# RMO 2016 practice board

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 $\angle A=60^\circ$.

Note by A Former Brilliant Member
4 years, 3 months ago

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## Comments

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Solution

1)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°

- 4 years, 3 months ago

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

- 4 years, 3 months ago

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Brilliant staff are working on this issue.

- 4 years, 3 months ago

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Hey Sharky post some problems (RMO level.)

- 4 years, 3 months ago

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

- 4 years, 3 months ago

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

- 4 years, 3 months ago

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

- 4 years, 3 months ago

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sorry https://postimg.org/image/ylrffpqh1/ https://postimg.org/image/e0cnnt8w5/ open the links

- 4 years, 3 months ago

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

- 4 years, 3 months ago

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Oh that's awesome use of trignometery!

- 4 years, 3 months ago

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See an even more efficient use of $trigonometry$. 1st Part of @Svatejas Shivakumar 's question :

$1.$ 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.

- 3 years, 9 months ago

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please explain how BO = 2 OD if AO = AH. Thanks!

- 3 years, 7 months ago

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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 $AH = 2OD$.

- 3 years, 7 months ago

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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 $\angle DEP = \angle PFD$.

- 4 years, 3 months ago

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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.

- 4 years, 3 months ago

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- 3 years, 9 months ago

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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 $\dfrac{BD}{CD}$ = $\dfrac{AB}{AC}$ => $\dfrac{BE}{CF}$ = $\dfrac{BD}{CD}$.

=> $\Delta EBD\sim \Delta FCD$.

=> $\angle BED$ = $\angle CFD$ =>$\angle DEP = \angle PFD$.

$KIPKIG.$

- 3 years, 9 months ago

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

- 3 years, 7 months ago

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yes ..like geogebra

- 3 years, 7 months ago

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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.

- 4 years, 3 months ago

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Yes you are right. Thanks for pointing it out.

- 4 years, 3 months ago

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

- 4 years, 3 months ago

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Draw $OD\perp BC$.

Just $Pythagoras$ then, $BD^{2}$ $+$ $OD^{2}$ = $BO^{2}$

=> $4$ $BD^{2}$ $+$ $4$ $OD^{2}$ = $4$ $BO^{2}$

=> $[2BD]^{2}$ $+$ $[2OD]^{2}$ = $4$ $BO^{2}$

=> $BC^{2}$ + $AH^{2}$ = $4$ $AO^{2}$.

- 3 years, 9 months ago

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Can you post the solution please ?

- 4 years, 3 months ago

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

- 4 years, 3 months ago

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How did you get AH^2 = 2RcosA ?

- 4 years, 3 months ago

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Its an identity.

- 4 years, 3 months ago

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Its AH*

- 4 years, 3 months ago

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Oh, I knew that but I forgot lol

- 4 years, 3 months ago

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

- 4 years, 3 months ago

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- 4 years, 3 months ago

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Thanks. But I left Olympiad mathematics forever.

- 4 years, 3 months ago

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

- 4 years, 3 months ago

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That's true.

- 4 years, 3 months ago

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Are you ok? Say no

- 4 years, 3 months ago

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Yes, I'm fine.

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.

- 4 years, 3 months ago

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

- 4 years, 3 months ago

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

- 4 years, 3 months ago

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yep, i gave it today :)

- 4 years, 3 months ago

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How was it?

- 4 years, 3 months ago

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pretty good, better than last time's

- 4 years, 3 months ago

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How was ur rmo?how many did get correct?

- 4 years, 3 months ago

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Mine is on 16th. How was your paper?

- 4 years, 3 months ago

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

- 4 years, 3 months ago

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From which region did u give RMO?

- 4 years, 3 months ago

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Gujarat

- 4 years, 3 months ago

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Please most the paper.

- 4 years, 3 months ago

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Please most the paper.

- 4 years, 3 months ago

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How to most a paper? 😂😂 XD

- 4 years, 3 months ago

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Lol I meant post.Please post your paper.

- 4 years, 3 months ago

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What is circumdiameter?

- 4 years, 3 months ago

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sorry it is circumcenter.

- 4 years, 3 months ago

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

- 4 years, 3 months ago

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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.

- 4 years, 3 months ago

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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.

- 4 years, 3 months ago

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I have a solution using trivial inequalities.

- 4 years, 3 months ago

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QM-AM is trivial.

- 4 years, 3 months ago

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Yes after QM-AM the result directly follows.

- 4 years, 3 months ago

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Alright.Post some problems.

- 4 years, 3 months ago

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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.

- 4 years, 3 months ago

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

- 4 years, 3 months ago

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Good choice from MATHEMATICAL OLYMPIAD TREASURES.

- 4 years, 3 months ago

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

As $k$ is integer $2k-2$ will also be an integer.

- 4 years, 3 months ago

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

- 4 years, 3 months ago

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Sorry but this is not a REGIONAL MATHEMATICAL OLYMPIAD level problem.

Please post difficult ones.

- 4 years, 3 months ago

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OK, sorry! I'll post IMO level probs next time.

- 4 years, 3 months ago

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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.

- 4 years, 3 months ago

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@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.

- 4 years, 3 months ago

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Determine all positive triplets of integers such that

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

- 4 years, 3 months ago

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

Find the largest real solution to this equation.

- 4 years, 3 months ago

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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

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

- 4 years, 3 months ago

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Please post again as you cannot edit that.

- 4 years, 3 months ago

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Sure I can! Mod powers! :P

- 4 years, 3 months ago

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But how you edited that?

- 4 years, 3 months ago

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With great skill (and a large screen)!

