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rmo aakash lecture 10, 15th oct 2026

Q1. Let a, b, c ∈ ℝ. Prove the inequalities: 3(ab + bc + ca) ≤ (a + b + c)² ≤ 3(a² + b² + c²) S1. First see the RHS one. Cauchy shwartz. (1^2 + 1^2 + 1^2)(a^2 + b^2 + c^2) >= (1.a + 1.b + 1.c)^2 Since (a1.b1 + a2.b2 + a3.b3)^2 <= (a1^2 + a2^2 + a3^2).(b1^2 + b2^2 + b3^2) And equality happens when: a1/b1 = a2/b2 = a3/b3 => a/1 = b/1 = c/1 => a = b = c Now the LHS one, we need to do effectively this: ab + bc + ca <= a^2 + b^2 + c^2 WLOG a <= b <= c Which is simple using rearrangement inequality. And in RE also a = b = c gives the equality.
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p-adic valuation and legendre's formula

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 V_p(n) = k => highest power of prime p in number 'n' is k. This is known as p-adic valuation. For e.g. V2(12) = 2 => p-adic valuation of 2 wrt 12 is 2. Also it's additive, ie V_p(m.n) = V_p(m) + V_p(n) V2(36) = V2(18.2) = V2(18) + V2(2) = 1 + 1 = 2 Legendre's formula:
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Bertrand's postulate

 for any integer k > 1, there will always be a prime p s.t. k < p < 2k.
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RMO aakash 24 sep 2026 lecture 5

Q1. Find all positive integers n for which 2^n + 7^n is a perfect square. S1. for n = 1, it's 9. Ok. for n even  2^n = (-1)^n mod 3 = 1 mod 3 7^n = 1^n mod 3 = 1 mod 3 2^n + 7^n = 2 mod 3 for all even 'n'. But any positive number square = 0,1 mod 3. So no even n is possible. for n odd and n>=3 2^n = 0 mod 4 3^n = (-1)^n = -1 mod 4 = 3 mod 4 2^n + 7^n = 3 mod 4 Any positive square  = 0,1 mod 4 So only answer is n = 1 Q2. Given five distinct integers, consider the ten differences formed by pairs of these numbers. (Note that some of these differences may be equal.) Determine the largest integer that is certain to divide the product of these ten differences, regardless of which five integers were originally given. S2. 1,2,3,4,5 => 288 = 2^5.3^2 Now show 288 will always divide any such product. 3k,3k+1,3k+2 3 buckets, 5 objects. No matter how you distribute, you will get 2 factors of 3. Because at least 1 bucket will have more than 1 object. Case 1: 1 bucket with 3 objects...
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in any triangle longer side corresponds to shorter median

Proof: Compute ma and mb using appollonius and equate. 2[ma^2 + a^2/4] = b^2 + c^2 __[1] 2[mb^2 + b^2/4] = a^2 + c^2 __[2] ma^2 - mb^2 = (b^2 - a^2)/2 - (a^2 - b^2)/4 = 3/4[b^2 - a^2] H.P.
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two similar triangles with same circumradius will be congruent

Yes. Why? In first triangle: a/SinA = 2R Next triangle: p/SinP = 2R Angles are same => sides are same H.P.
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rmo aakash lecture 8, 6th october - done

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Q1. In a convex quadrilateral ABCD , angle ABD = 30 , angle BCA = 75 , angle ACD = 25 , and CD = CB . Extend CB  to meet the circumcircle of triangle DAC  at E . Prove that CE = BD . S1. Q2. Let ABC be a triangle with circumcircle Gamma, and let G be the centroid of triangle ABC. Extend AG, BG, and CG to meet the circle Gamma again at A1, B1, and C1, respectively. Suppose that triangle ABC ~ A1B1C1, specifically with: angle BAC = angle A1B1C1,  angle ABC = angle A1C1B1,  and angle ACB= B1A1C1. Prove that ABC and A1B1C1 are equilateral triangles. S2. Step 1. Do angle chasing. Same chord same angle. Step 2. Use given angle equalities. A1 = C => 5 + 6 = 4 + 6 => 4 = 5 B1 = A => 1 + 4 = 1 + 2 => 4 = 2 C1 = B => 3 + 5 = 3 + 2 => 2 = 5 Now draw the modified figure: And all the medians intersect at Centroid G. We know that medians divide the triangle into 3 equal areas. [GAB] = [GBC] = [GCA] Each of these triangles as angle 2. [GAB] = GB.AB.sin(2)/2 [GBC] ...
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