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.

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## Comments

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TopNewest(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 \(a, b, a+b \geq 2\sqrt {ab}\).

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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)

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

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

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

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

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(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.

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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?

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It's 3+2 root 2. Not -

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Calculus is not allowed in RMO.

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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.

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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

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