# Demonstrating inequality.

This article heralds the release of version 2 of Geometry
Expressions by Saltire Software. You can download a demonstration
version for either Windows or Intel Macintosh at
www.geometryexpressions.com. This site details the changes made in
version 2. It also features examples of how to use the software and
includes an online User Manual. When I got really stuck on Saturday
evening, I sent an email. There was a helpful response on Sunday
morning--now that's service.

The partial screen dump printed below shows that the user interface remains largely unchanged. The tool icons are racked logically on the right of the desktop and they are duplicated in the pull-down menus. When using the software by myself, I use the icons, but when I am demonstrating to students, I think they have a better chance to see what is happening if I use the pull-down menus.

The exercise described here will work just as well using version 1. On the left I have summarised the traditional geometric construction that illustrates the inequality relating the arithmetic, geometric and harmonic means of two positive real numbers. Here we duplicate the demonstration using Geometry Expressions software.

The first step is to draw a circle centre C on a diameter AB. I chose to draw a line segment AB, found its midpoint and then drew the circle with centre C passing through B. However, because Geometric Expressions uses constraint-based geometry, we could just as easily have drawn a circle with a chord AB and constrained the chord to pass through the centre.

[FIGURE 1 OMITTED]

[FIGURE 2 OMITTED]

The advantage of the constraint system is better illustrated in the next step. We need to draw a half chord perpendicular to AB at a point E to meet the circle at D. Using traditional geometry packages we would need to construct a perpendicular through E, find the point of intersection with the circle and then join DE. To be tidy we would then hide the perpendicular.

Using this package we choose any point D on the circle and join it to any point E on the diameter as shown in Figure 2. We then select AB and DE using the shift key and click on the third constraint icon which forces the two line segments to be perpendicular, as shown in Figure 3.

Another powerful feature of this software is a built in algebra system. If we highlight a line segment and click the length constraint icon, the segment is allocated a variable.

For example, to tell the system that the line segment AE has a length x, we need to draw the segment AE using the line segment tool. To distinguish between AE and AB we aim the cursor near the centre of the line segment we want to select and then click. The selected line segment is then highlighted to reassure you that you have got it right. Then you click on the length constraint tool and the line segment is labelled a. This can be changed to x simply by typing x. Repeat with EB to define y.

[FIGURE 3 OMITTED]

[FIGURE 4 OMITTED]

Now here is the clever bit: go through the same procedure with the line segment CD but when the new variable is allocated, press the return key or pull down the Draw menu and select Expression (Figure 4). Immediately, a new drop-down menu asks if you want to back out, define the new variable in terms of previously defined variables, or relax previously defined constraints to ensure independence of the new variable (Figure 5). When you select the second option, the software adds the algebraic expression

[z.sub.0] [right arrow] x + y/2

and when you double-click, the z0 can be edited to a (Figure 6).

The drop-down menu makes the process more complicated but it clearly provides an early intervention mechanism to avoid further problems developing.

Now choose the line segment DE and repeat the procedure. Click on the length restraint icon and press return. Choose to define the new variable in terms of existing variables and then edit the expression z0 to be g as shown in Figure 7.

[FIGURE 5 OMITTED]

[FIGURE 6 OMITTED]

At this stage we should drag E left and right along the diameter to show that for all possible positions the value of a is greater than or equal to g. Thus illustrating the AG inequality which for two variables is:

x+y/2 [greater than or equal to] [square root of x.y]

This model can now be extended by drawing a line segment from E to a point F on the line segment CD. Now select both DC and EF and click on the third constraint icon which forces the line segments to be perpendicular. Draw the line segment DF and select it. Click on the length constraint icon, press return and define the length of DF in terms of x and y. This gives

[z.sub.0] [right arrow] 2xy/x + y

and by double clicking we can edit the z0 to h the harmonic mean, as shown in Figure 8.

The hypotenuse of triangle DEF is greater than or equal to the side DF. Therefore the geometric mean g is always greater than or equal to the harmonic mean h.

Draw a line segment EH where H is a point on ED. Use the shift key to highlight both EH and DF and then by clicking the congruence constraint icon we force EH to have the same length (h) as DF. As we move point E left and right, the relative positions of H and D make it clear that g [greater than or equal to] h for all values of x and y.

[FIGURE 7 OMITTED]

[FIGURE 8 OMITTED]

If we also compare the height of H and D with a point R at the top of the circle, we can illustrate the overall inequality a [greater than or equal to] g [greater than or equal to] h for all values of x and y.

