Clawson point

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In Euclidean geometry, the Clawson point is a special point in a triangle defined by the trilinear coordinates $tan α : tan β : tan γ$ ,^[1] where $α, β, γ$ are the interior angles at the triangle vertices $A, B, C$ . It is named after John Wentworth Clawson, who published it 1925 in the American Mathematical Monthly. It is denoted X(19) in Clark Kimberling's Encyclopedia of Triangle Centers.

The Clawson point, the orthocenter, the mittenpunkt and the Spieker center are collinear.^[1]

Construction 1

There are at least two ways to construct the Clawson point, which also could be used as coordinate free definitions of the point. In both cases you have two triangles, where the three lines connecting their according vertices meet in a common point, which is the Clawson point.

Construction 1: Clawson point as a homothetic center
  Reference triangle $△ ABC$

  Extended sides of $△ ABC$

  Excircles of $△ ABC$

  Orthic triangle $△ H A H B H C$ , and triangle $△ T A T B T C$ , internally tangent to each pair of excircles

  Perspective lines between $△ H A H B H C$ and $△ T A T B T C$ , which are similar; meet at the homothetic center $P$ (the Clawson point)

For a given triangle $△ ABC$ , let $△ H A H B H C$ be its orthic triangle and $△ T A T B T C$ the triangle formed by the outer tangents to its three excircles. These two triangles are similar and the Clawson point is their center of similarity, therefore the three lines $T A H A, T B H B, T C H C$ connecting their vertices meet in a common point, which is the Clawson point.^[2]^[3]

Construction 2

Construction 2: Clawson point as a center of perspective
  Reference triangle $△ ABC$

  Extended sides of $△ ABC$

  Circumcircle and excircles of $△ ABC$

  Triangle $△ A'B'C',$ formed by lines connecting pairs of intersection points between each excircle and the circumcircle

  Perspective lines between $△ ABC$ and $△ A'B'C',$ meeting at the perspective center $P$ (the Clawson point)

For a triangle $△ ABC$ , its circumcircle intersects each of its three excircles in two points. The three lines through those points of intersections form a triangle $△ A'B'C'.$ This triangle and $△ ABC$ are perspective triangles with the Clawson point being their perspective center. Hence the three lines $AA', BB', CC'$ meet in the Clawson point.^[1]

History

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