Tuesday, January 28, 2014

Geometry Problem 969: Equilateral Triangle, Rectangle, Common Vertex, Right Triangles Areas

Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to view the complete problem 969.

Online Geometry Problem 969: Equilateral Triangle, Rectangle, Common Vertex, Right Triangles Areas

13 comments:

  1. Let z(P) be the complex number representing P.
    Let z(C)=0, z(B)=−a, z(E)=−a−bi.

    Then
    z(F) = (−a−bi)(1/2 + √3/2 i) = (−1/2 a + √3/2 b) + (−√3/2 a − 1/2 b)i
    z(D) = (−√3/2 a − 1/2 b)i
    z(A) = −a + (−√3/2 a − 1/2 b)i

    Thus
    BC = a
    BE = b
    AE = √3/2 a − 1/2 b
    DF = 1/2 a − √3/2 b
    AF = 1/2 a + √3/2 b
    CD = √3/2 a + 1/2 b

    S₁ = 1/2 × AE × AF = √3/8 (a^2 − b^2) + 1/4 ab
    S₂ = 1/2 × BC × BE = 1/2 ab
    S₃ = 1/2 × CD × DF = √3/8 (a^2 − b^2) − 1/4 ab

    Hence,
    S₁ = S₂ + S₃

    ReplyDelete
  2. Use the area formula of triangle:
    A=c^2/(2[cotA+cotB])

    In this case, c^2/2 cancel out of the equation S2+S3=S1. What we are left with is a simple trigonometric identity which is always true.

    ReplyDelete
  3. Please if you want explain it more.
    than you.

    ReplyDelete
  4. using trigonometry should always be the last resort. There must be another, direct, proof, using geometric reasoning.

    ReplyDelete
  5. Fie E(0,e),F(f,o)atunci C(f/2+e√3/2; e/2+f√3/2)pentru a arata ca S₁ = S₂ + S₃ trebuie sa aratam ca
    AE.AF=FD.CD+BE.BC adica ef= (f/2+e√3/2-f)(e/2+f√3/2)+(e/2+f√3/2-e)(f/2+e√3/2) care este adevarata

    ReplyDelete
  6. Think about pythagoras triangle BEC and DFC

    ReplyDelete
  7. i think their might be wrong in the question'

    ReplyDelete
  8. Triangles EBC and FDC are congruent (RHS congruency)
    => BE = FD = x let, AE = AF = a − x where a is length of the side of square ABCD.
    By Pythagoras Theorem, S2 + S3 = ax/2 + ax/2 = ax
    S1 = (1/2)(a − x)^2

    Now notice that EF = CF = CE
    so (a − x)^2 + (a − x)^2 = a^2 + x^2
    so (a − x)^2 = 2ax
    so (1/2)(a − x)^2 = ax
    => S1 = S2 + S3

    ReplyDelete
    Replies
    1. To Abdul Problem 969:
      Please more details to your first statement:
      Triangles EBC and FDC are congruent (RHS congruency)
      Thanks

      Delete
  9. BC = CD, CE = CF, m(EBC) = 90° = m(FDC)

    ReplyDelete
    Replies
    1. To Abdul: In your statement BC = CD
      ABCD is a rectangle (not necessary a square) therefore BC is not equal CD
      Thanks.

      Delete
  10. Is there a simple geometric proof for this Antonio?

    ReplyDelete
  11. Let M be midpoint of EC so that AEMF and CDFM are cyclic quadrilaterals from which we can easily show that ADM is equilateral

    Let BE = a, AE = b, AF = c, FD = d and FE = x

    Apply Ptolemy to the cyclic quad AEMF we have

    Sqrt3/2 xb + x/2 c = x(c+d) from which

    b = (c+2d)/sqrt3.....(1)

    Similarly from cyclic quad CDFM we have upon using Ptolemy and simplifying as before

    a+b = (2c+d)/sqrt3 from which using (1) we have

    a =(c-d)/sqrt3......(2)

    So S 2 + S3 = 1/2a(c+d) + 1/2(a+b)d which upon substituting for a and b from (1) and (2)

    = (1/2sqrt3){(c-d)(c+d) + (2c+d)d} = (1/2sqrt3)(c+2d)c = bc/2 = S1

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete