Homework 3, due Tuesday, September 7

1. Areas of parallelograms.
(a) Draw a parallelogram, and label the vertices A, B, C, D (in order around the figure).
(b) Construct the diagonal AC.
(c) Pick a point P on AC. (Use the command "Point on Object" in the "Construct" menu.)
(d) Construct lines through P parallel to the two sides of the parallelogram ABCD.
(e) Measure the areas of the four smaller parallelograms into which ABCD is divided by the lines you just constructed.
(f) What do you see?

2. Points on lines.
(a) Draw any two lines x and y.
(b) Pick three points A, B, C on x and three points A', B', C' on y.
(c) Construct the six segments AB', AC', A'B, BC', A'C, B'C.
(d) Constuct the three intersection points: AB' with A'B, AC' with A'C, BC' with B'C.
(e) What do you see?

3. A triangle construction.
The original statement of this problem was not what I intended, but students found lots of interesting relations anyway! Here is what I meant to say, as explained in an email message to the class.
(a) Draw a triangle, and label the vertices A, B, C.
(b) Construct the midpoints of the three sides, and label them L, M, N.
(c) Construct the feet of the altitudes of the triangle ABC, and label them D, E, F.
(d) Construct the orthocenter H and the midpoints of the segments AH, BH, CH, and label them X, Y, Z.
(e) What do you see?

4. Quadrangles.
A quadrangle consists of four points A, B, C, D such that no three of them are collinear, together with the four segments AB, BC, CD, DA. For short we refer to "the quadrangle ABCD." Let P, Q, R, S be the midpoints of AB, BC, CD, DA, respectively. What can you say about the quadrangle PQRS? What is the relation between the area of PQRS and the area of ABCD?

5. The Pythagorean Theorem.
(a) State the Pythagorean Theorem.
(b) Find a proof of the Pythagorean Theorem. (It is OK to look a proof up in a book, or to ask someone to explain a proof to you, but be prepared to present the proof to the class.)
(c) Illustrate your proof with Geometer's Sketchpad.

6. Extra Credit.
Show how to construct a regular pentagon using "ruler and compass."
Explanation: This means you start with a segment and construct a regular pentagon, one of whose sides is the given segment, using only the following commands from the GSP "Construct" menu: Point At Intersection, Segment (or Line), Circle By Center + Point. (The following commands are allowable too, since they can be duplicated using only the three basic commands: Point At Midpoint, Perpendicular Line, Parallel Line, Angle Bisector, Circle By Center + Radius.)

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