Dynamic Geometry Module Lesson 1: What Is Dynamic Geometry?

Introduction

The term "dynamic" in mathematics refers to the ideas of motion and change. Dynamic Geometry is a new term coined in response to the new software packages such as Sketchpad and Cabri. These products act as a sort of electronic ruler and compass. What really sets a Sketchpad sketch apart from one that you might have made on a sheet of paper with the good old tools of our high school days is not just the accuracy of the constructions. It is the fact that the software remembers relationships among the various components of a construction --- that point P is the midpoint of the segment AB, that circle c has center O and goes through point X etc.

So, for instance, when you construct the circumcircle of a triangle (Open the file ex1_1.gsp found in the Examples folder in the download) you can move any of the vertices of the triangle and see that the circle still circumscribes the triangle. Try this by clicking on a vertex, holding down the mouse, and dragging the vertex around. Notice that you can make the center, O of the circumcircle either inside, outside, or on an edge of the triangle. Can you find a triangle where the circumcenter is at a vertex? What can you say about the triangles where the circumcenter is on one of the edges? These are the kinds of questions you can have students explore when they can see different instances of the same construction just by moving the initial objects (in this case the three vertices of the triangle).

 

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Points equidistant from a pair of points

Let's illustrate how we might guide a student to a simple discovery. Suppose we are trying to find the locus of all point X that are equidistant from points A and B.

We can start with a simple special case. Find all the points that are a given distance from both A and B. The points at distance d from A form a circle (center A and radius d) and the This means that the points at distance d from B form a circle (center B and radius d). Thus the points at distance d from both A and B will be the points of intersection of these two circles.

To illustrate this with a Sketchpad sketch (See file ex1_2.gsp in the Examples folder), you need the points A and B along with a segment whose length we can use for the distance d. On the sketch:

1. Select point A and the segment d,
2. Construct the Circle By Center+Radius,
3. Repeat this process, constructing the circle center B with radius d,
4. Select both these circles,
5. Construct the Point At Intersection of these circles.

Now we are in the position to ask, "What happens if we change the distance d?" To answer this:

1. Select the two points of intersection of the circles drawn previously,
2. Choose Trace Points from the Display menu,
3. Drag one end (either X or Y) of the segment to change the distance d.

When you finish dragging the endpoint around, Sketchpad will connect the traced points with a "curve." This curve looks for all the world like a straight line, or perhaps two pieces of a straight line. What is the relationship between this line and the segment AB? To make sure that this is not a coincidence, move either A, or B, or both, then repeat the process of dragging an endpoint of the segment. Do you still appear to get a straight line?

At this point you could construct the perpendicular bisector of AB and check try dragging an endpoint again. You should see the traced points lying along this perpendicular bisector. In some sense, this is a proof of the result:

The locus of points equidistant from the two distinct points A and B is the perpendicular bisector of the segment AB.

You, of course, could proceed with a proof of this fact. That can be illustrated by other Sketchpad sketches, but the software can't do everything!!

At this point, let us examine a more sophisticated version (See file ex1_3.gsp) of the sketch we just used, and see how it was constructed. Notice that, in this version:

• when X or Y are moved, the segment remains horizontal,
• there is a button you can double click to animate the changing of the distance d --- clicking anywhere in the sketch will halt the animation,
• the circles are drawn as dashed lines,
• the intersection points are traced in red, so the line connecting them at the end of the animation is also in red.

These are all quite simple enhancements. Try these steps on a new sketch.

• First choose the line tool and draw a horizontal line. (Try holding down the shift key while you draw the line if you have trouble making it horizontal.)
• Select the two points that define the line and Hide them.
• Place the point X on this line near the left end and draw a segment from X to another point on the line about 3/4 of the way across the screen.
• Select the original line and Hide it.
• Place a point Y on the segment showing and construct the segment XY.
• Label this segment d.
• Place the points A and B and construct the circles with centers at these points and with radius d.
• Construct the intersection points and change their color to dark red.
• Select the point Y and the longer of the two segments. Then choose Action Button / Animation ... from the Edit menu and click OK. This will produce an animation button. Drag this to the location you would like.
• Select the hand tool then double click on the animation button. This gives you a chance to edit the message in the button. You are limited to 33 characters.
• Select the longer of the two segments and its right endpoint and Hide them.

