Math IOQM

Q1. If b1,b2 ... .bn is a permutation of n positive real numbers a1,a2...an then find the minimum value of a1/b1 + a2/b2 ... an/bn. S1. Apply AM GM inequality to get answer = n. Q2. If a1 + a2 ... an = 1, ai > 0 for all i then find the minimum value of 1/a1 + 1/a2 ... 1/an. S2. AM >= GM >= HM HM = n/[1/a1 + 1/a2 .. 1/an] <= (a1 + a2 .. an)/n => n/X <= 1/n => X >= n^2 Q3. A,B are the A.M. and G.M. of two positive numbers a,b. Show that, B < (a-b)^2/8(A-B) < A S3. (a-b)^2 = (a+b)^2 - 4ab = (2A)^2 - (2B)^2 = 4(A^2 - B^2) So middle part becomes: 4(A-B)(A+B)/8(A-B) = (A+B)/2 B <= (A+B)/2 => B/2 <= A/2 which is correct. Similarly the other part. Q4. a,b,c,d are distinct positive real numbers in H.P. Prove that: 1) a + d > b + c 2) ad > bc S4. let 1/a = A-3D 1/b = A-D 1/c = A+D 1/d = A+3D 1/a * 1/d = A^2 - 9D^2 1/b * 1/c = A^2 - D^2 Clearly 1/ad < 1/bc since D^2 is positive and A^2 is also positive. => ad > bc Now: 1/a + 1/d = 1/b + 1/c =...

Q1. Let ABC be a given equilateral triangle. Denote the mid-points of sides BC, CA, AB respectively by A1, B1, C1. Three distinct parallel lines p, q,r are drawn through A1, B1, C1, respectively. Line p cuts B1C1 at A2; line q cuts C1A1 at B2; line r cuts A1B1 at C2. Prove that the lines AA2, BB2, CC2 are concurrent. S1. Q2. Let (P) be any point inside triangle ABC. Let A1,A2 in AP, B1,B2 in BP, C1,C2 in CP with BA1 || AB2 || CP, CB1 || BC2 || AP, AC1 || CA2 || BP Prove that [A1B1C1] = [A2B2C2] where ([K]) denotes the area of (K). S2.  Q3. In a triangle ABC, let (AD), (BE), (CF) be the medians. Prove that AB+BC+CA <= 4/3(AD+BE+CF). When does equality hold? S3. 2z + 2y > AB 2z + 2x > AC 2x + 2y > BC Add all: 4(x+y+z) > (AB + BC + AC)____[1] LHS = 4/3(AD + BE + CF) Equality will hold for a degenerate triangle when A,B,C are colinear. For e.g. let C lie on F, then CF = 0 and centroid also lies on F. 2z = AC, 2y = BC and x = 0 So [1] becomes: 2AC + 2BC + 0 >= 2AB And ...

Q1. How many two digit numbers have exactly 4 positive factors? (Here 1 and the number (n) are also considered as factors of (n).) S1. Only 2 possibilities: 1. p^3 where p is prime 2. p*q where p,q are prime. p^3 only one 3^3: 1 p*q we can construct 2*(5 to 47 primes): 13 3*(5 to 31): 9 5*(7 to 19): 5 7*(11 to 13): 2 Total: 30 = answer. Q2. Find the number of pairs ((a,b)) of natural numbers such that (b) is a 3-digit number, (a+1) divides (b-1), and (b) divides (a^2+a+2). S2. b-1 = k(a+1) b = k(a+1) + 1 k(a+1) + 1 divides a^2 + a +2 => ak + k + 1 divides k(a^2 + a +2) = 2k + ak + a^2.k = 2k + a(k + ak + 1 - 1) = 2k - a + a(k + ak + 1) => ak + k + 1 divides 2k - a Now 3 cases from here: 2k - a = 0 => a = 2k 2k - a > 0 => ak + k + 1 < 2k - a => a(k+1) -k + 1 < 0 => a(k+1) <= k-1 => a <= (k-1)/(k+1) which is not possible since a > 1 and RHS is < 1. 2k - a < 0 =>  ak + k + 1 < a - 2k => a(k-1) + 3k + 1 < 0 which is not possible ...

Q1. Let X = {1,2,...,n}). In how many ways can you choose (r) elements from (X) such that no two chosen elements are consecutive, where 0 <= r <= (n+1)/2. S1. Answer: (n-r+1)c(r) Solution here. Q2. In how many ways can three numbers in arithmetic progression (A.P.) be selected from the set (1,2...n)? S2. Method 1: a,b,c in A.P. => a,c are both even or both odd. b = (a+c)/2 So if we pick 2 even or 2 odd numbers, we will find a number between them to complete the A.P. Case 1: n is even. n = 2m. m odd and m even numbers. Pick 2 even: mC2 Pick 2 odd: mC2 Total: 2 * mC2 = m(m-1) = n/2(n/2-1) = n(n-2)/4 Case 2: n = 2m+1 m+1 odd and m even numbers. Total ways: (m+1)C2 + mC2 = 1/2[m.(m+1) + m.(m-1)] = m/2[2m] = m^2 = (n-1)^2/4 Method 2: When n is even: You can pick an A.P. with d = 1 as 1,2,3 2,3,4 .... n-2,n-1,n d = 1 => n-2 ways to pick the first term d = 2 => 1,3,5 2,4,6 ... n-4,n-2,n => n-4 ways .... For e.g. n=6, max d = 2(1,3,5 2,4,6) n = 8, max d = 3(1,4,7 2,5,8) n = 1...

Q1. Straight lines are drawn by joining (m) points on a straight line to (n) points on another line. No two lines drawn are parallel and no three lines are concurrent. How many total intersecting points are there? S1. Let's rephrase the problem as how many new intersection points are created? Solution 1: Let's create a quadrilateral by joining 2 vertices from each line. In this quadrilateral, 2 new intersection points are created. One by diagonals. And one by extending the 2 sides which we just created. Total ways to create a quadrilateral like this: Choose 2 vertices from one line and 2 from another. So, answer = 2 * mC2 * nC2 Solution 2: Total lines: mn Total intersection points: (mn)C(2) From this we need to subtract the intersection points which were already there. On each of the 'm' points, 'n' lines converge. So nC2 intersection points are made by them on a single point. In total m*nC2 such points. Similarly n*mC2 other points. So mnC2 - m*nC2 - n*mC2 = [m...

Q1. (ABC) is a triangle. (D, E, F) are arbitrary points on (BC, AC,) and (AB) respectively (or on their extensions). Draw the three circumcircles of triangles (AEF), (DBF), and (DEC). Prove that these three circles intersect at a single point (M). S1. This is known as Miquel's theorem and M is known as Miquel's point. Case 1. When AEF,DBF intersect at 2 different points : F and M. Proof 1: Angle AFM = theta => MFB = 180 - theta (supplementary angles) and AEM = 180 - theta (opposite angles in cyclic quadrilateral AEMF) => MEC = theta (supplementary) and BDM = theta (opposite in cyclic quadrilateral BDMF) => CDM = 180 - theta (supplementary) => CEM + CDM = 180 => CEMD is cyclic quadrilateral H.P. Proof 2: AFME is cyclic => FME = 180 - A = B + C Similarly, FMD = A + B Now angles around M should add up to 360. => FME + EMD + DMF = 360 => EMD = 360 - (B+C) - (A+B) =  So we showed that M is the Miquel's point. But could it have been F? I mean, is it possib...