Math IOQM

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...

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 ...

Q1. Find the number of ways to choose an ordered pair ((a,b)) of numbers from the set (1...10) such that (|a-b| <= 5). S1. Answer: 80 Method 1: Direct counting 1: 1,2,3,4,5,6 => count:6 Similarly the count is 6 for 2,3,4,5. So total: 30*2 = 60 pairs. Then for 6,7,8,9,10 there are 5,4,3,2,1 options. So 15*2 = 30 pairs. Total: 90 But we have counted same number pairs twice. So 90-10=80 = answer Method 2: faster and better: complementary counting Total: 10x10 = 100 Subtract invalid: 1: 7,8,9,10 2: 8,9,10 3: 9,10 4: 10 Total: 10*2 = 20 100-20 = 80 = Answer Q2. Suppose that in a poll made of 150 people, the following information was obtained: 70 of them read The Hindu , 80 read The Indian Express and 50 read Deccan Herald . 30 read both The Hindu and The Indian Express ; 20 read both The Hindu and Deccan Herald and 25 read both The Indian Express and Deccan Herald . Find at most how many of them read all the three. S2. Incorrect solution first: Initially I applied PIE without thi...

Q1. Ten 1's and ten 2's are written on a blackboard. In one turn, a player may erase any two figures. If the two figures erased are identical, they are replaced with a 2. If they are different, they are replaced with a 1. The first player wins if a 1 is left at the end, and the second player wins if a 2 is left. S1. Only P2 can win. Why? 1. There are total 19 moves since in every move the count of total numbers goes down by 1. At the end we will have a single number. 2. Let's say P1 plays the first move, then he will also play the 19th move. Let C1, C2 denote count of 1s and 2s at any given time. 3. For P1 to win, after the 18th move we need C1 = 1, C2 = 1. For P2 to win C1 = 0, C2 = 2 or C1 = 2, C2 = 0. 4. Let's focus on P1's win. That means that in the initial 18 moves both C1 and C2 should decrease by 9 each. 5. What happens in every move? Case 1: Two 1s replaced by 2 => C1 -= 2, C2 += 1 Case 2: Two 2s replaced by 2 => C1 += 0, C2 -= 1 Case 3: One 1 and One...

Q1. On a board, the numbers a, b, c, d, e, f are written in a circle (say counter-clockwise). You may increase two neighbouring numbers by 1. For which case is it impossible to equalize all of them by a sequence of such steps? Select any one option a = 4, b = 10, c = 5, d = 6, e = 9, f = 2 a = 3, b = 9, c = 7, d = 6, e = 9, f = 4 a = 4, b = 11, c = 7, d = 7, e = 10, f = 2 a = 7, b = 10, c = 5, d = 8, e = 9, f = 3 S1. First a partially correct approach which works in this case but will not work always. Final sum of all numbers is even. All the increments are also even.(multiple of 2) So the original sum of all the numbers should also be even. Only option 3 fails. Hence that's the answer. But that approach fails for: a=2,b=4,c=2,d=4,e=2,f=4 So the better way to solve it is this: Let there be 2 groups:  Group 1: a,c,e Group 2: b,d,f Each time we increment, we increment by 1 in each group. So the original sum in both the groups should be same. Else it won't work. Q2. On a 10x10 boa...

Q1. While applying Menelaus on a triangle ABC, can the traversal pass through one of its vertices? S1. No. In one of the fractions numerator would become zero. In another denominator. Q2. The points (X, Y) are taken on (CA, AB) respectively of triangle ABC. If (BX) and (CY) meet at (P) and AX/CX = BY/AY = 1/2 then find BP/PX. S2. Triangle XAB, Traversal YPC. Apply Menelaus. Answer: 3/4 Q3. In △ABC the points E, F, G are on AB, BC & CA respectively, such that AE/EB = BF/FC = CG/GA = 1/3. The (K, L, M) are the intersections of (AF) & (CE); (BG) & (AF); (CE) & (BG) respectively. If ([ABC] = 1), find ([KLM] = ?) S3. Here we need to use mass points on 2 Cevians at a time. We can't apply on all 3 together since they are not concurrent. Our aim is to find [KLM] like this:  [KLM] = [ABC] - [AKC] - [ALB] - [BMC] For [AKC], use mass points on cevians AF,CE to get the ratio AK:KF. Then [CKA]:[CKF] = AK:KF But [CKA] + [CKF] = [AFC] And [AFC]/[ABC] = FC/BC So [AFC] = 3/4 So...

