Tuesday, March 3, 2015

Geometry Problem 1092. Equilateral Triangle, Square, Circle, Tangent, 90 Degree, Sangaku

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

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Online Math: Problem 1092. Equilateral Triangle, Square, Circle, Tangent, 90 Degree, Sangaku.

4 comments:

  1. Observe that ΔABC is a 30°-60°-90° triangle, and so are ΔAHB, ΔHBD, ΔHMD.
    Also observe that HBDM is a kite.

    Let BN⊥AC at N which is the mid-point of GE.
    Let EP⊥BC at P which is the mid-point of DF.

    Let AG=a.
    Then
    AB = AH + HB = 2 AG + GH = (2+√3)a
    BC = 2 ED + DM = 3 AG + 2 GH = (3+2√3)a
    AE = AG + GH = (1+√3)a

    Hence, AE = BC − AB

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  2. Join the collinear points H, O, D.
    The three 60-30-90 Right Triangles AHG, HDB, HDM are congruent to one another.
    Let each side of the equilateral triangle be "a" and each side of the square be "b".
    So BD = DM = a - b. Also HB = HM = b, AH = 2AG = 2BD = 2DM = 2a - 2b.
    Follows AB = AH + HB = (2a - 2b) + b = 2a - b, AE = AG + GE = (a - b) + b = a
    while BC - AB = (BD + DC) - AB = [(a - b) + 2a] - [2a - b] = a.
    Hence AE = BC – AB.
    (Incidentally AB = BF and AE = FC)

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  3. Let BA meet FE in N. Let the side of the square be a and let AG = b. Easily Tr. EFC is isoceles with FC = a+b.

    Now H, O, D are collinear and by congruent Tr.s easily we can prove that HB = a and BD = b. So BC = 2a+3b.

    Now since < BNF = 30, Tr. AEN is isoceles and AN = a+b, AH = 2b and HB = a. So BN = 2a+3b = BC.

    So AN + AB = BC hence BC - AB = AN = AE.

    (Note that a = sqrt3 X b which I have avoided using to maintain clarity)

    Sumith Peiris
    Moratuwa
    Sri Lanka

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