Find the triangle(s) formed by nine arbitrary concyclic points on a circle

Question

I have read a lot about the nine point circle. Finding the nine point circle of a triangle is straightforward. But what about going the other way?

Let $w$ be a circle. Given nine arbitrary concyclic points on $w$, how can one find the triangle (if it exists) for which $w$ is the nine-point circle?

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In other words, let's say there exists a triangle, $t$, where

• $P$, $Q$, and $R$ are the midpoints of each side of $t$
• $P'$, $Q'$, and $R'$ are the foot of each altitude of $t$
• $P''$, $Q''$, and $R''$ are the midpoints of the line segments from each vertex of $t$ to the orthocenter
• $P$, $Q$, $R$, $P'$, $Q'$, $R'$, $P''$, $Q''$, and $R''$ all lie on circle, $c$

Given $P$, $Q$, $R$, $P'$, $Q'$, $R'$, $P''$, $Q''$, $R''$, and $c$, find $t$

Note: there may exist multiple triangles which fit the criteria for $t$, or there may exist no triangles which fit the criteria for $t$. Thus, there are two halves to the problem:

a) Determine of a triangle $t$ exists whose significant points are $P$, $Q$, $R$, $P'$, $Q'$, $R'$, $P''$, $Q''$, and $R''$

b) If $t$ exists, find the vertices of $t$

Answer 1

I've significantly re-worked this answer. For a previous version, see the Edit History.

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"The" nine points of a triangle fall into three types:

1. the three midpoints of its sides ($P$, $Q$, $R$ in the figure),
2. the three points ($P'$, $Q'$, $R'$) half-way between its orthocenter and its vertices (we'll call these "ortho-midpoints"), and
3. the three feet ($P''$, $Q''$, $R''$) of its altitudes.

These points naturally determine three concyclic right triangles:

By Thales' Theorem, each hypotenuse is necessarily a diameter of the nine-point circle. Each diameter has a triangle midpoint and an ortho-midpoint as its endpoints. The left-over points are the altitude feet, which determine the original triangle's orthic triangle. These elements happen to be related in an interesting way:

Fun Fact. The midpoint/ortho-midpoint diameters must be the perpendicular bisectors of the sides of the orthic triangle.

The figure shows the situation for diameter $\overline{PP'}$ and orthic triangle side $\overline{Q''R''}$:

For proof, we note that, since $\angle BQ''C$ and $\angle BR''C$ are right angles, Thales tells us that $Q''$ and $R''$ live on a circle with diameter $\overline{BC}$ (and, therefore, with center $P$). Similarly, with $G$ the orthocenter of $\triangle ABC$, we have right angles $\angle AQ''G$ and $\angle AR''G$, so that $Q''$ and $R''$ live on a circle with center $P'$. Thus, $\overline{Q''R''}$ is a chord common to the two circles, so that it must be perpendicular to, and bisected by, the line $\overline{PP'}$ connecting the centers. $\square$

The Fun Fact provides the "only if" portion of this characterization of viable nine-point sets:

Theorem. A set of nine distinct concyclic points is "the" nine-point set of some triangle if and only if it can partitioned into three diametric pairs $\{P,P'\}$, $\{Q,Q'\}$, $\{R,R'\}$ and a triad $\{P'', Q'', R''\}$ such that $$\overline{PP'}\perp\overline{Q''R''} \qquad \overline{QQ'}\perp\overline{R''P''} \qquad \overline{RR'}\perp\overline{P''Q''}$$

For the "if" part, we provide a construction a quartet of solution triangles. To begin, one can show (as @mathlove did in this answer)

Lemma. The incenter $G$ and excenters $A$, $B$, $C$ of $\triangle P''Q''R''$ form an orthocentric system such that $\triangle ABC$, $\triangle GBC$, $\triangle AGC$, and $\triangle ABG$ have altitude feet $P''$, $Q''$, $R''$. No other triangles have these altitude feet.

The four triangles determined by the orthocentric system share a common nine-point set, namely: the three altitude feet ($P''$, $Q''$, $R''$), as well as the three midpoints of $\overline{AB}$, $\overline{BC}$, $\overline{CA}$ (the midpoints of $\triangle ABC$), and the three midpoints of $\overline{GA}$, $\overline{GB}$, $\overline{GC}$ (the ortho-midpoints of $\triangle ABC$). The roles of some points change for $\triangle GBC$, $\triangle AGB$, $\triangle ABG$, but the point-set itself remains the same.

By the Fun Fact, the midpoints and ortho-midpoints determine diameters perpendicular to the sides of $\triangle P''Q''R''$. Such diameters are unique, so they must coincide the the Theorem's assumed pairs $\{P,P'\}$, etc, so that the midpoints and ortho-midpoints themselves coincide with the points $P$, $P'$, $Q$, $Q'$, $R$, $R'$. Thus, the given nine points are "the" nine points of $\triangle ABC$. $\square$

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So, given nine points, the Theorem tells us when the set is viable, and the Lemma tells us how to construct an orthocentric system that yields exactly four solution triangles.

Note that nine points can contain four diametric pairs, raising the possibility of additional solutions. What happens then?

In the case of four diametric pairs, there the altitude-feet triad must contain one of them (so that the other three remain in tact), making the orthic triangle a right triangle (by Thales, yet again). Consequently, we have two midpoint/ortho-midpoint diameters perpendicular to the legs, which makes them perpendicular to each other, while the third midpoint/ortho-midpoint diameter is perpendicular to the hypotenuse diameter. In other words: Those four diameters comprise two mutually-perpendicular pairs.

I'll leave it as an exercise to the reader to show that there is only one choice of altitude-foot triad, unless the diameters make angles of $30^\circ$ and $60^\circ$; in that case, there are two symmetric choice of triad, leading to two orthocentric systems, for a total of eight solution triangles.