VIEWS: 10 PAGES: 15 POSTED ON: 2/26/2012
Visual Mathematics Project #7 Compass & Straightedge Constructions From Wikipedia: The "compass" and "straightedge" of compass and straightedge constructions is an idealization of rulers and compasses in the real world: The compass can be opened arbitrarily wide, but (unlike some real compasses) it has no markings on it. It can only be opened to widths that have already been constructed, and it collapses when not used for drawing. The straightedge is infinitely long, but it has no markings on it and has only one edge, unlike ordinary rulers. It can only be used to draw a line segment between two points or to extend an existing line. Each construction must be exact. "Eyeballing" it (essentially looking at the construction and guessing at its accuracy, or using some form of measurement, such as the units of measure on a ruler) and getting close does not count as a solution. Stated this way, compass and straightedge constructions appear to be a parlour game, rather than a serious practical problem; but the purpose of the restriction is to ensure that constructions can be proven to be exactly correct, and is thus important to both drafting (design by both CAD software and traditional drafting with pencil, paper, straight-edge and compass) and the science of weights and measures, in which exact synthesis from reference bodies or materials is extremely important. One of the chief purposes of Greek mathematics was to find exact constructions for various lengths; for example, the side of a pentagon inscribed in a given circle. All compass and straightedge constructions consist of repeated application of five basic constructions using the points, lines and circles that have already been constructed. These are: Creating the line through two existing points Creating the circle through one point with centre another point Creating the point which is the intersection of two existing, non-parallel lines Creating the one or two points in the intersection of a line and a circle (if they intersect) Creating the one or two points in the intersection of two circles (if they intersect). For example, starting with the minimal state of a drawing, with just two distinct points, we can create a line or either of two circles. From the two circles, two new points are created at their intersections. Drawing lines between the two original points and one of these new points completes the construction of an equilateral triangle. Visual Mathematics Project #7 – Compass & Straightedge Constructions Assignment: Following the instructions contained in the following pages, create the following 13 compass and straightedge constructions on separate plain white paper. A compass will be checked out to you for this project. You will not receive a grade for the project until you have returned the compass. Both accuracy and neatness will be taken into account in grading. Construction Points 1. Construct the perpendicular bisector of a line segment, or construct the midpoint of a line /8 segment. 2. Given a point on a line, construct a perpendicular line through the given point. /8 3. Given a point not on the line, construct a perpendicular line through the given point. /8 4. Construct the bisector of an angle. /8 5. Construct an angle congruent to a given angle. /8 6. Construct a line through a given point, parallel to a given line. /8 7. Construct an equilateral triangle, or construct a 60º angle. /8 8. Divide a line segment into n congruent line segments. /8 9. Construct a line through a given point, tangent to a given circle. /8 10. Construct the center point of a given circle. /8 11. Construct a circle through three given points. /8 12. Circumscribe a circle about a given triangle. /4 13. Inscribe a circle in a given triangle. /8 Total /100 1. Construct the perpendicular bisector of a line segment. Or, construct the midpoint of a line segment. 1. Begin with line segment XY. X Y 2. Place the compass at point X. Adjust the compass radius so that it is more than (½)XY. Draw two arcs as shown here. X Y 3. Without changing the compass radius, place the A compass on point Y. Draw two arcs intersecting the previously drawn arcs. Label the intersection points X Y A and B. B 4. Using the straightedge, draw line AB. Label the intersection point M. Point M is the midpoint of line A segment XY, and line AB is perpendicular to line segment XY. X M Y B 2. Given point P on line k, construct a line through P, perpendicular to k. 1. Begin with line k, containing point P. k P 2. Place the compass on point P. Using an arbitrary k X P Y radius, draw arcs intersecting line k at two points. Label the intersection points X and Y. 3. Place the compass at point X. Adjust the compass radius so that it is more than (½)XY. Draw an arc as k shown here. X P Y 4. Without changing the compass radius, place the A compass on point Y. Draw an arc intersecting the k previously drawn arc. Label the intersection point A. X P Y 5. Use the straightedge to draw line AP. Line AP is perpendicular to line k. A k X P Y 3. Given point R, not on line k, construct a line through R, perpendicular to k. 1. Begin with point line k and point R, not on the line. R k 2. Place the compass on point R. Using an arbitrary R k radius, draw arcs intersecting line k at two points. Label X Y the intersection points X and Y. 3. Place the compass at point X. Adjust the compass R k radius so that it is more than (½)XY. Draw an arc as X Y shown here. 4. Without changing the compass radius, place the R k compass on point Y. Draw an arc intersecting the X Y previously drawn arc. Label the intersection point B. B 5. Use the straightedge to draw line RB. Line RB is perpendicular to line k. R k X Y B 4. Construct the bisector of an angle. 