Homework: Geometer's Sketchpad Homework
Up: Solutions Manual Table of Contents

Geometer's Sketchpad Homework
Solutions Manual

SOLUTIONS ARE IN BOLD

Work through the first part of Jim King's notes on Sketchpad. The more you do, the easier you will find labs using Sketchpad. At the very least, learn to do all of the items on the following list of basic Sketchpad commands:
1. Construct points, segments, lines, circles.
2. Label an object. Change the name of the object.
3. Construct a segment, line, or ray through two specified points.
4. Construct a circle given its center and a point on it. Construct a circle given its center and radius.
5. Construct the intersection of two objects.
6. Trace the locus of an object.
7. Construct the midpoint of a segment.
8. Given a line j and a point A, construct lines perpendicular and parallel to j through A.
9. Make and save a Script.
10. Translate, rotate, and dilate. Specifically, given a segment AB, find a point C such that AC = .3 (AB) such as in the following picture:
  A correctly constructed sketch is available here. [GSP Help]
  This was constructed by drawing line segment AB. Select point A only and choose from the menu Transform / Mark Center. Then select point B only. Choose Transform / Dilate... and type in 0.3 for the New Scale Factor.
11. Save a Sketch.
12. Hide an object. This is never explicitly mentioned in the homework, but if the sketch becomes too complicated, you may wish to hide objects to make it simpler. In particular, do not delete objects used in defining other objects. Only hide them.
The following questions are designed to give a general familiarity with Sketchpad. They can be done at any time in relationship to the Sketchpad labs.
Any three noncollinear points lie on a circle. In this exercise, you will use Sketchpad to construct the circle through three points. Turn in the sketches described below as "sketch1" and "sketch2", which is the completed picture. Briefly summarize the mathematics justifying what you did to make the completed picture.
This exercise uses Sketchpad to discover that the center of the desired circle is the circumcenter of the triangle with the three points as its vertices. A triangle's circumcenter is the point of intersection of the three perpendicular bisectors of the triangle's sides. The important property of the circumcenter is that it is equidistant from the three vertices, thus being the desired center of the circle.
1. Construct three noncollinear points. Use the Show Labels command on the Display menu so you can distinguish them. Save this as "sketch1".
  Eventually, you will construct the circle through these points. However, first think about all circles passing through two points.
2. Create a separate sketch ("sketch2") consisting of just two points, A and B. Where are the centers of circles through these two points?
  The answer to this is discovered in parts c and d.
3. Construct circles containing the two points A and B: Construct a line segment at the top of your sketch. This represents the radius of the circle containing A and B. Construct a circle with center A and radius given by the segment. Then construct a circle with center B and radius given by the segment. The intersection points of these two circles are possible centers of circles containing A and B with radius given by the segment. Think about why this is true.
  The circles centered at A and B were constructed to have the same radii so the distances from the centers of the circles to a point of intersection are equal. Thus a circle centered at one of these points of intersection containing point A will als o contain point B.
  Are there any other circles with this radius that go through A and B?
  There are only two circle with this radius - one centered at each intersection point of the circles.
  The intersections are at points F and G in the picture below. Now pick either F or G to be the center of a circle, and pick A to be a point on the circle. Notice that B is also on the circle. This supports the statement above about circles containing A and B.
  A sample of how "sketch2" should be at this point is available by clicking here. [GSP Help]
4. Look at all the possible centers of circles through A and B. To do this, select the points F and G, and pick the Trace Locus command from the Display menu. Vary the length of the segment. What does the set of centers for circles through A and B look like? (Think about this before you look at the next part.)
  A sample of how "sketch2" should be at this point is available by clicking here. [GSP Help]
  
