Math IOQM

  Now, BMP ~ BAE. Why? Since Angle B is same in both. BM/BA = BP/BE = 1/2. So SAS similarity criteria. So: MP/AE = 1/2 => x + z = 2MP => x = MP. But MP = AB/2 = (x+y)/2 => 2x = x + y => x = y. So ABE is equilateral and hence all angles are 60. So DAE = 60 as well. Q4. Solution: After applying the transformation 4 times you will get: 81M + 80 = 16N + 405 => 81(M-5) = 16(N-5) Minimum values of M,N when LHS  = RHS, M = 5, N = 5 and both sides become 0. Next equal values will be when M-5 = 16 and N-5 = 81 since there are no common factors in 81 and 16. So M + N = 10 Q5. Let a , b ∈ Z > 0 a,b\in\mathbb Z_{>0} with a 3 − b 3 − a b = 25 > 0    ⇒    a > b . a^3-b^3-ab=25>0\;\Rightarrow\; a>b. Put d = a − b ≥ 1 d=a-b\ge1 . Then a 3 − b 3 − a b = ( a − b ) ( a 2 + a b + b 2 ) − a b = d ( 3 b 2 + 3 b d + d 2 ) − b ( b + d ) = d 3 + ( 3 d − 1 ) a b . a^3-b^3-ab=(a-b)(a^2+ab+b^2)-ab = d(3b^2+3bd+d^2)-b(b+d) = d^3+(3d-1)ab. So d 3 + ( 3 d − 1 ) a b = 2...

Question: Consider 4 boxes, where each box contains 3 red balls and 2 blue balls. Assume that all 20 balls are distinct. In how many different ways can 10 balls be chosen from these 4 boxes so that from each box at least one red ball and one blue ball are chosen? Solution 1: We start by noting that each of the 4 boxes contains 3 red and 2 blue distinct balls. In any valid selection of 10 balls, from each box at least one red and one blue must be chosen. Step 1. Determine How Many Balls Are Chosen from Each Box Let x i x_i be the number of balls chosen from box i i (for i = 1 , 2 , 3 , 4 i=1,2,3,4 ). We have: x 1 + x 2 + x 3 + x 4 = 10 x_1+x_2+x_3+x_4 = 10 Since from each box at least one red and one blue must be chosen, we must choose at least 2 balls from each box: x i ≥ 2 for  i = 1 , 2 , 3 , 4. x_i \ge 2 \quad \text{for } i=1,2,3,4. Define y i = x i − 2 y_i = x_i - 2 (so y i ≥ 0 y_i \ge 0 ). Then: y 1 + y 2 + y 3 + y 4 = 10 − 8 = 2. y_1+y_2+y_3+y_4 = 10 - 8 = 2. The n...

Answer: 1