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A variable conic is inscribed in a given triangle. If the sum of the squares of its axes is constant, then the locus of its center is a circle whose center is the orthocenter of the triangle.

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    $\begingroup$ Welcome to Math.SE! The community prefers/expects questions to include something of what the asker knows of the problem. (What have you tried? Where did you get stuck? etc) This information helps answerers tailor their responses to best serve you, without wasting time (theirs or yours) explaining things you already know or talking over your head. (It also helps convince people that you aren't simply trying to get them to do your homework for you.) Since comments are easily overlooked, please edit your question to add details. $\endgroup$ – Blue Nov 25 '19 at 13:04
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Not an answer, just a start to stimulate some work on this interesting problem.

I'll show first of all how to construct an ellipse inscribed into a triangle. Given a triangle $ABC$, we can choose at will two tangency points (for instance $P$ on side $AB$ and $Q$ on side $BC$, as in figure below). If $M$ is the midpoint of $PQ$, then the center $O$ of the ellipse lies on line $BM$.

To construct the third tangency point, draw from $A$ a line parallel to $PQ$, intersecting line $BM$ at $L$. Tangency point $R$ is then the intersection between line $QL$ and side $AC$. We can then construct $N$, midpoint of $QR$: center $O$ of the ellipse is the intersection between lines $CN$ and $BM$.

Lines $OB$ and $OD$ (parallel to $PQ$) give the directions of a couple of conjugate diameters. Semi-diameters $OF$ and $OG$ can be found from the relations: $$ OF^2=PM\cdot OD,\quad OG^2=OM\cdot OB. $$ If $a$ and $b$ are the semi-axes of the ellipse, then the first theorem of Apollonius gives: $$ a^2+b^2=OF^2+OG^2=PM\cdot OD+OM\cdot OB. $$ One should then prove that the distance of $O$ from the orthocenter $H$ of $ABC$ depends only on the above quantity, but at the moment I have no idea of how to do that.

enter image description here

EDIT.

After some experiment with GeoGebra, I found for instance that $OH=\sqrt{a^2+b^2}$ when $ABC$ is a right triangle, for any position of $P$ and $Q$.

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  • $\begingroup$ +1. The key relation is shown in my answer's equation $(\star)$. My path to it isn't very satisfying, however. (The Apollonius relation was very helpful. Do you have a reference?) $\endgroup$ – Blue Dec 1 '19 at 11:17
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    $\begingroup$ @Blue Here's a reference to that theorem: cut-the-knot.org/Curriculum/Geometry/ConjugateDiameters.shtml $\endgroup$ – Intelligenti pauca Dec 1 '19 at 11:25
  • $\begingroup$ Nifty. I had a passing knowledge of conjugate diameters in a projective context, but I'm not sure I ever knew their metric properties. $\endgroup$ – Blue Dec 1 '19 at 11:31
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Here's an ugly coordinate algebra approach. I suspect there must be a much more straightfoward way to show the key result $(\star)$.


In $\triangle ABC$ let $D$, $E$, $F$ be the points of tangency of the conic as shown in the figure. (This argument is constructed with an ellipse in mind. I think that hyperbolas are properly covered simply by allowing parameters $\alpha$, $\beta$, $\gamma$ ---and therefore also the ratios in $(4)$--- to be negative.)

enter image description here

One can show that $\overline{AD}$, $\overline{BE}$, $\overline{CF}$ concur at a point; let $P$ be that point, and let $(\alpha:\beta:\gamma)$ be its barycentric coordinates, so that $$P = \frac{\alpha A+\beta B+\gamma C}{\alpha+\beta+\gamma} \tag{1}$$ then $$D = \frac{\beta B+\gamma C}{\beta+\gamma} \qquad E = \frac{\alpha A+\gamma C}{\alpha+\gamma} \qquad F = \frac{\alpha A+\beta B}{\alpha+\beta} \tag{2}$$ As @Aretino observes, the conic's center (here, $K$) lies on the line through $A$ and the midpoint of $\overline{EF}$; likewise, the line through $B$ and the midpoint of $\overline{FD}$, and the line through $C$ and the midpoint of $\overline{DE}$. If $K$ has barycentric coordinates $(\alpha':\beta':\gamma')$, then $$(\alpha':\beta':\gamma') = \left(\overline{\beta}+\overline{\gamma}: \overline{\gamma}+\overline{\alpha}: \overline{\alpha}+\overline{\beta}\right) \quad\text{where}\;\;\overline{x}:= 1/x\tag{3}$$

