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°

Log in to reply

can u post the solution for the 2nd part of problem 1

Log in to reply

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

Log in to reply

Log in to reply

Log in to reply

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.

Log in to reply

Log in to reply

Log in to reply

Log in to reply

I can't edit my comment, its "AHCM".

Log in to reply

Brilliant staff are working on this issue.

Log in to reply

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 \(\angle DEP = \angle PFD\).

Log in to reply

I have seen this question earlier in one of my books.

Its a good problem. As i have solution of it, i will not post this now. Let others try also.

Log in to reply

Below is the \(diagram\).

\(Const:\) 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 \(\dfrac{BD}{CD}\) = \(\dfrac{AB}{AC}\) => \(\dfrac{BE}{CF}\) = \(\dfrac{BD}{CD}\).

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

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

\(KIPKIG.\)

Log in to reply

How do you draw diagrams on the computer? Is there some tool you use?

Log in to reply

Log in to reply

Log in to reply

In \(\Delta ABC\), O is the circumcenter and H is the orthocenter. Prove that \(AH^2+BC^2=4AO^2\).

Log in to reply

Draw \(OD\perp BC\).

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

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

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

=> \(BC^{2}\) + \(AH^{2}\) = \(4\) \(AO^{2}\).

Log in to reply

Can you post the solution please ?

Log in to reply

i got it , its easy https://postimg.org/image/5d1sm6hm7/

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log 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.

Log in to reply

Yes you are right. Thanks for pointing it out.

Log in to reply

@Swapnil Das @Harsh Shrivastava @Ayush Pattnayak Please Help! I am a class 9 student and am Appearing for RMO. I am pretty intimidated by Geometry Problems.... I can make an accurate figure but I don't know how to proceed.(For example: This question by Brilliant Member) Please guide me on how to solve Geometry and geometrical proofs...... I know Theorems(like Menelaus' Ptolemy's, Sine rule, Co-sine rule) If I make it in INMO... You all will deserve the credit. Urgent Help Required!!

Log in to reply

Just read those theorems and get into actual problem solving, you may also try British mathematical Olympiad problems, they are also RMO level. No sort of 'magical' construction required for RMO. Best of luck!

Log in to reply

u just need some angle chasing and similarity for RMO.

Log in to reply

All INMO participants,please share ur marks.

Log in to reply

70-80.

Log in to reply

why dont u be active in slack?

Log in to reply

I have a lot of stuffs for FIITJEE. I do that only.

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

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.

Log in to reply

Me too.... gaharwar.02@gmail.com

Log in to reply

Me too.....ayushpattnayak2001@gmail.com

Log in to reply

Mine is rajdeep.ind24@gmail.com

Log in to reply

Ok i have sent you an invite You can check your email

By the way I am the co owner of the team and ayush is the owner

Log in to reply

I'm in, here's my email id alanj.33@cloud.com

Log in to reply

I meant icloud* there

Log in to reply

Log in to reply

Hello everybody,

RMO results are out.

Who are selected?

Log in to reply

Use trigonometry as there is a right triangle formed assuming centet

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

Log in to reply

What is your score in RMO?

Log in to reply

I gave GMO

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

At which website is the result of RMO declared?

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.

Log in to reply

this is north zone's (Delhi) problem

Log in to reply

could u do it?

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Please someone post this years rmo question paper.

Log in to reply

The papers are uploaded on AoPS.

Log in to reply

Please give me the link.Thanks.

Log in to reply

Log in to reply

Can someone post this year's problems?

Log in to reply

Mumbai region paper was really easy

Log in to reply

Could you post the question paper please?

Log in to reply

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

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Did anyone give RMO from north zone?

Log in to reply

Uttar Pradesh - Me

Log in to reply

At which center?

Log in to reply

Log in to reply

Best of luck everyone

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

\[\displaystyle \sum_{i=1}^{\left \lfloor \frac{k}{2} \right \rfloor + 1} \left ( \left \lfloor \dfrac {n_i}{2} \right \rfloor + 1 \right ) . \]

Log in to reply

Please post again as you cannot edit that.

Log in to reply

Sure I can! Mod powers! :P

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log 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.

Log in to reply

Determine all positive triplets of integers such that

\(\large\ {(x+1)}^{y+1} + 1 = {(x+2)}^{z+1}.\)

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.

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.

Log in to reply

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

Log in to reply

Good choice from MATHEMATICAL OLYMPIAD TREASURES.

Log in to reply

\((n-1)^3=(\sqrt[3]{k+\sqrt{k^2-1}} + \sqrt[3]{k-\sqrt{k^2-1}})^3 \\ n^3-3n^2=2k-21\)

As \(k\) is integer \(2k-2\) will also be an integer.

Log in to reply

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

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.

Log in to reply

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.

Log in to reply

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

I have a solution using trivial inequalities.

Log in to reply

Log in to reply

Log in to reply

Log in to reply

I am also giving the RMO this year, please help me out 😀

Log in to reply

What is circumdiameter?

Log in to reply

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!

Log in to reply

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

Log in to reply

Actually RMO happened today for most of the regions. Some are on 16th.

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Thanks. But I left Olympiad mathematics forever.

Log in to reply

Are you ok? Say no

Log 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.

Log in to reply

WHATT!!!!!!!!!!!!!

Log in to reply

Log in to reply

Here's the link to last year's RMO board

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/.

Log in to reply

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

Log in to reply

Even mine is on 16th

Log in to reply

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?

Log in to reply

Log in to reply

How was KVPY?

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?

Log in to reply

I got selected.

Log in to reply

Log in to reply

I have already started that. Check here.

https://brilliant.org/discussions/thread/inmo-2017-board/?ref_id=1273098

Log in to reply

Log in to reply

Log in to reply

Now it's ok, Harsh, I know you are gonna rock in other exams 😀😀

Log in to reply

Log in to reply

Log in to reply

So wait upto that and then solve INMO problems.

Log in to reply

I have RMO on 23rd.

Log in to reply