An extension of Poncelet's porism

A special case of Poncelet's porism states that whenever a triangle in inscribed in a circle and circumscribes another circle, there exist an infinite family of such triangles.

We can parametrise the possible inner circles with $\theta \in [0, \frac{\pi}{2})$ like so:

Let $\Gamma$ be the unit circle centred at $O = (0,0)$ . Call the point $(1,0) A$ . Pick a point $B$ on the upper half of $\Gamma$ , such that $\angle OAB = \theta$ . Reflect $B$ about the x-axis; call this point $C$ . Let the incircle of $\triangle ABC$ be $\gamma_\theta$ ; its radius is $r = 2 \sin \theta + \cos 2 \theta - 1$ , and it is centred at $(2 \sin \theta - 1, 0)$ .

Find a parametrisation for the infinite family of triangles inscribed in $\Gamma$ and circumscribing $\gamma_\theta$ . Then find the formula for the area of any given triangle, and determine when the area attains maxima/minima.

#Geometry

Easy Math Editor

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Comments

Thank you for this question. I found it fun to follow several different approaches. I only have a partial solution so far. I have used $d=2 \sin{θ}$ as the distance between the centres of $\mathbf{Γ}$ and $γ_{θ}$ .

Step 1. choose a parameterisation. I chose $t$ to be the angle between the x-axis and a line from $\mathbf{I}$ , the centre of $γ_{θ}$ (and the incentre of $\mathbf{ABC}$ ) to an arbitrary point on $\mathbf{Γ}$ : $\mathbf{{A}'}$ . The task is then to define a new triangle $\mathbf{{{A}'{B}'{C}'}}$ based on this parameter $t$ .

Step 2. Calculate $D$ , the distance from $\mathbf{I}$ to $\mathbf{Γ}$ along the line $\overline{\mathbf{I{A}'}}$ .

$D = \sqrt{{d^{2} cos^{2} t - d^{2}+1}} - d \cos{t}$

Step 3. Calculate $a$ , the distance from $\mathbf{{A}'}$ to $\mathbf{P_{1}}$ or $\mathbf{P_{2}}$ , the tangent points of $\mathbf{{A}'}$ on $γ_{θ}$ .

$a = \sqrt{{D^{2} - r^{2}}}$

Step 4. Calculate angle $\epsilon$ , the angle between $\overline{\mathbf{O{A}'}}$ and $\overline{\mathbf{I{A}'}}$ where $\mathbf{O}$ is the centre of $\mathbf{Γ}$ and the circumcentre of $\mathbf{ABC}$ .

$\epsilon = sin^{-1}(d sin{t})$

Step 5. calculate angle $\phi$ , the angle between $\overline{\mathbf{I{A}'}}$ and $\overline{\mathbf{P_{1}{A}'}}$ .

$\phi = sin^{-1}(\frac{D}{r})$

Step 6. Note that $\mathbf{P_{3}}$ , the third point of tangency between $\mathbf{{{A}'{B}'{C}'}}$ and $γ_{θ}$ lies at an angle $t + π - \epsilon$ from $\mathbf{I}$ . The angles $t_{2}$ and $t_{3}$ from $\mathbf{I}$ to $\mathbf{{B}'}$ and $\mathbf{{C}'}$ respectively bisect the angles $\mathbf{P_{1}IP_{3}}$ and $\mathbf{P_{2}IP_{3}}$ respectively so:

$t_{2} = t + \frac{3\pi}{4} - \frac{\phi}{2} - \frac{\epsilon}{2}$

$t_{3} = t + \frac{\pi}{4} + \frac{\phi}{2} - \frac{\epsilon}{2}$

Step 7. Calculate $D_{2}$ and $D_{3}$ , the distances from $\mathbf{I}$ to $\mathbf{{B}'}$ and $\mathbf{{C}'}$ respectively by the formula in step 2 (using $t_{2}$ and $t_{3}$ in place of $t$ ) and then calculate $b$ and $c$ using the same method as the calculation of $a$ .

Step 8. We now have a description of $\mathbf{{{A}'{B}'{C}'}}$ . The area is $r(a+b+c)$ and the perimeter is $2(a+b+c)$ . Area and perimeter are maximised when the " $-d cos{t}$ " terms in steps 2 and 7 are maximised. If $d$ is positive, this happens when $t=\pi$ and if $d$ is negative when $t=0$ . So for small $θ$ , triangle $\mathbf{ABC}$ is of larger area than any other $\mathbf{{{A}'{B}'{C}'}}$ . For large $θ$ , triangle $\mathbf{ABC}$ is smaller than any other $\mathbf{{{A}'{B}'{C}'}}$ with equality at $\theta=\frac{\pi}{6}$ where $\mathbf{ABC}$ is an equilateral triangle and every $\mathbf{{{A}'{B}'{C}'}}$ is of equal area and all have the greatest area of any triangle at any $θ$ .

Justin Travers - 12 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$