB\Delta IAB lies on the circumcircle of ΔABC\Delta ABC.

Sharky Kesa - 4 years, 1 month 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 - 4 years, 1 month ago

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@Sharky Kesa What!Really?!My solution is a page long!

Adarsh Kumar - 4 years, 1 month ago

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@Adarsh Kumar Q6 Q6, Q3 Q3

Adarsh Kumar - 4 years, 1 month ago

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

Adarsh Kumar - 4 years, 1 month ago

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@Adarsh Kumar ab(a+b)+ac(a+c)+bc(b+c)=ab(a+b)+ac(a+c)+bc(b+c)+ab(a+b)+ac(a+c)+bc(b+c)2=a2(b+c)+b2(a+c)+c2(a+b)+(a2b+b2c+a2c)+(ab2+bc2+ac2)2ab(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, ab2+bc2+ac23abcab^2+bc^2+ac^2 \ge 3abc

Therefore,

a2(b+c)+b2(a+c)+c2(a+b)+(a2b+b2c+a2c)+(ab2+bc2+ac2)2a2(b+c)+b2(a+c)+c2(a+b)+(a2b+b2c+a2c)+3abc2=a2(b+c)+b2(a+c)+c2(a+b)+ab(a+c)+bc(a+b)+ac(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} \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)

(a2(b+c)+ab(a+c))2+(b2(a+c)+bc(a+b))+2+(c2(a+b)+ac(b+c))2a3b(b+c)(a+c)+b3c(a+c)(a+b)+c3a(a+b)(a+c)=cycabab(b+c)(c+a).\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=ca=b=c

Hence proved.

A Former Brilliant Member - 4 years, 1 month ago

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@A Former Brilliant Member Woah!That is a big solution!Could you write the motivation please?

Adarsh Kumar - 4 years, 1 month ago

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Congratulations @Nihar Mahajan !

Adarsh Kumar - 4 years, 1 month ago

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

A Former Brilliant Member - 4 years, 1 month ago

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

Surya Prakash - 4 years, 1 month 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 - 4 years, 1 month ago

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Thanks! to Surya Prakash for this opportunity.

Siddharth Singh - 4 years, 1 month ago

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

Surya Prakash - 4 years, 1 month ago

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When was it held?

Anandhu Raj - 4 years ago

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

Shrihari B - 3 years, 10 months ago

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