I was immensely surprised and amused by the idea of the fourth side of a triangle that was introduced by B.F.Sherman in 1993. 'Sherman's Fourth Side of a Triangle' by Paul Yiu is available here. Naturally a question arises whether it is possible to determine the '4th vertex' of a triangle in a somewhat similar way?

Basically a vertex of the given triangle can be described as a point that is lying on its circumcircle. Then the middle point of the segment connecting this point to the Orthocenter must belong to the nine point circle of the triangle. (This holds true for any point that is lying on the circumcircle.) Finally it appears that we should just add another equation somehow linking our 'vertex' to the inscribed circle. In other words, there might exist some triangle center on the circumcircle that is also satisfying a certain condition related to the inscribed circle, so that eventually this point can be called the fourth vertex of the triangle.

Luckily I found in my files a construction of a triangle center X that in a sense satisfies these conditions:

  • It belongs to the circumcircle of ABC
  • The midpoint of the segment XX(4) belongs to the nine point circle.
  • The constriction of X primarily relies on the Incenter of the triangle (i.e. the inscribed circle)

Last but not least, this point X is not included in Kimberling's encyclopaedia:

A',B',C' is the circumcevian triangle with respect to the Incenter I. Lines AB and A'B' intersect at point C'', points B'', A'' are defined cyclically. Circumcircles for the triangles IRA, IRB, IRC were drawn. These 3 circles intersect the circumcircle at the points A''', B''', C'''. Finally A''A''',B''B''',C''C''' always cross each other at some point X that conveniently belongs to the circumcircle of the original triangle ABC.

enter image description here

Geogebra dynamic sketch.

It is highly speculative to call our point X the fourth vertex of a triangle, of course. Presumably this attribution will be outright dismissed or proven wrong. However I assume that perhaps a better fit for the 4th vertex of a triangle can be found? What would its construction be?

  • $\begingroup$ Looking at the picture and contemplating over my ability to contribute here, I came to think of the "Find X - Here it is"-joke. $\endgroup$ Nov 30, 2021 at 21:15

1 Answer 1


I propose an alternative "fourth vertex".

To paraphrase Sherman's result:

The "fourth side" ($w$) of a triangle is a chord of the circumcircle, is a tangent to the incircle ($\bigcirc I$), and is bisected by the nine-point circle ($\bigcirc N$). (The last aspect is equivalent to $w$ intersecting the nine-point circle at the foot of the perpendicular from circumcenter $K$.)

Thus, this "fourth side" fits a description that applies to the (standard) three; importantly, there is exactly one such "fourth side" that can do so.

enter image description here

My proposal:

The "fourth vertex" ($W$) of a triangle determines a line through the orthocenter ($H$) that intersects the nine-point circle at the foot of a perpendicular tangent to the incircle, and is such that the midpoint of $\overline{WH}$ is the "other" intersection with the nine-point circle. (The last part distinguishes $W$ from the other endpoint of the chord determined by $\overleftrightarrow{WH}$.)

Note. As observed in comments, point $W$ is Kimberling's triangle center X(1309).

As before, the ostensible "fourth vertex" fits a description that applies to the (standard) three; also, as it turns out, there is exactly one "fourth vertex" that can do so.

The simplicity of this proposed definition has some appeal, but what makes it particularly relevant is a fact the reader may have suspected from the images: The two "fourth" elements involve the same construction. The line of segment $w$ serves as the tangent to $\bigcirc I$ perpendicular to $\overline{WH}$ where they meet on $\bigcirc N$; that is, Sherman could have —and may have (I haven't checked)— noted:

$w$ intersects the nine-point circle again at the foot of the perpendicular from orthocenter $H$.

Proof isn't complicated (essentially all that is needed is to rotate some elements around the center of the nine-point circle), but getting too caught-up in the triangle context is unnecessarily limiting. It's better to see these "three elements and a spare" results as special cases of a broader "four elements" result.