- 4 years, 3 months ago

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What is that skill!?

- 4 years, 3 months ago

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Big iMac skills! :P :P

- 4 years, 3 months ago

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Best of luck everyone

- 4 years, 3 months ago

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Did anyone give RMO from north zone?

- 4 years, 3 months ago

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Uttar Pradesh - Me

- 4 years, 3 months ago

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At which center?

- 4 years, 3 months ago

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Meerut

- 4 years, 3 months ago

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Mumbai region paper was really easy

- 4 years, 3 months ago

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Could you post the question paper please?

- 4 years, 3 months ago

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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?

- 4 years, 3 months ago

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Post it on Slack.

- 4 years, 3 months ago

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Umm...how do I do that?

- 4 years, 3 months ago

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

- 4 years, 3 months ago

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You can do it on the browser.

- 4 years, 3 months ago

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Can I mail it to you and then you can post it?

- 4 years, 3 months ago

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Sure. sharkesa@gmail.com

- 4 years, 3 months ago

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I have sent it to you plz check

- 4 years, 3 months ago

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Please post the paper.

- 4 years, 3 months ago

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Can someone post this year's problems?

- 4 years, 3 months ago

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Please someone post this years rmo question paper.

- 4 years, 3 months ago

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The papers are uploaded on AoPS.

- 4 years, 3 months ago

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Please give me the link.Thanks.

- 4 years, 3 months ago

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

- 4 years, 3 months ago

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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.

- 4 years, 3 months ago

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this is north zone's (Delhi) problem

- 4 years, 3 months ago

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could u do it?

- 4 years, 3 months ago

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Then pls post the solution

- 4 years, 3 months ago

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Show that AB is independent of the choice of point P

- 4 years, 3 months ago

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Try using sine rule

- 4 years, 3 months ago

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It looks like Power of a Point, but Extended Sine Rule definitely works.

- 4 years, 3 months ago

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please add the solution

- 4 years, 3 months ago

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I won't give solution but the crux of this proof is to show that $AB$ is constant, irrespective of where $P$ is.

- 4 years, 3 months ago

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yeah. I have done it

- 4 years, 3 months ago

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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.

- 4 years, 3 months ago

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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

- 4 years, 2 months ago

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What is your score in RMO?

- 4 years, 2 months ago

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I gave GMO

- 4 years, 2 months ago

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

- 4 years, 2 months ago

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

- 4 years, 2 months ago

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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 25

- 4 years, 2 months ago

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Only 35. I don't think so.

At which website is the result of RMO declared?

- 4 years, 2 months ago

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Pls give for delhi region also.

- 4 years, 2 months ago

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Is RMO DELHI result out?

- 4 years, 2 months ago

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Yes

- 4 years, 2 months ago

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At which website?

- 4 years, 2 months ago

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Use trigonometry as there is a right triangle formed assuming centet

- 4 years, 1 month ago

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Hello everybody,

RMO results are out.

Who are selected?

- 4 years, 1 month ago

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@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.

- 4 years ago

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

- 4 years ago

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I meant icloud* there

- 4 years ago

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you can check ur mail now

- 4 years ago

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Mine is rajdeep.ind24@gmail.com

- 4 years ago

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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

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

- 3 years, 10 months ago

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

- 3 years, 3 months ago

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All INMO participants,please share ur marks.

- 3 years, 11 months ago

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do u want to join my INMO team?

- 3 years, 11 months ago

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70-80.

- 3 years, 11 months ago

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why dont u be active in slack?

- 3 years, 11 months ago

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I don't have time for these "SLACK" things.

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

- 3 years, 11 months ago

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oh...ok I am very sorry for disturbing u.

- 3 years, 11 months ago

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Its true. Kuch mazaa nahi aata slack chat pe.

- 3 years, 11 months ago

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

- 3 years, 11 months ago

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Initially i am only RMO qualifier. I am not INMO qualifier till the result is declared.

- 3 years, 11 months ago

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Which centre?

- 3 years, 11 months ago

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@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!!

- 3 years, 3 months ago

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u just need some angle chasing and similarity for RMO.

- 3 years, 3 months ago

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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!

- 3 years, 3 months ago

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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/.

- 4 years, 3 months ago

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

- 4 years, 3 months ago

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Even mine is on 16th

- 4 years, 3 months ago

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So wait upto that and then solve INMO problems.

- 4 years, 3 months ago

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How was your rmo :)?

- 4 years, 2 months ago

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Well it could not have been any more bad.

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?

- 4 years, 2 months ago

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Wth! Was the paper reallyl that tough? I solved 4 completely. I am on the boundary line of getting selected, since it's very difficult to get selected from my state, lot of competition here! 😅

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

- 4 years, 2 months ago

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Not sure about that rocking part :(

- 4 years, 2 months ago

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Did you give from Mumbai region?

- 4 years, 2 months ago

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Your Olympiad maths journey is not over. Congo :)

- 4 years, 1 month ago

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It might have been over if i was in a state like yours.

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?

- 4 years, 1 month ago

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Do you get selected in RMO?

I got selected.

- 4 years, 1 month ago

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Yes.Let's start the INMO Board!!

- 4 years, 1 month ago

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Oh congratulations.

I have already started that. Check here.

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

- 4 years, 1 month ago

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I think you should post a new note because that thread has died.

- 4 years, 1 month ago

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Ok i will post a fresh one.

- 4 years, 1 month ago

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Also,how's ya' iit prep going on?

How was KVPY?

- 4 years, 1 month ago

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I have RMO on 23rd.

- 4 years, 3 months ago

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