Comments on Math 444 Assignment Due Wednesday, November 07, 2001
There were 4 problems, worth 30 points total. Problem 1 was worth up to 10 points, 3 each for the counterexamples and 2 each for the descriptions of all quadrilaterals with the special midpoint quadrilaterals. Problems 2 and 3 were worth 5 points each. Problem 4 was worth 10, 6 points for the proofs and 4 for the 2 straightedge and compass constructions.
The median grade on this assignment was 17/30. One major source of points lost for some students was from proving that the statements in #1 were true when they were false. Another problem area was assuming the conclusion in #4.
Both statements were false. A counterexample for both is ABCD where A = (1,0), B = (0,1), C = (-2,0), D = (0,-2). Many other counterexamples were possible.
The part of the question that asks for a description of all the quadrilaterals that have midpoint quadrilateral as a rectangle or a rhombus can be answered as follows.
What is a valid counterexample?
There was some discussion of this in class, but it was hurried and not too clear, so this is an attempt to be more complete.
First, you only need one counterexample, but if you can find a lot of them, all the better. Your counterexample should convince an intelligent but skeptical reader that it is truly a correct example of a case when the theorem is false.
(1) A small, rough sketch is evidence that a counterexample exists and may suggest what it looks like, but since the angles and lengths are not exact, a sketch does not really spell out what the counterexample is without additional explanation. This could be marks of equal length in the figure, but probably should include at least a few words of explanation of what the example is. (On this assignment such a sketch earned partial credit.)
(2) A careful construction of some size with Sketchpad or with straightedge and compass is much more convincing, but again to be completely convincing it needs to be self-evident what the shape is (e.g. an isosceles trapezoid) or else it still needs some words explaining exactly what the figure is. (On this assignment such an example also got partial credit, but more than for (1).)
(3) An example on graph paper can be self-explanatory if it is crystal clear what the coordinates are (so it is clear that certain points are truly midpoints, certain lengths are equal, etc.). The example of Angela that was discussed in class is of this type. Such an example is a convincing counterexample.
(4) A counterexample can be either a crude sketch or a nice figure if there are words telling what it is. For example, a sketch of a quadrilateral with the words that "the diagonals are perpendicular but do not bisect each other" or "this is an equilateral triangle" would be such an example of descriptive words. This is a very good way to give a convincing counterexample.
This is an example of "anti-Thales" or two right triangles sharing a common angle, so if M is the midpoint of BC, N is the midpoint of AB, and O is the circumcenter, then triangle ABM is similar to triangle ANO. Combine this with Pythagoras to compute missing lengths, and you get the answer, which is
R = b^2/sqrt(4b^2 – a^2)
Other forms of the correct answer were OK also. Some students got lost in algebra or did not put in enough information to be able to solve for R.
One observation – think of examples: No one gave evidence of checking her/his answer in an example. Some simple examples could be checked exactly by hand; one could do others numerically by calculator or Sketchpad. This would have caught some mistakes. Some examples we already know well. For instance, if angle A is a right angle and the other two angles are 45 degrees, then the circumcenter is the midpoint of BC. If b = 1, then a = sqrt2 and r = (1/2)sqrt 2. Another easy example to check would be an equilateral triangle.
Most students got this right, but not all. Again this is what we called anti-Thales. One needs to realize that the statements "triangle AOB similar to triangle A'OB'" and "triangle AOB similar to triangle B'OA'" are not the same. Then just check what the constant of proportionality is, using algebra.
Again, it was possible to check against example, but it was less needed here than in #3.
In proving that there is a circle, by and large the difference between the correct proofs and the incorrect ones was this: in a correct proof, there will be a point P that turns out to be the center of the circle, but you have to tell exactly how this point is defined at the outset and the prove the other properties. Good examples of defining P correctly would be ONE of these:
An incorrect proof would typically take P to be the intersection of all 3 perpendicular bisectors. This is exactly the same as assuming there is a circle through A, B, C, D through P, so it is assuming the conclusion in advance. It misses the whole point, which is not so exotic: for a "random" set of 4 points, there is no circle through them. For some special arrangements of points, there is a circle but to know this, one has to explain how the points are arranged guarantees the existence of the circle. To do this, you have to start only with things that are true in general and then let the special relationships kick in, not the reverse.
There were many correct proofs, once it was set up. For example, if we take P to be the intersection of the perpendicular bisectors of AB and of BC, then P is the circumcenter of triangle ABC and we have PA = PB = PC = r. Then triangles PAB and PBC are congruent isosceles triangles by SSS. If we let x = angle PAB = angle PBA = angle PBC = angle PCB, then angle ABC = 2x = angle BCD, so angle PCD = 2x – x = x. Then triangle PCD is congruent to triangle PCB by SAS, since PC = PC, CD = CB and angle PCD = angle PCB. From this, r = PB = PD, so this means that D is also on the circle of radius r with center P.