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# Let $A_1 A_2 \dotsb A_{11}$ be a regular 11-gon inscribed in a circle of radius 2. Let $P$ be a point, such that the distance from $P$ to th

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Let $A_1 A_2 \dotsb A_{11}$ be a regular 11-gon inscribed in a circle of radius 2. Let $P$ be a point, such that the distance from $P$ to the center of the circle is 3. Find $PA_1^2 + PA_2^2 + \dots + PA_{11}^2.$

Jan 25, 2019

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Let $A_1 A_2 \dotsb A_{11}$ be a regular 11-gon inscribed in a circle of radius 2.

Let $P$ be a point, such that the distance from $P$ to the center of the circle is 3.

Find $PA_1^2 + PA_2^2 + \dots + PA_{11}^2.$

$$\text{Let A_n=\dbinom{x_a}{y_a}, ~x_a =2\cos\Big((n-1)\cdot \dfrac{360^{\circ}}{11} \Big), ~y_a =2\sin\Big((n-1)\cdot \dfrac{360^{\circ}}{11} \Big) , ~ n=1,2,\ldots 11 }\\ \text{Let P=\dbinom{x_p}{y_p}, ~x_p =3\cos(\beta), ~y_p =3\sin(\beta) } \\ \text{Let PA^2 = (x_p-x_a)^2 + (y_p-y_a)^2}$$

$$\begin{array}{|rcll|} \hline PA_1^2 &=& \left[ 3\cos(\beta) -2\cos\left(0\cdot \dfrac{360^{\circ}}{11} \right) \right]^2 + \left[ 3\sin(\beta) -2\sin\left(0\cdot \dfrac{360^{\circ}}{11} \right) \right]^2 \\ &=& 13 - 12\left[ \cos(\beta)\cos\left(0\cdot \dfrac{360^{\circ}}{11} \right) + \sin(\beta)\sin\left(0\cdot \dfrac{360^{\circ}}{11} \right)\right]\\\\ PA_2^2 &=& \left[ 3\cos(\beta) -2\cos\left(1\cdot \dfrac{360^{\circ}}{11} \right) \right]^2 + \left[ 3\sin(\beta) -2\sin\left(1\cdot \dfrac{360^{\circ}}{11} \right) \right]^2 \\ &=& 13 - 12\left[ \cos(\beta)\cos\left(1\cdot \dfrac{360^{\circ}}{11} \right) + \sin(\beta)\sin\left(1\cdot \dfrac{360^{\circ}}{11} \right)\right]\\\\ PA_3^2 &=& \left[ 3\cos(\beta) -2\cos\left(2\cdot \dfrac{360^{\circ}}{11} \right) \right]^2 + \left[ 3\sin(\beta) -2\sin\left(2\cdot \dfrac{360^{\circ}}{11} \right) \right]^2 \\ &=& 13 - 12\left[ \cos(\beta)\cos\left(2\cdot \dfrac{360^{\circ}}{11} \right) + \sin(\beta)\sin\left(2\cdot \dfrac{360^{\circ}}{11} \right)\right]\\\\ \ldots \\\\ PA_{11}^2 &=& \left[ 3\cos(\beta) -2\cos\left(10\cdot \dfrac{360^{\circ}}{11} \right) \right]^2 + \left[ 3\sin(\beta) -2\sin\left(10\cdot \dfrac{360^{\circ}}{11} \right) \right]^2 \\ &=& 13 - 12\left[ \cos(\beta)\cos\left(10\cdot \dfrac{360^{\circ}}{11} \right) + \sin(\beta)\sin\left(10\cdot \dfrac{360^{\circ}}{11} \right)\right]\\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline && PA_1^2 + PA_2^2 + PA_3^2 + \dots + PA_{11}^2 \\\\ &=& 11\cdot 13 \\ && -12\cdot \Big\{ \cos(\beta) \Big(\underbrace{ \cos(0\cdot \dfrac{360^{\circ}}{11}) + \cos(1\cdot \dfrac{360^{\circ}}{11}) + \cos(2\cdot \dfrac{360^{\circ}}{11}) + \ldots + \cos(10\cdot \dfrac{360^{\circ}}{11}) }_{=0} \Big) \\ && +\sin(\beta) \Big(\underbrace{ \sin(0\cdot \dfrac{360^{\circ}}{11}) + \sin(1\cdot \dfrac{360^{\circ}}{11}) + \sin(2\cdot \dfrac{360^{\circ}}{11}) + \ldots + \sin(10\cdot \dfrac{360^{\circ}}{11}) \Big) }_{=0} \Big\} \\\\ &=& 11\cdot 13 -12\cdot \Big( \cos(\beta)\cdot 0 +\sin(\beta)\cdot 0 \Big) \\ &=& 11\cdot 13 -12\cdot 0 \\ &=& 11\cdot 13 \\ &\mathbf{=}& \mathbf{143} \\ \hline \end{array}$$

$$PA_1^2 + PA_2^2 + \dots + PA_{11}^2 = \mathbf{143}$$

Jan 25, 2019