**Midpoint Polygons Revisited on the Internet**

In the Spring semester of 1999, Professor Thomas Banchoff of Brown University presented the midpoint polygons problem to the students in his Fundamental Problems of Geometry course, Math 104. He mentioned Rose Mary Zbiek's article, and challenged his students to go beyond the earlier investigations. The students discussed the midpoint problem on the interactive course webpage, as well as in class, and came up with a number of new results, including an explicit general formula for the area ratio. This paper is based on the results of that discussion.

## The Problem

## Midpoint Polygons

Rose Mary Zbiek asked her students to investigate midpoint pentagons, the figures obtained by joining the midpoints of a pentagon in the same order. What is the ratio of their perimeters? What about their areas? Computer experiments using Geometers' Sketchpad initially suggested that these ratios should be constant, but when attempts to prove this conjecture failed, further investigation led to the opposite conjecture, namely that this ratio depends on the shape of the pentagon. The object lesson was that technology can lead students to make conjectures, and it can also provide evidence to show that a conjecture is false.[RMZ]

This investigation formed the basis of an example in the Standards, in dialogue form, showing how students might discover preliminary theorems and then go on to investigate more general cases, make conjectures, and accumulate evidence to see whether a conjecture is likely to be true or false. [NCTM]

This investigation formed the basis of an example in the Standards, in dialogue form, showing how students might discover preliminary theorems and then go on to investigate more general cases, make conjectures, and accumulate evidence to see whether a conjecture is likely to be true or false. [NCTM]

## The Solution

Triangles: the base case

The first example, familiar to most students, concerns a triangle. The midpoint polygon divides the triangular region up into four congruent triangles so the area ratio is 1/4. Each edge of the midpoint triangle is half the length of its corresponding edge on the original, so the perimeter ratio is 1/2.

The first example, familiar to most students, concerns a triangle. The midpoint polygon divides the triangular region up into four congruent triangles so the area ratio is 1/4. Each edge of the midpoint triangle is half the length of its corresponding edge on the original, so the perimeter ratio is 1/2.

**Demonstration 1.**Area ratio for midpoint trianglesConvex Quadrilaterals

Area

For convex quadrilaterals, the midpoint polygon is a parallelogram with sides parallel to the diagonals (See Figure 1). The diagonals divide the original quadrilateral up into four triangles. If we consider one of these triangles, say ABO, it is clear that the portion of the midpoint polygon contained within the triangle (shaded red) has exactly half the area of the entire triangle (red and blue regions). Therefore the total area of the midpoint polygon, is 1/2 the area of the original quadrilateral

PerimeterThe perimeter question is more complicated. For a square, the midpoint polygon is a square with sides one over the square root of 2 times the sides of the original square, so the perimeter ratio is 1/sqrt(2), about .7071. More generally, for a rectangle with sides 2

Area

For convex quadrilaterals, the midpoint polygon is a parallelogram with sides parallel to the diagonals (See Figure 1). The diagonals divide the original quadrilateral up into four triangles. If we consider one of these triangles, say ABO, it is clear that the portion of the midpoint polygon contained within the triangle (shaded red) has exactly half the area of the entire triangle (red and blue regions). Therefore the total area of the midpoint polygon, is 1/2 the area of the original quadrilateral

**Figure 1.**Finding the area of a midpoint quadrilateralPerimeterThe perimeter question is more complicated. For a square, the midpoint polygon is a square with sides one over the square root of 2 times the sides of the original square, so the perimeter ratio is 1/sqrt(2), about .7071. More generally, for a rectangle with sides 2

*a*and 2*b*, the perimeter of the midpoint parallelogram isThe perimeter ratio is then,

so if

*a*= 3 and*b*= 4, the ratio is 5/7=.714285 and repeating. The ratio is not constant. In general, the length of the midpoint parallelogram is equal to the sum of the lengths of the two diagonals.**Demonstration 2.**Perimeter ratio for rectangles

It is interesting to calculate what the perimeter ratios can be for rectangles. Some experiment suggests that the ratio

since this is equivalent to,

which in turn is equivalent to.

The minimum is achieved for a square.At this point we can go in two directions. We can ask about non-convex quadrilaterals, and we can go further to ask about self-intersecting quadrilaterals. Or we can move up to pentagons.

