Math IOQM

Distance of (x0,y0) from ax + by + c = 0 formula is: d = (ax0 + by0 + c)/sqrt(a^2 + b^2) Proof: Let the perpendicular from (x0,y0) fall on the line at (x1,y1). ax1 + by1 + c = 0 Slope of the perpendicular line = (y1 - y0)/(x1- x0) = b/a Let y1 - y0 = kb x1 - x0 = ka d^2 = (x1 - x0)^2 + (y1 - y0)^2 = k^2(a^2 + b^2) Now ax1 + by1 + c = 0 => a(ka + x0) + b(kb + y0) + c = 0 => k = -(ax0 + by0 + c)/(a^2 + b^2) =>  d^2 = (ax0 + by0 + c)^2/(a^2 + b^2) H.P. Distance between 2 parallel lines ax + by + c1 = 0  and  ax + by + c2 = 0 formula is: d = |c1 - c2|/sqrt(a^2 + b^2) Proof: Let (x0,y0) lie on the first line =>  ax0 + by0 + c1 = 0 Distance of (x0,y0) from the second line is: (ax0 + by0 + c2)/sqrt(a^2 b+2) = |c1 - c2|/sqrt(a^2 + b^2) = answer.

Q1. Let X be the midpoint of the side AB of △ABC. Let Y be the midpoint of CX. Let BY cut AC at Z. Prove that AZ=2ZC. Q2. ABC is a equilateral triangle with vertex A fixed and B moving in a given straight line. Find the locus of C. Q3. Let  A be one of the two points of intersection of two circles with centres  X and  Y. The tangents at  A to these two circles meet the circles again at  B,C. Let the point  P be located so that  PXAY is a parallelogram. Show that  P is the circumcentre of triangle  ABC. Q4: Let ABC be a triangle and h be the altitude from A to BC. Prove: (b+c)^2 >= a^2 + 4h^2 Q5: The interior of a quadrilateral is bounded by the graphs of  ( x + a y ) 2 = 4 a 2 ( x + a y ) 2 = 4 a 2  and  ( a x − y ) 2 = a 2 ( a x − y ) 2 = a 2 , where  a a  is a positive real number. What is the area of this region in terms of  a a , valid for all  a > 0 a > 0  ? Q6. Three equally spaced p...

Q1) Solve the cubic equation (9x^3 - 27x^2 + 26x - 8 = 0), given that one of the root of this equation is double the other. Q2) If the product of two roots of the equation (4x^4 - 24x^3 + 31x^2 + 6x - 8 = 0) is 1, find all the roots. Q3 Obtain a polynomial of lowest degree with integral coefficient, whose one of the zeros is sqrt{5} + sqrt{2}. Solution 1: Assume a,2a are the roots. Put the values in the given equation and equate them. You will get a cubic in 'a' with one root as 0. a can't be 0 since that would mean -8 = 0 Other roots will be 2/3,13/21 Putting a = 2/3 in the original equation works out nicely and 13/21 doesn't quite fit it. So roots will be 2/3,4/3,1 = answer. Solution 2: Solution 3: Simplest way to solve such question is to start with x = given root = sqrt(5) + sqrt(2) Square both sides: x^2 = 7 + 2sqrt(10) x^2 - 7 = 2.sqrt(10) Again square to eliminate the sqrt on the right side and now you will have a degree 4 polynomial with all integer coefficients...

Q1). Andy and Bethy are at same point. Andy leaves at 1:30 toward north at a steady 8 miles/hr speed. Bethy leaves at 2:30, toward east at a steady 12 miles/hr speed. At what time they will be exactly the same distance away from their starting point? Q2). A box contains 10 pounds of a nut mix i.e., 50% peanuts, 20% cashews, 30% almonds. A 2nd nut mix containing 20% peanuts, 40% cashews, 40% almonds is added to the box resulting in a new nut mix i.e., 40% peanuts. How many pounds of cashews are now in the box? Q3). How many isosceles triangles are there with positive area whose side lengths are all positive integers and whose longest side has length 2025? Q4) 1, 2, 1, 2, 3, 2, 1, 2, 3, 4, 3, 2, 1, 2, 3, 4, 5, 4, 3, 2, 1, 2, 3, 4, 5, 6, 5, 4, 3, 2, 1, 2, ... What is the 2025th term? Q5) Suppose a and b are real numbers. When the polynomial (x^3 + x^2 + ax + b) is divided by (x - 1), the remainder is 4, and when divided by (x - 2), the remainder is 6. b - a = ? Q6) The sequence (1, x, y, ...

Proof: p = a.x q = b.x gcd(p,q) = k Using Bezout's identity: k = s.p + r.q = s.a.x + r.b.x = x(s.a + r.b) => x divides k H.P.

Prove that: If x divides y, then (a^x - 1) divides (a^y - 1) where a,x,y are positive integers and a > 1. Proof: We know that a^x - 1 can be written as (a - 1)(1 + a + a^2 ... a^(x-1)) Let y = kx Then a^y - 1 = a^(kx) - 1 = (a^x)^k - 1 Let u = a^x then a^y - 1 = u^k - 1 = (u - 1)(1 + u + u^2 ... u^(k-1)) = (a^x - 1)(...) H.P. The result above can be used to prove that: gcd(a^m - 1, a^n - 1) = a^(gcd(m,n)) - 1 Proof: Let gcd(m,n) = g Then using Bezout's identity, there exist x,y such that mx + ny = g => a^g = a^mx.a^ny____________[1] Also, since g divides m and n a^g - 1 divides a^m - 1 and a^n - 1 So a^g -1 is a common divisor. Now we need to prove that it is greatest common divisor. Let gcd(a^m - 1, a^n - 1) = D Then a^m - 1 mod D = 0 => a^m = 1 mod D and a^n = 1 mod D => (a^m)^x = 1 mod D and (a^n)^y = 1 mod D => a^g = 1 mod D using [1] => D divides (a^g -1) So D divides (a^g -1) which is a common divisor of (a^m - 1) and (a^n - 1) while D is their gcd. GCD can ...

Q. Let S be the sum of the base 10 logarithms of all the proper divisors of 1000000. What is the integer nearest to S? Solution: Before that prove that product of divisors of an integer N is P = N^(d(N)/2) where d(N) = number of divisors of N. Let there be k divisors of N which are 1 = d1, d2, ... dk = N. So k = d(N). If we multiply them pairwise d1.dk = N = d2.d(k-1) = dk.d1 Multiply all together. (d1.d2..dk)^2 = N^k = P^2 => P = N^(k/2) H.P. Now: S = log(d1) + log(d2)... log(dk) = log(d1.d2...dk) = log(N^(k/2)) = (k/2).log(N) = (1/2).k.6 = 3.k k = number of divisors = 7*7 So S = 49*3 = 147 But from this we need to subtract log of improper divisor(number itself). Answer = 147 - 6 = 141 Q. For k≥2, show each of the following: (a)  n=2^(k−1) satisfies the equation σ(n)=2n−1. (b) If 2^k −1 is prime, then  n=2^(k−1)(2^k−1) satisfies the equation σ(n)=2n. (c) If 2^k −3 is prime, then  n=2^(k−1)(2^k −3) satisfies σ(n)=2n+2. Solution: (a) Do the sum of the GP 1 + 2 + ... 2^(...