Let be the vertices of a regular octagon that lie on a circle of radius 2. Let P be a point on the circle and let denote the distance between the points P and for . If P varies over the circle, then the maximum value of the product is

Question

Let $A_{1}, A_{2}, A_{3}, \dots, A_{8}$ be the vertices of a regular octagon that lie on a circle of radius 2. Let P be a point on the circle and let $P A_{i}$ denote the distance between the points P and $A_{i}$ for $i = 1, 2, \dots, 8$ . If P varies over the circle, then the maximum value of the product $P A_{1} \cdot P A_{2} \dots P A_{8}$ is [2023]

Smart Achievers · Accepted Answer

(512)

$In ∆ A_{1} O P, ∠ O A_{1} P = ∠ O P A_{1} = \frac{π}{2} - \frac{θ}{2}$

$\frac{P A_{1}}{2} = \frac{\sin θ}{\sin (\frac{π}{2} - \frac{θ}{2})} = 2 \sin \frac{θ}{2} \Rightarrow P A_{1} = 4 \sin (\frac{θ}{2}) = x_{1}$ (say)

$P A_{8} = 4 \sin (\frac{π}{8} + \frac{θ}{2}) = x_{8}$ $[∵ ∠ P A_{8} O = \frac{π}{8} + \frac{θ}{2}]$

$P A_{7} = 4 \sin (\frac{π}{4} + \frac{θ}{2}) = x_{7}, P A_{6} = 4 \sin (\frac{3 π}{8} + \frac{θ}{2}) = x_{6}$

Similarly

$P A_{2} = 4 \sin (\frac{ϕ}{2}) = x_{2}, P A_{3} = 4 \sin (\frac{π}{8} + \frac{ϕ}{2}) = x_{3}$

$P A_{4} = 4 \sin (\frac{π}{4} + \frac{ϕ}{2}) = x_{4}, P A_{5} = 4 \sin (\frac{3 π}{8} + \frac{ϕ}{2}) = x_{5}$

$Now, P A_{1} \cdot P A_{2} \dots P A_{8} =$

$P = 4^{8} \sin (\frac{θ}{2}) \sin (\frac{3 π}{8} + \frac{ϕ}{2}) . \sin (\frac{π}{8} + \frac{θ}{2}) \sin (\frac{π}{4} + \frac{θ}{2})$ $\sin (\frac{π}{4} + \frac{ϕ}{2}) \sin (\frac{π}{8} + \frac{ϕ}{2}) . \sin (\frac{3 π}{8} + \frac{θ}{2}) \sin (\frac{ϕ}{2})$

$= 4^{8} {\sin \frac{θ}{2} . \cos \frac{θ}{2} . \sin (\frac{π}{8} + \frac{θ}{2}) \cos (\frac{π}{8} + \frac{θ}{2}) . \sin (\frac{π}{4} + \frac{θ}{2}) \cos (\frac{π}{4} + \frac{θ}{2}) \sin (\frac{3 π}{8} + \frac{θ}{2}) \cos (\frac{3 π}{8} + \frac{θ}{2})}$

$= 4^{8} {\frac{\sin θ \sin (\frac{π}{4} + θ) \sin (\frac{π}{4} + θ) \sin (\frac{3 π}{4} + θ)}{2^{4}}}$

$= 4^{6} {\sin θ \cos θ \sin (\frac{π}{4} + θ) \cos (\frac{π}{4} + θ)}$

$= 4^{6} {\frac{\sin 2 θ \sin (\frac{π}{2} + 2 θ)}{4}}$

$= 4^{5} \frac{\sin (4 θ)}{2} = 2^{9} \sin 4 θ$

$P is maximum when \sin 4 θ = 1 \Rightarrow θ = \frac{π}{8}$

$P_{\max} = 2^{9} = 512$

Solution

Want Us to Email you About Special Offers & Updates?

Solution

Q. Let A1,A2,A3,…,A8 be the vertices of a regular octagon that lie on a circle of radius 2. Let P be a point on the circle and let PAi denote the distance between the points P and Ai for i=1,2,…,8. If P varies over the circle, then the maximum value of the product PA1·PA2⋯PA8 is [2023]

Want Us to Email you About Special Offers & Updates?