Hello everyone!

As many of us are preparing for RMO (Regional Mathematics Olympiad), let us start posting problems and help each other prepare. Everyone is more than welcome to post problems or post the solutions to problems.

Here is a problem to start with:

In \(\Delta ABC\), \(O\) is the circumcenter and \(H\) is the orthocenter. If \(AO=AH\), prove that \(\angle A=60^\circ\).

Also, if the circumcircle of \(\Delta BOC\) passes through H, prove that \(\angle A=60^\circ\).

## Comments

Sort by:

TopNewestSolution1)Extend BO to meet at the circumcircle of the ∆ at M.Also extend CH to meet at AB at N and AH meet BC at K.Join AM and MC.

Now observe that angle ANC = angle BAM = 90°.This implies AM || CN.

Similarly,AH || CM.

This implies,AHMC is a parallelogram.

Now, AH = MC(=OA because AH=AO.)

Thus,OMC is an equilateral triangle with angle MOC =60°.

angle BOC = 180° - angle MOC = 120°.

This implies angle BAC = 60° – Harsh Shrivastava · 9 months, 3 weeks ago

Log in to reply

– Neel Khare · 9 months, 2 weeks ago

can u post the solution for the 2nd part of problem 1Log in to reply

– Harsh Shrivastava · 9 months, 2 weeks ago

I will try to solve it and if I succeed ,I'll post the solution.Log in to reply

– Neel Khare · 9 months, 2 weeks ago

i got it here is the solution [url=https://postimg.org/image/e0cnnt8w5/][img]https://s18.postimg.org/e0cnnt8w5/IMG_0100.jpg[/img][/url]Log in to reply

– Neel Khare · 9 months, 2 weeks ago

sorry https://postimg.org/image/ylrffpqh1/ https://postimg.org/image/e0cnnt8w5/ open the linksLog in to reply

– Neel Khare · 9 months, 2 weeks ago

the 1 st part can be solved much much easily AO=AH R=2RcosA 2cosA=1 cosA=1/2 A=60 done!!Log in to reply

@Svatejas Shivakumar 's question :

See an even more efficient use of \(trigonometry\). 1st Part of\(1.\) Draw \(OD\perp BC\).

\(AO\) = \(AH\) => \(BO\) = \(2OD\) => \(\cos\) \(\angle BOD\) = \(1/2\) => \(\angle BOD\) = \(60\) => \(\angle BOC\) = \(120\) => \(\angle A\) = \(60\).

Second part :

\(2\) Quad. \(BHOC\) is cyclic.

=> \(\angle BHC\) = \(\angle BOC\)

=> \(180 - \angle A\) = \(2 \angle A\)

=> \(\angle A\) = \(180 / 3\) = \(60\).

I know, as expected it was a \(non-trigonometric\) one. – Rohit Camfar · 3 months, 3 weeks ago

Log in to reply

– Yash Mehan · 1 month ago

please explain how BO = 2 OD if AO = AH. Thanks!Log in to reply

– Rohit Camfar · 1 month ago

That's just one of the Euler line properties. In a triangle \(\Delta ABC\), if \(O\) is the circumcenter, \(H\) is the orthocentre and \(D\) is the foot of perpendicular from \(O\) on \(BC\) then by a well known result \(AH = 2OD\).Log in to reply

– Harsh Shrivastava · 9 months, 2 weeks ago

Oh that's awesome use of trignometery!Log in to reply

– Harsh Shrivastava · 9 months, 3 weeks ago

I can't edit my comment, its "AHCM".Log in to reply

– Sharky Kesa · 9 months, 3 weeks ago

Brilliant staff are working on this issue.Log in to reply

– Harsh Shrivastava · 9 months, 3 weeks ago

Hey Sharky post some problems (RMO level.)Log in to reply

In a triangle \(ABC\) the point \(D\) is the intersection of the interior angle bisector of \(\angle BAC\) with side \(BC\). The line through \(A\) that is perpendicular to \(AD\) intersects the circumcircle of triangle \(ABC\) for a second time at point \(P\). A circle through points \(A\) and \(P\) intersects line segment \(BP\) internally in \(E\) and line segment \(CP\) internally in \(F\).

Prove \(\angle DEP = \angle PFD\). – Sharky Kesa · 9 months, 3 weeks ago

Log in to reply

Its a good problem. As i have solution of it, i will not post this now. Let others try also. – Priyanshu Mishra · 9 months, 3 weeks ago

Log in to reply

\(Const:\) Let \(\odot APE\) be the circle passing through \(A\) and \(P.\)[where \(E\) is a point on \(BP\).] Join \(AE\), \(AF.\)

\(Solution\): Let \(\angle BAD\) = \(\angle CAD \) = \(\theta\).

