6.1. In an isosceles triangle \(ABC\), one of the angles is equal to the difference of the other two, and one of the angles is twice another. The angle bisectors of angles \(A\), \(B\), and \(C\) intersect the circumcircle of the triangle at points \(L\), \(O\), and \(M\) respectively. Find the area of triangle \(LOM\), if the area of triangle \(ABC\) is 8. If the answer is not an integer, round it to the nearest integer.
Answer: 11. Solution. Let in triangle $ABC$ the angle $\alpha$ be equal to the difference of angles $\beta-\gamma$. Since $\alpha+\beta+\gamma=180^{\circ}$, then $\beta-\gamma+\beta+\gamma=180^{\circ}$ and therefore $\beta=90^{\circ}$. If $\beta=90^{\circ}$ is twice one of the angles $\alpha$ or $\gamma$, then the triangle will be isosceles, which contradicts the condition. Let for definiteness $\alpha>\gamma$. Then $\beta=90^{\circ}, \alpha=60^{\circ}, \gamma=30^{\circ}$.

By the inscribed angle theorem:

$\angle MLC = \angle MBC = 45^{\circ}, \angle CLO = \angle CAO = 30^{\circ}$. Therefore, $\angle MLO = 75^{\circ}$. Similarly,

$\angle LOM = \angle LOA + \angle AOM = \angle LCA + \angle ABM = 15^{\circ} + 45^{\circ} = 60^{\circ}$,

![](https://cdn.mathpix.com/cropped/2024_05_06_1415bcd4b27ed98bd836g-05.jpg?height=569&width=596&top_left_y=2177&top_left_x=1324)
$\angle OML = \angle OMB + \angle BML = \angle OAB + \angle BCL = 30^{\circ} + 15^{\circ} = 45^{\circ}$.

Thus, the angles of triangle $LOM$ are $\alpha_{1}=75^{\circ}, \beta_{1}=60^{\circ}, \gamma_{1}=45^{\circ}$.

Using the sine rule, we get the formula for the area of a triangle inscribed in a circle of radius $R: S=\frac{1}{2} a b \sin \gamma=\frac{1}{2} \cdot 2 R \sin \alpha \cdot 2 R \sin \beta \cdot \sin \gamma=2 R^{2} \sin \alpha \sin \beta \sin \gamma$.

Therefore, $\frac{S_{LOM}}{S_{ABC}}=\frac{2 R \cdot \sin \alpha_{1} \sin \beta_{1} \sin \gamma_{1}}{2 R \cdot \sin \alpha \sin \beta \sin \gamma}=\frac{\sin 75^{\circ} \sin 60^{\circ} \sin 45^{\circ}}{\sin 90^{\circ} \sin 60^{\circ} \sin 30^{\circ}}=\sqrt{2} \cdot \sin 75^{\circ}$. Since $\sin 75^{\circ}=\sin \left(45^{\circ}+30^{\circ}\right)=\frac{\sqrt{6}+\sqrt{2}}{4}$, we get: $\frac{S_{LOM}}{S_{ABC}}=\frac{\sqrt{3}+1}{2}$.

Therefore, $S_{LOM}=4 \sqrt{3}+4 \approx 11$.