sábado, 17 de junio de 2017

Generalization of a Sangaku problem

This is a generalization of a sangaku problem.


  • How to make the construction?
  • What is the formula for the radius of the two congruent circles?
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2 comentarios:

  1. Let s=(a+b+c)/2, sa=(-a+b+c)/2, sb=(a-b+c)/2 and sc=(a+b-c)/2

    Point P ( =Pa) must have trilinear coordinates:
    Pa = 1 : (sqrt(a*s)-sb)/b : (sqrt(a*s)-sc)/c

    The radii of both circles is:
    ra =| (s-sqrt(a*s))*tan(A/2)|

    The incircle of BCP has center:
    Ia = a-sqrt(a*s) : (sqrt(a*s)*(a-c)-SC)/b : (sqrt(a*s)*(a-b)-SB)/c

    The other circle has center:
    Ja = -(b+c-sqrt(a*s))/(a-sqrt(a*s)) : 1 : 1

    Best regards.
    César Lozada

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  2. Note that this solution requires that one circle be inscribed into the triangle BCP and the other be inscribed into the quadrilateral bounded by lines AB, AC, BP and CP. Please change the order of the vertices in the triangle of the figure (counter-clockwise), with A as the top one.

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