1. Objective:

B. Historical Perspective (Explanation of how history fits into the lesson.)

3. Prerequisites:

I propose using this unit at the end of the year as a creative way to review for the final exam. Students should have studied regular polygons, symmetry, similar triangles and circle theorems. However, it could be used as a one day stand-alone lesson by eliminating the introductory activity, the details of the preliminary propositions and the assigned constructions.

4. Materials needed to teach the lesson:

Students will need a compass and a straightedge.

5. Outline of Procedures:

1. Optional introductory activity: Scatter Theorems to review for a final exam.
2. Class discussion on how to construct an inscribed regular pentagon.
3. The constructions used to follow Euclid’s construction of an inscribed regular pentagon.
4. Understanding the golden ratio
5. Book ii, Proposition 11
6. Book iv, Proposition 10.
7. Book ii, Proposition 2
8. Demonstration of the construction of a regular pentagon using Book iv,

9. Assign these constructions:

1. Given a regular pentagon, construct the circumscribed circle.
2. Given a regular pentagon, construct the inscribed circle.
3. Given a circle, circumscribe a regular pentagon about the circle.

1. Equal chord have equal arcs and its converse
2. The measure of an inscribed angle is half the measure of the arc intercepted.
3. Vertical angles are congruent.
4. Parallel lines imply alternate interior angles are congruent and its converse.
5. Parallel lines imply same-side interiors angles are supplementary and its converse.
6. The sum of the interior angles of a polygon.
7. The sum of the exterior angles of a polygon.
8. The exterior angle of a triangle equals the sum of the remote interior angles.
9. Similar triangle theorems
10. Congruent triangle theorems
11. Isosceles triangle theorems
12. Isosceles trapezoid theorems
13. Rhombus theorems

1. The bisector of one angle of the pentagon
2. Two radii
3. A tangent line
4. The medians, altitudes, or angle bisectors of any triangle

Distribute a geometry sketch, such as the inscribed regular pentagon below, to each student. Each student should study the sketch and write five geometry theorems that are embodied in the sketch.

Next group the students in threes or fours and have each group compose a list of its best ten theorems. The idea is to state theorems that other groups overlook.

A good copy with no changes will be collected at the end of the activity.

A spokesperson for each group will read the theorems to the class, one theorem per turn. Group1 will state a theorem and the other groups will indicate if that theorem is on their list. If another group has it on their list, all groups with it on their list must cross it off. If no other group has it on their list, group 1 circles it for a point. Group 2 then reads a theorem and the other groups check for it on their list. The process repeats until every theorem has been read or crossed off.

The winner is the group with the most circled theorems.

To take the game to a higher level, try round two. Instruct each group to add up to three lines or line segments and create a new list of theorems that depend on them.

1. Class discussion on how to construct an inscribed regular pentagon.

Whether the introductory activity is used or not, students should be able to offer ideas on what must be known to inscribe the regular pentagon. A wonderful dialogue on angles and triangles that can be constructed should occur. The end result will hopefully be that help is needed! This opens the door for revisiting Euclid’s Elements.

2. The constructions used to follow Euclid’s construction of an inscribed regular pentagon.

1. An isosceles triangle with base angles double the vertex angle:

b) An inscribed triangle equiangular to a given triangle.

1. Understanding the Golden Ratio

Just as pi is the constant ratio of the circumference of a circle to its diameter, the Golden Ratio is a constant equal to (1 + Ö 5)/2 or the reciprocal of (-1 + Ö 5)/2 or approximately 1.618. It is believed that the students of Pythagoras first discovered this ratio. It is further supposed that Plato (-429 to –347) "began the study of the ‘golden section’ as a subject in itself" (Heath, vol. 2, 99)

The golden rectangle is a rectangle divided into a square and a rectangle that is similar to the original rectangle. The ratio of its length to its width is the golden ratio. Since this rectangle is "pleasing to the eye, " it is often found in art, architecture, nature and the human body (Bass, 498).

The isosceles triangle needed for the pentagon contains this special ratio. The leg divided by the base equals the golden ratio. In fact as you look at the sketch in Scatter Theorems handout, AD/DC, AT/TD, and AP/PT all equal the golden ratio. Most geometry textbooks refer to the golden rectangle, but few discuss the 36-72-72 triangle containing the golden ratio.

 

 

 

2. Book II, Proposition 11: To cut a given straight line so that the rectangle contained by the whole and one of the segments is equal to the square on the remaining segment.(Heath, vol.1,402-403)

This construction allows us to divide a segment such that the ratio of the given segment to the "remaining segment" is equal to the golden ratio. I will call this dividing a segment into the golden ratio property.