Then \(\angle PAC\) = \(\angle PBC\) = \(\angle PCB\) = \(90\) - \(\theta\) & \(\angle BPC\) = \(\angle EPF\) = \(\angle EAF\) = \(2\theta.\)

Now, \(\angle BAC\) = \(\angle EAF\) => \(\angle BAE\) = \(\angle CAF\). Also, \(\angle ABE\) = \(\angle ACF\)

=> \(\Delta ABE\sim \Delta ACF\) => \(\dfrac{AB}{AC}\) = \(\dfrac{BE}{CF}\) ( Similarity properties )

Now, \(\angle EBD\) = \(\angle FCD\) = \(90\) - \(\theta\) , \(\dfrac{BE}{CF}\) = \(\dfrac{AB}{AC}\). But also \(\dfrac{BD}{CD}\) = \(\dfrac{AB}{AC}\) => \(\dfrac{BE}{CF}\) = \(\dfrac{BD}{CD}\).

=> \(\Delta EBD\sim \Delta FCD\).

=> \(\angle BED\) = \(\angle CFD\) =>\(\angle DEP = \angle PFD\).

\(KIPKIG.\) – Rohit Camfar · 3 months, 3 weeks ago

Log in to reply

– Yash Mehan · 1 month ago

How do you draw diagrams on the computer? Is there some tool you use?Log in to reply

– Ayush Rai · 1 month ago

yes ..like geogebraLog in to reply

Log in to reply

In \(\Delta ABC\), O is the circumcenter and H is the orthocenter. Prove that \(AH^2+BC^2=4AO^2\). – Svatejas Shivakumar · 9 months, 3 weeks ago

Log in to reply

Just \(Pythagoras\) then, \(BD^{2}\) \(+\) \(OD^{2}\) = \(BO^{2}\)

=> \(4\) \(BD^{2}\) \(+\) \(4\) \(OD^{2}\) = \(4\) \(BO^{2}\)

=> \([2BD]^{2}\) \(+\) \([2OD]^{2}\) = \(4\) \(BO^{2}\)

=> \(BC^{2}\) + \(AH^{2}\) = \(4\) \(AO^{2}\). – Rohit Camfar · 3 months, 3 weeks ago

Log in to reply

– Alan Joel · 9 months, 2 weeks ago

Can you post the solution please ?Log in to reply

– Neel Khare · 9 months, 2 weeks ago

i got it , its easy https://postimg.org/image/5d1sm6hm7/Log in to reply

– Alan Joel · 9 months, 2 weeks ago

How did you get AH^2 = 2RcosA ?Log in to reply

– Priyanshu Mishra · 9 months, 2 weeks ago

Its an identity.Log in to reply

– Alan Joel · 9 months, 2 weeks ago

Its AH*Log in to reply

– Alan Joel · 9 months, 2 weeks ago

Oh, I knew that but I forgot lolLog in to reply

Well, this first part of the question can't be right! (Below is a summary of why)

If \(AO=OH\), \(H\) must also be on the circumcircle of \(ABC\), from which we get the triangle being right-angled, and \(H\) is on the vertex with right angle. Nothing else can be gathered from the given information.

Perhaps you meant \(AO=AH\), which makes sense. – Sharky Kesa · 9 months, 3 weeks ago

Log in to reply

– Svatejas Shivakumar · 9 months, 3 weeks ago

Yes you are right. Thanks for pointing it out.Log in to reply

All INMO participants,please share ur marks. – Ayush Pattnayak · 5 months, 3 weeks ago

Log in to reply

– Priyanshu Mishra · 5 months, 2 weeks ago

70-80.Log in to reply

– Ayush Rai · 5 months, 2 weeks ago

why dont u be active in slack?Log in to reply

I have a lot of stuffs for FIITJEE. I do that only. – Priyanshu Mishra · 5 months, 2 weeks ago

Log in to reply

– Rajdeep Das · 5 months, 2 weeks ago

Which centre?Log in to reply

– Ayush Rai · 5 months, 2 weeks ago

oh...ok I am very sorry for disturbing u.Log in to reply

– Priyanshu Mishra · 5 months, 2 weeks ago

Its true. Kuch mazaa nahi aata slack chat pe.Log in to reply

– Ayush Rai · 5 months, 2 weeks ago

Its not chatting.I invited bcoz ur an INMO qualifier and u can help us solve problems that are posted.Log in to reply