  The centers of the circles through A and B would lie on the perpendicular bisector of the segment connecting A and B.
  General Explanation:
  We see from the picture that all centers of circles through A and B lie on a line. Also, the midpoint between A and B is on the same line since it is a center of a circle through A and B. Since the points F and G are reflections across the segment AB, w e must have a perpendicular line by symmetry.
  Analytic Proof:
  Introduce a coordinate system such that segment AB is on the x-axis and the midpoint of segment AB is at the origin. Let (p,0) be the coordinate of B. Then (-p,0) is the coordinates of A. Let F (x,y) by any point that is the center of a circle through A and B. We will show that F must line on the y-axis, thus being on the perpendicular bisector of segment AB (on the x-axis).
  We know that the distance from A to F must be equal to the distance from B to F. Using the distance formula (actually squared) gives
  (-p - x)^2 + (0 - y)^2 = (p - x)^2 + (0 - y)^2
  p^2 + 2px + x^2 = p^2 - 2px + x^2
  2px = -2px x = 0
  Thus the coordinates of F are (0,y), so F is on the y-axis.
5. Construct this line of possible centers: make a segment with endpoints A and B. Construct the Point at Midpoint of the segment. Select the midpoint and segment, and construct the Perpendicular Line.
  The final version of "sketch2" is available by clicking here. [GSP Help]
6. You may want to find the perpendicular bisector of the segment between two points again. Make a script to do it in one of the following two ways:
  1. Select everything in the sketch. Select Make Script from the Work menu. Save the script.
  2. Start with two new points. Select New Script from the File menu. On the Script window, hit the record button. Do all the steps to create the perpendicular bisector. Hit the Stop button. Save the script. Select two new points, and try playing it back.
  The script to create a perpendicular bisector is available by clicking here. [GSP Help]
7. Go back to sketch1. For each pair of points, there is a line of centers of circles through the pair of points. Construct these lines using your script.
  A sample of how "sketch1" should be at this point is available by clicking here. [GSP Help]
8. How do you know that the three lines intersect at a point? (State a theorem from geometry.) This intersection point is the center of the circle through the three points.
  The following theorem is directly out of a high school geometry textbook.
  This point of intersection is called the circumcenter, which is the center of the circle passing through points A, B, and C (which we proved in part d).
9. Construct the circle through the three points.
  The final version of "sketch1" is available by clicking here. [GSP Help]
Given two points A and B, what is the set of points P such that AP + BP = a constant K? ("AP" is my shorthand for "the distance from point A to point P." Also, K must be greater than the distance AB, or there are no points P.) In this exercise, you use Sketchpad to look at this set and formulate a conjecture about its shape. (Do you remember what the shape will be from previous geometry courses?)
This is the definition of an ellipse with foci A and B.
Turn in your final sketch. Briefly describe the mathematics of how you constructed it. Write your conjecture for the shape traced in part d. Answer the questions in part e.
1. Construct two points A and B. Construct a segment CD at the top of your sketch. The distance CD is the constant K. Construct a point E on this segment CD.
2. Notice that CE + ED = constant K. Use this to construct P as follows; construct the set of points distance CE from A. Construct the set of points distance ED from B.
  A circle centered at A with radius CE is the set of points distance CE from A. To do this, first hide segment CD. Then construct segments CE and ED. Finally, select segment CE and point A and go to Construct / Circle By Center+Radius. Likewise, construct the circle centered at B with radius ED.
3. The points of intersection of these two circles (F and G) have the property we wanted to study. Construct the points F and G, the intersection of the circles.
  A sample of what the sketch should be at this point is available by clicking here. [GSP Help]
4. Trace Locus of F and G as you vary E. What does this set look like?
  The traced locus lie on an ellipse which is available by clicking here. [GSP Help]
5. Experiment: move A and B around. Change the constant K (by moving D). Now trace the locus in part d again. How does the value of K effect the set of points you trace? What would happen if K were exactly distance AB?
  For values of K near AB (the distance from A to B), the eccentricity of the ellipse is near 1 (but less than 1).
  By increasing the constant K (making segment CD longer), the eccentricity of the ellipse formed will decrease closer to 0, meaning it looks more like a circle. The eccentricity of the ellipse could only be 0 if A and B were the same point. In this case th e ellipse would be a circle.
  When the constant K is equal to the distance AB, the ellipse collapses into line segment AB. This is because the two circles intersect at only one location (F = G). The circles are tangent to one another in this case.
6. By now you should have a conjecture about the traced set. Can you prove your conjecture? Place the origin at the midpoint of the line segment between A and B. Write equations involving the coordinates for points in the traced set. Write an equation describing the shape that you conjecture. Are the these equivalent?
  The sketch of this last part is here. [GSP Help]
  
  Proof:
  To derive the equation of the set of points P, let the segment AB lie on the x-axis with its midpoint being at the origin. For any point P(x,y) we know that AP + PB = K. Let (c,0) be the coordinates of B. Then the coordinates of A is (-c,0). By the distance formula we have
  
  squareroot[(x + c)^2 + (y - 0)^2] + squareroot[(x - c)^2 + (y - 0)^2] = K.
  By simplification this reduces to
  
  (4K^2 - 16c^2)x^2 - 4(Ky)^2 = (K^2 - 4c^2)K^2.
  By letting b^2 = (K/2)^2 - c^2 this becomes
  
  (4x^2)/(K^2) + (y^2)/(b^2) = 1.
  Finally, if we let a = K/2 we get the standard equation for an ellipse,
  
  (x^2)/(a^2) + (y^2)/(b^2) = 1.
Given a point A and a line j, what is set of points P such that for some e< 1, PA = e (distance from P to j) ? In this exercise, you use Sketchpad to look at this set. Refer to the following picture.

