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 θ[0,π2)\theta \in [0, \frac{\pi}{2}) like so:

Let Γ\Gamma be the unit circle centred at O=(0,0)O = (0,0). Call the point (1,0)A(1,0) A. Pick a point BB on the upper half of Γ\Gamma, such that OAB=θ\angle OAB = \theta. Reflect BB about the x-axis; call this point CC. Let the incircle of ABC\triangle ABC be γθ\gamma_\theta; its radius is r=2sinθ+cos2θ1r = 2 \sin \theta + \cos 2 \theta - 1, and it is centred at (2sinθ1,0)(2 \sin \theta - 1, 0).

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

Note by Jake Lai
4 months, 2 weeks ago

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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=2sinθd=2 \sin{θ} as the distance between the centres of Γ\mathbf{Γ} and γθγ_{θ}.

Step 1. choose a parameterisation. I chose tt to be the angle between the x-axis and a line from I\mathbf{I}, the centre of γθγ_{θ} (and the incentre of ABC\mathbf{ABC}) to an arbitrary point on Γ\mathbf{Γ} : A\mathbf{{A}'}. The task is then to define a new triangle ABC\mathbf{{{A}'{B}'{C}'}} based on this parameter tt.

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

D=d2cos2td2+1dcostD = \sqrt{{d^{2} cos^{2} t - d^{2}+1}} - d \cos{t}

Step 3. Calculate aa, the distance from A\mathbf{{A}'} to P1\mathbf{P_{1}} or P2\mathbf{P_{2}}, the tangent points of A\mathbf{{A}'} on γθγ_{θ}.

a=D2r2a = \sqrt{{D^{2} - r^{2}}}

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

ϵ=sin1(dsint)\epsilon = sin^{-1}(d sin{t})

Step 5. calculate angle ϕ\phi, the angle between IA\overline{\mathbf{I{A}'}} and P1A\overline{\mathbf{P_{1}{A}'}}.

ϕ=sin1(Dr)\phi = sin^{-1}(\frac{D}{r})

Step 6. Note that P3\mathbf{P_{3}}, the third point of tangency between ABC\mathbf{{{A}'{B}'{C}'}} and γθγ_{θ} lies at an angle t+πϵt + π - \epsilon from I\mathbf{I}. The angles t2t_{2} and t3t_{3} from I\mathbf{I} to B\mathbf{{B}'} and C\mathbf{{C}'} respectively bisect the angles P1IP3\mathbf{P_{1}IP_{3}} and P2IP3\mathbf{P_{2}IP_{3}} respectively so:

t2=t+3π4ϕ2ϵ2t_{2} = t + \frac{3\pi}{4} - \frac{\phi}{2} - \frac{\epsilon}{2}

t3=t+π4+ϕ2ϵ2t_{3} = t + \frac{\pi}{4} + \frac{\phi}{2} - \frac{\epsilon}{2}

Step 7. Calculate D2D_{2} and D3D_{3}, the distances from I\mathbf{I} to B\mathbf{{B}'} and C\mathbf{{C}'} respectively by the formula in step 2 (using t2t_{2} and t3t_{3} in place of tt) and then calculate bb and cc using the same method as the calculation of aa.

Step 8. We now have a description of ABC\mathbf{{{A}'{B}'{C}'}}. The area is r(a+b+c)r(a+b+c) and the perimeter is 2(a+b+c)2(a+b+c). Area and perimeter are maximised when the "dcost-d cos{t}" terms in steps 2 and 7 are maximised. If dd is positive, this happens when t=πt=\pi and if dd is negative when t=0t=0. So for small θθ, triangle ABC\mathbf{ABC} is of larger area than any other ABC\mathbf{{{A}'{B}'{C}'}}. For large θθ, triangle ABC\mathbf{ABC} is smaller than any other ABC\mathbf{{{A}'{B}'{C}'}} with equality at θ=π6\theta=\frac{\pi}{6} where ABC\mathbf{ABC} is an equilateral triangle and every ABC\mathbf{{{A}'{B}'{C}'}} is of equal area and all have the greatest area of any triangle at any θθ.

Justin Travers - 3 months, 2 weeks ago

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