Sunday, March 13, 2016

Geometry Problem 1199: Equilateral Triangle, Square, Altitude, Circle, Incircle, Inradius, Metric Relations

Geometry Problem. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.

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Geometry Problem 1199: Equilateral Triangle, Square, Altitude, Circle, Incircle, Inradius, Metric Relations, Online College School

4 comments:

  1. Let DF = p, GH =q and AG = m

    So BD = sqrt3 -1 and from similar triangles p = 1 - sqrt(1/3)

    and p/1 = (1-q)/q from which q = (6+sqrt3)/11 upon simplifying the algebra

    m^2 = q^2 + 1 and so simplifying again m = (1/11) sqrt(149-10sqrt3)

    x = S(AGC)/s = (1/2)q X 2/{(1/2) (2m+2)} = 2q / (2m+2)

    Substituting for q and m and again simplifying the algebra

    x = (5 - sqrt3)/{11 + sqrt(149-10sqrt3)}

    I have not tried to simplify this unwieldy expression but it can be done by rationalizing the denominator.

    Sumith Peiris
    Moratuwa
    Sri Lanka

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    Replies
    1. Dear Sumith,
      Please review the algebra simplification
      Thanks

      Delete
    2. From my proof above x = q/(1+m) which upon simplifying gives

      (6+sqrt3)/(11+ sqrt(160+12sqrt3))

      Thanks Antonio

      Delete
  2. 배덕락(Bae deokrak)

    We see that two triangles BDF, CEF are 30-60-90 triangle and so
    CE=1, EF=1/sqrt3, FC 2/sqrt3, DF=1-1/sqrt3 , BD=sqrt3 –1 and BF=2sqrt3 –1
    Furthermore, since two triangles FDG and AHG are similar, we have
    DF/DG=AM/(1-DG) <=> DG=(5-sqrt3)/11 <=> HG= 1-DG=(6+sqrt3)/11
    The area of triangle ACG =[tri_ACG]= HG= x(1+GA) =x(1+sqrt{1+HG^2}), that is,
    x=HG/(1+sqrt{1+HG^2}). where HG=(6+sqrt3)/11.

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