It is also possible to build a separate graphing section using the locus facility. I removed the line segment EH and instead drew a line segment OH equal in length to AE and drew a perpendicular at H. On this perpendicular I drew line segments HL = DF, HK = ED and HJ = CD. As E is moved left and right on AB, the line segment HJ also moves left and right and should allow the points J, K and L to trace graphs of the three means. They do--but not in the way I expected.

I sent an email to Philip Todd at Saltire Software and within a few hours received a reply describing the different trace and locus systems used and the modifications he had made to the model, as shown in Figure 9. From the locus curves he found the equations of the curves and we can see that the arithmetic mean generates a linear locus, the geometric mean generates a parabola and the harmonic mean a hyperbola. This is high power use by a man who knows his software and well illustrates some of the powerful new features of Geometric Expressions version 2.

[FIGURE 9 OMITTED]

The ancient approach

In Figure 1, a point E is free to move along the diameter of a circle centre C and radius a. The lengths of AE and EB are designated the variables x and y. Thus the diameter of the circle is (x + y) units and the radius is half of that. Therefore the radius has a length equal to the arithmetic mean of x and y.

CD = a = x+y/2

Similarly, the length of EC is half the difference between x and y.

Triangle DEC is right angled at E and so using Pythagoras' Theorem

[(x+y).sup.2]/4 = [DE.sup.2] + [(x - y).sup.2]/4

From which we find that DE is equal to the geometric mean of x and y.

DE = g = [square root xy]

If we consider the triangle DEC to have a base EC and an altitude of DE, the area will be half EC x DE. But if we consider it to have base DC we get an area of half DC x EF.

Thus EF x DC = EC x DE giving us that

EF = [absolute value of x - y] [square root of xy]/x + y

Triangle DEF is right angled at F and so (again using Pythagoras)

[([square root xy]).sup.2] = [DF.sup.2] + xy [(x - y/x + y).sup.2]

From which we find that DF is equal to the harmonic mean of x and y. Thus we obtain

2/h = 1/x + 1/y

and therefore DF = h = -2xy/x + y

The hypotenuse is the longest side of a right-angled triangle, so we know from triangle DEC that DC > DE and from triangle DEF that DE > DF. In the case where E and C coincide we must allow equality. Therefore DC [greater than or equal to] DE [greater than or equal to] DF which demonstrates the inequality that a [greater than or equal to] g [greater than or equal to] h.

Hartley Hyde

cactus.pages@internode.on.net

The partial screen dump printed below shows that the user interface remains largely unchanged. The tool icons are racked logically on the right of the desktop and they are duplicated in the pull-down menus. When using the software by myself, I use the icons, but when I am demonstrating to students, I think they have a better chance to see what is happening if I use the pull-down menus.

The exercise described here will work just as well using version 1. On the left I have summarised the traditional geometric construction that illustrates the inequality relating the arithmetic, geometric and harmonic means of two positive real numbers. Here we duplicate the demonstration using Geometry Expressions software.

The first step is to draw a circle centre C on a diameter AB. I chose to draw a line segment AB, found its midpoint and then drew the circle with centre C passing through B. However, because Geometric Expressions uses constraint-based geometry, we could just as easily have drawn a circle with a chord AB and constrained the chord to pass through the centre.

[FIGURE 1 OMITTED]

[FIGURE 2 OMITTED]

The advantage of the constraint system is better illustrated in the next step. We need to draw a half chord perpendicular to AB at a point E to meet the circle at D. Using traditional geometry packages we would need to construct a perpendicular through E, find the point of intersection with the circle and then join DE. To be tidy we would then hide the perpendicular.

Using this package we choose any point D on the circle and join it to any point E on the diameter as shown in Figure 2. We then select AB and DE using the shift key and click on the third constraint icon which forces the two line segments to be perpendicular, as shown in Figure 3.

Another powerful feature of this software is a built in algebra system. If we highlight a line segment and click the length constraint icon, the segment is allocated a variable.

For example, to tell the system that the line segment AE has a length x, we need to draw the segment AE using the line segment tool. To distinguish between AE and AB we aim the cursor near the centre of the line segment we want to select and then click. The selected line segment is then highlighted to reassure you that you have got it right. Then you click on the length constraint tool and the line segment is labelled a. This can be changed to x simply by typing x. Repeat with EB to define y.

[FIGURE 3 OMITTED]

[FIGURE 4 OMITTED]

Now here is the clever bit: go through the same procedure with the line segment CD but when the new variable is allocated, press the return key or pull down the Draw menu and select Expression (Figure 4). Immediately, a new drop-down menu asks if you want to back out, define the new variable in terms of previously defined variables, or relax previously defined constraints to ensure independence of the new variable (Figure 5). When you select the second option, the software adds the algebraic expression

[z.sub.0] [right arrow] x + y/2

and when you double-click, the z0 can be edited to a (Figure 6).