Lemma. Suppose tangents at $A$, $B$, $C$ of $\bigcirc P$ meet $\bigcirc Q$ at feet ($A_+$, $B_+$, $C_+$) of perpendiculars from a common point $R_+$. Then there is a unique point $D$ of $\bigcirc P$ such that the tangent at $D$ meets $\bigcirc Q$ at the foot ($D_+$) of the perpendicular from $R_+$.

Moreover, reflecting $R_+$ in $Q$ gives a point $R_-$ such that the feet ($A_-$, $B_-$, $C_-$, $D_-$) of perpendiculars to the tangents lie on $\bigcirc Q$.

(In the triangle context, $\bigcirc P$ and $\bigcirc Q$ are respectively the incircle and nine-point circle; lines $\overleftrightarrow{A_+A_-}$, etc, contain the "four sides"; and points $R_+$ and $R_-$ are the circumcenter and orthocenter.)

enter image description here

The "moreover" follows from recognizing that a $180^\circ$ rotation about $Q$ effectively creates four inscribed rectangles. This rotational symmetry guarantees that sides of these rectangles concur iff the opposite sides concur.

The main part of the Lemma can be proven with some light coordinate bashing. For instance, taking $P=(0,0)$ and $R_+=(r,0)$, defining angles $\alpha:=\angle R_+PA$, $\beta:=\angle R_+PB$, $\gamma:=\angle R_+PC$, $\delta:=\angle R_+PD$, and letting the radius of $\bigcirc P$ be $p$, then $$A_+ = R_+ + (p - r \cos\alpha) (\cos\alpha,\sin\alpha), \qquad B_+=\cdots, \qquad C_+=\cdots, \qquad D_+=\cdots$$ We find that $A_+$, $B_+$, $C_+$, $D_+$ are concyclic iff $$2 p \sin\sigma = r \left(\;\sin(\sigma-\alpha) +\sin(\sigma-\beta) + \sin(\sigma-\gamma) + \sin(\sigma-\delta)\;\right) \tag{1}$$ where $\sigma:=\frac12(\alpha+\beta+\gamma+\delta)$. Condition $(1)$ determines any one angle from the other three; eg, we can write $$\begin{align} &\;\phantom{-}\cos\tfrac12\delta\left(\; 2 p \sin\tau - r ( \sin(\tau-\alpha) + \sin(\tau-\beta) + \sin(\tau-\gamma) + \sin\tau) \;\right) \\[0.5em] = &-\sin\tfrac12\delta \left(\; 2 p \cos\tau - r (\cos(\tau-\alpha) + \cos(\tau-\beta) + \cos(\tau-\gamma) - \cos\tau)\;\right) \end{align} \tag{1'}$$ where $\tau:=\frac12(\alpha+\beta+\gamma)$. This identifies $\delta/2$ up to a half-turn, hence $\delta$ up to a full turn, guaranteeing the unique $D$ claimed in the Lemma. $\square$

It is perhaps worth noting that (barring degeneracies) $(1)$ allows for determining distance $r=|PR_+|$ to make any chosen collection of distinct angles (ie, any chosen collection of points $A$, $B$, $C$, $D$ on $\bigcirc P$) "work" to give the concurrencies shown in the figure.

  • $\begingroup$ @A.Zakharov, are you sure? My reading of the Yiu paper you link is that X(953) is "the other endpoint of the chord determined by $\overleftrightarrow{WH}$". $\endgroup$ Aug 16, 2021 at 13:25
  • 1
    $\begingroup$ @PeterTaylor Sherman's line is the perpendicular to the line X(1309)X(953) drawn through the midpoint of the segment X(4)X(953). The "other" intersection with the nine-point circle is the midpoint of the segment X(1309)X(4). So X(1309) is the point W, indeed. And finally X(953) is the Miquel point of the sidelines of ABC and the Sherman line. (Angel Montesdeoca, July 24, 2019) What will be the Miquel point of the quadrilateral ABCX(1309) then?.. $\endgroup$
    – A.Zakharov
    Aug 16, 2021 at 14:24
  • 1
    $\begingroup$ @A.Zakharov, aha, yes. Kimberling's Encyclopedia says "X(1309) lies on the circumcircle and these lines: 4,953", which would indeed make it W. $\endgroup$ Aug 16, 2021 at 14:38

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