Golden Section in Two Equilateral Triangles, II

Quang Tuan Bui

November 7, 2011

ABC and AMN are two equilateral triangles where M is midpoint of side BC. Arc 60° centered at B passing through A, C intersects side MN by golden ratio.

Construct two circles: circle A(P) centered at A passing through P and circle C(A) centered at C passing through A. These two circles intersect at two points. Let F denote the one that is in the same side of AC as N.

Suppose D is symmetric to A in M. Two triangles ABP and ACF are congruent because AP = AF (the radius of A(P)) and BA = BP = CA = CF (the radius of C(A)). Therefore ΔACF is obtained by a 60° rotation R(A, 60°) of ΔABP around A. This shows that ΔAPF is equilateral.

The rotation R(A, 60°) also maps ΔAMP onto ΔANF; therefore,

(1) MP = NF

and ∠FNA = ∠PMA = ∠MAN = 60°, implying

(2) FN||AD.

∠APD = 120° and ∠APF = 60°, so that
(3) D, P, F are collinear.

From (2) and (3) two triangles PMD, PNF are similar. Therefore:

NF/PN = MD/PM
MP/PN = NF/PN by (1)

Now we can calculate:

x = MP/PN = MD/PM = AM/PM = NM/PM
  = (NP + PM)/PM = NP/PM + 1 = PN/MP + 1

or x = 1/x + 1. From this x = φ, the Golden ratio.

(There is another proof for the same construction.)

Fibonacci Numbers

  1. Ceva's Theorem: A Matter of Appreciation
  2. When the Counting Gets Tough, the Tough Count on Mathematics
  3. I. Sharygin's Problem of Criminal Ministers
  4. Single Pile Games
  5. Take-Away Games
  6. Number 8 Is Interesting
  7. Curry's Paradox
  8. A Problem in Checker-Jumping
  9. Fibonacci's Quickies
  10. Fibonacci Numbers in Equilateral Triangle
  11. Binet's Formula by Inducion
  12. Binet's Formula via Generating Functions
  13. Generating Functions from Recurrences
  14. Cassini's Identity
  15. Fibonacci Idendtities with Matrices
  16. GCD of Fibonacci Numbers
  17. Binet's Formula with Cosines
  18. Lame's Theorem - First Application of Fibonacci Numbers

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