+0  
 
+1
604
1
avatar

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
 #1
avatar+25528 
+8

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 \[PA_1^2 + PA_2^2 + \dots + PA_{11}^2.\]

 

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

 

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

 

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

 

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

 

laugh

 Jan 25, 2019

18 Online Users

avatar
avatar
avatar