Two parallel lines are tangent to an ellipse. Between those two lines, every line parallel to those two intersects the ellipse in two points.

The precise midpoint between those two points lies exactly on the line connecting the two points of tangency.

My question here is whether that last proposition is true for every pair of parallel tangents only for the ellipse and no other shape.

PS: For present purposes, let's define "tangent" as "touching but not crossing the boundary." Then an oblique line through the corner of a rectangle is a "tangent", and we see that the rectangle is NOT an example of another shape with the property of interest.

PPS: ok, Let's refine the statement of the problem a bit. Consider a closed bounded set with non-empty interior in the plane. Suppose it is strictly convex, i.e. every point between two of its points is one of its interior points. This entails that a line that intersecting its boundary but not its interior intersects it at only one point. Call such a line a tangent line. It follows that for every tangent line, there is exactly one other tangent line parallel to it. Suppose that for every line parallel to those two and between them, the midpoint of the intersection of that line with our closed bounded convex set is on the line connecting the two points of tangency.

Does it follow that our closed bounded set is the convex hull of an ellipse?

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    $\begingroup$ Does the property hold for any two parallel tangents of the shape? $\endgroup$ Aug 20, 2020 at 5:07
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    $\begingroup$ What about a rhombus ? $\endgroup$
    – Spectre
    Aug 20, 2020 at 5:19
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    $\begingroup$ I think you have to restrict to convex shape with smooth boundary. Or a rectangle is another example. $\endgroup$ Aug 20, 2020 at 5:22
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    $\begingroup$ Just to mention it: If we replace "Between those two lines, every line parallel to those two [tangents] intersects the ellipse in two points." with "Every secant line parallel to those two [tangents] intersects the ellipse in two points.", then the property holds for hyperbolas. Likewise, an appropriately-massaged variant holds for parabolas, as well. ... That said, I suspect OP's use of "between" is intentional, and speaks to a specific interest in bounded curves. $\endgroup$
    – Blue
    Aug 20, 2020 at 5:44
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    $\begingroup$ @g.kov : This doesn't come close to working with a Reuleaux triangle. $\qquad$ $\endgroup$ Aug 21, 2020 at 18:06

1 Answer 1


Every closed convex piecewise-differentiable curve with the given tangent property is an ellipse.

Proof: The problem is affine, in the sense that if a curve has the given property then so does any affine transformation of it. So, starting with a pair of tangents at the widest extent of the curve, use a rotation to make the tangents vertical and a shear to bring the curve to $\mathcal{C}$ whose line of symmetry is the $x$-axis.

enter image description here $\hspace{2cm}\mapsto\hspace{2cm}$ enter image description here

Now take the horizontal pair of tangents on $\mathcal{C}$, meeting it at two points one vertically above the other. Translate it so this vertical line is the $y$ axis. Then $\mathcal{C}$ is symmetric about both the $x$ and $y$ axes. Scaling along these axes brings their intercepts to $1$. Every other point has radius at most $1$, by the way the original tangents were chosen.

Proposition 1. $\mathcal{C}$ is balanced, i.e., $x\in \mathcal{C}\implies -x\in\mathcal{C}$.

This follows directly from the symmetry along the two perpendicular axes.

Hence given any pair of tangents, the line joining the points of contact passes through the origin.

Proposition 2. The curve is differentiable.

Join opposite corners by a line through the origin. Then $\mathcal{C}$ would have equal distances from this line along two sets of parallel lines, which gives a contradiction.

$\hspace{4cm}$enter image description here

Proposition 3. Any point on $\mathcal{C}$ with radius $1$ has a perpendicular tangent.

A point with the maximum radius $r(\theta)=1$ must have $r'=0$.

Proposition 4. If $OA$ and $OB$ have radii of $1$ then so does their angle bisector $OC$.

The tangent parallel to $AB$ touches the curve at some point $C$. The line $OC$ cuts $AB$ in half by hypothesis and is thus the median and angle bisector of $AOB$, and perpendicular to $AB$. Thus $\mathcal{C}$ is symmetric about $OC$ and so the tangent at $C$ is perpendicular to $OC$.

$\hspace{3cm}$enter image description here

Let the tangent at $C$ meet the tangent at $A$ at the point $P$. Consider tangents parallel to $AC$ and the line $Q'OQ$ joining the opposite tangents. This line passes through the midpoint of $AC$ by hypothesis. In the limit, nearby points $A'$ on $AP$ and $C'$ on $CP$ with $A'C'$ parallel to $AC$ are also bisected by $OQ$ since $AP$ and $CP$ are tangents to $\mathcal{C}$. But this means that $OQ$ is the median of $APC$, and thus $Q$ is on $OP$. Since $OAPC$ is a cyclic quadrilateral with diameter $OP$, the bisected chord $AC$ is perpendicular to $OP$ and so $OC=OA=1$.

Proposition 5. $\mathcal{C}$ is a circle.

Since the $x$ and $y$ intercepts have radius $1$, one can keep taking the angle bisectors, forming a dense set of points of radius $1$. By continuity, all points have the same radius.

Hence the original curve is an affine transformation of a circle, namely an ellipse.

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    $\begingroup$ In the paragraph before Proposition 1 you wrote "Then $\mathcal C$ is symmetric about both the $x$ and $y$ axes." How did you reach that conclusion based only on the curve having the given propperties? $\endgroup$ Aug 23, 2020 at 17:56
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    $\begingroup$ For any two tangents, the shape is symmetric along a skew-reflection. $\mathcal{C}$ is not the original curve but a rotated and sheared version. However it still satisfies the property of the question and thus the reflection property. $\endgroup$ Aug 23, 2020 at 18:35
  • $\begingroup$ ok, Now we (or at least I) face the question of what to make of Raffaele's comment under the question. $\qquad$ $\endgroup$ Aug 25, 2020 at 19:09
  • $\begingroup$ $\ldots\,$and some numerical work and graphics seems to turn up a counterexample to Raffaele's claim$\,\ldots\qquad$ $\endgroup$ Aug 25, 2020 at 19:38
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    $\begingroup$ @MichaelHardy The first paragraph is a technique - to 'simplify' a problem by transformations. The rest I thought of in bed over three dawns :) $\endgroup$ Aug 26, 2020 at 5:38

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