Math IOQM

 Q1. S1. First by trial and error. 2^(1/2) > 1. 2^(1/2) < 3^(1/3). Why? Raise both to powers 6. 2^3 < 3^2 => 8 < 9 correct. So 3^(1/3) is biggest so far. Now 3^(1/3) > 4^(1/4) why? Raise both to powers 12. 3^4 > 4^3 => 81 > 64. True. So looks like it could be 3^(1/3) since it's bigger than both its neighbors. Second by calculus. How to differentiate x^(1/x) and find its critical points? y = x^(1/x) =>  ln(y) = f(y) = (1/x).ln(x) To differentiate f(y) w.r.t. 'x' use chain rule. d(f(y))/dx = d(f(y))/dy . dy/dx d(f(y))/dy = 1/y => LHS = dy/dx.1/y Now differentiate RHS wrt 'x'. Using quotient rule. What is quotient rule? f(x) = u(x)/v(x) => f'(x) = [v(x).u'(x) - u(x).v'(x)]/(v(x))^2 So f'(RHS) = [x.1/x - ln(x).1]/x^2 for it to be 0: 1 - ln(x) = 0 => x = e So critical point is at 'e'. At x < e slope > 0 At x > e slope < 0 => There is a maxima at 'e'. But x has to be an integer. So compa...

Prove that Circumcenter(O),Centroid(G) and Orthocenter(H) in a triangle are colinear - also known as Euler's line - and OG/GH = 1/2. Proof: Simplest way to see it in action is in a right triangle. In a right triangle: |\ | \ |   \  O | G  \ |       \ ------- H It's clear that OGH is a straight line. Why? Orthocenter is simply the intersection of base and height. Circumcenter is the midpoint of hypotenuse and HO is the median hence G lies on HO. Centroid divides the median in 2:1 hence GH = 2OG. Now, a formal proof. In a triangle ABC, we know that AH = 2OM where M is midpoint of BC . Also: AH || OM why? Since AH is perpendicular to BC by definition. BC is a chord in the circumcircle and O is the center so OM will be perpendicular bisector of BC. Hence AH || OM And of course A,G,M are colinear. Now we claim that triangles AGH and MGO are similar. Why? AH = 2MO AG = 2MG Now the traversal AGM intersects the || lines AH and OM => angles HAG and OMG ...

Q1. Let O be the Circumcentre of triangle ABC. Find the value of  angle BAC + angle OBC in degrees. S1. Angle OBC = angle OCB since OB = OC in triangle OBC. 2OBC = OBC + OCB = 180 - 2A => OBC = 90 - A angle BAC = A OBC + BAC = 90 deg = answer Q2. Triangle ( AMC ) is isosceles with ( AM = AC ). Medians ( MV ) and ( CU ) are perpendicular to each other, and ( MV = CU = 12 ). What is the area of ( \triangle AMC )? S2. Let the centroid be G. It will divide the triangle into 3 smaller ones with same areas => GCA,GMC,GMA Area of MGC is easy to find since GM = 2/3*12 = 8 = CG and it is a right triangle. [MGC] = 1/2 * 8 * 8 = 32 [AMC] = 3*32 = 96 = Answer Q3. In a ( \triangle ABC ), let ( O, G, H ) denote its circumcenter, centroid, and orthocenter respectively. If GH/OH = m/n, where ( m, n ) are relatively prime positive integers, find ( m + n ). S3. Assume that ABC is not equilateral else it will become 0/0. Since this question doesn't mention the type of triangle we can use a quic...

Prove that in a triangle ABC, AH = 2OM where H is the Orthocenter, O is the Circumcenter and M = midpoint of BC. Using  Sylvester's Triangle Theorem We know that OH = OA + OB + OC so assume Circumcenter O as the origin of the vector space. H = A + B + C AH = OH - OA = B + C OM = (OB + OC)/2 = (B + C)/2 proof here . => AH = 2OM H.P.

Prove that  OM = (OB + OC)/2 where OM,OB,OC are vectors. M is midpoint of BC. For any O. Proof: OM  = OB + BM = OB + (BC)/2 BC = OC - OB => OM = OB + OC/2 - OB/2 = (OB+OC)/2 H.P.

Prove that H = A + B + C in triangle ABC where Circumcenter O is the origin for the vector space and H,A,B,C are position vectors of the Orthocenter and vertices A,B,C respectively. Proof: For position vectors a,b,c: |a| = |b| = |c| = R = circumradius And a.a = b.b = c.c = R^2 Let a vector h = a + b + c. We will show that line segment from any vertex to H is perpendicular to the opposite side. AH = OH - OA = h - a = a + b + c - a = b + c BC = OC - OB = c - b Compute dot product of AH and BC AH.BC = (b + c) ( c - b) = b.c + c.c - b.b - c.b = c.c - b.b + b.c - c.b Since b.c = c.b AH.BC = c.c - b.b = R^2 - R^2 So AH is perpendicular to BC. Similarly for other vertices and their opposite sides. Since H lies on each altitude, it's the orthocenter.