[Cambridge/J78/II/6] An isosceles triangle $ABC$ with $AB = AC$ is inscribed in a fixed circle, centre $O$, whose radius is 1 unit. Angle $BOC = 2\theta$, where $\theta$ is acute. Express the area of the triangle $ABC$ in terms of $\theta$, and hence show that, as $\theta$ varies, the area is a maximum when the triangle is equilateral.

1. Find the base length of the triangle using the given circle and angle.

Let $M$ be the midpoint of $BC$. By symmetry, since $AB=AC$, the line $AM$ is the perpendicular bisector of $BC$ and passes through the centre $O$ of the circle.
In right-angled $\triangle OBM$, $\angle BOM = \frac{1}{2}\angle BOC = \theta$.
With the radius of the circle $OB = 1$, we have:
$BM = OB \sin \theta = \sin \theta$
Therefore, the base of the triangle is $BC = 2\sin \theta$.

2. Express the area of the triangle in terms of $\theta$.

To maximise the area, vertex $A$ must lie on the major arc $BC$. Because $\theta$ is acute, $2\theta < 180^\circ$, meaning the centre $O$ lies between $A$ and $M$.
The distance from $O$ to $M$ is:
$OM = OB \cos \theta = \cos \theta$
The height of $\triangle ABC$ is $AM = AO + OM = 1 + \cos \theta$.
The area of the triangle is:
$\text{Area} = \frac{1}{2} \times BC \times AM$
$\text{Area} = \frac{1}{2} (2\sin \theta)(1 + \cos \theta) = \sin \theta(1 + \cos \theta)$

3. Differentiate the area function to find stationary points.

To find the maximum area, we differentiate the area with respect to $\theta$ (where $\theta$ must be in radians for calculus):
$\frac{d(\text{Area})}{d\theta} = \frac{d}{d\theta}(\sin \theta + \sin \theta \cos \theta)$
Using the product rule for the second term:
$\frac{d(\text{Area})}{d\theta} = \cos \theta + (\cos \theta)(\cos \theta) + (\sin \theta)(-\sin \theta)$
$\frac{d(\text{Area})}{d\theta} = \cos \theta + \cos^2 \theta - \sin^2 \theta$
Using the identity $\sin^2 \theta = 1 - \cos^2 \theta$:
$\frac{d(\text{Area})}{d\theta} = \cos \theta + \cos^2 \theta - (1 - \cos^2 \theta) = 2\cos^2 \theta + \cos \theta - 1$

4. Set the derivative to zero and solve for $\theta$.

At a stationary point, $\frac{d(\text{Area})}{d\theta} = 0$:
$2\cos^2 \theta + \cos \theta - 1 = 0$
$(2\cos \theta - 1)(\cos \theta + 1) = 0$
Since $\theta$ is acute ($0 < \theta < \frac{\pi}{2}$), $\cos \theta > 0$, so $\cos \theta + 1 \neq 0$.
Therefore, we must have:
$2\cos \theta - 1 = 0 \implies \cos \theta = \frac{1}{2}$
$\theta = \frac{\pi}{3}$

5. Verify that this value of $\theta$ gives a maximum area.

Using the second derivative test:
$\frac{d^2(\text{Area})}{d\theta^2} = \frac{d}{d\theta}(2\cos^2 \theta + \cos \theta - 1)$
$\frac{d^2(\text{Area})}{d\theta^2} = 4\cos \theta(-\sin \theta) - \sin \theta = -\sin \theta(4\cos \theta + 1)$
When $\theta = \frac{\pi}{3}$, both $\sin \frac{\pi}{3} > 0$ and $\cos \frac{\pi}{3} > 0$.
Therefore, $\frac{d^2(\text{Area})}{d\theta^2} < 0$, which confirms that the area is a maximum when $\theta = \frac{\pi}{3}$.

6. Show that the triangle is equilateral when the area is maximum.

When $\theta = \frac{\pi}{3}$, the angle at the centre is $\angle BOC = 2\theta = \frac{2\pi}{3}$.
The angle subtended at the circumference by the minor arc $BC$ is:
$\angle BAC = \frac{1}{2}\angle BOC = \frac{\pi}{3}$
Since $\triangle ABC$ is isosceles with $AB = AC$, the base angles are equal:
$\angle ABC = \angle ACB = \frac{\pi - \angle BAC}{2} = \frac{\pi - \frac{\pi}{3}}{2} = \frac{\pi}{3}$
Since all three angles of $\triangle ABC$ are $\frac{\pi}{3}$ (or $60^\circ$), the triangle is equilateral.