Sunday, January 07, 2007

Golden Triangle

In today's blog, I will talk about the properties a golden triangle. A golden triangle is an isoceles triangle that has two of its sides equal to the golden ratio.

I will show how the properties of the golden triangle enable us to conclude that:

cos (2π/5) = √5 - 1/4

and

sin (2π/5) = √5 + √5/8

First, here is the definition of a golden triangle:

Definition 1: golden triangle










A golden triangle is any isoceles triangle ABC with AB ≅ AC and AB/BC = AC/BC = (1 + √5)/2.

This is none other than a 36 °-72 °-72° triangle as show by:

Theorem 1: A 36 °-72 °-72° triangle is a golden triangle













Proof:

(1) Let ABC be a 36 °-72 °-72° triangle where ∠ BAC = 36 ° and ∠ ABC, ∠ ACB= 72 °

(2) It is an isoceles triangle with AB ≅ AC since congruent angles imply congruent sides (see Corollary 1, here)

(3) Let D be a point on AC such that ∠ CBD ≅ ∠ BAC.

(4) Since triangles add up to 180 ° (see Lemma 4, here), it is clear that ∠ BDC ≅ ∠ ABC

(5) This means that triangle CBD is similar to triangle CAB (see here for review of similar triangles) and it is also an isoceles triangle for the same reason (see step #2 above)

(6) Now, triangle ADB is also an isoceles triangle since ∠ CAB ≅ ∠ ABD since:

(a) ∠CAB is 36 °

(b) ∠ ABC is 72 °

(c) ∠ DBC is 36 ° from step #5 above.

(d) So this gives us that ∠ ABD = ∠ ABC - ∠ DBC = 72 ° - 36 ° = 36 °.

(7) From step #5, we have AC/BC = BC/DC (see Lemma 3, here)

(8) Let p be the measurement of AC ≅ AB

(9) Let q be the measurement of BD ≅ BC ≅ AD

(10) It is clear that DC = AC - AD = p - q

(11) So AC/BC = BC/DC gives us:

p/q = q/p-q

(12) If we let:

a = q
b = p-q

(13) Then step #11 gives us:

(a+b)/a = a/b = q/(p-q) = p/q

(14) Using (a+b)/a = a/b (see details here) gives us:

p/q = (1 + √5)/2.

QED

Corollary 1.1: Additional Properties

cos (2π/5) = (√5 - 1)/4

sin (2π/5) = √5 + √5/8











Proof:

(1) Let ABC be a a 36 °-72 °-72° triangle where ∠ BAC = 36 ° and ∠ ABC, ∠ ACB= 72 °

(2) Let D be a midpoint on BC such that BD ≅ CD

(3) ∠ ABC = 72 ° = 360/5 = 2π/5 (see here for review of radians)

(4) We can see that ∠ ADB and ∠ ADC are right angles since:

(a) AB ≅ AC since the bottom angles are congruent (see Corollary 1, here)

(b) This means triangle ABD ≅ triangle ACD by side-side-side (see Postulate 2, here)

(c) This then gives us that ∠ ADB and ∠ ADC are congruent.

(d) Since they are on the same line ∠ ADB + ∠ ADC = 180 ° (see Postulate 1, here) which shows that they must be right angles.

(5) By Theorem 1 above, we know that AB/BC = (1 + √5)/2

(6) If we assume that the measurement of AB = 1, then BC = 2/(1 + √5) = 2(1 - √5)/(1 - 5) = -2(1 - √5)/4 = (√5 - 1)/2

(7) Since AD = (1/2)AB:

cos(2π/5) = AD/AB = (√5 - 1)/4 [See here for review of cosine if needed]

(8) From the Pythagorean Theorem (see Theorem 1, here),

(AB)2 = (BD)2 + (AD)2

so that:

BD = √(AB)2 - (AD)2

(9) (AD)2 = [√5 - 1)/4]2 = [5 - 2√5 + 1]/16 = (3 - √5)/8

(10) BD = √(1)2 - (AD)2 =

= √(8 - 3 + √5)/8 =

= √(5 + √5)/8

(11) sin(2π/5) = (AD)/(AB) = (AD)/1 = √(5 + √5)/8 [See here for a review of sine if needed]

QED

References

No comments :