The circumcenter of the equilateral triangle | vedclass

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(A) Let the vertices of the equilateral triangle be $P_i = (a \cos 	heta_i, b \sin 	heta_i)$ for $i = 1, 2, 3$. Since the triangle is equilateral,it

Vedclass · Accepted Answer

$\frac{1}{2}\left[\frac{3r^2}{a^2}+\frac{3s^2}{b^2}-1\right]$

Vedclass · Answer

(A) Let the vertices of the equilateral triangle be $P_i = (a \cos 	heta_i, b \sin 	heta_i)$ for $i = 1, 2, 3$. Since the triangle is equilateral,its circumcenter $(r, s)$ is also its centroid. Thus,$r = \frac{a}{3} \sum \cos 	heta_i$ and $s = \frac{b}{3} \sum \sin 	heta_i$. This implies $\sum \cos 	heta_i = \frac{3r}{a}$ and $\sum \sin 	heta_i = \frac{3s}{b}$. Squaring and adding these,we get $(\sum \cos 	heta_i)^2 + (\sum \sin 	heta_i)^2 = \frac{9r^2}{a^2} + \frac{9s^2}{b^2}$. Expanding the left side: $3 + 2(\cos(	heta_1-	heta_2) + \cos(	heta_2-	heta_3) + \cos(	heta_3-	heta_1)) = \frac{9r^2}{a^2} + \frac{9s^2}{b^2}$. Dividing by $6$,the average is $\frac{1}{3} \sum \cos(	heta_i-	heta_j) = \frac{1}{3} \left[ \frac{1}{2} \left( \frac{9r^2}{a^2} + \frac{9s^2}{b^2} - 3 ight) ight] = \frac{1}{2} \left[ \frac{3r^2}{a^2} + \frac{3s^2}{b^2} - 1 ight]$.

The circumcenter of the equilateral triangle having the three points $\theta_1, \theta_2, \theta_3$ lying on the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ as its vertices is $(r, s)$. Then the average of $\cos(\theta_1-\theta_2)$,$\cos(\theta_2-\theta_3)$ and $\cos(\theta_3-\theta_1)$ is

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