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We can think of an incircle and three excircles of a triangle. So the question is : do exellipses exist? Inellipses do exist, and there are a number of studies on it such as the Steiner Ellipse and Marden's Theorem. However, I couldn't find any information on the 'exellipse' ; the ellipse that behaves like an excircle.

Am I getting the name wrong? I expected it to be 'exellipse', but it seems that the word does not exist at all. Also, I would like to find out whether the properties of the incircle and the excircle still holds when it comes to inellipses and 'exellipses'.

The following picture expresses both the inellipse and the 'exellipse' of a triangle. (Sorry for bad resolution.)

Image

Vue
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    See, for instance Wikipedia's "Circumconic and inconic" entry. Once you generalize circles to ellipses, it's not long before parabolas and hyperbolas get added to the mix; at that point, the "in/ex" distinction starts to fade away, leaving only the essential property of tangency with the side-lines of the triangle, and we just use "inconic" as the catch-all term, regardless of how the curves relate to the interior of the triangle. – Blue Feb 21 '22 at 12:17
  • @Blue understood. Are there any papers or theorems related to the picture? – Vue Feb 21 '22 at 13:29
  • @Vue You could export your picture to clipboard in Geogebra, and then paste to any other graphical software and save as different file types you wanted. – Ng Chung Tak Feb 22 '22 at 00:21

2 Answers2

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Using skew (axes) transformation,

$$(x',y')=(x+y\cos \omega,y\sin \omega)$$

You could always generate a two-parameter family of ellipse touching the three lines defining $\Delta ABC$,

$$ \left( \frac{x}{\lambda a}+\frac{y}{\mu b}-1 \right)^2=\frac{4xy}{ab} \left( \frac{1}{\lambda}-1 \right) \left( \frac{1}{\mu}-1 \right)$$

which touches the axes at $A(\lambda a,0)$ and $B(0,\mu b)$ and the line $\dfrac{x}{a}+\dfrac{y}{b}=1$.

For touching ellipse, $$(1-\lambda)(1-\mu)(\lambda+\mu-\lambda \mu)>0$$

In particular, $\lambda,\mu \in (0,1)$ gives inscribed ellipse.

enter image description here

Ng Chung Tak
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  • Thanks, is there a paper about it? Or did you do it by yourself? Since I may need official reference. – Vue Feb 22 '22 at 16:48
  • The technique is quite standard, you may look up some old texts for analytical geometry in library or digital archive in the internet. You can rederive on you own from $$(\alpha x+\beta y-1)^2=\lambda x y$$ and checking tangency with discriminant and so on. – Ng Chung Tak Feb 22 '22 at 22:07
  • You may search different answers from mine or other users and feel free to cite my own answer. Click "cite" under the answer to see the link. – Ng Chung Tak Feb 22 '22 at 22:14
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You can choose at will tangency points $L$ and $M$ on the prolongations of $AB$ and $AC$: tangency point $R$ is then uniquely determined by the proportion $RC:RB=FC:FB$, where $F$ is the intersection of line $BC$ with line $LM$. This follows from a nice property of conics:

If two tangents be drawn to a conic, any third tangent is harmonically divided by the two tangents, their chord of contact, and the point in which it touches the curve.

The centre $O$ of the ellipse can then be found as the intersection of $CH$ and $BK$, where $H$ and $K$ are the midpoints of $RL$ and $RM$.

Note that for some positions of $L$ and $M$, lines $CH$ and $BK$ become parallel (in that case we get a parabola) or meet on the opposite side of line $BC$, in which case the ex-inscribed conic is no more an ellipse but a branch of hyperbola.

enter image description here

EDIT.

Construct point $A'$ such that $ABA'C$ is a parallelogram. If the conic is a parabola, then one can prove that contact chord $LM$ passes through $A'$. This gives a useful criterion for the kind of conic we are going to obtain: if $A$ and $A'$ lie on opposite sides of line $LM$ then the conic is an ellipse, if they lie on the same side then the conic is a hyperbola.

Intelligenti pauca
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