Q0. In a ||gram ABCD, does the diagonal AC bisect the angle DAB? Q1. M,N,D are midpoints of AB,AC,BC. X,Y are midpoints of NC,ND. Find [MNXY]/[ABC]. S1. Let [ABC] = f => [AMN] = f/4 => [BCNM] = 3f/4 [CND] = f/4 => [NYX] = f/16 [MNXY] = [NYX] + [MNY] => we need to find [MNY] now. [BDNM] = [BCNM] - [NDC] = f/2 [BDM] = [NDM] = f/4 [MNY] = [DMY] = f/8 => [MNXY] = [NYX] + [MNY] = f/16 + f/8 = 3f/16 Q2. ABCD is a parallelogram. X divides AB in the ratio 3:2. Y divides CD in the ratio 4:1. XY intersects AC at Z. Find (AZ : AC). S2.

 1. Show that length of direct common tangent of 2 non intersecting circles(touching is fine) is sqrt(d^2 - (r1-r2)^2) and that of the transverse tangent is sqrt(d^2 - (r1+r2)^2) where 'd' is the distance between the centers of those 2 circles. 2. Now show that if the 2 circles touch each other than DCT length = 2.sqrt(r1r2). 3.  If two circles with radii (a) and (b) touch each other externally. Let (c) be the radius of a circle that touches these two circles as well as a common tangent of these two circles. Prove 1/sqrt(c) = 1/sqrt(a) + 1/sqrt(b)

 Q1. In △ABC, AD, BE and CF are concurrent lines. P, Q, R are points on EF, FD, DE such that DP, EQ and FR are concurrent. Prove that AP, BQ and CR are also concurrent. S1. Prerequisite: Standard and trigonometric form of Ceva's theorem. Standard form gives us: AF/FB * BD/DC * CE/EA = 1 and EP/PF * FQ/QD * DR/RE = 1 Using the trigonometric form, we need to prove that: Sin(CAP)/Sin(PAB) * Sin(ABQ)/Sin(QBC) * Sin(BCR)/Sin(RCA) = 1 Look at triangles EAP and PAF and compute their area ratio: [EAP]/[PAF] = AE * AP * Sin(CAP)/AF*AP*Sin(PAB) = EP/PF Similarly do for others. Multiply both sides of all 3. You will get your desired relation. H.P. Q2. If (X) and (Y) are variable points on the sides (CA, AB) of (\triangle ABC) such that CX/XA + AB/AY = 1 prove that (XY) passes through a fixed point. S2. CX/XA > 0 since everything is positive. => AB/AY < 1 => AY > AB => Y lies on AB extended towards B. => AB = AY - BY => AB/AY = 1 - BY/AY From here: we have 2 cases for X...

 Q1. The internal angle bisector of ∠A of △ABC meets BC at P and (b = 2c) in the usual notation. Prove that ((9AP^2 + 2a^2)) is an integral multiple of (c^2). S1. Let PC = n, BP = m, AP = d We need to prove: 9d^2 + 2a^2 = k.c^2 where 'k' is an integer. c/m = b/n by Angle Bisector theorem. => 2m = n => 3m = a => m = a/3, n = 2a/3 Using Stewart's theorem : man + dad = bmb + cnc = m.4c^2 + nc^2 = a(mn + d^2) = c^2 (4m + n) => a(n^2/2 + d^2) = c^2(3n) => a(4a^2/18 + d^2) = c^2.2a/3 => 4c^2/3 = (2a^2/9 + d^2) => 12c^2 = (2a^2 + 9d^2) H.P. Q2. In (\triangle ABC), in the usual notation, the area is bc/2 sq. units. (AD) is the median to (BC). Prove that Angle ABC = (Angle ADC)/2 S2. Area = 1/2.bc.sin(A) => A = 90 deg. So D is the circumcenter. Chord AC subtends angle ADC at center and ABC at circumference. H.P. Q3. Let C1 be any point on (AB) of triangle ABC. Draw CC1 meeting AB at C1. The lines through A and B parallel to CC1 meet BC produced and AC produce...

Q1) Let Γ be a semicircle with diameter AB. The point (C) lies on the diameter (AB) and points (E) and (D) lie on the arc (BA), with (E) between (B) and (D). Let the tangents to (\Gamma) at (D) and (E) meet at (F). Suppose that (\angle ACD = \angle ECB). Prove that (\angle EFD = \angle ACD + \angle ECB). S1: Let's quickly get an intuition about it. If the point C is same as the center then CDFE is a cyclic quadrilateral since the angles at D and E are both 90. Angle at C is 180 - 2.theta so the angle at F is 2.theta which is what we need to show. H.P. So it works for the case when the point C is same as the center. Now, let's try the general case. Now ODFE is the cyclic quadrilateral. => Angle DFE = 180 - DOE So proving that DFE = ACD + ECB = 2.theta is same as proving that: DOE = 180 - 2.theta = DCE But if we need to prove that DOE = DCE then we need to prove that E,D,C,O are concyclic since the chord DE is subtending the same angles at O,C. Now look at the original problem...

