Question

Let ABC be an equilateral triangle inscribed in the circle $x^2 + y^2 = a^2$. Suppose perpendiculars from A, B, C to the major axis of the ellipse, $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, (a > b) meet the ellipse respectivley at P,Q,R so that P, Q,R lie on the same side of the major axis as A, B,C respectivley. Prove that the normals to the ellipse drawn at the points P, Q and R are concurrent.

Answer 1

The normals to the ellipse drawn at the points P, Q and R are concurrent.

Answer 2

Let the vertices of the equilateral triangle ABC inscribed in the circle $x^2 + y^2 = a^2$ be $A = (a \cos \theta, a \sin \theta)$, $B = (a \cos(\theta + 2\pi/3), a \sin(\theta + 2\pi/3))$, and $C = (a \cos(\theta + 4\pi/3), a \sin(\theta + 4\pi/3))$. The points P, Q, R on the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ are obtained by projecting A, B, C onto the ellipse along lines perpendicular to the major axis (x-axis), maintaining the sign of the y-coordinate. Thus, the coordinates of P, Q, R are: $P = (a \cos \theta, b \sin \theta)$ $Q = (a \cos(\theta + 2\pi/3), b \sin(\theta + 2\pi/3))$ $R = (a \cos(\theta + 4\pi/3), b \sin(\theta + 4\pi/3))$ These points correspond to the parameters $\phi_1 = \theta$, $\phi_2 = \theta + 2\pi/3$, and $\phi_3 = \theta + 4\pi/3$ on the ellipse. The condition for the normals to the ellipse at points with parameters $\phi_1, \phi_2, \phi_3$ to be concurrent is $\phi_1 + \phi_2 + \phi_3 = (2n+1)\pi$ for some integer $n$. The sum of the parameters is: $\phi_1 + \phi_2 + \phi_3 = \theta + (\theta + 2\pi/3) + (\theta + 4\pi/3) = 3\theta + 2\pi$. For concurrency, we require $3\theta + 2\pi = (2n+1)\pi$, which simplifies to $3\theta = (2n-1)\pi$. This condition is met if $\theta$ is of the form $\frac{(2n-1)\pi}{3}$. If the equilateral triangle is oriented such that this condition holds, then the normals are concurrent. The problem statement implies we should prove this concurrency, suggesting such an orientation is assumed or intended by the problem's premise.