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I am an artist and want a more formulaic way of finding where to place perspective circles (ellipses) inside perspective squares in 3-point perspective (quadrilaterals where none of the sides are parallel). I asked another question regarding quadrilaterals where two sides are parallel (trapezoids) here.

Here is an example of an ellipse inside a quadrilateral. Points E2 and D2 are the Foci. Point H2 is the 'perspective' center of the quadrilateral.

The four tangent points must be the perspective centers of each side (Line I2, U1 & the line intersecting H2 that isn't I2, U1).

A 3 point quadrilateral with an ellipse inside

How do I find where to place an ellipse inside a scalene quadrilateral such that the ellipse is tangent to the 4 perspective centers of each side of the quadrilateral?

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  • $\begingroup$ Are tangency points given? $\endgroup$
    – Intelligenti pauca
    Commented Sep 10, 2020 at 6:52
  • $\begingroup$ In the cases I am dealing with I will know the tangency points. $\endgroup$
    – Audus
    Commented Sep 10, 2020 at 7:08

1 Answer 1

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Let $ABCD$ be the given (convex) quadrilateral, and $P$, $E$, $Q$ the tangency points of the ellipse on sides $AD$, $DC$, $CB$ respectively (see figure below).

The line joining the intersection point of two tangents with the midpoint of their tangency points, passes through the centre of the ellipse. If $G$ and $K$ are the midpoints of $PE$ and $QE$, we can then find the centre $O$ of the ellipse as the intersection between lines $DG$ and $CK$.

Once the centre has been found, we can construct point $F$ reflecting $E$ about $O$, thus finding a first diameter $EF$ of the ellipse. The diameter conjugate to $EF$ is $LM$, parallel to tangent $CD$ and with $$ OL=OM={PH\cdot EO\over\sqrt{EO^2-HO^2}}, $$ where $H$ is the intersection of $EF$ with the line through $P$ parallel to $CD$.

Having constructed a pair of conjugate diameters $EF$ and $LM$, we can find the axes of the ellipse following the construction given here.

enter image description here

EDIT.

If you don't need to construct the axes of the ellipse, but just want to find some other points on it, then it is easy to construct the four points of the ellipse lying on the lines joining the center of the ellipse with the vertices of the quadrilateral. If $S$, for instance, is the intersection of the ellipse with $OD$, then $OS$ is the mean proportional between $OD$ and $OG$: $$ OS^2=OD\cdot OG. $$

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  • $\begingroup$ Thank you. This technique of finding the midpoint of the ellipse is so useful! $\endgroup$
    – Audus
    Commented Sep 11, 2020 at 5:06
  • $\begingroup$ Is there a rule and compass method for determining points M and L $\endgroup$
    – cdosborn
    Commented Oct 7, 2023 at 4:29
  • $\begingroup$ @cdosborn Of course: construct $\sqrt{EO^2-HO^2}$ using Pythagoras' theorem and then you can construct $OL$ and $OM$ as a fourth proportional $\endgroup$
    – Intelligenti pauca
    Commented Oct 7, 2023 at 12:20
  • $\begingroup$ @cdosborn This could also be of interest: math.stackexchange.com/questions/3821917/… $\endgroup$
    – Intelligenti pauca
    Commented Oct 7, 2023 at 13:57
  • $\begingroup$ I will try this ty! $\endgroup$
    – cdosborn
    Commented Oct 8, 2023 at 15:45

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