Math IOQM

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...