Neka je a = n − 1 a=n-1 a = n − 1 b = n b=n b = n c = n + 1 c=n+1 c = n + 1 α < β < γ \alpha<\beta<\gamma α < β < γ
1 ∘ 1^\circ 1 ∘ β = 2 α \beta=2\alpha β = 2 α
cos α = sin 2 α 2 sin α = sin β 2 sin α = b 2 a = n 2 ( n − 1 ) , \cos\alpha=\frac{\sin2\alpha}{2\sin\alpha}=\frac{\sin\beta}{2\sin\alpha}=\frac b{2a}=\frac n{2(n-1)}, cos α = 2 sin α sin 2 α = 2 sin α sin β = 2 a b = 2 ( n − 1 ) n , cos α = b 2 + c 2 − a 2 2 b c = n + 4 2 ( n + 1 ) . \cos\alpha=\frac{b^2+c^2-a^2}{2bc}=\frac{n+4}{2(n+1)}. cos α = 2 b c b 2 + c 2 − a 2 = 2 ( n + 1 ) n + 4 . Iz jednadžbe n 2 ( n − 1 ) = n + 4 2 ( n + 1 ) \dfrac n{2(n-1)}=\dfrac{n+4}{2(n+1)} 2 ( n − 1 ) n = 2 ( n + 1 ) n + 4 n = 2 n=2 n = 2
2 ∘ 2^\circ 2 ∘ γ = 2 α \gamma=2\alpha γ = 2 α
cos α = sin 2 α 2 sin α = sin γ 2 sin α = n + 1 2 ( n − 1 ) , \cos\alpha=\frac{\sin2\alpha}{2\sin\alpha}=\frac{\sin\gamma}{2\sin\alpha}=\frac{n+1}{2(n-1)}, cos α = 2 sin α sin 2 α = 2 sin α sin γ = 2 ( n − 1 ) n + 1 , cos α = b 2 + c 2 − a 2 2 b c = n + 4 2 ( n + 1 ) . \cos\alpha=\frac{b^2+c^2-a^2}{2bc}=\frac{n+4}{2(n+1)}. cos α = 2 b c b 2 + c 2 − a 2 = 2 ( n + 1 ) n + 4 . Iz jednadžbe n + 1 2 ( n − 1 ) = n + 4 2 ( n + 1 ) \dfrac{n+1}{2(n-1)}=\dfrac{n+4}{2(n+1)} 2 ( n − 1 ) n + 1 = 2 ( n + 1 ) n + 4 n = 5 n=5 n = 5
3 ∘ 3^\circ 3 ∘ γ = 2 β \gamma=2\beta γ = 2 β
cos β = sin 2 β 2 sin β = sin γ 2 sin β = n + 1 2 n , \cos\beta=\frac{\sin2\beta}{2\sin\beta}=\frac{\sin\gamma}{2\sin\beta}=\frac{n+1}{2n}, cos β = 2 sin β sin 2 β = 2 sin β sin γ = 2 n n + 1 , cos β = a 2 + c 2 − b 2 2 a c = n 2 + 2 2 ( n 2 − 1 ) . \cos\beta=\frac{a^2+c^2-b^2}{2ac}=\frac{n^2+2}{2(n^2-1)}. cos β = 2 a c a 2 + c 2 − b 2 = 2 ( n 2 − 1 ) n 2 + 2 . Iz jednadžbe n + 1 2 n = n 2 + 2 2 ( n 2 − 1 ) \dfrac{n+1}{2n}=\dfrac{n^2+2}{2(n^2-1)} 2 n n + 1 = 2 ( n 2 − 1 ) n 2 + 2 n 2 − 3 n − 1 = 0 n^2-3n-1=0 n 2 − 3 n − 1 = 0 n 1 , 2 = 3 ± 13 2 n_{1,2}=\dfrac{3\pm\sqrt{13}}2 n 1 , 2 = 2 3 ± 13
Dakle, jedino rješenje je trokut sa stranicama duljina 4, 5 i 6.
