RMO 2014 (Rajasthan region)

Let ABCD be an isosceles trapezium having an incircle; let AB and CD be the parallel sides and let CE be the perpendicular from C on to AB. Prove that AB, CCE and CD are in GP.
If $x$ and $y$ are positive real numbers, prove that $4x^{4}+4y^{3}+5x^{2}+y+1\geq 12xy$ .
Determine all pairs $m$ > $n$ of positive integers such that $1=gcd(n+1,m+1)=gcd(n+2,m+2)=......=gcd(m,2m-n)$
What is the minimal area of right-angled triangle whose inradius is 1 unit?
Let ABC be an acute-angled triangle and let I be its incentre. Let the incircle of triangle ABC touch BC in D. The incircle of the triangle ABD touches AB in E, the incircle of triangle ACD touches BC in F. Prove that B, E, I, F are concyclic.
In the adjacent figure, can the numbers 1,2,3,...,18 be placed, one on each line segment such that the sum of the numbers on the three line segments meeting at each point is divisible by 3.

Image Question 6

#Algebra #Geometry #GreatestCommonDivisor(GCD/HCF/GCF) #RMO #Minimumvalue

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

(2). By AM-GM, $4y^{3}+y \geq 4y^{2}, 4x^{4}+1 \geq 4x^{2}, 4x^{2}+5x^{2}+4y^{2}=9x^{2}+4y^{2} \geq 12xy$ .

This is using the fact that for all positive $a, b, a+b \geq 2\sqrt {ab}$ .

Joel Tan - 6 years, 6 months ago

OM*G. This was so easily simplified by you :( :( :( :( ...I did 30-40% of the same using AM-GM...... Felt that the powers were disproportionately large and left it midway :/ :'( (sob)

Krishna Ar - 6 years, 6 months ago

Never! I gave RMO 2014, Rajasthan region. Paper was completely different but the number of questions. Why?

Satvik Golechha - 6 years, 6 months ago

I got the same paper. :D How many did you solve? (How can you appear when in 12th, BTW?)

Krishna Ar - 6 years, 6 months ago

I didnt appeard! My brother did! I took paper frm him!

Pranjal Jain - 6 years, 6 months ago

Oh, I see :P. Which class is he in? How many did he do?

Krishna Ar - 6 years, 6 months ago

@Krishna Ar – He is in 9th. Attempted RMO just for some exposure. I didn't asked him how much he did as I know he won't tell anyone!

Pranjal Jain - 6 years, 6 months ago

@Pranjal Jain – -No Issues- :)

Krishna Ar - 6 years, 6 months ago

@Krishna Ar – How much are you scoring in NSEs?

Pranjal Jain - 6 years, 6 months ago

@Pranjal Jain – I didn't write them this year because I had to write another test on that day. Anyway, I wasn't prepared upto 12th level in any of the subjects :/ , So there wasn't any point in giving them. What about you?

Krishna Ar - 6 years, 6 months ago

@Krishna Ar – NSEC=180, NSEA=144, NSEP=105

Pranjal Jain - 6 years, 6 months ago

@Pranjal Jain – Impressive !!!!!!!!! :D

Krishna Ar - 6 years, 6 months ago

@Krishna Ar – Not at all!! NSEC was worth 200+, No comments about astro and NSEP changed its pattern. I was prepared for subjective ones. Well just waiting for results. Just some hopes in NSEC and NSEA.

Pranjal Jain - 6 years, 6 months ago

(3). Note that gcd (m-n, n+i)=1, i=1, 2, ..., m-n. Otherwise, gcd (m-n+n+i, n+i)=gcd (m+i, n+i) > 1. But one of n+i must be a multiple of m-n. Let it be n+x. Then gcd(n+x, m+x)= (n+x, m-n) is at least m-n. So m-n=1.

Note that gcd (n+1, m+1)=gcd (n+1, n+2)=1. Hence all pairs are in the form (x, x+1) for an integer x.

Joel Tan - 6 years, 6 months ago

For number 4, The minimum area of right triangle is (3 - 2 sqrt(2)) square units.

Using the relationship of inradius, semiperimeter, and area of triangles, by letting a and b be the sides of the triangle and c be the hypotenuse, (a + b + c)/2 = ab/2 implying the relationship. By Pythagorean theorem, a + b - ab = -sqrt(a^2 + b^2). After manipulations, it implies that ab + 2 = 2b + 2a. We solve for a in terms of b, where a = (2b - 2)/(b - 2) and ab/2 = (2b^2 - 2b)/2(b - 2). I am trying to get a non-calculus solution here, but using derivatives, the minimum area is the given answer above...

P.S. Somehow confused because the function relating the area and one side has actually no minimum as graphed... What might the actual answer be?

John Ashley Capellan - 6 years, 6 months ago

It's 3+2 root 2. Not -

Krishna Ar - 6 years, 6 months ago

Calculus is not allowed in RMO.

Pranjal Jain - 6 years, 6 months ago

FYI, it is allowed but not required. All the sums can be solved without calculus. However, if one wants to use calculus, one is allowed to do so.

Mohnish Chakravarti - 6 years, 6 months ago

@Mohnish Chakravarti – Well, I really didn't knew this! I still doubt it...

Pranjal Jain - 6 years, 6 months ago

I got the answer as 3-2sqrt2 without using calculus. We'll get a quadratic equation in the side and hence the area taking both sides as equal. I got the minimum side as 2-sqrt2

easha manideep d - 5 years, 6 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$