Math IOQM

Q1. Two friends, Marco and Ian, are talking about their ages. Ian says, "My age is a zero of a polynomial with integer coefficients." Having seen the polynomial p(x) Ian was talking about, Marco exclaims, "You mean, you are seven years old? Oops, sorry I miscalculated! p(7) = 77 and not zero." "Yes, I am older than that," Ian's agreeing reply. Then Marco mentioned a certain number, but realizes after a while that he was wrong again because the value of the polynomial at that number is 85. Ian sighs, "I am even older than that number." Determine Ian's age. S1. Let the root be 'a'. p(a) = 0 p(7) = 77 p(b) = 85 and b > 7. a > b > 7. For a polynomial with integer coefficients and x,y as integers: x-y | p(x) - p(y) So a-7 | p(a) - p(7) a-7 | -77 So a-7 can be 1,7,11,77 => a can be {8,14,18,84} b - 7 | 8 => b-7 can be 1,2,4,8 => b can be {8,9,11,15} a-b| 85 => a-b can be 1,5,17,85 Only option which works is a = 14,...

Q1. Prove that if the two angle bisectors of a triangle are equal, then the triangle is isosceles. ( Steiner-Lehmus Theorem ) S1. So, let's see what are we trying to prove. BE = CF and angle B = angle C We know that length of angle bisector in triangle is given by the below formula. BE = 2ac.cos(B/2)/(a+c) Similarly, CF = 2ab.cos(C/2)/(a+b) make the equal: c.cos(B/2)/(a+c) = b.cos(C/2)/(a+b) cos(B/2)/cos(C/2) = ab + bc/ac + bc We will prove by contradiction. WLOG, Angle B > Angle C => b > c also => cos(B/2) < cos(C/2) (since B/2 and C/2 are both < 90). So LHS < 1 But RHS > 1 why? Because  ab + bc > ac + bc (since b > c) And if Angle C < Angle B, again LHS > 1 and RHS < 1. So only option is that both are equal. H.P.

  [ABC] = a.c.sin(B)/2 [ABC] = [ABE] + [EBC] [ABE] = x.c.sin(B/2)/2 [EBC] = a.x.sin(B/2)/2 x = ac.sin(B)/sin(B/2).(a+c) x = ac/(a+c) * 2cos(B/2) BE = 2ac.cos(B/2)/(a+c)

Q1. Determine all solutions in the positive integers of 18x+5y=48 S1. y = (48-18x)/5 48-18x = 0 mod 5 3 = 3x mod 5 x = 1 mod 5 x = 1 => y = 6 x >= 6 not possible else y will become negative. what if I had done x = (48 - 5y)/18 in that case 48 = 5y mod 18 => 12 = 5y mod 18 Now, the thing is that y mod 18 can range from 0 to 17, so my trial and error will take much longer. So for these kind of equations it's always better to go for a smaller modulus, as we did in the first approach. Q2. 5x+3y=52. S2. x = (52 - 3y)/5 => 2 = 3y mod 5 => y = 4 mod 5 y = 4,9,14 => x = 8, 5, 2 Q3. Find all residues r with 0≤r≤7 such that x^2 ≡r(mod8) has a solution. S3. 0,1,4 Q5. A single bench section at a school event can hold either 7 adults or 11 children. When N bench sections are connected end to end, an equal number of adults and children together will occupy all the bench space. What is the least possible positive integer value of N? S5. a + c = N 7a = 11c 7,11 are primes => s...

Q1. S1. QR || AC by MPT. => line BHE is perpendicular to QR. In triangle AHB, using MPT, QP || BH. => QR is perpendicular to QP => angle PQR = 90 H.P. Q2. ABCD is a quadrilateral in which AB = AD . The bisectors of ∠BAC and ∠CAD intersect the sides BC and CD at E and F , respectively. Prove that (EF ||BD). S2. Using Angle bisector theorem: Triangle ABC: AB/BE = AC/EC Triangle ADC: AD/DF = AC/FC => AB/AC = BE/EC = AD/AC = DF/FC (since AB = AD) Now in triangle BDC: EF divides the sides BC and DC in same ratio => EF || BD H.P.

  Q1) Let (ABCD) be a rectangle such that (BC = 3AB). (P) and (Q) are points on the side (BC) such that BP = PQ = QC. Show that \angle DBC + \angle DPC = \angle DQC. S1. Q2. In a quadrilateral ABCD, given that angle A + angle D = 90. Pr. th. AC^2 + BD^2 = AD^2 + BC^2. S2. BC^2 = OB^2 + OC^2 AD^2 = OA^2 + OD^2 BC^2 + AD^2 = OB^2 + OC^2 + OA^2 + OD^2 Look at RHS: OA^2 + OC^2 = AC^2 OB^2 + OD^2 = BD^2 H.P. Q3.  In ( \triangle ABC ), (BM) and (CN) are perpendiculars from (B) and (C) respectively on a line passing through (A). If (L) is the midpoint of (BC), prove that [ ML = NL. ] S3. Method 1: Let LP be perpendicular to the same line passing through A. LP || BM || CN According to Intercept theorem, if 3 parallel lines cut 2 equal segments from a traversal, they will do the same with any other traversal. So MP = PN Now we will show triangle LMP congruent to LNP. LP is common side. Angle P is 90 in both. MP = PN. => LM = LN. Method 2: co-ordinate geometry Let the line thro...