Math IOQM

Theorem: If a1x1 + a2x2 + .... an.xn = K then this Diophantine equation has a solution if d | K where d = gcd(a1,a2...an). We can note the similarities with Bezout's lemma here. Practice: Which of these Diophantine equations have integer solutions? 1. 21x + 15y = 17 2. 7x + 17y = 27 Theorem: If ax + by = c and d | c, where d = gcd(a,b), then this Diophantine equation has infinitely many solutions, and the solutions are of the form x = x0 + (b/d).k, y = y0 - (a/d).k, where k is integer and (x_0, y_0) are particular solutions of (ax + by = c); and x0, y0 are integers.

Q1. Solve for x: 32x = 79 mod 1225 S1. 32x = 79 mod 25 => 7x = 4 mod 25, since 7*3 = 21 = -4 m 25 => 7*-3 = 4 m 25 => x = 22 m 25 32x = 79 mod 49 => 32x = 30 m 49 => 16x = 15 m 49, since 16.4 = 64 = 15 m 49 => x = 4 m 49 Use CRT now: x = p.25 + q.49 m 25 22 = q.-1 m 25 => q = -22 = 3 mod 25 m 49 4 = p.25 mod 49 => p = 8 mod 49, since 25.8 = 200 = 4 mod 49 x = 8.25 + 3.49 mod 1225 = 347 mod 1225

Q1. There is a 21-sided regular polygon with vertices (A1, A2 .... A21). Triangles are formed by joining these vertices. a) How many of these triangles are acute-angled ? b) How many of these triangles are right-angled ? c) How many of these triangles are obtuse-angled ? d) How many of these triangles are equilateral ? e) How many of these triangles are isosceles ? S1.  Q2.  S2. Total: 11!/6!5! Routes via CD: 5!/2!3!*5!3!2! Answer: [1] - [2]

Q1. Two friends, Marco and Ian, are talking about their ages. Ian says, "My age is a zero of a polynomial with integer coefficients." Having seen the polynomial p(x) Ian was talking about, Marco exclaims, "You mean, you are seven years old? Oops, sorry I miscalculated! p(7) = 77 and not zero." "Yes, I am older than that," Ian's agreeing reply. Then Marco mentioned a certain number, but realizes after a while that he was wrong again because the value of the polynomial at that number is 85. Ian sighs, "I am even older than that number." Determine Ian's age. S1. Hint: (a-b) divides (a^n - b^n) for all integer 'n'. Let the root be 'a'. p(a) = 0 p(7) = 77 p(b) = 85 and b > 7. a > b > 7. For a polynomial with integer coefficients and x,y as integers: x-y | p(x) - p(y) So a-7 | p(a) - p(7) a-7 | -77 So a-7 can be 1,7,11,77 => a can be {8,14,18,84} b - 7 | 8 => b-7 can be 1,2,4,8 => b can be {8,9,11,15} a-b| 85 ...

Q1. Prove that if the two angle bisectors of a triangle are equal, then the triangle is isosceles. ( Steiner-Lehmus Theorem ) S1. So, let's see what are we trying to prove. BE = CF and angle B = angle C We know that length of angle bisector in triangle is given by the below formula. BE = 2ac.cos(B/2)/(a+c) Similarly, CF = 2ab.cos(C/2)/(a+b) make the equal: c.cos(B/2)/(a+c) = b.cos(C/2)/(a+b) cos(B/2)/cos(C/2) = ab + bc/ac + bc We will prove by contradiction. WLOG, Angle B > Angle C => b > c also => cos(B/2) < cos(C/2) (since B/2 and C/2 are both < 90). So LHS < 1 But RHS > 1 why? Because  ab + bc > ac + bc (since b > c) And if Angle C < Angle B, again LHS > 1 and RHS < 1. So only option is that both are equal. H.P. A slippery slope: If you go a different way, you will miss the proof! It's quite tricky. c.cos(B/2)/(a+c) = b.cos(C/2)/(a+b) from here: b/c = sinB/sinC => cos(B/2)/(a+c) = sinB/sinC * cos(C/2)/(a+b) sinB = 2sin(B/2)cos(B/2) ...

  [ABC] = a.c.sin(B)/2 [ABC] = [ABE] + [EBC] [ABE] = x.c.sin(B/2)/2 [EBC] = a.x.sin(B/2)/2 x = ac.sin(B)/sin(B/2).(a+c) x = ac/(a+c) * 2cos(B/2) BE = 2ac.cos(B/2)/(a+c)