Q1. The diagonals of a parallelogram (ABCD) intersect at (O). A line through (O) intersects (AB) at (X) and (DC) at (Y). Another line passing through (O) intersects (AD) at (P) and (BC) at (Q). Prove that (XQYP) is a parallelogram. S1. Triangles OAX and OCY are congruent. Why? OA = OC Angle CYO = AXO Angle AOX = YOC => OX = OY => O bisects XY Similarly O bisects PQ. So in quadrilateral XQYP, the diagonals bisect each other. Hence it's a ||gram. H.P. So it gives us a property about ||gram that a line passing through diagonal midpoint and ending on opposite sides is also bisected at the diagonal midpoint. Q2. Prove that the feet of perpendiculars drawn from the vertices of a parallelogram onto its diagonals are the vertices of another parallelogram. S2. Let the feet of perpendiculars from D,B on AC be Q,P. We will try to show that OP = OQ so that we can show that O bisects PQ. Similarly, O will also bisect the other line joining the other 2 feet of perpendiculars. And in that c...

Q1). A number N in base 10, is 503 in base b and 305 in base b+2. What is the product of the digits of N? S1. N = 248 Answer = 64 Q2).  Let (ABCD) be a rectangle, in which (AB + BC + CD = 20) and (AE = 9), where (E) is the midpoint of the side (BC). Find the area of the rectangle. S2. 2x + y = 20______[1] 81 = x^2 + y^2/4______[2] We just need to find xy. Square the first equation and you will 4x^2 + y^2, replace it using 2 and you will get x.y = 19. Q3). What is the least positive integer by which 2^5. 3^6. 4^3. 5^3.6^7 should be multiplied so that the product is a perfect square? S3. Write it as 2^(5+6+7).3^(6+7).5^3 = 2^18.3^13.5^3 So the answer is 15. Q4). Find the number of integer solutions to ||x| - 2020| < 5 S4. -5 < |x| - 2020 < 5 2015 < |x| < 2025 => x has 18 possible integer solutions when |x| = 2016 to 2024 Q5). The product (55 * 60 * 65) is written as the product of five distinct positive integers. What is the least possible value of the largest of th...

 Q1. ABC is a triangle and D and E are interior points of the sides AB and BC respectively such that: AD/DB = 1/3 CE/EB = 3 If AE and CD intersect at F, find CF/FD. S1. Approach 1: Using mass points : AD/DB = 1/3 => A has higher mass since it's closer to D. Let mA = 3 mB = 1 => mC = 1/3 => mE = 4/3 Now mF = 3 + 4/3 = 13/3 and it matches 4 + 1/3 = 13/3 So we have assigned masses correctly. => CF/FD = 4/(1/3) = 12 Approach 2: Menelaus theorem Typically you would notice that if we have a solution using mass points, we can also solve it using Ceva's theorem or Menelaus' theorem. Here, in triangle BDC, AFE is the traversal intersecting all sides. BE/EC * CF/FD * DA/AB = 1 => 1/3 * CF/FD * 1/4 = 1 => CF/FD = 12 = Answer Q2.  L and M are the mid-points of the diagonals BD and AC respectively of the quadrilateral ABCD. Through D, draw DE equal and parallel to AB. Show that EC || LM EC = 2LM S2. It's straightforward with co-ordinate geometry. A (0,0) B(x1,0) C(...

 Q1. In quadrilateral ABCD, the diagonals AC and BD meet at O. Suppose the four triangles AOB,BOC,COD and DOA are equal in area, prove that ABCD is a parallelogram. S1. [AOB] = [COD] => 1/2.OA.OB.Sin(AOB) = 1/2.OC.OD.Sin(COD) Angles AOB and COD are same. => OA.OB = OC.OD Similarly, OA.OD = OC.OB Multiply both => OA = OC and OB = OD => O bisects both diagonals. If the diagonals bisect each other, the quadrilateral is a ||gram. Why? Let the quadrilateral be A (0,0), B(x,0), C(a,b) D(p,q) O = Midpoint of AC = (a/2,b/2) = Midpoint of BD = ((p+x)/2,q/2) b/2 = q/2 => b = q => CD || AB and a = p+x Slope of AD = q/p Slope of BC = b/(a-x) = q/p => AD || BC Opposite sides are || => ABCD is a ||gram. H.P. Q2. In a parallelogram  A B C D A BC D , a point  P P on the segment  A B A B is taken such that  A P A B = A B A P ​ =   61 2022 2022 61 ​  and a point  Q Q on the segment  A D A D is taken such that  A Q A D = 61 2065...