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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 circumcenter.) 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.

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 this 4th vertex of a triangle can be found? What would its construction be?

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.

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?