Turn in your final sketch. Briefly describe the mathematics of how you constructed it. Answer the questions in part f.
This exercise also creates an ellipse with eccentricity e, although in a very different manner than in the previous problem.
1. Construct a point A and a line j.
2. Construct a point D not on j. Construct a line m parallel to j through D.
3. You need a segment that has length equal to the distance from j to n. Construct a line n perpendicular to j through D. Construct the intersection point E between n and j. The distance ED is the distance from j to m. Do you see why?
4. You want a segment length e times ED. Thus use dilation: choose E and D. In the Transform menu, select the Mark Vector command. Now use the Mark Center command. Now use the dilate command. Dilate by fixed ratio e (the picture uses e=.8) to 1. This will give you a point D' such that ED' =e ED.
5. Make a circle around A with radius given by ED'.
6. Finally, the intersection F and G of the circle with line m is the set in question. Trace the locus of intersection points as you move D. What do you think this set is? How would the set change if you used a different number e?
  When point D moves, F and G trace out an elipse.
  A value of e close to 1 makes the traced ellipse have an eccentricity close to1 - in other words is appears stretched out. A value of e close to 0 makes the traced ellipse have an eccentricity close to 0 - in other words it is "close" to a circle. When e=0, only point A is traced. When e=1, the traced points lie on a parabola.
  The finished sketch is here. [GSP Help]
A person is standing 9 feet up on a 12 foot ladder. The ladder slides down the wall. What is the path that the person follows while falling? Here is a picture of the situation. I have hidden some things, so don't just try to reproduce this picture.
Turn in your final sketch. Briefly describe the mathematics of how you constructed it. Answer the question in part d.
Once again, the person on the slidding ladder is on an elipse.
1. Construct the line which is the wall. Construct a point on the wall. Construct the floor, a line perpendicular to the wall through the point. Construct another point L on the wall. This is the point the ladder hits the wall.
2. Off to the side, draw a segment the length of the ladder. Use this segment and point L to construct the ladder. Hint: Remember that the ladder is a segment with one endpoint L. The ladder is always the same length. Its other endpoint L' is on the floor. Thus the other endpoint is on a circle with center L. What is the radius of the circle?
  Use the segment drawn in this step to be the length of the ladder as the radius of the circle centered at L. Then find one point of intersection of the circle and the floor. This is L'. Connect L and L' to form the ladder.
  A sample of what the sketch should be at this point is available here. [GSP Help]
3. Construct the person. Hint: the distance from L to the person is .25 times the distance LL'. Do you understand why?
  
  The person is 9 feet up the 12 foot ladder, or .25 time the distance LL' from point L. To construct this point where the person is standing on the ladder, first select point L. Then choose Transform / Mark Center "L". Now select point L'. Choose Transform / Dilate... and type in 0.25 for the New Scale Factor. The resulting image is the point where the person is on the ladder.
  A sample of what the sketch should be at this point is available here. [GSP Help]
4. Trace the locus of the person as you move L. Does the set of traced points look like part of an object which is familiar to you? What do you suppose the shape would be?
  Surprise! This traced ellipse is in the final sketch available here. [GSP Help]
  
  Proof:
  It is fairly easy to prove that the path taken by the person is on an ellipse for any length of the ladder. Let d be the length of the ladder. Impose a coordinate system so that the wall is the y-axis and the floor is x-axis. Let the person on the ladder have coordinates (x,y), and let t be the angle the ladder makes with the floor. Then cos t = x/(d/4) and sin t = y/(3d/4). (Remember the person is 3/4 of the way up the ladder.) By the Pythagorean identity, (cos t )^2 + (sin t )^2 = 1, we get x^2/(d/4)^2 + y^2/(3d/4)]^2 = 1. This is the equation of an ellipse with major axis length 3d/2 and minor axis length d/2. In the case of the 12 foot ladder, the equation is x^2/3^2 + y^2/9^2 = 1.
Simpson's line: In an above question, you constructed a circle through three points, known as its circumcircle. Connect these points to form a triangle. Pick a point p on the circumcircle; through p, construct the perpendiculars to the sides of your triangle. What can you say about the feet of these perpendiculars? This is called Simpson's line. (The foot of the perpendicular is the point at which the perpendicular line intersects the line containing the side of the triangle.) What happens when you trace Simpson's line as you vary p? Your writeup should reference a sketch. (Reference "A Little Geometry," Mathematica in Education, Vol.3, No.1, Winter, 1994.)

The feet of the perpendiculars are collinear. They lie on what is called Simpson's line. If we trace Simpson's line while varying P, we get an interesting envelope of lines called a deltoid.
The final sketch is available here. [GSP Help]
Pythagorean theorem. Use Sketchpad to illustrate the Pythagorean theorem. Your writeup should reference a sketch. You may need to include text for explanation.
There are many ways to illustrate the Pythagorean theorem. One method is by constructing squares extending from the sides of a right triangle and measuring the areas of these squares. From this you can easily see that the Pythagorean theorem is verified using the sketch available here. [GSP Help]
Given any triangle consider the following configuration of the following nine points:
- the midpoints of the sides,
- the feet of the altitudes, and
- the midpoints of the segments joining the orthocenter to the vertices.
Construct the points, guess on what curve the points lie and construct this curve. Test your conjecture by moving your original triangle around in a variety of ways.
Your writeup should reference a sketch. For more information, consult the Penguin Dictionary of Curious and Interesting Geometry, by David Wells.

This construction makes what is known as the Nine-Point Circle.
Definition: The orthocenter is the point where the altitudes of a triangle intersect.
The final sketch is available here. [GSP Help]