Verified Solution • 2025–26 Session moderate

QJEEAdv23Ma7579: JEE Advanced 2023 Mathematics - Class 11 JEEAdv23Ma7579 Solution: Let be the vertices of a regular octagon that lie on a circle…

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 PAiP A_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×⋯⋯×PA8P A_1 \times P A_2 \times \cdots \cdots \times P A_8 , is :

Step-by-Step Verified Solution

A1,A2,A3,…,A8A_1, A_2, A_3, \ldots, A_8 vertices of a regular octagon lying on a circle of radius 2 .

Let say, Z=(2)(1)1/8Z=(2)(1)^{1 / 8}

amp;⇒Z8=28×1amp;⇒Z8−28=0\begin{aligned}& \Rightarrow Z^8=2^8 \times 1 \\\\& \Rightarrow Z^8-2^8=0\end{aligned}

amp;⇒Z=2,2α,2α2,2α3,…,2α7;α=ei2π8amp;⇒Z8−28=(Z−2)(Z−2α)(Z−2α2)(Z−2α3)…(Z−2α7)amp;⇒∣Z8−28∣=∣Z−2∣∣Z−2α∣….∣Z−2α7∣amp; But ∣Z8+(−28)∣≤∣Z∣8+28\begin{aligned} & \Rightarrow Z=2,2 \alpha, 2 \alpha^2, 2 \alpha^3, \ldots, 2 \alpha^7 ; \alpha=e^{i \frac{2 \pi}{8}} \\\\ & \Rightarrow Z^8-2^8=(Z-2)(Z-2 \alpha)\left(Z-2 \alpha^2\right)\left(Z-2 \alpha^3\right) \ldots\left(Z-2 \alpha^7\right) \\\\ & \Rightarrow\left|Z^8-2^8\right|=|Z-2||Z-2 \alpha| \ldots .\left|Z-2 \alpha^7\right| \\\\ & \text { But }\left|Z^8+\left(-2^8\right)\right| \leq|Z|^8+2^8\end{aligned}

⇒∣Z−2∣∣Z−2α∣…∣Z−2α7∣amp;≤∣Z∣8+28amp;≤28+28amp;≤29\begin{aligned} \Rightarrow|Z-2||Z-2 \alpha| \ldots\left|Z-2 \alpha^7\right| & \leq|Z|^8+2^8 \\\\ & \leq 2^8+2^8 \\\\ & \leq 2^9\end{aligned}

⇒Max⁡(PA1⋅PA2…PA8)=29\Rightarrow \operatorname{Max}\left(P A_1 \cdot P A_2 \ldots P A_8\right)=2^9

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