[1] Through every point P inside the deltoid pass three tangents to it. These are the Wallace lines {W

[2] The sides {RS,ST,TR} are respectively orthogonal to the Wallace lines {W

These facts are proved in Orthopole2.html . Triangle RST is called an

Inversely, for every system of points {T,R,S} on the circumcircle satisfying this relation, the corresponding triangle TRS is an S-triangle i.e. the Wallace lines {W

The proof of the first assertion follows trivially from the way the Wallce line, W

To prove the inverse note that the condition implies the orthogonality of RS and W

This property has a very important corollary since the relation on the three arcs is symmetric with respect to the two triangles ABC and RST.

If a Wallace line W

This is

Triangle RST is an S-triangle of ABC if and only if the Wallace lines {W,W'} of the two triangles for every point Q of their circumcircle are parallel.

This follows from the arc relation of the previous section. For this draw two parallel lines {AA',TT'} and the orthogonals to {BC,RS} respectively from {A',T'}. Use the arc relation to show that these lines intersect at a point Q of the circumcircle. This implies that the Wallace lines {W,W'} of the two triangles with respect to Q are parallel. The argument can be reversed, even in a stronger form to prove:

[1] The Wallace line of T w.r to RST is the altitude parallel to the Wallace line W

[2] The orthocenters {H,H'} of ABC and RST are symmetric with respect to P.

[3] The Wallace lines of points {A,B,C} with respect to triangle RST pass also through P, the point defining RST.

First property follows from the fact that the altitude of every triangle is the Wallace line of the vertex on that altitude. Thus, this altitude and W

The symmetry of the orthocenters is a consequence of the other two properties and the fact that HT is bisected by the corresponding Wallace line W

This equilateral triangle was constructed and discussed in Deltoid.html . It results by taking first {A'',B'',C''} to be the intersections of the medial lines respectively to {BC,CA,AB} with the circle lying on the same side with the opposite vertices {A,B,C}. Then taking on the oriented arcs {A''A,B''B,C''C} points {A',B',C'} dividing these arcs in ratio 1:2. An easy angle chasing argument shows that these points define indeed an equilateral triangle and that the arc relation holds true: AA' + BB' + CC' = 0.

Since two different equilaterals inscribed in the same circle cannot satisfy the above condition we conclude that:

This property follows immediately from the way these deltoids are generated as hypocycloids discussed in Deltoid.html . The orthopoles {X,Y} are the contact points of the parallel Wallace lines of Q with respect to the two triangles. Hence they are corresponding points under this translation.

There is a particular position of Q on the circumcircle for which the two corresponding parallel Wallace lines with respect to ABC and RST

[2] Goormaghtigh, R.

[3] Honsberger, R.

[4] Lalesco, T.

[5] Ramler, J. O.

[6] Ramler, J. O.

Orthopolar.html

Orthopole.html

Orthopole2.html

SimsonProperty.html

SteinerLine.html

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