Question

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

• A. $\frac{1}{2}[\frac{3r^2}{a^2}+\frac{3s^2}{b^2}-1]$
• B. $\frac{3}{2}[\frac{r^2}{a^2}+\frac{s^2}{b^2}]$
• C. $\frac{1}{3}[\frac{r^2}{a^2}+\frac{s^2}{b^2}]$
• D. $\frac{1}{3}[\frac{r^2}{a^2}+\frac{s^2}{b^2}+\frac{rs}{ab}]$

Hint:

For an equilateral triangle inscribed in an ellipse, the circumcenter and centroid coincide. This fact allows you to relate the coordinates of the center $(r,s)$ to the sum of the coordinates of the vertices, which in turn can be related to the eccentric angles.

Answer 1

The Correct Option is A

Step 1: Relate the circumcenter to the centroid.
For an equilateral triangle, the circumcenter and the centroid coincide. The vertices are given by their eccentric angles $\theta_1, \theta_2, \theta_3$ on the ellipse. The coordinates of the vertices are $P_1(a\cos\theta_1, b\sin\theta_1)$, $P_2(a\cos\theta_2, b\sin\theta_2)$, and $P_3(a\cos\theta_3, b\sin\theta_3)$. The coordinates of the centroid $(r,s)$ are the average of the coordinates of the vertices. \[ r = \frac{a(\cos\theta_1+\cos\theta_2+\cos\theta_3)}{3} \implies \frac{3r}{a} = \sum \cos\theta_i. \] \[ s = \frac{b(\sin\theta_1+\sin\theta_2+\sin\theta_3)}{3} \implies \frac{3s}{b} = \sum \sin\theta_i. \]

Step 2: Use a known identity relating the sums of sines and cosines.
We square and add the two equations from Step 1. \[ \left(\frac{3r}{a}\right)^2 + \left(\frac{3s}{b}\right)^2 = (\cos\theta_1+\cos\theta_2+\cos\theta_3)^2 + (\sin\theta_1+\sin\theta_2+\sin\theta_3)^2. \] Let's expand the right side (RHS): RHS $= (\sum\cos^2\theta_i + 2\sum\cos\theta_1\cos\theta_2) + (\sum\sin^2\theta_i + 2\sum\sin\theta_1\sin\theta_2)$. RHS $= \sum(\cos^2\theta_i+\sin^2\theta_i) + 2\sum(\cos\theta_1\cos\theta_2 + \sin\theta_1\sin\theta_2)$. Using identities $\cos^2\theta+\sin^2\theta=1$ and $\cos(A-B)=\cos A\cos B+\sin A\sin B$: RHS $= (1+1+1) + 2[\cos(\theta_1-\theta_2) + \cos(\theta_2-\theta_3) + \cos(\theta_3-\theta_1)]$.

Step 3: Formulate the final equation.
Equating the left and right sides: \[ \frac{9r^2}{a^2} + \frac{9s^2}{b^2} = 3 + 2[\cos(\theta_1-\theta_2) + \cos(\theta_2-\theta_3) + \cos(\theta_3-\theta_1)]. \]

Step 4: Solve for the average of the cosine terms.
The average of the cosine terms is $\frac{\cos(\theta_1-\theta_2) + \cos(\theta_2-\theta_3) + \cos(\theta_3-\theta_1)}{3}$. Let this be `Avg`. The sum in the brackets is $3 \times \text{Avg}$. \[ \frac{9r^2}{a^2} + \frac{9s^2}{b^2} = 3 + 2(3 \times \text{Avg}) = 3 + 6 \times \text{Avg}. \] Now, we solve for `Avg`: \[ 6 \times \text{Avg} = \frac{9r^2}{a^2} + \frac{9s^2}{b^2} - 3. \] Divide by 6: \[ \text{Avg} = \frac{1}{6} \left( \frac{9r^2}{a^2} + \frac{9s^2}{b^2} - 3 \right) = \frac{3}{6} \left( \frac{3r^2}{a^2} + \frac{3s^2}{b^2} - 1 \right). \] \[ \text{Avg} = \frac{1}{2}\left[\frac{3r^2}{a^2}+\frac{3s^2}{b^2}-1\right]. \]