數學知識交流---PolyU math 6~7Jan2012

6JanLet 5A = 5tan⁻¹ (1/7) , 2B = 2tan⁻¹ (3/79) , ( A , B ∈ (0 , π/2) )then tan A = 1/7 , tan B = 3/79 , ∴tan 2A = 2tanA / (1 - tan²A) = 2(1/7) / (1 - 1/7²) = 7/24tan 3A = tan (2A + A) = (7/24 + 1/7) / (1 - (7/24)(1/7)) = 73/161tan 5A = tan (2A + 3A) = (7/24 + 73/161) / (1 - (7/24)(73/161)) = 2879/3353;tan 2B = 2(3/79) / (1 - (3/79)²) = 237/3116 ∴ tan (5A + 2B)= (2879/3353 + 237/3116) / (1 - (2879/3353) (237/3116))= 1= tan (π / 4)So N = 4

7JanSince sum of roots = - m - 1 , so m must be an integer.△ = (m+1)² - 4(2m+1) = m² - 6m - 3 must be a perfect square.Let m² - 6m - 3 = n²m² - 6m + 9 - 12 = n²(m - 3)² - n² = 12(m + n - 3) (m - n - 3) = 12 = 1 x 12 = 2 x 6 = 3 x 4 (m + n - 3) + (m - n - 3) = 2m - 6 = 1 + 12 = 2 + 6 = 3 + 4So the maximum value of 2m - 6 = 13The maximum value of m = 19/2 (rejected)
The true maximum value of 2m - 6 = 2 + 6The true maximum value of m = 7.


2011-11-03 01:44:17 補充：
For 6Jan :

The reason of 5A + 2B = π/4 but not 5π/4 as follow :

tan 2A = 7/24 > 0 and 2A < π ,
∴ 2A < π/2
A < π/4

tan 3A = 73/161 > 0 and 3A < 3π/4 ,
∴ 3A < π/2
A < π/6

tan 5A = 2879/3353 > 0 and 5A < 5π/6 ,
∴ 5A < π/2
A < π/10


tan 2B = 237/3116 > 0 and 2B < π ,
∴ 2B < π/2
B < π/4

2011-11-03 01:44:22 補充：
Therefore 5A + 2B < 5π/10 + 2π/4 = π ,
Hence 5A + 2B = π/4 only.