 

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Areas

Three of the four sketches for this section are actually from the examples that are distributed with Sketchpad itself. The first sketch for this section (See file ex1_4a.gsp) illustrates the formula for the area of a parallelogram. We are assuming that students believe the formula for the area of a rectangle. Area is actually a hard thing to define precisely and the formula for the area of a rectangle is often taken as one of the axioms of area.

This sketch presents a framework for experimentation and discovery, but it does not give any proof of the result it is looking for. The next sketch (See file ex1_4b.gsp) addresses the same issue, the formula for the area of a parallelogram, but it actually includes a proof a text box beneath the sketch. Obviously there are times when one approach is preferred over the other and vice versa.

The two sketches also vary somewhat in the number of objects that are hidden or shown. Remember that if you leave a point or a segment visible, the user can drag it. This movement may affect the way the sketch appears in a manner that does not contribute to its overall effect. The next sketch (See file ex1_5.gsp) illustrates the formula for the area of a trapezoid. The final sketch (See file ex1_6.gsp) illustrates the formula for the area of a triangle. You should experiment with all of these examples. Notice how nicely the sketches behave when the indicated point is dragged. The parallel lines don't move and just the right concept is illustrated. What happens if you move some of the other points in the sketches? These may seem like very simple sketches. Indeed, they are not too complicated, but it does take some practice to get them to behave as nicely as these do. You should try to replicate these sketches. If you are having trouble getting them to work properly, you might try going to the original and using Show All Hidden from the Display menu. This will show you what other objects were constructed and then hidden to make the sketch. It is not always easy to see what is going on, but it may help. If you have messed up the original, don't worry. Just close it without saving and open it again in its original condition! Be sure to contact us at dynamic@mtl.math.uiuc.edu if you have trouble here. The techniques you discover here will be very helpful in creating good sketches of your own later.

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Homework

Complete the exercises in Lesson 1 Homework and submit them to the MTL handin system.

Dynamic Geometry Module Lesson 2: Ratios and Dilations

Ratios

Suppose we have a point X on a segment AB. This point divides the segment into two pieces: AX and XB. It is often important to know the ratio of the lengths of these two segments. this is sometimes referred to as the ratio into which X divides the segment AB. There are two common ways to measure this:

 

You can compute the ratio Length(AX)/Length(XB), which describes the relative sizes of the two pieces. If X is close to A this ratio is close to 0. As X moves along the segment towards B, the ratio gets increases. When X is at the midpoint of AB the ratio is exactly 1. As X gets closer and closer to B the ratio gets larger and larger, without bound.

 

You could also compute the ratio Length(AX)/Length(AB). This measures what fraction of the whole segment is taken up by AX. When X is close to A, this ratio is close to 0. Again, as X moves toward B the ratio increases. This time, however the ratio is 1/2 when X is at the midpoint and is 1 when X is at B.

 

Let us take some time to see how to display these ratios effectively in Sketchpad.

 

1. Open a new sketch. Use the Line tool to draw a horizontal, dashed line, then hide the two defining points for the line.

    

2. Use the segment tool to draw a segment with both endpoints on this line. Make this segment thick and blue. Label the endpoints A and B.

    

3. Construct a point on the segment AB and label it X. Then draw the segment AX and make it thick and red. At this stage your sketch should look like You should be able to slide X back and forth along AB and see the red segment AX expand and contract.

    

4. At this stage we can measure one of the ratios. We really don't need the dashed horizontal line any more so you can select and hide it. Then, holding the shift key down, select first the segment AX and then the segment AB. Choose Ratio from the Measure menu. You should see the ratio displayed. The first segment selected gives the numerator and the second the denominator so the order in which you select the segments is crucial. Make sure that you have the correct ratio displayed.

    

5. To get the first ratio we discussed, you need to construct the segment XB (make it thick and green. Then, holding down the shift key, select first the segment AX and then the segment XB. Note that clicking on either of these segments could actually select segment AB. If this happens, just click again and you should get the next alternative, which will be the segment you desire. Be sure to save this file. You will need it later in the lesson.

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Dilations

A dilation is a transformation that contracts (or expands) all points towards (away from) a given central point by a fixed ratio. Thus, to specify a dilation you have to specify the central point and the ratio. If the ratio is less than 1 then points will be pulled in towards the central point and all distances will shrink by the ratio. In other words, if two points P and Q were distance 6 apart and you applied a dilation with center C and ratio 1/3 then P would go to a point P' and Q would go to a point Q' where now the distance from P' to Q' would be 2.

