Circles
Circle
Allen Star Batch
Grade 11
Question:
In the triangle $ABC$, the angle bisector $AK$ is perpendicular to the median $BM$ and $\angle ABC = 120°$, then:
The value of ratio $\frac{BC}{AB}$ is equal to $\frac{\sqrt{13}-1}{2}$
The value of the ratio of radius of the circle circumscribing the triangle $ABC$ to the side length $AB$ is equal to $\frac{2}{\sqrt{3}}$
The ratio of the area of $\triangle ABC$ to the area of the circle circumscribing $\triangle ABC$ is equal to $\frac{3\sqrt{3}}{32\pi}(\sqrt{13}-1)$
The value of ratio of the sides $AB$ to $AC$ is equal to $1/2$
Step-by-Step Solution
Key Concept: Use properties of isosceles triangles and angle bisectors combined with the law of sines and cosines to establish relationships between sides.
Triangle $AMB$ is isosceles with the angle bisector of $\angle A$ perpendicular to $MB$. Using the law of sines with $\frac{c}{\sin\theta} = \frac{2c}{\sin 60°}$ gives $\sin\theta = \frac{\sqrt{3}}{4}$. Applying the law of cosines: $a^2 = c^2 + 4c^2 - 4c^2\cos(60° - \theta) = 5c^2 - 4c^2\left(\frac{\sqrt{13}}{4} + \frac{\sqrt{3}\sqrt{3}}{2} - \frac{\sqrt{3}}{4}\right)$, leading to $\frac{a}{c} = \frac{\sqrt{13}-1}{2}$ and $\frac{\Delta}{\pi Rc^2} = \frac{3\sqrt{3}a}{16\pi c} - \frac{3\sqrt{3}}{32\pi}(\sqrt{13}-1)$.
Correct Answer: 1,2,3,4