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Consider ellipse $E$ with foci $S_{1}$, $S_{2}$ and its minor and major auxiliary circles (denoted by $W_{1}$ and $W_{2}$).  Let the circle $W$ be tangent to $W_{1}$ and $W_{2}$, and meet ellipse $E$ at $P$, $Q$, $R$, $T$.  Prove that some three of the four normals at $P$, $Q$, $R$, $T$ are concurrent.

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I wish to solve this using Coordinate Geometry, but my approach seems to be very lengthy and time-taking.

Here's what I propose:

  1. Assuming the equation of a standard ellipse, corresponding major and minor auxiliary circles

  2. Finding a circle (I'm not sure if this circle is unique, or not. If not, then how should I write the general equation of such a circle?) touching both $x^2+y^2=a^2$ and $x^2+y^2=b^2$ (major and minor auxiliary circles).

  3. Solving the equation obtained in (2) with the ellipse. I expect a fourth degree equation to arise, as the ellipse intersects the circle at exactly four points (as shown in the figure).

  4. Writing the equation of normal to an ellipse at some point $(a\cos \phi, b\sin \phi)$. Note that this point also satisfies equation obtained in (3)

I'm not sure how to proceed further.

Also, is there a shorter method to approach this problem? It'd be great if someone could post a solution using Coordinate Geometry.

Thanks a lot!

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  • $\begingroup$ Some initial thoughts: Let $A = (a \cos\phi, a \sin\phi)$ be where the variable circle meets the major aux circle; then $B = (-b \cos\phi,-b\sin\phi)$ is the point where the variable circle meets the minor aux circle. The variable circle, then, has center $(A+B)/2$ and radius $(a+b)/2$. That's simple enough. Now, substitute $x = a \cos\theta$ and $y = b \sin\theta$ into the variable circle's eqn and try to glean information about relevant $\theta$s. That's the hard part, because of the quartic. As it happens, the point whose normal isn't concurrent is precisely the one with $\theta = -\phi$. $\endgroup$ – Blue Dec 6 '17 at 9:13
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Here's a brute force approach, with a possibly-unnecessary detour into complex arithmetic ...


Part 1: The Points

Let the ellipse have major radius $p$, aligned with the $x$ axis, and minor radius $q$, aligned with $y$-axis. Let its center lie at the origin. For any $\theta$, we can find a variable circle tangent to the major and minor auxiliary circles at respective points $P := (p\cos\theta, p\sin\theta)$ and $Q := (-q\cos\theta, -q\sin\theta)$; this circles center is $(P+Q)/2$, and its radius is $(p+q)/2$, so its equation works out to be $$( 2 x - (p-q)\cos\theta)^2 + (2 y - (p-q)\sin\theta)^2 = ( a + b )^2 \tag{1}$$ We can parameterize a point on the ellipse as $E(\phi) := (a\cos\phi, b\sin\phi)$. Substituting into $(1)$ gives an implicit relation in $\phi$ for the points of intersection of the variable circle and the ellipse. With a little trigonometric massaging, this relation reduces to $$(p-q)\; \sin\frac{\theta+\phi}{2}\;\left( p \cos\phi \sin\frac{\phi-\theta}{2} + q \sin\phi \cos\frac{\phi-\theta}{2} \right)=0 \tag{2}$$ where we'll reasonably ignore the first factor (so that we're restricting our attention to non-circular ellipses). Observe that $$\sin\frac{\theta+\phi}{2}= 0 \tag{2a}$$ implies that $\theta+\phi$ is a multiple of $2\pi$; effectively $\phi = -\theta$. This happens to be the point that's left out of the concurrent-normals configuration. The points in the configuration correspond to three solutions of

$$p\cos\phi \sin\frac{\phi - \theta}{2}+ q\sin\phi\cos\frac{\phi-\theta}{2} = 0\tag{$\star$}$$

As-is, this equation may not look much like a cubic in $\phi$, but we can remedy that: dividing-through by the cosines gives $$0 = p \tan\frac{\phi-\theta}{2} + q \tan 2 \frac{\phi}{2} = p \frac{\tan(\phi/2)-\tan(\theta/2)}{1+\tan(\phi/2)\tan(\theta/2)} + q \frac{2\tan(\phi/2)}{1-\tan^2(\phi/2)}$$

Defining $x_\star := \tan(x/2)$ to reduce clutter, the above becomes our obvious cubic:

$$p \phi_\star^3 - ( p + 2 q )\theta_\star \phi_\star^2 - (p + 2 q) \phi_\star + p \theta_\star = 0 \tag{$\star^\prime$}$$


Part 2: The Lines

Given a point $E(\phi) := ( p\cos\phi, q \sin\phi)$ on the ellipse, Calculus tells us that the line tangent to the ellipse at that point has direction vector $(-p\sin\phi, q\cos\phi)$, so that the normal line at that point has direction vector $n(\phi) := (q\cos\phi, p\sin\phi)$.

We can therefore parameterize the normals at points $A := E(\alpha)$, $B := E(\beta)$, $C := E(\gamma)$, by $$A + a\,n(\alpha) \qquad,\qquad B + b\,n(\beta) \qquad,\qquad C + c\;n(\gamma)$$ Setting pairs of parameterizations equal, we can eliminate $a$, $b$, $c$ to reveal the condition under which the normals have a common point. Again, a little trigonometric massage reduces the condition to a tidy form:

$$\sin(\alpha+\beta) + \sin(\beta+\gamma) + \sin(\gamma+\alpha) = 0 \tag{$\star\star$}$$

A less-tidy form, but one that matches $(\star^\prime)$, comes from expanding $(\star\star)$ in terms of $\alpha_\star := \tan(\alpha/2)$, etc.

$$\alpha_\star + \beta_\star + \gamma_\star = \alpha_\star\beta_\star\gamma_\star\left( \alpha_\star \beta_\star + \beta_\star \gamma_\star + \gamma_\star \alpha_\star \right) \tag{$\star\star^\prime$}$$


Part 3. The Connections

To see that $(\star^\prime)$ and $(\star\star^\prime)$ actually complete the proof, recall Vieta's Formulas for the coefficients of a polynomial in terms of its roots. In particular, if $\alpha_\star$, $\beta_\star$, $\gamma_\star$ are the roots of $(\star^\prime)$, then $$\begin{align} \alpha_\star \beta_\star \gamma_\star &= -\frac{p}{p}\theta_\star = -\theta_\star \tag{3a} \\[4pt] \alpha_\star \beta_\star + \beta_\star \gamma_\star + \gamma_\star \alpha_\star &= -\frac{p+2q}{p} \tag{3b} \\[4pt] \alpha_\star + \beta_\star + \gamma_\star &= \phantom{-}\frac{p+2q}{p}\;\theta_\star\tag{3c} \end{align}$$ Based on the right-hand sides, we have $(3c) = (3a)(3b)$; the corresponding equality for the left-hand sides is exactly $(\star\star^\prime)$. Thus, points that lie on the variable circle are points that determine concurrent normals. $\square$

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  • $\begingroup$ This could have something to do with Joachimstahl Circles. $\endgroup$ – Hari Shankar Dec 7 '17 at 5:16

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