Wednesday, September 13, 2017

Geometry Problem 1345: Three Equilateral Triangles, Center, Area, Common Vertex, 60 Degrees

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

Details: Click on the figure below.

Geometry Problem 1345: Three Equilateral Triangles, Center, Area, Common Vertex, 60 Degrees.

3 comments:

  1. Trigonometry Solution

    Let AB = a, OE = b and CE = c and
    let < ACE = α.

    S1 = √3b2/4 …(1) and
    S2 = a2/(4√3)…(2)
    being 1/3rd of S(ABC) = (√3a2/4)/3

    S3 = S(ADC) – S(ACE) – S(CDE)
    S3 = ½ ac.sin(α+60) - √3c2/4 – ½ ac.sinα
    S3 = ½ ac.(1/2 sinα + √3/2 cosα – sinα) - √3c2/4
    S3 = ½ ac(cos30.cosα – sin30.sinα) - √3c2/4
    S3 = ½ ac cos(α+30) - √3c2/4…(3)

    Now from ∆OCE, cos(α+30) = (a2/3 + c2 – b2)/(2ac/√3) …(4)

    Substituting in (3),
    S3 = √3/4. (a2/3 + c2 – b2) - √3c2/4
    S3 = a2/(4√3) – √3b2/4
    So S3 = S2 – S1 from (1) and (2).

    If S1 > S2, then S3 = S1 – S2.

    So generally S3 = IS2 – S1I

    Sumith Peiris
    Moratuwa
    Sri Lanka

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  2. Do u have a pure geometry solution Antonio?

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    Replies
    1. Here is a pure geometry solution.

      See diagram here.

      Let OE = r and O’ be the symmetric of O with respect to AC :
      - ΔOCE and ΔO’CD are SAS congruent => O’D = OE = r

      Draw the circles C(O,r) and C’(O’,r) with radius r and centers O and O’ :
      - AD intercepts C’(O’,r) in I and D
      - BE extended intercepts C(O,r) in J and E, and AD in G

      Let EH be the altitude of ΔAED

      Consider R, the rotation by 60° about vertex C.
      From the problem statement, it follows that:
      - the images by R of C(O,r), B, J and E are C’(O’,r), A, I and D, respectively, and AD is that of BE
      - G is on the circumcircle of ABC => OG = OB
      - Hence ΔEOJ and ΔGOB are isosceles and share the same base line
      => EG = (BG – EJ) /2 = BJ = AI

      Now:
      - EH = AI.√3/2 and AI.AD =│AO’ - r│.(AO’ + r) = │AO^2 - r^2│ = │(AB^2)/3 - r^2│
      - S3 = EH.AD/2 = │(AB^2)/3 - r^2│ .√3/4 = │S2 – S1│

      QED

      Note:
      - In the spirit of Euclid’s Elements, we could also get the results derived from the use of the concept of rotation by those derived from suitable similarities and congruencies of pairs of triangles such as BEC and ACD, KAG and KBC (where K is the intersection of BE and AC) etc.
      - As for the power of A with respect to circle C', the property was known by Euclid for points outside as well as inside the circle (Book 3, prop 35 to 37).

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