– Priyanshu Mishra · 5 months, 2 weeks ago

Initially i am only RMO qualifier. I am not INMO qualifier till the result is declared.Log in to reply

– Ayush Rai · 5 months, 3 weeks ago

do u want to join my INMO team?Log in to reply

@Svatejas Shivakumar, @Harsh Shrivastava, @Ayush Pattnayak @Alan Joel @Racchit Jain @rajdeep das @naitik sanghavi and all other RMO aspirants ,I invite you'll to my RMO,INMO team. Those who are interested can give their email id over here. – Ayush Rai · 6 months, 1 week ago

Log in to reply

– Ayush Pattnayak · 4 months, 2 weeks ago

Me too.....ayushpattnayak2001@gmail.comLog in to reply

– Rajdeep Das · 6 months, 1 week ago

Mine is rajdeep.ind24@gmail.comLog in to reply

By the way I am the co owner of the team and ayush is the owner – Neel Khare · 6 months, 1 week ago

Log in to reply

– Alan Joel · 6 months, 1 week ago

I'm in, here's my email id alanj.33@cloud.comLog in to reply

– Alan Joel · 6 months, 1 week ago

I meant icloud* thereLog in to reply

– Ayush Rai · 6 months, 1 week ago

you can check ur mail nowLog in to reply

Hello everybody,

RMO results are out.

Who are selected? – Priyanshu Mishra · 7 months, 2 weeks ago

Log in to reply

Use trigonometry as there is a right triangle formed assuming centet – Biswajit Barik · 7 months, 2 weeks ago

Log in to reply

Hey!! Did anyone give GMO? Or RMO on 16th October. If yes please tell how many were you able to do, and what should be the expected cutoff – Racchit Jain · 9 months ago

Log in to reply

– Sayantan Saha · 8 months, 3 weeks ago

What is your score in RMO?Log in to reply

– Racchit Jain · 8 months, 3 weeks ago

I gave GMOLog in to reply

– Racchit Jain · 8 months, 3 weeks ago

However, I know marks of some of my friends from different regions, which region are you asking for?Log in to reply

– Rajdeep Das · 8 months, 3 weeks ago

Pls give for delhi region also.Log in to reply

– Priyanshu Mishra · 8 months, 3 weeks ago

Is RMO DELHI result out?Log in to reply

– Rajdeep Das · 8 months, 3 weeks ago

YesLog in to reply

– Priyanshu Mishra · 8 months, 3 weeks ago

At which website?Log in to reply

– Sayantan Saha · 8 months, 3 weeks ago

I am from WB region. I want to know how high the scores of rmo had gone in Delhi this year.Log in to reply

– Racchit Jain · 8 months, 3 weeks ago

The highest marks in Delhi that I know of is 35 out of 60 otherwise everyone is getting less than 15. The cutoff should be around 20 I think, but not more than 25Log in to reply

At which website is the result of RMO declared? – Priyanshu Mishra · 8 months, 3 weeks ago

Log in to reply

Given are two circles w1, w2 which intersect at points X, Y . Let P be an arbitrary point on w1. Suppose that the lines PX, PY meet w2 again at points A,B respectively. Prove that the circumcircles of all triangles PAB have the same radius. – Sayantan Saha · 9 months, 2 weeks ago

Log in to reply

– Sayantan Saha · 9 months, 2 weeks ago

this is north zone's (Delhi) problemLog in to reply

– Rajdeep Das · 9 months, 2 weeks ago

could u do it?Log in to reply

– Rajdeep Das · 9 months, 2 weeks ago

Then pls post the solutionLog in to reply

– Racchit Jain · 9 months, 2 weeks ago

Show that AB is independent of the choice of point PLog in to reply

– Racchit Jain · 9 months, 2 weeks ago

Try using sine ruleLog in to reply

– Sharky Kesa · 9 months, 2 weeks ago

It looks like Power of a Point, but Extended Sine Rule definitely works.Log in to reply

– Sayantan Saha · 9 months, 2 weeks ago

please add the solutionLog in to reply

– Sharky Kesa · 9 months, 2 weeks ago

I won't give solution but the crux of this proof is to show that \(AB\) is constant, irrespective of where \(P\) is.Log in to reply

