Neka su stranice trokuta a = n a=n a = n b = n + 1 b=n+1 b = n + 1 c = n + 2 c=n+2 c = n + 2 a a a α \alpha α c c c γ \gamma γ
Vrijedi γ = 2 α \gamma=2\alpha γ = 2 α
Iz poučka o sinusu slijedi
n sin α = n + 2 sin γ = n + 2 2 sin α cos α , \frac n{\sin\alpha}=\frac{n+2}{\sin\gamma} =\frac{n+2}{2\sin\alpha\cos\alpha}, sin α n = sin γ n + 2 = 2 sin α cos α n + 2 , pa je
cos α = n + 2 2 n . \cos\alpha=\frac{n+2}{2n}. cos α = 2 n n + 2 . Iz poučka o kosinusu slijedi
n 2 = ( n + 1 ) 2 + ( n + 2 ) 2 − 2 ( n + 1 ) ( n + 2 ) cos α = ( n + 1 ) 2 + ( n + 2 ) 2 − 2 ( n + 1 ) ( n + 2 ) ⋅ n + 2 2 n . \begin{aligned} n^2&=(n+1)^2+(n+2)^2-2(n+1)(n+2)\cos\alpha\\ &=(n+1)^2+(n+2)^2-2(n+1)(n+2)\cdot\frac{n+2}{2n}. \end{aligned} n 2 = ( n + 1 ) 2 + ( n + 2 ) 2 − 2 ( n + 1 ) ( n + 2 ) cos α = ( n + 1 ) 2 + ( n + 2 ) 2 − 2 ( n + 1 ) ( n + 2 ) ⋅ 2 n n + 2 . a nakon množenja s n n n
n 3 = n ( n + 1 ) 2 + n ( n + 2 ) 2 − ( n + 1 ) ( n + 2 ) 2 = n 3 + 2 n 2 + n + n 3 + 4 n 2 + 4 n − ( n + 1 ) ( n 2 + 4 n + 4 ) = 2 n 3 + 6 n 2 + 5 n − n 3 − 4 n 2 − 4 n − n 2 − 4 n − 4. \begin{aligned} n^3&=n(n+1)^2+n(n+2)^2-(n+1)(n+2)^2\\ &=n^3+2n^2+n+n^3+4n^2+4n-(n+1)(n^2+4n+4)\\ &=2n^3+6n^2+5n-n^3-4n^2-4n-n^2-4n-4. \end{aligned} n 3 = n ( n + 1 ) 2 + n ( n + 2 ) 2 − ( n + 1 ) ( n + 2 ) 2 = n 3 + 2 n 2 + n + n 3 + 4 n 2 + 4 n − ( n + 1 ) ( n 2 + 4 n + 4 ) = 2 n 3 + 6 n 2 + 5 n − n 3 − 4 n 2 − 4 n − n 2 − 4 n − 4. Rezultat sređivanja je kvadratna jednadžba
n 2 − 3 n − 4 = 0. n^2-3n-4=0. n 2 − 3 n − 4 = 0. Jedino moguće rješenje je n = 4 n=4 n = 4
Tada su stranice trokuta a = 4 a=4 a = 4 b = 5 b=5 b = 5 c = 6 c=6 c = 6
Rotacijom ovog trokuta oko stranice c c c
Polumjer baze je visina danog trokuta. Izračunajmo prvo sin α \sin\alpha sin α
cos α = n + 2 2 n = 6 8 = 3 4 , \cos\alpha=\frac{n+2}{2n}=\frac68=\frac34, cos α = 2 n n + 2 = 8 6 = 4 3 , sin α = 1 − 9 16 = 7 4 . \sin\alpha=\sqrt{1-\frac9{16}}=\frac{\sqrt7}{4}. sin α = 1 − 16 9 = 4 7 . Kako je sin α = r 5 \sin\alpha=\frac r5 sin α = 5 r
r = 5 7 4 c m . r=\frac{5\sqrt7}{4}\ \mathrm{cm}. r = 4 5 7 cm . Obujam dobivenog tijela je
V = r 2 π ⋅ x 3 + r 2 π ⋅ ( 6 − x ) 3 = r 2 π ⋅ 6 3 = 5 2 ⋅ 7 16 ⋅ π ⋅ 2 = 175 π 8 c m 3 . \begin{aligned} V&=\frac{r^2\pi\cdot x}{3}+\frac{r^2\pi\cdot(6-x)}{3} =\frac{r^2\pi\cdot6}{3}\\ &=5^2\cdot\frac7{16}\cdot\pi\cdot2 =\frac{175\pi}{8}\ \mathrm{cm}^3. \end{aligned} V = 3 r 2 π ⋅ x + 3 r 2 π ⋅ ( 6 − x ) = 3 r 2 π ⋅ 6 = 5 2 ⋅ 16 7 ⋅ π ⋅ 2 = 8 175 π cm 3 . 