**Original Discussion Here**

**Pedagogy**

Using technology in the classroom can often present old problems in new ways, and in some cases that can lead to a chance to extend classical and traditional topics in new directions. The midpoint polygon problem is a good example of this phenomenon, and in this article, we attempt to explore several different ideas that arise in this way.At one level, we want to illustrate how an Internet-based course can foster interaction of students with instructors and with other students by encouraging class members to investigate complicated problems in ways that would be difficult in a general discussion or in individual homework exercises. We do this by presenting, in unedited form, a series of responses from students in an upper level course in geometry open to those who have completed the calculus sequence and one semester of linear algebra. The course was elected by students in mathematics, physics, and computer science, and it was specifically recommended for those interested in secondary school teaching.

We also want to introduce some natural extensions of standard approaches in the geometry of polygons to admit wider classes of geometrically interesting objects. For example many presentations of the goemetry of polygons treat almost exclusively polygons that are convex. Some will deal with non-convex polygons, but most avoid any treatment of polygons that intersect themselves. Introducing interactive geometric demonstrations changes all that. If the student is allowed to move a particular vertex in a polygon with more than three vertices, then it is a difficult programming problem to keep him or her from moving that vertex so as to create something non-convex, let alone self-intersecting. A curious student will ask whether or not certain formulas and relationships still hold even when the polygon is not convex or when it self-intersects. Although such investigations fall outside traditional treatments, they arise very naturally when computer programs are used.

As it happens, most commonly used geometry technology can easily handle the area of general polygons, given by a sequence of points in the plane, without ever considering convexity or self-intersection. The way they do this is by using formulas that assign to each triangle an "algebraic area", which is the usual area if the triangle is oriented in a counterclockwise matter, and the negative of that area if the triangle is oriented clockwise.

For example, if a triangle has its first vertex at the vertical axis and the second vertex further above on the vertical axis, then a choice of third vertex determines a positive area if that vertex is to the left of the axis and a negative area if the vertex is to the right. More generally, the oriented line from the first vertex through the second divides the plane into a left side of the line and a right side. The area of a triangle is positive if the third vertex is on the left side and negative if it is on the right side.

What is the advantage of considering both positive and negative areas for triangles? The first one comes when we consider the area of a non-convex quadrilateral. On diagonal lies on the interior of the figure and divides the quadrilateral into two non-overlapping triangles. The sum of the (positive) areas of these two triangles will give the area of the region determined by the quadrilateral. On the other hand, the second diagonal expresses the area of this quadrilateral region as the difference between the area of a large triangle and the area of a triangle contained inside it. If the large triangle is traversed in a counterclockwise way, counted positively, then the second is traversed in a clockwise way and its area is counted negatively.

Note that in this way of counting areas, if we traverse a convex quadrilateral in a counterclockwise manner, then the area is positive as the sum of the areas of two non-overlapping triangles traversed in a counterclockwise manner, while if the quadrilateral is traversed in a clockwise manner, then so are the two triangles and the algebraic area of the quadrilateral is negative, given by the sum of the negative areas of the two triangles.

Standard geometric drawing programs like the Geometer's Sketchpad and the applet used to illustrate the pentagon problem in the NCTM Standards use this method to find areas. If the total area turns out to be negative, then the program reports the absolute value of the oriented area as the area.

This does not cause problems for quadrilaterals, even when the polygon intersects itself, but it does cause problems for pentagons, where the algebraic area of the midpoint polygon can be negative even when the area of the original pentagon is positive.

Self-intersecting quadrilaterals arise naturally when a student is encouraged to explore with a geometry program and moving a vertex causes one edge to cross another, resulting in an "hourglass shape". We may think of this shape as consisting of two triangular regions, one traversed in a counterclockwise manner and the other in a clockwise manner. The algebraic area will be the algebraic sum of these two signed areas. It can be zero if both parts of the hourglass have the same absolute area. If the algebraic area is positive, because the area of the triangle traversed counterclockwise is greater than that of the area traversed clockwise, then if we traverse the quadrilateral in the opposite direction, the roles will be reversed and the algebraic sum will be negative.

We can also express the algebraic area of the hourglass quadrilateral by introducing a "diagonal", for example joining the first vertex to the third. We can then compute the algebraic areas of the two triangles, one positive and one negative. The overlapping region of these two triangles is counted once positively and once negatively, and the leftover parts are the two triangular pieces treated in the previous paragraph.