 

Open the first sample sketch for this lesson (Open file ex2_1.gsp) and experiment with the results of a dilation. This sketch has defined a dilation centered at C with ratio determined by AX/XB. The three blue figures are the originals and the results of dilating them are colored in magenta. Try moving X to get different ratios and moving C to give the original figures a different position relative to the center. You can also change the shape and size of the original figures.

You should be able to make some simple conjectures about the relationship of a figure to its image after a dilation. For instance:

• A circle always dialtes to a ......

   

• A segment always dilates to a .......

   

• A triangle always dilates to a ......

   

When you consider how you would complete these conjectures, fill in both the kind of figure (segment, circle, ellipse rectangle etc.), the shape (similar to the original, larger angles than the original etc.), and the size (relative to the original).

Now let's see if we can create an interesting dilation.

 

1. Open the sketch you saved from earlier in the lesson. Draw a segment at an angle across the sketch. Label the endpoints P and Q. We are going to construct a point Y on PQ so that the ratio PY/YQ is exactly the same as the ratio AX/XB. We will do this by using a dilation to "pull" Q down towards P.

    

   

    

2. Start by setting the center of the dilation at P: use the selection tool and double click on P. You should briefly see a sort of bulls-eye appear around P.

    

3. Select Q and then choose Dilate... from the Transform menu. A dialog box will appear. This lets you set the dilation ratio. It is clear where to enter the ratio you want. In our situation, however (as is often the case), we don't know what the ratio is. We want to use a quantity that is determined by the configuration to set the dilation ratio.

   Notice that the dialog box refers to the ratio in terms of the New distance and the Old distances. In other words, the ratio of PY to PQ. But this is what we want to be the same as the ratio of AX to AB. This is a quantity we have already measured.

    

4. Click on the measurement of the ratio AX/AB in the sketch. The dialog box should change to say that you are now dilating by a Marked Ratio. Click OK. You should see a new point on the segment PQ. Label this point Y.

    

5. Slide X along AB and observe Y slide along PQ. Change the slope and length of PQ and slide X some more.

    

Save this sketch. You will need it in your homework assignment.

 

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Homework

Complete the exercises in Lesson 2 homework and submit them to the MTL handin system.

Dynamic Geometry Module Lesson 3: Ceva's Theorem

Discovering Ceva's Theorem

This result deals with line segments that go from one vertex of a triangle to a point on the opposite side. In tribute to the Italian mathematician Giovanni Ceva, these are sometimes called cevians of the triangle. In the figure below, AW, AX, BY, and CZ (in red) are all examples of cevians for the triangle ABC (in blue).

 

Although the term cevian may be new to you, the concept is certainly not. You have seen examples. For instance, the medians, the altitudes, and the angle bisectors are examples of cevians.

 

You have probably noticed that, in each of these examples, the three cevians all go through a single point. Ceva's Theorem gives a condition that determines whether or not three cevians from the three vertices of a triangle will have this concurrency property.

Open the next sketch (See file ex3_1.gsp). This shows a generic triangle along with three cevians, one from each vertex. We have also included the ratios into which the endpoints of the cevians divide their corresponding sides of the triangle. Use this sketch to perform the following experiments. In each case, record the values for the ratios CX/XB, AY/YC, and BZ/ZA.

 

1. Set X and Y to be as close as you can make them to the midpoints of BC and CA respectively. This means that the corresponding ratios should be as close as you can make them to 1.000. Now slide Z until CZ goes through the intersection point of AX and BY. In other words, until the three cevians are concurrent. What is the value of the final ratio BZ/ZA?

    

2. Move X and Y so that they are each about 1/3 of the way along their segments. That means the ratios will be as close as you can get them to 0.500. Now slide Z until the cevians are concurrent. What is the value of BZ/ZA?

    

3. Move X so that it is as close as you can get to the midpoint of CB. This means that the ratio CY/YB should be as close as you can get to 1.000.

    

   1. Move Y so that the ratio AY/YC is as close as you can get to 2.00. Then slide Z until the cevians are concurrent. What is the value of BZ/ZA?

       

   2. Move Y so that the ratio AY/YC is as close as you can get to 3.00. Then slide Z until the cevians are concurrent. What is the value of BZ/ZA?

       

   3. Move Y so that the ratio AY/YC is as close as you can get to 4.00. Then slide Z until the cevians are concurrent. What is the value of BZ/ZA?

       

   Can you see any pattern here?

    

4. Select the three measurements and then use Calculate... from the Measure menu to compute the product of these three ratios. Return to the sketch and move the vertices and the cevians into different positions, but make sure that the cevians are concurrent. What is the value of the product of the three ratios in each case?

