Math IOQM

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...

Q1. There is a 5x10 chocolate bar. P1 and P2 are 2 players cutting it through the grid lines. In a single move you can take either a complete row or a complete column. Whoever picks up a 1x1 piece first, wins. S1. Since one of the dimensions is even there is a guaranteed winner here: Player 1. If both dimensions were odd there won't be any guaranteed winner. So player 1 will split it into 2 equal halves of 5x5. And from there will simply mirror whatever player 2 does on the symmetric half. But as soon as player 2 reduces 1 dimension of a grid to 1, player 1 picks up that piece and wins.  Q2. S2. 1. For the 2x1 domino case Player 2 will win as explained in Q4 here . 2. For the 3x1 Tromino case, again player 2 will win using the same reflection strategy as done for 2x1 domino. In both 1. and 2. the 2x1, 3x1 pieces cannot overlap their own reflection across the center. But in 3. it's possible. Remember that reflection across center formula here is 11-r,11-c 3.  Q3. "The nos. ...

  Where 'n' is a positive integer. For e.g.

Q1. Find the number of ways in which 30 marks can be allotted to 8 questions if each question carries at least 2 marks? S1. x1 + ... x8 = 30-16 = 14 Stars and bars, n = 14, k = 8 14+8-1C8-1 = 21C7 Q2. In an exam the maximum marks for each of three papers is 'n' and that for fourth paper is  2n. Find the number of ways in which a student can get (3n) marks. S2. Using negative binomial expansions :

 Q1.  A. How many shortest paths from X to Y? B. From AB? S1. A.  7R4U => 11!/7!4! = 330 B. 5!/3!2! * 5!/3!2! = 100 Q2. In a box, there are 10 balls: 4 red, 3 black, 2 white, 1 yellow. In how many ways can a child select 4 balls out of these 10 balls? S2. If all balls are distinct, answer is simply 10C4  = 210 If balls of same color are identical, which is typically the case, then: All 4 same: 1 3 Same, 1 different: 2*3C1 = 6 2 same, 2 same: 3C2 = 3 2 Same, 1 diff, 1 diff: 3C1*3C1 = 9 All 4 different: 1 Total: 20 Q3. There are three papers of 100 marks each. Then find the no. of ways a student can get 150 s.t. he scores at least 60% in two papers. (only integer marks are given) S3. Let' say the marks in first 2 papers are x + 60, y + 60 and in third paper z marks. So: 60 + x + 60 + y + z = 150 => x + y + z = 30 But we could have chosen any 2 papers. So there are 3C2 ways to choose the 2 papers. So we will multiply the final answer with 3. Now each of x,y,z can ...