Prvo rješenje.
Neka je a = A B ‾ a=\overline{AB} a = A B △ A B S \triangle ABS △ A B S x x x △ A B C \triangle ABC △ A B C r + x r+x r + x
Vrijedi
x = r cos α 2 . x=r\cos\frac\alpha2. x = r cos 2 α . Za površine danih trokuta vrijedi
( r + x ) a 2 x a 2 = 2 3 + 3 3 . \frac{\frac{(r+x)a}{2}}{\frac{xa}{2}}=\frac{2\sqrt3+3}{3}. 2 x a 2 ( r + x ) a = 3 2 3 + 3 . Nakon sređivanja tog izraza slijedi
r x + 1 = 2 3 3 + 1 , x r = 3 2 3 . \frac rx+1=\frac{2\sqrt3}{3}+1, \qquad \frac xr=\frac3{2\sqrt3}. x r + 1 = 3 2 3 + 1 , r x = 2 3 3 . Odatle je
cos α 2 = 3 2 . \cos\frac\alpha2=\frac{\sqrt3}{2}. cos 2 α = 2 3 . Tada je α = 60 ∘ \alpha=60^\circ α = 6 0 ∘
Drugo rješenje.
Površina manjeg trokuta △ A B S \triangle ABS △ A B S
P 1 = 1 2 r 2 sin α , P_1=\frac12r^2\sin\alpha, P 1 = 2 1 r 2 sin α , a površina većeg trokuta △ A B C \triangle ABC △ A B C
P 2 = P 1 + 2 P △ A S C = 1 2 r 2 sin α + r 2 sin ( 180 ∘ − α 2 ) = 1 2 r 2 ⋅ 2 sin α 2 cos α 2 + r 2 sin α 2 = r 2 sin α 2 ( cos α 2 + 1 ) . \begin{aligned} P_2&=P_1+2P_{\triangle ASC} =\frac12r^2\sin\alpha+r^2\sin\left(180^\circ-\frac\alpha2\right)\\ &=\frac12r^2\cdot2\sin\frac\alpha2\cos\frac\alpha2+r^2\sin\frac\alpha2\\ &=r^2\sin\frac\alpha2\left(\cos\frac\alpha2+1\right). \end{aligned} P 2 = P 1 + 2 P △ A S C = 2 1 r 2 sin α + r 2 sin ( 18 0 ∘ − 2 α ) = 2 1 r 2 ⋅ 2 sin 2 α cos 2 α + r 2 sin 2 α = r 2 sin 2 α ( cos 2 α + 1 ) . Sada je
P 1 : P 2 = 1 2 r 2 sin α r 2 sin α 2 ( cos α 2 + 1 ) = sin α 2 cos α 2 sin α 2 ( cos α 2 + 1 ) . P_1:P_2 =\frac{\frac12r^2\sin\alpha}{r^2\sin\frac\alpha2\left(\cos\frac\alpha2+1\right)} =\frac{\sin\frac\alpha2\cos\frac\alpha2}{\sin\frac\alpha2\left(\cos\frac\alpha2+1\right)}. P 1 : P 2 = r 2 sin 2 α ( cos 2 α + 1 ) 2 1 r 2 sin α = sin 2 α ( cos 2 α + 1 ) sin 2 α cos 2 α . ⟹ cos α 2 cos α 2 + 1 = 3 2 3 + 3 ⟹ α = 60 ∘ . \Longrightarrow\quad\frac{\cos\frac\alpha2}{\cos\frac\alpha2+1} =\frac3{2\sqrt3+3} \quad\Longrightarrow\quad \alpha=60^\circ. ⟹ cos 2 α + 1 cos 2 α = 2 3 + 3 3 ⟹ α = 6 0 ∘ . 