    

By this stage you should have discovered:

 

Ceva's Theorem:

If the three cevians AX, BY and CZ are concurrent, then the product of the ratios (CX/XB) (BZ/ZA) (AY/YC) = 1.

 

The ratios are taken in the same "order," that is, either all counterclockwise (as in the statement of the theorem above) or, equivalently, all clockwise. In the clockwise case the condition would be

(BX/XC) (CY/YA) (AZ/ZB) = 1.

 

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The Extended Version of Ceva's Theorem

First we should consider what we mean by the ratio in which a point divides a segment if the point is not actually inside the segment. This may sound kind of silly, but consider the following diagram.

where X is on the line determined by A and B but is not between them. Although you might well say that X does not divide the segment at all, you can still make perfectly good sense of the ratio AX/XB. In the picture below, the red segment is equal in length to AX and the blue on to XB

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You can open a Sketchpad version of this ( See file ex3_2.gsp). Once you have done that, drag the point X along the line and observe the changes in the ratio.

Now we can create a sketch that will check Ceva's Theorem for more general cevians. Open a New Sketch. In this sketch:

 

1. draw three blue lines to form a triangle. Label the vertices of this triangle A, B, and C,
2. construct points X (on BC) and Y (on AC) then construct two dashed red lines AX and BY,
3. construct the intersection point, O, of these two lines,
4. construct the dashed red line CO and construct its intersection point, Z, with the line AB,
5. construct the thick green segments CX, XB, BZ, ZA, AY, and YC and use them to measure the ratios CX/XB, BZ/ZA, and AY/YC,
6. select these ratios and use them to calculate the product (CX/XB) (BZ/ZA) (AY/YC).

Now you can drag X and Y along their respective lines and the three cevians should remain concurrent. What happens to the product of the ratios?

 

In other words, Ceva's Theorem is still true when the cevians do not all lie in the interior of the triangle.

 

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Proving Ceva's Theorem

To prove Ceva's Theorem in the case where all the cevians lie in the interior of the triangle, we have to show that if the Cevians are concurrent then the product of the ratios is 1. Use the point of concurrency, O, to form the three triangles (colored red, green, and blue in the figure below) AOC, COB, and BOA.

 

From your homework for Lesson 2, you know, for instance, that the ratio Area(COB)/Area(AOC) is equal to the ratio BZ/ZA. Using this, and similar equalities, you can complete the proof. This is one of your homework exercises.

 

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Homework

Complete the exercises in Lesson 3 homework and submit them to the MTL handin system.

Dynamic Geometry Module Lesson 4: Circles and Tangents

A Tangent Line to One Circle

One of your prerequisite homework exercises was to construct a line tangent to a circle C through a point P outside the circle. The key here was to recall that the radial line (in red in the figure below) from the center, O, through the point of contact, X, is perpendicular to the tangent line (in blue).

 

Let's see how to create a dynamic sketch that should induce a student to discover how to use this fact to come up with the construction of the tangent line. This is another great use of the dynamic features of Sketchpad.

Open a new sketch and perform the following steps:

 

1. draw a circle and a point, P, outside that circle,
2. create a point A anywhere in the plane,
3. draw the line through A and the center, O, of the circle,
4. draw the line perpendicular to this, though P,
5. construct the point of intersection, Q, of these two lines.

 

You have constructed a line, PQ, perpendicular to the radial line AO. Moreover, for this choice of A, Q is the only point that will make PQ perpendicular to the radial line AO. The only problem is that this line PQ is almost certainly not a tangent line! To find the tangent line, we need to choose a different radial line and hence a different perpendicular.

Try dragging point A, to get a new radial line, until Q sits right on the circle. At this point PQ will be the tangent line we are looking for. Unfortunately, this doesn't tell us how to make an accurate construction that will create the correct position of Q.

Now select Q and choose Trace Point from the Display menu. Then drag the point A completely around circle C. What is the locus traced by Q during this motion? What points does it go through? Is OP a diameter of this locus and why?

The answers to these questions should lead to the appropriate construction:

 

1. construct the segment OP,
2. construct the midpoint of this segment,
3. construct the circle centered at this midpoint and going through O and P,
4. find the intersection points of this circle with the original circle,
5. construct the lines through P and each of these intersection points,
6. select and hide all the objects that are not needed in the final figure.

If you try this construction you will get both the tangent lines to C through P.