Let $M$ be the midpoint of $\overline{EF}$ and let $L$ be the point on $\overline{CA}$ such that $\overline{KL}\parallel\overline{ME}$. With some symbol-crunching, we find $$\frac{|KL|}{|ME|}=\frac{|KA|}{|MA|}=\frac{ (\overline{\alpha}+\overline{\beta})(\overline{\alpha}+\overline{\gamma})}{\overline{\alpha}(\overline{\alpha}+\overline{\beta}+\overline{\gamma})}\qquad \frac{|KM|}{|MA|}=\frac{\overline{\beta}\overline{\gamma}}{\overline{\alpha}(\overline{\alpha}+\overline{\beta}+\overline{\gamma})} \tag{4}$$ We also have $$\begin{align} |ME|^2 &= \overline{\alpha}^2\;\frac{a^2(\overline{\alpha}+\overline{\beta})(\overline{\alpha} + \overline{\gamma}) +b^2 (\overline{\beta}-\overline{\gamma})(\overline{\alpha} + \overline{\beta}) +c^2(\overline{\gamma}-\overline{\beta})(\overline{\alpha} + \overline{\gamma})}{4(\overline{\alpha}+\overline{\beta})^2(\overline{\alpha} + \overline{\gamma})^2} \\[4pt] |MA|^2 &= \overline{\alpha}^2\;\frac{ - a^2(\overline{\alpha}+\overline{\beta})(\overline{\alpha}+\overline{\gamma}) + b^2 (2\overline{\alpha}+\overline{\beta}+\overline{\gamma})(\overline{\alpha} + \overline{\beta}) + c^2 (2\overline{\alpha}+\overline{\beta}+\overline{\gamma})(\overline{\alpha} + \overline{\gamma})}{ 4(\overline{\alpha}+\overline{\beta})^2(\overline{\alpha}+\overline{\gamma})^2 } \end{align}\tag{5}$$

where $a:=|BC|$, $b:=|CA|$, $c:=|AB|$. These calculations are relevant, with another nod to @Aretino (and Apollonius), because $$\begin{align} s^2 &:= (\text{maj.radius})^2 + (\text{min.radius})^2 \\[4pt] &= |KL||ME|+|KM||KA| \\[4pt] &= \frac{|KL|}{|ME|}|ME|^2 + \frac{|KM|}{|MA|}\frac{|KA|}{|MA|} |MA|^2 \\[4pt] &= \frac{a^2\overline{\alpha}+b^2\overline{\beta}+c^2\overline{\gamma}}{4(\overline{\alpha}+\overline{\beta}+\overline{\gamma})} -\frac{a^2\overline{\beta}\overline{\gamma} +b^2\overline{\gamma}\overline{\alpha} +c^2\overline{\alpha}\overline{\beta}}{4(\overline{\alpha}+\overline{\beta}+\overline{\gamma})^2} \tag{6} \end{align}$$

To relate this with the locus of the conic's center, we can position $\triangle ABC$ with its orthocenter $O$ at the origin; eg, $$A = 2r\,(0,\cos A) \quad B = 2r \cos B\,(-\sin C, -\cos C) \quad C = 2 r \cos C\,(\sin B,-\cos B) \tag{7}$$ where $r$ is the circumradius of $\triangle ABC$. Then, since $$K = \frac{\alpha'A+\beta'B+\gamma'C}{\alpha'+\beta'+\gamma'} \tag{8}$$ with $\alpha'$, $\beta'$, $\gamma'$ from $(3)$, we find, after much more symbol-crunching, that $$k^2 := |OK|^2 = 4r^2 - \sum_{cyc} a^2 \frac{(\overline{\alpha}+2\overline{\beta}+\overline{\gamma} )(\overline{\alpha}+\overline{\beta}+2\overline{\gamma}) }{4 (\overline{\alpha}+\overline{\beta}+\overline{\gamma})^2} \tag{9}$$

Therefore,

$$k^2 - s^2 = 4r^2 - \tfrac12\left(a^2+b^2+c^2\right) \tag{$\star$}$$

Since the right-hand side of $(\star)$ is a constant, we have that the variable conic's center has a fixed distance from the orthocenter if and only if the sum of the squares of the conic's radii is itself fixed.

Note that this doesn't say that the locus of the center is a full circle about the orthocenter. For that, we must convince ourselves that (barring degeneracies) any point is the center of some tangent conic. But this is clear: relation $(3)$ tells us that, given a center with barycentric coordinates $(\alpha':\beta':\gamma')$, we can solve for the coordinates of a corresponding $P$ to get $$(\alpha:\beta:\gamma)= \left(\frac1{-\alpha'+\beta'+\gamma'}:\frac1{\alpha'-\beta'+\gamma'}:\frac1{\alpha'+\beta'-\gamma'}\right)$$ Cevian lines through this $P$ meet the side-lines of $\triangle ABC$ at the points of tangency for the conic. $\square$

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  • $\begingroup$ Fine work! I'd be surprised if a simpler proof would appear. $\endgroup$ – Intelligenti pauca Dec 1 '19 at 11:35

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