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Problem 460 from Wernick's corpus: Mc, Hb, Tc


Problem

Given a point Mc, a point Hb and a point Tc, construct the triangle ABC.

Status

ArgoTriCS says that the problem is solvable.

Illustration

Figure of construction

Solution


  1. Using the point Mc and the point Tc, construct a line c (rule W02);

    % DET: points Mc and Tc are not the same

  2. Using the point Hb and the point Mc, construct a circle k(Mc,A) (rule W06);

    % NDG: points Hb and Mc are not the same

  3. Using the circle k(Mc,A) and the line c, construct a point A and a point B (rule W04);

    % NDG: line c and circle k(Mc,A) intersect

  4. Using the point Hb and the point A, construct a line b (rule W02);

    % DET: points Hb and A are not the same

  5. Using the point A, the point B, the point Tc and the line c, construct a point T`c (rule W19);

    % NDG: points A and B are not the same points B and Tc are not the same points B and midpoint([A,Tc]) are not the same

  6. Using the point Tc and the point T`c, construct a circle k_over(Tc,T`c) (rule W09);

    % NDG: points Tc and T`c are not the same

  7. Using the circle k_over(Tc,T`c) and the line b, construct a point Cwa and a point C (rule W04);

    % NDG: line b and circle k_over(Tc,T`c) intersect

animation
Construction in GCLC language

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