Wednesday, June 23, 2010

Triangulation in the pentagon

As everyone knows, a pentagon is a 5-sided polygon, while a regular pentagon has all five sides the same length. As shown in the first figure (left), one can generate a pentagon starting with an isosceles triangle whose apical angle α is equal to 360/5 = 72°. Successive rotations of the triangle about the apex by this angle generate a pentagon with the apex as the center.

The other two angles of the triangle are equal to each other (β) and measure (180 - α)/2 = 54°. The total angle at each vertex (θ) is twice that or 108°.

Another calculation (which is really the same one) considers the orientation of a person we imagine walking around the perimeter of the figure. After one complete tour, the walker will have rotated a total of 360°. At each vertex, the amount turned is the same, 360/5 = 72°. The angle at the vertex is the supplement of that, 108°. Both approaches to calculating the angles will work for any regular polygon.

Next, consider the chords of the pentagon, as shown in the same figure, right panel---lines that connect vertices which are not adjacent. Using the rotational and reflective symmetry of the regular pentagon, it is easy to show by similar triangles that chord QT in the second figure is parallel to edge RS. A similar relationship holds for each of the other chords and a corresponding edge.

Label the smaller angle between a chord and adjacent edge as γ. We can show that the two chords originating from each vertex divide the vertex angle θ into three equal parts. Hence the label γ on angle SQT in the figure.

Perhaps the simplest method to show this is algebraic. The triangle PQT is isosceles. The apical angle QPT is a vertex angle of the pentagon so it equals 108°. Thus, the two remaining angles (γ) each equal (180 - 108)/2 = 36°. But TQS equals θ - twice γ, or 108° - 2*36°, which is also equal to 36°. We have:

 `α = 72°β = 54°θ = 2 * β = 108°γ = 1/3 θ = 36°`

Because their base angles are equal, PQT and PQU are similar triangles (second figure, left). So the angle PUQ (labeled δ) is the same as the vertex angle of the pentagon, and is also equal to RUT. This means that the inner five-sided figure is a regular pentagon. (The five-fold rotational symmetry also makes this clear).

There is a special relationship between the parts of each chord, which are labeled a and b in the right panel.

We label the sides of the inner pentagon as b, and the other parts of each chord (which are equal to each other), as a. Recall that QT is parallel to RS. By symmetry then, PT is parallel to QS, and PT is also parallel to VS. This means that the quadrilateral PVST is a parallelogram. Opposing sides of a parallelogram are equal, so we deduce that the length of each edge is equal to a plus b. The parallelogram is actually a rhombus, with all four sides equal.

Now, ignore for a moment the labels a and b. (But remember the result about the rhombus). The two triangles QTW and RSW (next figure, left) are similar because each is an isosceles triangle with base angle γ and apical angle theta;, the vertex angle of the pentagon. We can mentally set the scale of the pentagon so an edge has length 1. Relabel the segment WS with length a as 1/φ and the length of an entire chord as d.

By similar triangles, the ratio of d to 1 (triangle QTW) is equal to the ratio of 1 to 1/φ (triangle RSW).

 `d/1 = 1/(1/φ) = φ`

But QS = QT, so

 `1 + 1/φ = d1 + 1/φ = φ`

Multiply by φ and rearrange terms:

 `φ2 - φ - 1 = 0`

Using the quadratic formula and taking the positive root, we solve for φ:

 `φ = 1/2(1 + √5)`

This is the golden ratio (explaining our choice of symbols!). That is:

 `a + b a----- = - = φ a b`

To construct a regular pentagon, we need to find the center, labeled O in the next figure, below. O is also the center of the circumscribing circle. Knowing that all the vertices lie on that circle, together with the edge length, makes it possible to construct the pentagon.

Finding the center by construction starting from the pentagon is pretty easy. Bisect each of two sides, and draw the corresponding altitudes to the opposite vertex. These altitudes cross at the center.

I found two construction methods to go from a circumscribing circle to the pentagon.

Start with the red circle in the figure, centered at O, containing all five vertices of the pentagon we want to construct. Draw the diameter from vertex P and then the perpendicular diameter to that. Bisect this radius to find point X.

At this point, in the first method, draw the dotted purple circle with radius PX and center X to find point Y. Now draw the dotted magenta circle with radius PY and center P to find point T at the intersection with the red circle. That gives one side of the pentgon, PT.

I found it here under Interactive exercises > Ruler & compass > level 4.

The second method is from wikipedia. Starting from the point X, draw the line PX. Bisect the angle formed with the horizontal diameter and extend the line of angle bisector to where it meets the vertical diameter. Draw the perpendicular to the diameter through this point. This is the chord connecting Q and T.

It is not clear to me yet why either one works. The first construction is connected to the golden ratio result, but I haven't figured it out. When I do, I'll let you know.

1 comment:

Richard Careaga said...

Geometry hasn't been this clear since the first week of 10th grade, after which it started to become obscure