1. Let point P be the vertex of the angle. Place the Q compass on point P and draw an arc across both sides of the angle. Label the intersection points Q and R. P R 2. Place the compass on point Q and draw an arc across the interior of the angle. Q P R 3. Without changing the radius of the compass, place it on point R and draw an arc intersecting the one Q W drawn in the previous step. Label the intersection point W. P R 4. Using the straightedge, draw ray PW. This is the bisector of QPR. Q W P R 5. Construct an angle congruent to a given angle. 1. To draw an angle congruent to A, begin by drawing a ray with endpoint D. A D 2. Place the compass on point A and draw an C arc across both sides of the angle. Without changing the compass B radius, place the A compass on point D and draw a long arc D E crossing the ray. Label the three intersection points as shown. 3. Set the compass so that its radius is BC. C Place the compass on point E and draw an arc F intersecting the one B drawn in the previous A step. Label the intersection point F. D E 4. Use the straightedge to draw ray DF. C EDF BAC F B A D E 6. Given a line and a point, construct a line through the point, parallel to the given line. 1. Begin with point P and line k. P k 2. Draw an arbitrary line through point P, intersecting line k. Call the intersection point Q. P Now the task is to construct an angle with vertex P, congruent to the angle of intersection. Q k 3. Center the compass at point Q and draw an arc intersecting both lines. Without changing the P radius of the compass, center it at point P and draw another arc. Q k 4. Set the compass radius to the distance between the two intersection points of the first arc. Now P center the compass at the point where the R second arc intersects line PQ. Mark the arc Q k intersection point R. 5. Line PR is parallel to line k. P R Q k 7. Given a line segment as one side, construct an equilateral triangle. This method may also be used to construct a 60 angle. 1. Begin with line segment TU. T U 2. Center the compass at point T, and set the compass radius to TU. Draw an arc as shown. T U 3. Keeping the same radius, center the compass V at point U and draw another arc intersecting the first one. Let point V be the point of intersection. T U 4. Draw line segments TV and UV. Triangle TUV is V an equilateral triangle, and each of its interior angles has a measure of 60. T U 8. Divide a line segment into n congruent line segments. In this example, n = 5. 1. Begin with line segment AB. It will be divided into five congruent line segments. A B 2. Draw a ray from point A. Use the compass C to step off five uniformly spaced points along the ray. Label the last point C. A B 3. Draw an arc with the compass centered at C point A, with radius BC. Draw a second arc with the compass centered at point B, with radius AC. Label the intersection point D. Note that ACBD is a parallelogram. A B D 4. Use the compass to step off points along line C segment DB, using the same radius that was used for the points along line segment AC. A B D 5. Use the straightedge to connect the C corresponding points. These line segments will be parallel. They cut line segments AC and DB into congruent segments. Therefore, A B they must also cut line segment AB into congruent segments. D 9. Given a circle, its center point, and a point on the exterior of the circle, construct a line through the exterior point, tangent to the circle. 1. Begin with a circle centered on point C. Point P is on the exterior of the circle. P C 2. Draw line segment CP, and construct point M, the midpoint of line segment CP. (For the construction of the midpoint, refer to the perpendicular bisector construction, on page C P M 1.) 3. Center the compass on point M. Draw a circle R through points C and P. It will intersect the P other circle at two points, R and S. C M S 4. Points R and S are the tangent points. Lines PR R and PS are tangent to the circle centered on P point C. S 10. Construct the center point of a given circle. 1. Begin with a circle, but no center point. 2. Draw chord AB. A B 3. Construct the perpendicular bisector of A chord AB. Let C and D be the points where it intersects the circle. (Refer to the construction of a perpendicular C bisector, on page 1.) D B 4. Chord CD is a diameter of the circle. Construct point P, the midpoint of diameter CD. Point P is the center point of the circle. (Refer to the construction of the midpoint of a line segment, on page 1.) C P D 11. Given three noncollinear points, construct the circle that includes all three points. 1. Begin with points A, B, and C. A C B 2. Draw line segments AB and BC. A C B 3. Construct the perpendicular bisectors of line A C P segments AB and BC. (Refer to the perpendicular bisector construction, on page 1.) Let point P be the intersection of the perpendicular bisectors. B 4. Center the compass on point P, and draw the circle through points A, B, and C. A C P B 12. Given a triangle, circumscribe a circle. 1. Begin with triangle STU. T S U 2. If a circle is circumscribed around the triangle, T then all three vertices will be points on the circle, S so follow the instructions above, for construction of a circle through three given points. U 13. Given a triangle, inscribe a circle. 1. Begin with triangle KLM. L K M 2. Construct the bisectors of K and L. L (Refer to the angle bisector construction, on page 4.) Let point Q be the intersection of the two angle bisectors. Q K M 3. Construct a line through point Q, L perpendicular to line segment KL. Let point R R be the point of intersection. (Refer to the construction of a perpendicular line Q through a given point, on page 3.) K M 4. Center the compass on point Q, and draw a L circle through point R. The circle will be R tangent to all three sides of a triangle. Q K M constructions courtesy of http://whistleralley.com/construction/reference.htm