Golden Ratio and Midpoints

In a 2011 article, Jo Niemeyer, offered an elegant way of constructing the Golden Ratio with three equal segments, their midpoints and a pair of perpendicular lines.

For a proof, drop a perpendicular B₃H from B₃ to A₁A₂.

Assume for convenience that all three line segments are of length 2. Then in right triangle A₃B₃H, A₃B₃ = 2, and B₃H = 1/2 (as a midline in ΔA₁A₂B₂).

By the Pythagorean theorem,

(A₃H)² = (A₃B₃)² - (B₃H)² = 4 - 1/4 = 15/4,

so that A₃H = √15/2.

On the other hand, in right triangle A₁A₂B₂, A₂B₂ = 2, and A₁B₂ = 1, making A₁A₂ = √3 and A₁H = √3/2.

It follows that A₁A₃ = A₁H + A₃H = √3(√5 + 1)/2, and, finally, A₁A₃ / A₁A₂ = [√3(√5 + 1)/2] / √3 = φ.

References

1. Jo Niemeyer, A Simple Construction of the Golden Section, Forum Geometricorum, Volume 11 (2011) 53

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
   • Getting a Scout out of Desert
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

Golden Ratio

1. Golden Ratio in Geometry
2. Golden Ratio in an Irregular Pentagon
3. Golden Ratio in a Irregular Pentagon II
4. Inflection Points of Fourth Degree Polynomials
5. Wythoff's Nim
6. Inscribing a regular pentagon in a circle - and proving it
7. Cosine of 36 degrees
8. Continued Fractions
9. Golden Window
10. Golden Ratio and the Egyptian Triangle
11. Golden Ratio by Compass Only
12. Golden Ratio with a Rusty Compass
13. From Equilateral Triangle and Square to Golden Ratio
14. Golden Ratio and Midpoints
15. Golden Section in Two Equilateral Triangles
16. Golden Section in Two Equilateral Triangles, II
17. Golden Ratio is Irrational

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