– Sayantan Saha · 9 months, 2 weeks ago

yeah. I have done itLog in to reply

– Sayantan Saha · 9 months, 2 weeks ago

Two circles C1 and C2 intersect each other at points A and B. Their external common tangent (closer to B) touches C1 at P and C2 at Q. Let C be the reflection of B in line PQ. Prove that angleCAP = angleBAQ. Can you convince me what this reflection does mean.Log in to reply

Please someone post this years rmo question paper. – Harsh Shrivastava · 9 months, 2 weeks ago

Log in to reply

– Svatejas Shivakumar · 9 months, 2 weeks ago

The papers are uploaded on AoPS.Log in to reply

– Harsh Shrivastava · 9 months, 2 weeks ago

Please give me the link.Thanks.Log in to reply

– Naitik Sanghavi · 9 months, 2 weeks ago

I have posted Gujarat rmo paper here,https://brilliant.org/discussions/thread/rmo-2016-gujarat-region/?ref_id=1272714Log in to reply

Can someone post this year's problems? – Sharky Kesa · 9 months, 2 weeks ago

Log in to reply

Mumbai region paper was really easy – Racchit Jain · 9 months, 2 weeks ago

Log in to reply

– Kush Singhal · 9 months, 2 weeks ago

Could you post the question paper please?Log in to reply

– Racchit Jain · 9 months, 2 weeks ago

Yeah sure, but I use the app and I don't know how to post an image here, can you give me your email and I'll mail it to you?Log in to reply

– Sharky Kesa · 9 months, 2 weeks ago

Post it on Slack.Log in to reply

– Racchit Jain · 9 months, 2 weeks ago

Umm...how do I do that?Log in to reply

– Racchit Jain · 9 months, 2 weeks ago

It's asking me too get the app can't I do it using the browser only?Log in to reply

– Sharky Kesa · 9 months, 2 weeks ago

You can do it on the browser.Log in to reply

– Racchit Jain · 9 months, 2 weeks ago

Can I mail it to you and then you can post it?Log in to reply

– Sharky Kesa · 9 months, 2 weeks ago

Sure. sharkesa@gmail.comLog in to reply

– Harsh Shrivastava · 9 months, 2 weeks ago

Please post the paper.Log in to reply

– Racchit Jain · 9 months, 2 weeks ago

I have sent it to you plz checkLog in to reply

Did anyone give RMO from north zone? – Rajdeep Das · 9 months, 2 weeks ago

Log in to reply

– Rishik Jain · 9 months, 2 weeks ago

Uttar Pradesh - MeLog in to reply

– Rajdeep Das · 9 months, 2 weeks ago

At which center?Log in to reply

– Rishik Jain · 9 months, 2 weeks ago

MeerutLog in to reply

Best of luck everyone – Rajdeep Das · 9 months, 2 weeks ago

Log in to reply

Suppose that \(k, n_1, \ldots, n_k\) are variable positive integers satisfying \(k \geq 3\), \(n_1 \geq n_2 \geq \ldots \geq n_k \geq 1\), and \(n_1 + n_2 + \ldots + n_k = 2016\).

Find the maximal value of

\[\displaystyle \sum_{i=1}^{\left \lfloor \frac{k}{2} \right \rfloor + 1} \left ( \left \lfloor \dfrac {n_i}{2} \right \rfloor + 1 \right ) . \] – Sharky Kesa · 9 months, 3 weeks ago

Log in to reply

– Priyanshu Mishra · 9 months, 3 weeks ago

Please post again as you cannot edit that.Log in to reply

– Sharky Kesa · 9 months, 3 weeks ago

Sure I can! Mod powers! :PLog in to reply

– Priyanshu Mishra · 9 months, 3 weeks ago

But how you edited that?Log in to reply

– Sharky Kesa · 9 months, 3 weeks ago

With great skill (and a large screen)!Log in to reply

– Priyanshu Mishra · 9 months, 3 weeks ago

What is that skill!?Log in to reply

– Sharky Kesa · 9 months, 3 weeks ago

Big iMac skills! :P :PLog in to reply

\(\large\ \frac{3}{x-3}+\frac{5}{x-5}+\frac{17}{x-17}+\frac{19}{x-19}=x^2-11x-4\).