A great advantage of this method is that it does not make any difference how we divide a given polygon up into a number of triangles. The sum of the algebraic areas of those triangles will always give the sum of the areas of the parts traversed counterclockwise minus the absolute areas of the parts traversed clockwise.

This fundamental intuition is at the heart of the computation of so-called "line integrals" which student will encounter in their third semester of calculus. When we teach this subject, often this is the first time that students are exposed to the concepts of oriented area for smooth curves, without an appreciation of the fact that the same concept works for polygons.

There are formulas for the area of a triangle in terms of the coordinates of the vertices that automatically give the algebraic area. For example, to find the algebraic area of a triangle with first vertex at (0,0), second vertex at (a,b) and third vertex at (c,d), we may use the formula ad - bc. This will be positive if the slope d/c of the line from (0,0) through (c,d) is greater than the slope b/a of the line from (0,0) through (a,b), since d/c > b/a implies that ad > bc and ad - bc > 0. In terms of vectors, if the line along (a,b) rotates in a counterclockwise direction to get to the line along (c,d), then the triangle from (0,0) to (a,b) to (c,d) has positive area, and if it rotates in a clockwise direction, the algebraic area of the triangle is negative.

For those familiar with vectors in physics, another way of expressing this algebraic area is to think of the three vertices as lying in a plane in ordinary three-dimensional space, and computing the cross-product between the vectors (a,b,0) and (c,d,0) to get (0,0,ad-bc). The algebraic area is then obtained by taking the dot product of this vector with (0,0,1). If the rotation from (a,b,0) to (c,d,0) is counterclockwise, then the cross-product points upward and the dot product with (0,0,1) is positive. If the rotation is negative, then the cross-product points in the direction opposite to (0,0,1) and the algebraic area is negative.

A more thorough treatment of this concept of algebraic area can be found in various textbooks in linear algebra. Most secondary school courses will not go far into such subjects, but it can be helpful to students to get an idea of what lies ahead.

We also want to introduce some natural extensions of standard approaches in the geometry of polygons to admit wider classes of geometrically interesting objects. For example many presentations of the goemetry of polygons treat almost exclusively polygons that are convex. Some will deal with non-convex polygons, but most avoid any treatment of polygons that intersect themselves. Introducing interactive geometric demonstrations changes all that. If the student is allowed to move a particular vertex in a polygon with more than three vertices, then it is a difficult programming problem to keep him or her from moving that vertex so as to create something non-convex, let alone self-intersecting. A curious student will ask whether or not certain formulas and relationships still hold even when the polygon is not convex or when it self-intersects. Although such investigations fall outside traditional treatments, they arise very naturally when computer programs are used.

As it happens, most commonly used geometry technology can easily handle the area of general polygons, given by a sequence of points in the plane, without ever considering convexity or self-intersection. The way they do this is by using formulas that assign to each triangle an "algebraic area", which is the usual area if the triangle is oriented in a counterclockwise matter, and the negative of that area if the triangle is oriented clockwise.

For example, if a triangle has its first vertex at the vertical axis and the second vertex further above on the vertical axis, then a choice of third vertex determines a positive area if that vertex is to the left of the axis and a negative area if the vertex is to the right. More generally, the oriented line from the first vertex through the second divides the plane into a left side of the line and a right side. The area of a triangle is positive if the third vertex is on the left side and negative if it is on the right side.

What is the advantage of considering both positive and negative areas for triangles? The first one comes when we consider the area of a non-convex quadrilateral. On diagonal lies on the interior of the figure and divides the quadrilateral into two non-overlapping triangles. The sum of the (positive) areas of these two triangles will give the area of the region determined by the quadrilateral. On the other hand, the second diagonal expresses the area of this quadrilateral region as the difference between the area of a large triangle and the area of a triangle contained inside it. If the large triangle is traversed in a counterclockwise way, counted positively, then the second is traversed in a clockwise way and its area is counted negatively.

Note that in this way of counting areas, if we traverse a convex quadrilateral in a counterclockwise manner, then the area is positive as the sum of the areas of two non-overlapping triangles traversed in a counterclockwise manner, while if the quadrilateral is traversed in a clockwise manner, then so are the two triangles and the algebraic area of the quadrilateral is negative, given by the sum of the negative areas of the two triangles.