 

 

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A Tangent Line to Two Circles

In this section we want to start with two circles, each one outside the other. We are looking for a line that is tangent to both these circles. If you try to draw the picture, you will see that there are essentially two different ways that this can happen. The line can touch each of the circles, with both of the circles on the same side of the line (the blue line in the figure below), or it can touch each of the circles with the circles on opposite sides of the line the red line in the figure below).

 

The first case (the blue line) is called an external common tangent for the two circles, while the other case (the red line) is called an internal tangent. In the rest of this lesson, we will only work with external common tangents. You will create similar constructions with internal common tangents in the homework exercises.

First, let us try a few experiments to try to draw the common external tangents to a pair of circles and see if they give us any hint as to how we might find a construction. Begin by opening a new sketch and drawing two circles, C and D, each outside the other. Pick any point P in the plane and draw the two tangents to circle C through P.

 

Now move P until the two lines are tangent to D as well as to C. Can you see anything about where P must be in relation to the two original circles? Try moving the circles and repeating to see if your idea still seems to hold.

Open a new sketch and draw two circles as before. Construct points X and Y on circle C and then draw the radial segments from the center of C to both X and Y. Draw lines perpendicular to each of these segments at X and Y respectively. By construction these must be tangent to C.

 

Move X and Y until you have found (at least approximately) the two common external tangents. What can you say about this configuration? Do the common tangents intersect? If so, do they intersect in a point on the line between the centers of the circles? If you knew the point of contact, X, of one common tangent, how could you find the point of contact, Y, of the other? Think of the symmetry of the picture.

If we have an external common tangent, then the two radii, one for each circle, out to the points of contact, must both be perpendicular to this line. This means that they must be parallel.

 

We could try to find the common tangent, then, by drawing parallel radii and joining their endpoints. Open a new sketch and draw the two circles as before. Draw a radius OX for circle C. Then draw a parallel line through the center O' of D and find its intersection, Y, with D so that you can draw a radius O'Y that is parallel to OX. Then construct the line joining X and Y.

 

This is probably not the common tangent that we are looking for. We will find that by changing the position of X. Drag X around circle C until you appear to have the common tangent.

We now have three different ways to get a pretty good approximation for the common tangent line. Unfortunately, none of these has yet given us a precise construction. In the final sketch you created, select the line XY and choose Trace Line from the Display menu. Now repeat the process of dragging X around circle C. What do you notice about the images of the line XY? Do they all go through a single point? If they do, then how could you find that point? Once you know how to locate this point, how can you use it to find the common external tangent(s)?

Answers to these questions give you a method for finding the external tangents.

 

1. Draw parallel radii OX for circle C and O'Y for circle D.
2. Join X and Y with a line.
3. Join the centers O and O' with a line.
4. Construct the point of intersection, P, of these two lines.
5. Construct the tangent line from P to either C or D using the construction method from earlier in this lesson.

The point of intersection of the two external common tangents is called the external center of similitude of the two circles.

Practice this construction several times until you are sure that you have it mastered. Do not hesitate to contact us for help if you are not sure about any step. It is essential that you understand this. Also, be sure that you hide all auxiliary lines and points. At the end of the construction the only new objects on your sketch should be the to tangent lines, their intersection point and their points of contact with the circles.

 

At this stage, it would be good practice to create a custom tool for this construction. Open a New Sketch from the File menu and draw two circles as usual. Complete the entire sketch, constructing the common tangents, and hiding the appropriate auxiliary constructions. Once the sketch is complete, do an Edit, Select All command, and then click on the Custom Tool button, selecting Create New Tool. Name the new tool "external center" and click on the box for Show Script View. You will now see a step by step listing of each of the actions you have taken to create the external centers of similitude. Check the beginning of your script to see what the Given requirements are for this script. This list will depend on how you did your construction, but it will probably contain at least the two circles and their centers. As described in the section on custom tools in Module 4, when you want to replicate this construction for any other two circles, you use the Custom Tool button, select the "external center" tool, and ask it to Show Script View. Then in the script view window, click on All Steps. This method will work particularly well as you do Lesson 4, Excercise 2 in the homework, where you need to create three different centers of similitude in the same drawing.

When you want to use the script, you need to select the corresponding objects on a sketch, in the order they appear in the Given list, and the click either PLAY or FAST on the script. Be sure to save your script. Then try it out a few times to make sure that you know how to use it and to make sure that it works correctly. There is no good way to edit scripts since they are just recordings of actions. If you need to make a correction or alteration, you need to start over again with a new recording of the script.