Find the largest real solution to this equation. – Priyanshu Mishra · 9 months, 3 weeks ago

Log in to reply

Determine all positive triplets of integers such that

\(\large\ {(x+1)}^{y+1} + 1 = {(x+2)}^{z+1}.\) – Priyanshu Mishra · 9 months, 3 weeks ago

Log in to reply

@Everyone

Find the smallest positive number \(\lambda\), such that for any complex numbers \({z_1},{z_2},{z_3} \in \{z\in \mathbb C \big| |z| < 1\}\), if \(z_1+z_2+z_3 = 0\), then \( \left|z_1z_2 +z_2z_3+z_3z_1\right|^2+\left|z_1z_2z_3\right|^2 < \lambda\).

Solve this and provide the solution. – Priyanshu Mishra · 9 months, 3 weeks ago

Log in to reply

3) Let \(k\) be an integer and let

\[n=\sqrt[3]{k+\sqrt{k^2-1}} + \sqrt[3]{k-\sqrt{k^2-1}}+1\]

Prove \(n^3 - 3n^2\) is an integer. – Sharky Kesa · 9 months, 3 weeks ago

Log in to reply

– Harsh Shrivastava · 9 months, 3 weeks ago

Hint : Use the fact that if a+b+c = 0 then \(a^3+b^3+c^3 = 3abc\)Log in to reply

– Priyanshu Mishra · 9 months, 3 weeks ago

Good choice from MATHEMATICAL OLYMPIAD TREASURES.Log in to reply

As \(k\) is integer \(2k-2\) will also be an integer. – Akshat Sharda · 9 months, 3 weeks ago

Log in to reply

– Priyanshu Mishra · 9 months, 3 weeks ago

Good but take \(n\) on RHS and see that \(a + b + c = 0\), so \(a^3 + b^3 + c^3 = 3abc\) and we are done.Log in to reply

– Sharky Kesa · 9 months, 3 weeks ago

Firstly, your final statement is incorrect. Secondly, you have put no working. Sorry, but this is a null solution.Log in to reply

Please post difficult ones. – Priyanshu Mishra · 9 months, 3 weeks ago

Log in to reply

– Sharky Kesa · 9 months, 3 weeks ago

OK, sorry! I'll post IMO level probs next time.Log in to reply

Prove that \[\dfrac{a^2+b^2+c^2}{d^2} >\dfrac{1}{3} \], where a,b,c,d are the sides of a quadrilateral. – Harsh Shrivastava · 9 months, 3 weeks ago

Log in to reply

– Sharky Kesa · 9 months, 3 weeks ago

This question I have done before (I'm pretty sure it was an application of QM-AM), so I'm leaving it as an exercise for everyone else.Log in to reply

– Harsh Shrivastava · 9 months, 3 weeks ago

I have a solution using trivial inequalities.Log in to reply

– Sharky Kesa · 9 months, 3 weeks ago

QM-AM is trivial.Log in to reply

– Harsh Shrivastava · 9 months, 3 weeks ago

Alright.Post some problems.Log in to reply

– Svatejas Shivakumar · 9 months, 3 weeks ago

Yes after QM-AM the result directly follows.Log in to reply

I am also giving the RMO this year, please help me out 😀 – Alan Joel · 9 months, 3 weeks ago

Log in to reply

What is circumdiameter? – Harsh Shrivastava · 9 months, 3 weeks ago

Log in to reply

– Svatejas Shivakumar · 9 months, 3 weeks ago

sorry it is circumcenter.Log in to reply

@Vaibhav Prasad @Kalash Verma @Nihar Mahajan @Adarsh Kumar @Akshat Sharda @AkshayYadav @Swapnil Das @Rajdeep Dhingra @Anik Mandal @Lakshya Sinha @Abhay Kumar @Priyanshu Mishra @Dev Sharma @Sharky Kesa and everyone! – Harsh Shrivastava · 9 months, 3 weeks ago

Log in to reply

– Nihar Mahajan · 9 months, 2 weeks ago

Sorry guys, I wasn't active on brilliant (and might not be for a period of time) Best luck for your rmosLog in to reply

– Svatejas Shivakumar · 9 months, 2 weeks ago

Actually RMO happened today for most of the regions. Some are on 16th.Log in to reply

