Last weekend was incredibly busy, as I, along with many others wrote two competitive exams: the AMTI Inter NMTC, Second Level, and the KVPY

First, the AMTI Second Level

This is the paper:

a) If $x$ and $y$ are positive reals such that $x^{2014}+y^{2014}=1$, prove that $(\displaystyle\sum^{1007}_{k=1}\frac{1+x^{2k}}{1+x^{4k}})(\displaystyle\sum^{1007}_{k=1}\frac{1+y^{2k}}{1+y^{4k}})<\frac{1}{(1-x)(1-y)}$

b) The angles of a triangle are in arithmetical progression. The altitudes of this triangle are also in arithmetical progression. Show that the triangle is equilateral.

$ABC$ is an acute angle triangle in which the three sides are unequal. $L,M,N$ are respectively the midpoints of the sides $BC,CA,AB$. The perpendicular bisectors of the sides $AB,AC$ intersect $AL$ at $D,E$ respectively. $BD, CE$ cut at $F$ inside the triangle. Show that $A,M,N,F$ are concyclic.

$S$ is a set of all positive integers, such that for any two distinct members of $S$, $a$ and $b$, $\frac{a+b}{GCD(a,b)}$ is also a member of $S$. Find and describe all such finite sets.

a) Find all real number triples $(x,y,z)$ which satisfy $3(x^2+y^2+z^2)=1, x^2y^2+y^2z^2+z^2x^2=xyz(x+y+z)^3$

Ab) Show that all the numbers $1$ to $16$ can be written in a line, but not in a circle, so that the sum of any two adjacent numbers is a perfect square.

A certain number is the product of three different prime factors, the sum of whose squares is $2331$. There are $7560$ number (including unity) which are less than the number and prime to it. The sum of all its divisors (including unity and the number) is $10560$. Find the number.

A rectangular parallelepiped is given, such that its intersection with a plane is a regular hexagon. Prove that the rectangular parallelepiped is a cube

A triangle $ABC$ is given. The midpoints of the sides $AC$ and $AB$ are $B_1$ and $C_1$ respectively. The in centre of the $\triangle ABC$ is $I$. The lines $B_1I$ and $C_1I$ meet the sides $AB,AC$ at $C_2,B_2$ respectively. If the areas of $\triangle ABC$ and $\triangle AB_2C_2$ are equal, find the measure of the angle $BAC$

If $(a+b)$ and $(a-b)$ are relatively prime integers for some natural numbers $a,b$, find the greatest common divisor of $2a+(1+2a)(a^2-b^2)$ and $2a(a^2+2a-b^2)(a^2-b^2)$

I would like to know how other Brilliant people who qualified did in their levels

The other exam was the KVPY

I found the exam rather tough. How did you do?

After checking the answer key, I am getting around 45. So, unfortunately, I will not be qualifying.

But how did you do? Post a comment about your KVPY experience or AMTI experience.

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1 vote

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## Comments

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TopNewestFor the third question, is the answer {a,a(a-1)} where a runs from 3 to infinity

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AMTI was tough.

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What was the Junior paper like?

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Not too hard, Not too trivial. Lots of lengthy problems with one being improperly worded. It definitely wasnt a good experience. Maybe you can look at Siddhart G's set which consists of few problems from the paper. How many did you solve?

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You don't say! The KVPY Math was waaay tougher than previous years. But maybe to counter that, the Biology section was strictly of 10th level. Oh and BTW, ask me no questions and I shall tell you no lies!

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Agreed. The 2 marks questions were fine though the Geometry in 1 markers put me off. BTW, why no questions-no lies ? :P And...how much are you expecting? (Was this why, no question-no lies? _

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Anyone know what is the use of NMTC round 2?

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No use. Only Rank is given+Certificate of Merit/Excellence

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What do you think the cutoffs will be? I'm probably getting ~40. Haven't really checked. Went horribly. They made us sit in benches meant for 4th grade students. EDIT:- Did you take the 11th one or the 12th one? I took the 11th one.

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I took the 11th one as well.

Last year's cutoff was 53 or so. I think they are reducing the cutoff this year. Maybe it will be 50.

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I am hoping they reduce the cutoff to 40. xD Everyone I know said their paper went horrible. Can't find anyone who actually said their paper went well.

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Wait, are you a 11th grader? :O

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Yes. Why the :O?

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The KVPY shortlist and the AMTI second level results have come out. I have qualified in both. I had also seen @Aditya Raut on both lists

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Whoa ! I didn't know this!!! Thank you very much for letting me know. What's the website?

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Never mind, I found it in google search

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What region are from, for RMO?

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Yeah.I saw, this is pretty cool!. How many did you solve in NMTC? Your rank is awesome.

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I think I attempted all of them, and got 6 right

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How did you do RMO?

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Very useful

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For how much was the test conducted?

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10 crores

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4(B) For the numbers to be arranged in a circle all the numbers must have at least to other distinct numbers from the set so that their sum is a perfect square. But here 8 has only 1 partner which is 1 and 16 has only one partner which is 9. So these to need to be at the terminal points of the line and so this set can't be arranged into a circle.

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For the 4th apply muirhead

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Root of cubic equation

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AMTI is coming up.....

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whats ans for 7 and 8

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