Monday, May 19, 2008

Elearn Geometry Problem 63

See complete Problem 63
Regular Heptagon, side and diagonals. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.


  1. Let AB = CD = DE = s, AC = CE = c, AD = AE = d.
    Applying Ptolemy's Theorem to cyclic quadrilateral ACDE, we obtain cs + ds = cd.
    Then s(c + d) = cd, and so 1/s = (c + d)/cd = 1/d + 1/c.

    Therefore, in regular heptagon ABCDEFG, 1/s = 1/c + 1/d

  2. Apply inversion on A with any inversion radius (A -> A' ; B -> B' ...etc)
    Then we can rewrite the statement by proving A'B'=A'C'+A'D'.

    The inversion image is easy : AB,AC,AD remains unchanged while the circumcircle becomes a line B'C'D' not passing through A.
    By the conformal properties, ∠BAC=∠B'A'C'=∠C'A'D'=∠A'B'C'=180/7

    Construct a point E' on A'B' such that A'E' = A'C', so we simply need to prove E'B'=A'D'.
    This is actually quite obvious.

    So ΔB'C'E'=ΔA'C'D', and hence E'B'=A'D


  3. Extend segments AB and DC to intersect outside of the circle in I (see graph).
    ABCD cyclic quadrilateral implies <IBC = <ADC = <BAD = <ICB = 2π/7, so ΔIAD and ΔIBC are similar and IB/BC = IA/AD (1).
    Consider ΔCAE and ΔIAD : they are equal (isoceles with same base length d and same base angles 2π/7). So IA = CA = c and from (1) IB = s*c/d.
    Also, IA = AB+IB = s+IB so IB = c-s.
    So the following holds :
    c-s = s*c/d
    c = s+s*c/d
    Dividing both terms by s*c : 1/s = 1/c + 1/d QED

  4. Hello, the graph supporting my proof can be found there: