# vismathproject7-compassandstraightedge by gegeshandong

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```									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
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
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

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