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Problem 256 from Connelly's corpus: Ea, Hb, Mc


Problem

Given a point Ea, a point Hb and a point Mc, 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 Ea, construct a circle k(Ea,A) (rule W06);

    % NDG: points Hb and Ea 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(Ea,A), the circle k(Mc,A), the point Hb, the point Ea and the point Mc, construct a point A (rule W08);

    % NDG: circles k(Ea,A) and k(Mc,A) intersect

    % DET: circles k(Ea,A) and k(Mc,A) are not the same points Hb and A must be different

  4. Using the point A and the point Ea, construct a point H (rule W01);

  5. Using the point Mc and the point A, construct a point B (rule W01);

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

    % DET: points Hb and A are not the same

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

    % DET: points Mc and A are not the same

  8. Using the circle k(Ea,A), the line c, the point Ea and the point A, construct a point Hc (rule W05);

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

    % DET: points A and Hc must be different

  9. Using the point H and the point Hc, construct a line hc (rule W02);

    % DET: points H and Hc are not the same

  10. Using the line b and the line hc, construct a point C (rule W03);

    % NDG: lines b and hc are not parallel

    % DET: lines b and hc are not the same

animation
Construction in GCLC language

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