# 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.

Three equal segments A_{1}B_{1}, A_{2}B_{2}, A_{3}B_{3} are positioned in such a way that the endpoints B_{2}, B_{3} are the midpoints of A_{1}B_{1}, A_{2}B_{2} respectively, while the
endpoints A_{1}, A_{2}, A_{3} are on a line perpendicular to A_{1}B_{1}.

In this arrangement, A_{2} divides A_{1}A_{3} in the golden ratio, namely, _{1}A_{3} / A_{1}A_{2} = φ.

For a proof, drop a perpendicular B_{3}H from B_{3} to A_{1}A_{2}.

Assume for convenience that all three line segments are of length 2. Then in right triangle A_{3}B_{3}H, _{3}B_{3} = 2,_{3}H = 1/2_{1}A_{2}B_{2}).

By the Pythagorean theorem,

(A_{3}H)² = (A_{3}B_{3})² - (B_{3}H)² = 4 - 1/4 = 15/4,

so that A_{3}H = √15/2.

On the other hand, in right triangle A_{1}A_{2}B_{2}, A_{2}B_{2} = 2, and _{1}B_{2} = 1,_{1}A_{2} = √3 and _{1}H = √3/2.

_{1}A_{3} = A_{1}H + A_{3}H = √3(√5 + 1)/2, and, finally, _{1}A_{3} / A_{1}A_{2} = [√3(√5 + 1)/2] / √3 = φ.

### References

### Fibonacci Numbers

- Ceva's Theorem: A Matter of Appreciation
- When the Counting Gets Tough, the Tough Count on Mathematics
- I. Sharygin's Problem of Criminal Ministers
- Single Pile Games
- Take-Away Games
- Number 8 Is Interesting
- Curry's Paradox
- A Problem in Checker-Jumping
- Fibonacci's Quickies
- Fibonacci Numbers in Equilateral Triangle
- Binet's Formula by Inducion
- Binet's Formula via Generating Functions
- Generating Functions from Recurrences
- Cassini's Identity
- Fibonacci Idendtities with Matrices
- GCD of Fibonacci Numbers
- Binet's Formula with Cosines
- Lame's Theorem - First Application of Fibonacci Numbers

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