CONSTRUCTION STEPS:

Begin with segment AB and construct square ABDC. Construct the midpoint of AC and label it E. Draw BE. Draw a circle with center E and radius EB. Extend CA through A intersecting circle E at point F. Construct square FGHA on AF. The area of HBDK is equal to the area of square FGHA. AB/AH will equal the golden ratio.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

GOLDEN RATIO JUSTIFICATION:

Let AB = r. Then AE = r/2. Using the Pythagorean Theorem,

BE = (Ö 5 r)/2. But radii of circle E, BE and EF, are equal and FA = AH in square FGHA. Since EF – AE = FA, FA = AH. So AH = (Ö 5 r)/2 – r/2 = (-1 + Ö 5) (r/2). Dividing by AB which is r, gives us AH/AB = (-1 + Ö 5)/2. Therefore its reciprocal AB/AH = (1 + Ö 5)/2 which is the golden ratio.

3. Book IV, Proposition 10: To construct an isosceles triangle having each of the angles at the base double of the remaining one.(Heath, vol.2,96-97)

We now recognize that this will be a 36-72-72 triangle. However, Euclid did not refer to degrees. He wrote in terms of right angles or in this case double an angle. Euclid understood that constructing an isosceles triangle with sides forming the golden ratio would result in the angles at the base being double the remaining angle.

CONSTRUCTION STEPS:

Begin with segment AB and locate point C so that the rectangle with length and width, AB and BC is equal in area to the square on AC. (This means you would have to repeat the construction that yielded point H of procedure 6.)

Draw circle A with radius AB. Mark off chord BD = AC. Triangle ABD is the desired isosceles triangle with legs AB = AD and base BD since AB/BD equals the golden ratio.

 

 

 

 

 

 

 

 

 

 

 

Euclid’s justification of this construction involves a rigorous proof, but one that high school students can follow in theory.

Using a circle circumscribed about triangle ACD, Euclid argues that angle BDC equals angle DAC because they both equal half of arc CD. Adding angle CDA to both angles makes angle BDA equal to angle BCD (exterior angle = remote interior angle DAC + angle CDA.) Therefore angle BCD = angle BDA and they also equal angle ABD because AB = AD. With the base angles of triangle BCD equal, BD = DC. But since BD was constructed equal to AC, AC=BD=DC. This makes triangle ADC isosceles and base angle DAC = base angle CDA. Again, the exterior angle BCD = angle CAD(or BAD) + angle CDA. Hence, angle BCD = double angle BAD. This concludes the argument that angle ABD is double angle BAD in isosceles triangle ABD.

A closer look at this isosceles triangle

verifies that AB/BD = (1 + Ö 5)/2.

Triangle ABD is similar to triangle DBC.

AB/BD = BD/BC = AD/DC

x/(r-x) = r/x yields x = (-1 + Ö 5) (r/2)

and r/x = (1 + Ö 5)/2, the golden ratio.

4. Book IV, Proposition 2: To inscribe a triangle equiangular with a given triangle in a given circle.(Heath, vol.1,81-83)

CONSTRUCTION STEPS:

Given triangle DEF and a circle. Construct a line PQ touching the circle at point A. (We would say tangent at A.)

Construct an angle QAC equal to angle DEF and construct angle PAB equal to angle DFE. C and B are the points where the sides of the angles intersect the circle.

Triangle ABC is the desired triangle inscribed in the given circle.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

JUSTIFICATION:

Angle ABC = angle QAC because they subtend equal arcs. Similarly angle ACB = angle PAB. This leaves angle BAC to equal angle FDE. Hence, triangle ABC is equiangular to triangle DEF.

5. Demonstrate the construction of a regular pentagon using Book IV, Proposition 11. In a given circle to inscribe an equilateral and equiangular pentagon.(Heath,vol.2,101-102)

JUSTIFICATION:

Historically, Euclid understood the golden ratio from the Pythagoreans. He gives us a construction (ii,11) which creates two segments with the golden ratio property. He constructs an isosceles triangle with the base angle double the vertex angle (iv,10) by building it on a segment with the golden ratio property. He understands that this triangle will yield (iv,11) an equilateral and equiangular pentagon inscribed in a given circle. He does not use similar triangles directly. However, he does construct the isosceles triangle inscribed in the given circle equiangular to a given triangle. We know the triangles are similar because the angles are equal.

In today’s classroom we begin by introducing the Golden Ratio and point out its existence in the regular pentagon. Working backwards, we show how to construct segments with the golden ratio property using Euclid’s Elements and justify it algebraically. Using our SSS Congruence theorem we can construct an isosceles triangle using those segments. By constructing a central angle in the given circle equal to one of the base angles of our isosceles triangle, we have determined the required 72-degree arc. The chord of that arc is the side of our inscribed regular pentagon.

9. To extend the lesson, assign these constructions:

1. Given a regular pentagon, construct the circumscribed circle.
2. Given a regular pentagon, construct the inscribed circle.
3. Given a circle, circumscribe a regular pentagon about the circle.

1. The perpendicular bisectors of the sides are concurrent at the center.
2. The same point will be the center of the inscribed pentagon. The radius is the perpendicular distance from that point to a side.
3. To begin construct a 72-degree central angle and the tangents at the endpoints of these radii. A variety of approaches can be used to complete the pentagon.

Constructions were drawn with Texas Instruments’ Cabri Geometry II Software.