You should notice that as long as one circle is not completely inside the other and the circles do not have the same radius, then the construction works properly. What happens if the circles have the same radius? Is there a common tangent line(s)? What happens if one circle is inside the other?

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Monge's Theorem

This theorem has to do with what happens when we have three circles. Open the sample sketch for this lesson (See file ex4_1.gsp). It shows three circles c1, c2, and c3. The red lines are common tangents to c1 and c2, the blue lines the common tangents to c2 and c3, and the green lines the common tangents to c1 and c3. You can move the centers of any of the circles and adjust their radii by moving the points labeled A, B, or C.

Label the center of similitude of c1 and c2 as X, the center of similitude of c2 and c3 as Y and the center of similitude of c1 and c3 as Z. Do you notice any relationship among these three points X, Y, and Z? Move the configuration and see if this still seems to be the case.

The three centers of similitude appear to be collinear (lined up on a single straight line). How would you check that? Move the configuration and see if your check still holds. This is Monge's Theorem:

 

For three disjoint circles of unequal radii with no one circle contained in any other, the three external centers of similitude for the three pairs of circles are collinear.

 

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Homework

Complete the exercises in Lesson 4 homework and submit them to the MTL handin system.

Dynamic Geometry Module Lesson 5: The Center of Things

Centers We Know and Love

The medians of a triangle all intersect in a point (the centroid of the triangle). The same is true of the angle bisectors (the incenter), the altitudes (the orthocenter) and the perpendicular bisectors of the sides (the circumcenter). These are all examples of important "centers" for a triangle.

Use Sketchpad to construct each of these centers.

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Napoleon's Configuration

This is also referred to as the Toricelli Configuration. It consists of an equilateral triangle drawn outward from each side of a triangle. You can see an example in the accompanying sketch for this lesson (See file ex5_1.gsp).

In this sketch, P, Q, and R are the vertices of the equilateral triangles and X, Y, and Z are their centroids (which also happen to be the circumcenters, incenters and orthocenters!). Napoleon's Theorem states that the centers X, Y, and Z form an equilateral triangle. Join these points and measure the sides of the resulting triangle to verify this. Be sure to move the vertices around and see that this property holds in all cases.

Now delete those segments and join each center to the vertex of the original triangle that is opposite it. That is, draw the lines AX, BY, and CZ. Notice that these lines meet in a point. This point is called the First Napoleon Center for the triangle. Is it always inside the original triangle? If not, can you determine under what conditions it will be outside?

Delete these lines and construct the lines AP, BQ, and CR. These also meet in a point. This is called the First Brocard Point for the triangle. Is it always inside the triangle? If not, can you determine when it is outside? Construct this intersection point and label it O. Measure the angles AOB, BOC, and COA. What do you find? Does this hold when you change the original triangle?

 

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The Gergonne Point

Open a new sketch and draw an arbitrary triangle ABC. Draw the inscribed circle and construct the point, X, where this circle touches BC; Y where this circle touches AC; Z where this circle touches AB. Then draw the lines AX, BY, and CZ. They meet in the Gergonne Point.

 

Save this sketch. You will need it for your homework.

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The Nagel Point

Let X be the point where the ex-circle opposite A touches BC; Y the point where the ex-circle opposite B touches AC; Z the point where the ex-circle opposite C touches AB.

 

The Nagel Point is the intersection of the lines AX, BY, and CZ. Beneath the figure for the Gergonne Point, create a figure for the Nagel point. Save the resulting sketch.

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And There's More!

There are almost as many such centers as you can imagine. The homework exercises ask you to draw diagrams for a few more and to invent some of your own!

For instance, here are some interesting centers:

 

1. The intersections of the symmedians. A symmedian is the line you get by reflecting a median across the angle bisector at its vertex. This is the Lemoine Point.

    

   

    

2. The angle bisectors of the triangle formed by the midpoints of the sides of a triangle meet at the Spieker Center of the triangle.

    

   

    

3. The orthic triangle is the triangle formed by the feet of the altitudes. The tangential triangle is the triangle formed by the tangents to the circumcircle at the vertices. These two new triangles have parallel sides! When corresponding vertices are joined, the resulting lines meet is a point. This point, sad to say, has no particular name.

    

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Homework

Complete the exercises in Lesson 5 homework and submit them to the MTL handin system.

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ex5_1_gsp.zip	999 bytes