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# RMO 2016 Karnataka Region(16th Oct)

1. Let $$ABC$$ be a triangle and $$D$$ be the mid-point of $$BC$$. Suppose the angle bisector of $$\angle ADC$$ is tangent to the circumcircle of triangle $$ABD$$ at $$D$$. Prove that $$\angle A=90^{\circ}$$.

2. Let $$a,b,c$$ be three distinct positive real numbers such that $$abc=1$$. Prove that $$\dfrac{a^3}{(a-b)(a-c)}+\dfrac{b^3}{(b-c)(b-a)}+\dfrac{c^3}{(c-a)(c-b)} > 3$$.

3. Let $$a,b,c,d,e,d,e,f$$ be positive integers such that $$\dfrac a b < \dfrac c d < \dfrac e f$$. Suppose $$af-be=-1$$. Show that $$d \geq b+f$$.

4. There are $$100$$ countries participating in an olympiad. Suppose $$n$$ is a positive integer such that each of the $$100$$ countries is willing to communicate in exactly $$n$$ languages. If each set of $$20$$ countries can communicate in at least one common language, and no language is common to all $$100$$ countries, what is the minimum possible value of $$n$$?

5. Let $$ABC$$ be a right-angled triangle with $$\angle B=90^{\circ}$$. Let $$I$$ be the incentre if $$ABC$$. Extend $$AI$$ and $$CI$$; let them intersect $$BC$$ in $$D$$ and $$AB$$ in $$E$$ respectively. Draw a line perpendicular to $$AI$$ at $$I$$ to meet $$AC$$ in $$J$$, draw a line perpendicular to $$CI$$ at $$I$$ to meet $$AC$$ at $$K$$. Suppose $$DJ=EK$$. Prove that $$BA=BC$$.

6.(a). Given any natural number $$N$$, prove that there exists a strictly increasing sequence of $$N$$ positive integers in harmonic progression.

(b). Prove that there cannot exist a strictly increasing infinite sequence of positive integers which is in harmonic progression.

Note by Svatejas Shivakumar
1 week ago

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@Svatejas Shivakumar @Sharky Kesa Can either of you help me verify the phrasing of Q4? Is it "exactly one common language" or "at least one common language"?

It makes a huge difference in appoaching the problem. Staff · 2 days, 11 hours ago

It is atleast one common language. · 2 days, 10 hours ago

Thanks. I've edited the question accordingly. Staff · 2 days, 3 hours ago

Anyone with 6 th done · 3 days, 7 hours ago

Both of $$6$$'s subdivions are available in Brilliant.

Grasp here and here · 3 days ago

20 is the answer for the 4th one · 4 days, 10 hours ago

Is the following solution correct for P2?

WLOG $a>b>c$. Multiplying both sides by $(a-b)(b-c)(c-a)$, it suffices to prove $\sum a^3(b-c) \ge 3(a-b)(b-c)(c-a)$.

Expanding both sides and simplifying , this reduces to prove $\sum (a^3b+3a^2b) \ge \sum( ab^3+3ab^2)$. Using $abc=1$ this simplifies to show $$\sum \frac{a^2+3a}{c} \ge \sum \frac{a^2+3a}{b}.$$

But this is just rearrangement on $a^3+3a>b^3+3b>c^3+3c$ and $1/a < 1/b < 1/c$. · 4 days, 23 hours ago

solutions are now uploaded on resonance. · 5 days, 10 hours ago

It was just 5 minutes after the exam was over LOL · 4 days, 23 hours ago

what is the answer of the 4 th one · 5 days, 21 hours ago

The same question paper was asked in jharkand . Being in eleventh is solving three correctly , enough to qualify ? · 5 days, 21 hours ago

Has anyone perfectly solved Q.4?? · 6 days, 7 hours ago

Comment deleted 2 days ago

The Farey sequence relates to question 3. Staff · 2 days, 22 hours ago

Yeah, sorry, looked at the wrong question. Question 4 looks like Pigeonhole Principle. · 2 days, 15 hours ago

For q2 since a,b,c are distinct WLOG a>b>c.Take a=c+y,b=c+x,and replace a,b with the above values in lhs and on solving you get finally lhs as 3c+x+y which is nothing but a+b+c,now apply am-gm,to prove it. · 6 days, 7 hours ago

Terms are [a^3/{3(a-b)(a-c)}]+a-b/(3^{5/2})+a-c/(3^{5/2})>=a/3 · 6 days, 17 hours ago

Better use the formatting guide to write the expression.

$$\frac{a^{3}}{3(a-b)(a-c)}+\frac{a-b}{3^{5/2}}+\frac{a-c}{3^{5/2}} \geq \frac{a}{3}$$

That looks better and much understandable too! · 6 days, 7 hours ago

*a-b is also in bracket.sane with a-c · 6 days, 17 hours ago

Terms can be negative but they are eventually cancelled so we can apply am-gm · 6 days, 17 hours ago

@Svatejas Shivakumar Could you post the solution to problem 6? · 6 days, 23 hours ago

I am from jharkhand.I have solved q1,2,5,6.solution for q4 please · 1 week ago

You are in which class and did you got dps ranchi as exam centre ? · 5 days, 21 hours ago

I am from Jamshedpur,in class 11 · 4 days, 10 hours ago

Same paper was for chhattisgarh also! · 1 week ago

Apply am -gm for these a^3/3(a-b)(a-c)+(a-b)/3^(5/2)+(a-c)/3^(5/2) Which is>=a/3 Now write same expressions for b and c and add all three We get in rhs- a+b+c/3>=1(am-gm) Hence proved :) · 1 week ago

You cannot apply AM GM Its not given that a-b,b-c,c-a are +ve · 1 week ago

Comment deleted 1 week ago

Yeah you are right · 1 week ago

But still denominator of second term is -ve

BTW,Rank In Technothlon? · 1 week ago

can u put brackets please · 1 week ago

you can solve the second one by using am gm inequality · 1 week ago

Can you post the solution? Also when does equality occur? · 1 week ago

Which questions did u do? · 1 week ago

Attempted all. Got 1,5,6 completely. Don't know the answer of 4. Did some progress in 1 and 2. · 1 week ago

I got 3 questions fully correct and 1 question 25% corrrect · 1 week ago

Dont u think q-1 is halwa? NCERT Question!!! · 1 week ago

Really!! Though it was very easy but don't think it could be an NCERT question. · 6 days, 7 hours ago

Yes even 5 was very easy. · 1 week ago

Can you please post the complete solution of Q.5?? · 6 days, 7 hours ago

Yes very easy everyone was able to solve it. Did you do Q.3? · 1 week ago

Yeah did it Though I spent 55 mins on this one · 6 days, 11 hours ago

Ok it took me also around 30-40 mins. How was RMO? · 6 days, 8 hours ago

I gave both RMO and GMO

Both went good · 4 days, 11 hours ago

It is better to evaluate it to be a+b+c · 1 week ago

It is better if u factorise given expression It is a+b+c · 1 week ago