Standard geometric drawing programs like the Geometer's Sketchpad and the applet used to illustrate the pentagon problem in the NCTM Standards use this method to find areas. If the total area turns out to be negative, then the program reports the absolute value of the oriented area as the area.

This does not cause problems for quadrilaterals, even when the polygon intersects itself, but it does cause problems for pentagons, where the algebraic area of the midpoint polygon can be negative even when the area of the original pentagon is positive.

Self-intersecting quadrilaterals arise naturally when a student is encouraged to explore with a geometry program and moving a vertex causes one edge to cross another, resulting in an "hourglass shape". We may think of this shape as consisting of two triangular regions, one traversed in a counterclockwise manner and the other in a clockwise manner. The algebraic area will be the algebraic sum of these two signed areas. It can be zero if both parts of the hourglass have the same absolute area. If the algebraic area is positive, because the area of the triangle traversed counterclockwise is greater than that of the area traversed clockwise, then if we traverse the quadrilateral in the opposite direction, the roles will be reversed and the algebraic sum will be negative.

We can also express the algebraic area of the hourglass quadrilateral by introducing a "diagonal", for example joining the first vertex to the third. We can then compute the algebraic areas of the two triangles, one positive and one negative. The overlapping region of these two triangles is counted once positively and once negatively, and the leftover parts are the two triangular pieces treated in the previous paragraph.

A great advantage of this method is that it does not make any difference how we divide a given polygon up into a number of triangles. The sum of the algebraic areas of those triangles will always give the sum of the areas of the parts traversed counterclockwise minus the absolute areas of the parts traversed clockwise.

This fundamental intuition is at the heart of the computation of so-called "line integrals" which student will encounter in their third semester of calculus. When we teach this subject, often this is the first time that students are exposed to the concepts of oriented area for smooth curves, without an appreciation of the fact that the same concept works for polygons.

There are formulas for the area of a triangle in terms of the coordinates of the vertices that automatically give the algebraic area. For example, to find the algebraic area of a triangle with first vertex at (0,0), second vertex at (a,b) and third vertex at (c,d), we may use the formula ad - bc. This will be positive if the slope d/c of the line from (0,0) through (c,d) is greater than the slope b/a of the line from (0,0) through (a,b), since d/c > b/a implies that ad > bc and ad - bc > 0. In terms of vectors, if the line along (a,b) rotates in a counterclockwise direction to get to the line along (c,d), then the triangle from (0,0) to (a,b) to (c,d) has positive area, and if it rotates in a clockwise direction, the algebraic area of the triangle is negative.

For those familiar with vectors in physics, another way of expressing this algebraic area is to think of the three vertices as lying in a plane in ordinary three-dimensional space, and computing the cross-product between the vectors (a,b,0) and (c,d,0) to get (0,0,ad-bc). The algebraic area is then obtained by taking the dot product of this vector with (0,0,1). If the rotation from (a,b,0) to (c,d,0) is counterclockwise, then the cross-product points upward and the dot product with (0,0,1) is positive. If the rotation is negative, then the cross-product points in the direction opposite to (0,0,1) and the algebraic area is negative.

A more thorough treatment of this concept of algebraic area can be found in various textbooks in linear algebra. Most secondary school courses will not go far into such subjects, but it can be helpful to students to get an idea of what lies ahead.

## Java Demonstrations

We have provided several demonstrations to help explain the solution to the problem, and to facilitate exploration of midpoint polygons. It should be noted that these demonstrations calculate the

Area of midpoint triangles

Perimeter of midpoint polygons for rectangles

Area of midpoint quadrilaterals

Area maximum for midpoint pentagons

Infinite area ratios for non-convex pentagons

Another nice demonstration can be found on the NCTM standards site, though it calculates the

*oriented*area, which can be negative. If you are unfamiliar with the concept of oriented area, you might want to read the short introduction that we have prepared.Area of midpoint triangles

Perimeter of midpoint polygons for rectangles

Area of midpoint quadrilaterals

Area maximum for midpoint pentagons

Infinite area ratios for non-convex pentagons

Another nice demonstration can be found on the NCTM standards site, though it calculates the

*absolute value*of the oriented area. Although we are taught that "area is always positive", using the oriented area allows us to make a more powerful statement about midpoint polygons in the most general case.