Math IOQM

  First of all, let's figure out whether we want ordered triplets or unordered ones. In combinatorics, if the question says "number of ways" it usually means ordered. If they wanted unordered triplets, they would have said: "Find the number of sets of 3 integers..." "Find the number of unordered combinations of 3 factors..." “in non-decreasing order”, “up to permutation”, “as an unordered product”. etc. Once it has been established that we want ordered triplets, it calls for stars and bars . 10^6 = 2^6 * 5^6 Let 10^6 = N then: N = a * b * c Here a,b,c are 3 distinct buckets in which 6 identical objects have to be distributed(for powers of 2), and similarly for powers of 5. So number of ways to distribute powers of 2 = (6 + 3 - 1)C(3-1) = 8C2 = 28 Similarly for powers of 5 = 8C2 = 28 Total: 28*28 = 784 But this number includes the cases where 1 or more factors could be 1. Let's count them: Cases where at least one factor is 1: A = set of triplets a,...

Every regular n-gon has a circumcircle. And formula for the circumradius : R = a/2.sin(pi/n) So a regular hexagon is the only regular n-gon where circumradius  = side length.

  Solution: Part 1: (AEHZ) is cyclic quadrilateral since opposite angles HZA and HEA sum to 180. Its diameter is AH since it subtends 90 deg at circumference. N is the center since it's the midpoint of AH.   (NZMDE) are also concyclic since all these fall on 9-point circle of ABC. EZ is the common chord of these 2 circles and hence the radical axis. Angle NDM = 90 => MN is a diameter of this 9-point circle. Hence the center of this will lie on MN. So both the centers lie on the line MN. Line joining centers is perpendicular to the radical axis. H.P. Part 2: MN is perpendicular to EZ as both the centers lie on MN. The line containing both the centers will bisect the chord EZ in circle (AEHZ). In triangle AEZ, MN is the perpendicular bisector of EZ. So using Arc midpoint lemma MN and AI will meet on the circumcircle of AEZ and that point is K. Using the same lemma again MN and AI' will meet on the circumcircle again at L. K,N,L are colinear hence KL is a diameter. Hence KL =...

In a triangle ABC, the vertices B,C make these angles: 1. At Incenter, angle BIC = 90 + A/2 2. At Orthocenter, angle BHC = B + C 3. At Circumcenter, angle BOC = 2A

  Proof video

In a circle where a chord BC subtends angles at A1,A2 on the circumference. Let H1,H2 be orthocenters of A1BC and A2BC. Then A1H1 = A2H2 Why? Since A1H1 = 2R.cosA1 A2H2 = 2R.cosA2 Circumradius is same and angles are also same.

 https://math.stackexchange.com/questions/589877/why-would-the-reflections-of-the-orthocentre-lie-on-the-circumcircle

Center of the nine-point circle lies on the Euler's line and it's the midpoint of Orthocenter and Circumcenter. Radius of the nine-point circle is half of the Circumradius of the original triangle. This proof uses the facts that OM1, i.e. distance from Circumcenter O to line BC(i.e. its midpoint) is R.cosA whereas AH = 2R.cosA where R = circumradius and H = Orthocenter. Proof video .

Why is AH = 2RcosA in a triangle ABC and OM1 = R.cosA. R = circumradius H = orthocenter O = circumcenter M1 = midpoint of BC. Proof for OM1 = R.cosA: Since M1 is the midpoint of BC, OM1 will be perpendicular to BC. BC is a chord of the circumcircle and hence it will subtend angle 2A at the center which will be bisected by OM1. In triangle OM1C, OC = R and OC.cos(ROM1) = OC.cos(2A/2) = R.cosA. Hence proved. Proof for  AH = 2R.cosA

 https://en.wikipedia.org/wiki/Miquel%27s_theorem Source: A beautiful journey through Olympiad

Radical axis questions. Facts used: 1. 3 Pairwise Radical axes of 3 circles are concurrent. 2. Common chord of 2 circles is their R.A.(radical axis) 3. Radical axis is all those points whose powers w.r.t. 2 circles are same. Power of a point P wr.t. a circle = PT^2 if PT is a tangent or PT1 * PT2 if a line drawn from P intersects circle at T1,T2. Question 1: Bulgarian Mathematical Olympiad 1996. Source . Question 2: Iranian Geometry Olympiad 2021. Source . Note that the same reasoning would work even if the ordering is AHG and not AGH.