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← JEE Advanced 2023 · Paper 2
MathematicsInteger+4Question 11 of 51

JEE Advanced 2023 · Paper 2, Mathematics question 11

Let A1,A2,A3,…,A8A_1, A_2, A_3, \ldots, A_8 be the vertices of a regular octagon that lie on a circle of radius 2. Let PP be a point on the circle and let PAiPA_i denote the distance between the points PP and AiA_i for i=1,2,…,8i = 1, 2, \ldots, 8. If PP varies over the circle, then the maximum value of the product PA1⋅PA2⋯PA8PA_1\cdot PA_2\cdots PA_8, is _____.

Answer and worked solution

Official answer: 512

Place the circle at the origin, with Ak=2ωkA_k=2\omega^k where ω=e2πi/8\omega=e^{2\pi i/8}. Let P=2eiθP=2e^{i\theta}.

Product of distances. Since ∏k(z−2ωk)=z8−28\prod_k(z-2\omega^k)=z^8-2^8,
PA1⋯PA8=∣P8−28∣=28∣e8iθ−1∣.PA_1\cdots PA_8=|P^8-2^8|=2^8\left|e^{8i\theta}-1\right|.

Maximum. ∣e8iθ−1∣≤2|e^{8i\theta}-1|\le2, with equality when e8iθ=−1e^{8i\theta}=-1. This is when PP is midway between two vertices.

max⁡=2⋅28=512.\max=2\cdot2^8=512.