Circles
Circle
star_batch_jee_advanced_2025
Grade None
Question:
Let $A, B, C, D$ lie on a line such that $AB = BC = CD = 1$. The points $A$ and $C$ are also joined by a semicircle with $AC$ as diameter and $P$ is a variable point on this semicircle such that $\angle PRD = 0, 0 \leq \pi \leq \pi$. Let $R$ is the region bounded by arc $AP$, the straight line $PD$ and line $AD$
The maximum possible area of region $R$ is $\frac{2\pi + 3\sqrt{3}}{6}$
If '$L$' is the perimeter of region '$R$', then $L$ is equal to $3 + \pi - \theta + \sqrt{5 - 4\cos\theta}$
The maximum possible area of region $R$ is $\frac{2\pi - 3\sqrt{3}}{6}$
If '$L$' is the perimeter of region '$R$', then $L$ is equal to $3 + \pi - \theta + \sqrt{5 + 4\cos\theta}$
Step-by-Step Solution
Key Concept: Optimize enclosed area by expressing it as a function of parameter $\theta$ and using calculus to find critical points.
The area enclosed is $A = \frac{1}{2}(1)^2(\pi - \theta) + \frac{1}{2}(2)\sin\theta = \frac{\pi}{2} - \frac{\theta}{2} + \sin\theta$. Taking the derivative: $\frac{dA}{d\theta} = -\cos\theta - \frac{1}{2}$, which equals zero when $\cos\theta = -\frac{1}{2}$, giving $\theta = \frac{\pi}{3}$. The maximum area is $A_{\max} = \frac{\pi}{3} + \frac{\sqrt{3}}{2} = \frac{2\pi + 3\sqrt{3}}{6}$.
Correct Answer: 1,2