Math IOQM

  Stewart's theorem Proof: Drop a perpendicular from A to BC at M. S.t. BM + MD = BD and CM - MD = CD AB^2 =  AM^2 + BM^2 => AM^2 = AB^2 - BM^2 = c^2 - (BD - MD)^2 = c^2 - (m - MD^2) Similarly: AM^2 = AC^2 - MC^2 = b^2 - (n - MD)^2 Now equate both to eliminate MD^2. Then multiply first equation by 'n', second by 'm' to eliminate the MD term. You will get the proof.

Q1. Two sides of a triangle are 10 cm and 5 cm in length, and the length of the median to the third side is 6.5 cm. If the area of the triangle is 6.sqrt(p) cm^2 then find the value of 'p. S1. By Apollonius theorem, third side is 9. Now use heron's formula to get answer p = 14 Q2. Two sides of a triangle are sqrt(3) units and sqrt(2) units. The medians to these two sides are mutually perpendicular. Prove that the third side has an integer measure. S2. (2x)^2 + y^2 = 1/2 = 4x^2 + y^2 (2y)^2 + x^2 = 3/4 5x^2 + 5y^2 = 5/4 4x^2 + 4y^2 = 1 = (2x)^2 + (2y)^2  = BC^2 => BC = 1 H.P. Q3. ABC is a triangle in which Angle B = 2 (Angle C). D is a point on BC such that AD bisects Angle BAC and AB = CD. Prove that Angle BAC = 72 deg. S3.

Prove that every n s.t. gcd(n,10) = 1 has a multiple which is made up of only 9's Proof: Let us pick one 'n'. When we divide 10^1, 10^2, 10^3 .. with 'n' there can be at max 'n' different remainders. So there must exist integers a,b with a>b s.t. 10^a and 10^b have same remainders when divided by 'n'. So 10^a - 10^b is divisible by 'n'. 10^b(10^(a-b) - 1) is div by n. gcd(n,10) = 1 => 10^(a-b) - 1 is div by 'n' and this is entirely made up of 9s. H.P.

 Q52 (Intro): Determine all positive integers n such that n has a multiple whose digits are nonzero. Q52 (Advanced)

1) ABCD is a parallelogram. Through C, a straight line RQ is drawn outside the figure and AP, BQ, DR are drawn ⟂ to RQ. Show that: DR + BQ = AP . S1) [RMO] Key idea: DR || BQ so they can form a trapezium. Since LHS is DR + BQ which also appears in Trapezium midline theorem we will try to use Trapezium midline theorem which says this: If AB || CD in trapezium ABCD, then line EF joining midpoints of AD and BC is || to AB,CD and also EF = (AB + CD)/2. Let O be intersection of diagonals AC and BD in the ||gram ABCD. O is their midpoint. Let OM be perpendicular to RQ. DR || BQ and DBQR is a trapezium. OM || DR || BQ and OM goes through midpoint of BD => M is the midpoint of RQ. OM = (BQ + DR)/2 We are almost there now. If we could show that OM = AP/2, we are done. OM || AP In Triangle APC, it's a strong hint to use Triangle midpoint theorem. O is midpoint of AC and OM || AP => M is midpoint of PC. => OM = AP/2 H.P. Another solution: Drop a perpendicular DO from D to AP. DRPO is...

1. Show that 10^(2n-1) + 1 is divisible by 11 for all natural number 'n'. Solution: 10 = -1 mod 11 10^(2n-1) = (-1)^(2n-1) mod 11 = -1 mod 11 => 10^(2n-1) + 1 = 0 mod 11 H.P. 2. Show that 11^(n+2) + 12^(2n+1) is divisible by 133 for all natural 'n'. Solution: 133 = 7*19 So LHS should be div by both 7 and 19 Try 7 first: 11 mod 7 = 4, 12 mod 7 = 5 LHS = 4^(n+2) + 5^(2n+1) 5 mod 7 = -2 => 5^2 mod 7 = 4 mod 7 LHS = 16.4^n + 5.4^n = 21.4^n mod 7 = 0 mod 7 Now do the same with 19. LHS = 121.11^n + 12.12^2n 12^2n = 144^n 144 = 11 mod 19 144^n = 11^n mod 19 121 = 7 mod 19 LHS = 7.11^n + 12.11^n = 19.11^n = 0 mod 19 H.P. 3. Find all integers a such that the quadratic expression (x+a)(x+1991) + 1 can be factored as (x+b)(x+c), where b and c are integers. Solution: Since -b is a root of RHS =>  (-b+a)(-b+1991) = -1 Case 1: (-b+a) = 1 (-b+1991) = -1 => b = 1992, a = 1993 Case 2: (-b+a) = -1 (-b+1991) = 1 => b = 1990, a = 1989 Answer: a = 1989, 1993.

Q1) Prove that the line segment joining the midpoints of the diagonals of a trapezium is parallel to each of the parallel sides and is equal to half the difference of these sides. Q2) Prove that the midpoint of the hypotenuse of a right-angled triangle is equidistant from all its vertices.