The drop-down menu makes the process more complicated but it clearly provides an early intervention mechanism to avoid further problems developing.

Now choose the line segment DE and repeat the procedure. Click on the length restraint icon and press return. Choose to define the new variable in terms of existing variables and then edit the expression z0 to be g as shown in Figure 7.

[FIGURE 5 OMITTED]

[FIGURE 6 OMITTED]

At this stage we should drag E left and right along the diameter to show that for all possible positions the value of a is greater than or equal to g. Thus illustrating the AG inequality which for two variables is:

x+y/2 [greater than or equal to] [square root of x.y]

This model can now be extended by drawing a line segment from E to a point F on the line segment CD. Now select both DC and EF and click on the third constraint icon which forces the line segments to be perpendicular. Draw the line segment DF and select it. Click on the length constraint icon, press return and define the length of DF in terms of x and y. This gives

[z.sub.0] [right arrow] 2xy/x + y

and by double clicking we can edit the z0 to h the harmonic mean, as shown in Figure 8.

The hypotenuse of triangle DEF is greater than or equal to the side DF. Therefore the geometric mean g is always greater than or equal to the harmonic mean h.

Draw a line segment EH where H is a point on ED. Use the shift key to highlight both EH and DF and then by clicking the congruence constraint icon we force EH to have the same length (h) as DF. As we move point E left and right, the relative positions of H and D make it clear that g [greater than or equal to] h for all values of x and y.

[FIGURE 7 OMITTED]

[FIGURE 8 OMITTED]

If we also compare the height of H and D with a point R at the top of the circle, we can illustrate the overall inequality a [greater than or equal to] g [greater than or equal to] h for all values of x and y.

It is also possible to build a separate graphing section using the locus facility. I removed the line segment EH and instead drew a line segment OH equal in length to AE and drew a perpendicular at H. On this perpendicular I drew line segments HL = DF, HK = ED and HJ = CD. As E is moved left and right on AB, the line segment HJ also moves left and right and should allow the points J, K and L to trace graphs of the three means. They do--but not in the way I expected.

I sent an email to Philip Todd at Saltire Software and within a few hours received a reply describing the different trace and locus systems used and the modifications he had made to the model, as shown in Figure 9. From the locus curves he found the equations of the curves and we can see that the arithmetic mean generates a linear locus, the geometric mean generates a parabola and the harmonic mean a hyperbola. This is high power use by a man who knows his software and well illustrates some of the powerful new features of Geometric Expressions version 2.

[FIGURE 9 OMITTED]

The ancient approach

In Figure 1, a point E is free to move along the diameter of a circle centre C and radius a. The lengths of AE and EB are designated the variables x and y. Thus the diameter of the circle is (x + y) units and the radius is half of that. Therefore the radius has a length equal to the arithmetic mean of x and y.

CD = a = x+y/2

Similarly, the length of EC is half the difference between x and y.

Triangle DEC is right angled at E and so using Pythagoras' Theorem

[(x+y).sup.2]/4 = [DE.sup.2] + [(x - y).sup.2]/4

From which we find that DE is equal to the geometric mean of x and y.

DE = g = [square root xy]

If we consider the triangle DEC to have a base EC and an altitude of DE, the area will be half EC x DE. But if we consider it to have base DC we get an area of half DC x EF.

Thus EF x DC = EC x DE giving us that

EF = [absolute value of x - y] [square root of xy]/x + y

Triangle DEF is right angled at F and so (again using Pythagoras)

[([square root xy]).sup.2] = [DF.sup.2] + xy [(x - y/x + y).sup.2]

From which we find that DF is equal to the harmonic mean of x and y. Thus we obtain

2/h = 1/x + 1/y

and therefore DF = h = -2xy/x + y

The hypotenuse is the longest side of a right-angled triangle, so we know from triangle DEC that DC > DE and from triangle DEF that DE > DF. In the case where E and C coincide we must allow equality. Therefore DC [greater than or equal to] DE [greater than or equal to] DF which demonstrates the inequality that a [greater than or equal to] g [greater than or equal to] h.

Hartley Hyde

cactus.pages@internode.on.net

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Title Annotation: | Calculator And Computer Technology User Service |
---|---|

Author: | Hyde, Hartley |

Publication: | Australian Mathematics Teacher |

Geographic Code: | 8AUST |

Date: | Mar 22, 2011 |

Words: | 1541 |

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