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Problem 451 from Connelly's corpus: Ec, Hb, Mb


Problem

Given a point Ec, a point Hb and a point Mb, construct the triangle ABC.

Status

ArgoTriCS says that the problem is solvable.

Illustration

Figure of construction

Solution


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

    % DET: points Hb and Mb are not the same

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

    % NDG: points Hb and Ec are not the same

  3. Using the circle k(Ec,C), the line b, the point Ec and the point Hb, construct a point C (rule W05);

    % NDG: line b and circle k(Ec,C) intersect

    % DET: points Hb and C must be different

  4. Using the point C and the point Ec, construct a point H (rule W01);

  5. Using the point Mb and the point C, construct a point A (rule W01);

  6. Using the point Hb and the point H, construct a line hb (rule W02);

    % DET: points Hb and H are not the same

  7. Using the point H and the point A, construct a line ha (rule W02);

    % DET: points H and A are not the same

  8. Using the circle k(Ec,C), the line ha, the point Ec and the point H, construct a point Ha (rule W05);

    % NDG: line ha and circle k(Ec,C) intersect

    % DET: points H and Ha must be different

  9. Using the point C and the point Ha, construct a line a (rule W02);

    % DET: points C and Ha are not the same

  10. Using the line hb and the line a, construct a point B (rule W03);

    % NDG: lines hb and a are not parallel

    % DET: lines hb and a are not the same

animation
Construction in GCLC language

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