– Nihar Mahajan · 9 months, 2 weeks ago

yep, i gave it today :)Log in to reply

– Harsh Shrivastava · 9 months, 2 weeks ago

Please most the paper.Log in to reply

– Nihar Mahajan · 9 months, 2 weeks ago

How to most a paper? 😂😂 XDLog in to reply

– Harsh Shrivastava · 9 months, 2 weeks ago

Lol I meant post.Please post your paper.Log in to reply

– Svatejas Shivakumar · 9 months, 2 weeks ago

How was it?Log in to reply

– Naitik Sanghavi · 9 months, 2 weeks ago

How was ur rmo?how many did get correct?Log in to reply

– Harsh Shrivastava · 9 months, 2 weeks ago

Please most the paper.Log in to reply

– Svatejas Shivakumar · 9 months, 2 weeks ago

Mine is on 16th. How was your paper?Log in to reply

– Naitik Sanghavi · 9 months, 2 weeks ago

3-4 correct ,This time 2 question were very easy, so may be cutoff will go high!Log in to reply

– Rajdeep Das · 9 months, 2 weeks ago

From which region did u give RMO?Log in to reply

– Naitik Sanghavi · 9 months, 2 weeks ago

GujaratLog in to reply

– Nihar Mahajan · 9 months, 2 weeks ago

pretty good, better than last time'sLog in to reply

– Swapnil Das · 9 months, 3 weeks ago

Thanks. But I left Olympiad mathematics forever.Log in to reply

– Nihar Mahajan · 9 months, 2 weeks ago

Are you ok? Say noLog in to reply

Number Theory, Euclidean Geometry, Classical Inequalities...Do they have any important application in my life? No, never. On the other hand, what I learn for the physics Olympiads will certainly have a huge impact on my future. Moreover, MOs make me slow, which is very harmful for these upcoming 3 years of my life. I'll be learning Math of relevant context like Calculus and stuff for PhOs, which will keep me away from MOs as well as keep my interest for math always alive. – Swapnil Das · 9 months, 2 weeks ago

Log in to reply

– Harsh Shrivastava · 9 months, 3 weeks ago

WHATT!!!!!!!!!!!!!Log in to reply

– Swapnil Das · 9 months, 3 weeks ago

That's true.Log in to reply

Here's the link to last year's RMO board – Harsh Shrivastava · 9 months, 3 weeks ago

Log in to reply

RMO is over now. So no need to look at that. Now we should prepare for INMO.

I have posted a sample of 6 questions here. You can practice that and post more questions there also.

https://brilliant.org/discussions/thread/inmo-2017-board/. – Priyanshu Mishra · 9 months, 2 weeks ago

Log in to reply

– Harsh Shrivastava · 9 months, 2 weeks ago

It is not over in many regions, mine is on 16th.Log in to reply

– Alan Joel · 9 months, 2 weeks ago

Even mine is on 16thLog in to reply

– Nihar Mahajan · 9 months ago

How was your rmo :)?Log in to reply

Very very bad. :(

Can do only one question, I succumbed to exam pressure :(

Though I could solve 3 out of remaining 5 question at home myself.

I know this this is a lame excuse but 😭

My Olympiad maths is officially over.

Sorry for long reply, wbu? – Harsh Shrivastava · 9 months ago

Log in to reply

– Nihar Mahajan · 7 months, 2 weeks ago

Your Olympiad maths journey is not over. Congo :)Log in to reply

How was KVPY? – Harsh Shrivastava · 7 months, 2 weeks ago

Log in to reply

But i think we should do such math when we are free 'coz we enjoy olympiad maths.

Can you please suggest me some resources for inmo level geometry and some important topics in geometry to be studied? – Harsh Shrivastava · 7 months, 2 weeks ago

Log in to reply

I got selected. – Priyanshu Mishra · 7 months, 2 weeks ago

Log in to reply

– Harsh Shrivastava · 7 months, 2 weeks ago

Yes.Let's start the INMO Board!!Log in to reply

I have already started that. Check here.

https://brilliant.org/discussions/thread/inmo-2017-board/?ref_id=1273098 – Priyanshu Mishra · 7 months, 2 weeks ago

Log in to reply

– Harsh Shrivastava · 7 months, 2 weeks ago

I think you should post a new note because that thread has died.Log in to reply

– Priyanshu Mishra · 7 months, 2 weeks ago

Ok i will post a fresh one.Log in to reply

Now it's ok, Harsh, I know you are gonna rock in other exams 😀😀 – Nihar Mahajan · 9 months ago

Log in to reply

– Racchit Jain · 8 months, 3 weeks ago

Did you give from Mumbai region?Log in to reply

– Harsh Shrivastava · 9 months ago

Not sure about that rocking part :(Log in to reply

– Priyanshu Mishra · 9 months, 2 weeks ago

So wait upto that and then solve INMO problems.Log in to reply

– Ayush Pattnayak · 9 months, 1 week ago

I have RMO on 23rd.Log in to reply