Assignment #5
Nicole Mosteller
EMAT 6680
All of the links below are to scripts to be used with Geometer's Sketchpad.
With each link is a picture of the finished sketch as well as a description of how the sketch is to be created.
Script #1: The Centroid.
The Centroid of a triangle is the point of intersection of the three medians of a triangle.
Point G is the centroid of DABC.
Script #2: The Orthocenter.
The orthocenter of a triangle is the point of intersection of a triangle's three altitudes.
Point H is the orthocenter for DABC.
Script #3: The Circumcenter.
The circumcenter is the point of intersection of the perpendicular bisectors of the sides of a triangle.
Point C is the circumcenter of DABM.
Script #4: The Circumcircle.
The circumcircle of a triangle is the circle that has the circumcenter as its center
and that goes through the vertices of DABM.
Script #5: The Incenter.
The incenter is the point of intersection of the angle bisectors of the triangle.
Point I is the circumcenter of DABC.
Script #6: The Incircle.
The incircle of a triangle is the circle that has the incenter as its center
and that is tangent to all sides of DABC.
Script #7: The Medial Triangle.
The Medial Triangle is constructed inside a given DABC with vertices at the midpoints of each side of DABC.
DDEF is a medial triangle.
Script #8: The Orthocenter-Midsegment Triangle.
The orthocenter-midsegment triangle is constructed using a couple of the previous scripts.
First, construct the orthocenter (point H) of the triangle (DABC).
Second, connect this orthocenter to the vertices of DABC with segments AH, BH, and CH.
The orthocenter-midsegment triangle have vertices that are the midpoints of these segments.
DDEF is the orthocenter-midsegment triangle of DABC.
Script #9: The Orthic Triangle.
The orthic triangle is also constructed using a previous construction.
First, construct the altitudes from each vertex of DABC.
Each altitude intersects the side opposite the vertex from which it is drawn.
These intersections are labeled with Points D, E, and F.
DDEF is the orthic triangle of DABC.
Script #10: The Pedal Triangle.
Script #11: Center of the Nine Point Circle.
The Nine-Point circle for any triangle (DABC) passes through the three mid-points of the sides (pts. D, E, and F),
the three feet of the altitudes (pts. G, J, and I), and the three mid-points of the segments
from the respective vertices to orthocenter (pts. K, L, and M).
After locating the nine points for the circle, the next point to locate is the center of the circle.
To find the center of the circle, find the perpendicular bisectors of two segments whose endpoints are points on the circle.
The intersection (pt. P) of these two perpendicular bisectors is the center of the Nine-Point Circle.
Script #12: Nine Point Circle.
The Nine-Point circle for any triangle (DABC) passes through the three mid-points of the sides (pts. D, E, and F),
the three feet of the altitudes (pts. G, J, and I), and the three mid-points of the segments
from the respective vertices to orthocenter (pts. K, L, and M).
After locating the center of the circle (see previous construction), construct the Nine-Point Circle with
center at point P going through all nine points constructed.
Script #13: Trisecting a segment.
To trisect segment AB, first pick a point C off the segment, and draw line AC.
Using circles with radii equal to segment CA, find four points that are equidistant from each other
(begin with pt. A, then pt. C, then two more...). The midsegment theorem can be extented to show that if three parallel
lines are equidistant (as constructed on line AC),
then these lines will continue to be equidistant as they intersect segment AB.
Segment AD = Segment DE = Segment EB.
Script #14: Equilateral Triangle, given a side.
To construct equilateral DABC given Segment AB, two circles are needed.
The first circle is centered at pt. A with radius AB, and the second circle is centered at pt. B with radius BA.
These circles are congruent which makes the following radii congruent: AB = BC = CA.
DABC is equilateral.
Script #15: Square, given a side.
Comparable to the equilateral triangle construction,
the square construction similarly uses two circles as well as two lines perpendicular to the given segment AB.
quad. ABCD is a square.
Script #16: Isosceles Triangle, given a base and altitude.
Given the base AB and the altitude CD, isosceles DABF is constructed after finding the perpendicular bisector
of the base (line EF) and constructing a circle whose center is pt. E with radius CD.
DABF is isosceles.
Script #17: Triangle Centers (H, G, C, and I).
Script #18: Trianlge Centers with Euler Line.
Script #19: Locus of vertex of a fixed angle that subtends a fixed segment.
Script #20: Divide a segment AB into two parts that form a golden ratio.
To divide segment AB into pieces that form the golden ratio, I constructed a 36-72-72 isosceles triangle using
segment AB as one of the legs of the triangle. To construct this triangle, I used the Rusin-Pentagon Construction
found on the Math Forum website. The angle bisector of the 72 degree angle opposite side AB
will split the segment AB so that the ratio of these pieces is the golden ratio.
Script #21: Pentagon, given a radius.
To perform this construction, I received assistance from the Rusin-Pentagon Construction
found on the Math Forum website. This construction gave me the basis for constructing a 72-degree angle.
The remainder of the construction required constructing this angle as needed to achieve the pentagon.
Script #22: Pentagon, given a side.
To perform this construction, I received assistance from the Rusin-Pentagon Construction
found on the Math Forum website. This construction gave me the basis for constructing a 72-degree angle.
The remainder of the construction required constructing this angle as needed to achieve the pentagon.
Script #23: Hexagon, given a side.
Script #24: Octagon, given a side.
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