MathDB

5

Part of 2013 India Regional Mathematical Olympiad

Problems(5)

Indian RMO- Paper 2

Source: Problem 4

12/11/2013
Let ABCABC be a triangle with ∠A=90∘\angle A=90^{\circ} and AB=ACAB=AC. Let DD and EE be points on the segment BCBC such that BD:DE:EC=1:2:3BD:DE:EC = 1:2:\sqrt{3}. Prove that ∠DAE=45∘\angle DAE= 45^{\circ}
geometryAngle Chasingsimilar triangles
Indian RMO- Paper 2

Source: Problem 5

12/11/2013
Let n≥3n \ge 3 be a natural number and let PP be a polygon with nn sides. Let a1,a2,⋯ ,ana_1,a_2,\cdots, a_n be the lengths of sides of PP and let pp be its perimeter. Prove that a1p−a1+a2p−a2+⋯+anp−an<2\frac{a_1}{p-a_1}+\frac{a_2}{p-a_2}+\cdots + \frac{a_n}{p-a_n} < 2
limitinequalities
Indian RMO - Paper -4[5]

Source:

12/12/2013
In a triangle ABCABC, let HH denote its orthocentre. Let PP be the reflection of AA with respect to BCBC. The circumcircle of triangle ABPABP intersects the line BHBH again at QQ, and the circumcircle of triangle ACPACP intersects the line CHCH again at RR. Prove that HH is the incentre of triangle PQRPQR.
geometrygeometric transformationreflectioncircumcircleincenterhomothetyReflections
Indian RMO - Paper 3

Source: RMO- Problem 5

12/11/2013
Let ABCABC be a triangle which it not right-angled. De fine a sequence of triangles AiBiCiA_iB_iC_i, with i≥0i \ge 0, as follows: A0B0C0A_0B_0C_0 is the triangle ABCABC and, for i≥0i \ge 0, Ai+1,Bi+1,Ci+1A_{i+1},B_{i+1},C_{i+1} are the reflections of the orthocentre of triangle AiBiCiA_iB_iC_i in the sides BiCiB_iC_i,CiAiC_iA_i,AiBiA_iB_i, respectively. Assume that ∠Am=∠An\angle A_m = \angle A_n for some distinct natural numbers m,nm,n. Prove that ∠A=60∘\angle A = 60^{\circ}.
geometrygeometric transformationreflectiongeometry unsolved
GCD and LCM Inception

Source: Indian RMO 2013 Mumbai Region Problem 5

2/1/2014
Let a1,b1,c1a_1,b_1,c_1 be natural numbers. We define a2=gcd⁡(b1,c1),        b2=gcd⁡(c1,a1),        c2=gcd⁡(a1,b1),a_2=\gcd(b_1,c_1),\,\,\,\,\,\,\,\,b_2=\gcd(c_1,a_1),\,\,\,\,\,\,\,\,c_2=\gcd(a_1,b_1), and a3=lcm⁡(b2,c2),        b3=lcm⁡(c2,a2),        c3=lcm⁡(a2,b2).a_3=\operatorname{lcm}(b_2,c_2),\,\,\,\,\,\,\,\,b_3=\operatorname{lcm}(c_2,a_2),\,\,\,\,\,\,\,\,c_3=\operatorname{lcm}(a_2,b_2). Show that gcd⁡(b3,c3)=a2\gcd(b_3,c_3)=a_2.
number theorygreatest common divisorleast common multiplenumber theory unsolved