Trigonometry – Problem 30: determine the shape of the triangle

Equilateral Triangle Verification

In $\triangle ABC$, sides $a, b, c$ are in arithmetic progression. The circle with diameter $AC$ has area $2\pi$. If the area of $\triangle ABC$ is $2\sqrt{3}$, determine the shape of the triangle.

Circle with diameter $AC = b$ has area $2\pi$:

$$\pi\left(\frac{b}{2}\right)^2 = 2\pi \Rightarrow b^2 = 8 \Rightarrow b = 2\sqrt{2}$$

Since $a, b, c$ are in AP:

$$2b = a + c \Rightarrow a + c = 4\sqrt{2}$$

Using Law of Cosines:

$$\cos B = \frac{a^2 + c^2 - b^2}{2ac} = \frac{(a+c)^2 - 2ac - b^2}{2ac}$$

$$= \frac{32 - 2ac - 8}{2ac} = \frac{24 - 2ac}{2ac} = \frac{12}{ac} - 1$$

Area: $S = \frac{1}{2}ac\sin B = 2\sqrt{3}$, so $ac\sin B = 4\sqrt{3}$.

Using $\sin^2 B + \cos^2 B = 1$ with $\sin B = \frac{4\sqrt{3}}{ac}$:

$$\left(\frac{4\sqrt{3}}{ac}\right)^2 + \left(\frac{12}{ac} - 1\right)^2 = 1$$

Let $t = ac$:

$$\frac{48}{t^2} + \frac{144}{t^2} - \frac{24}{t} + 1 = 1$$

$$\frac{192}{t^2} = \frac{24}{t} \Rightarrow t = 8$$

So $ac = 8$ and $a + c = 4\sqrt{2}$.

Solving: $a = c = 2\sqrt{2}$.

Since $a = b = c = 2\sqrt{2}$, it's an equilateral triangle.

Equilateral triangle