If ABCD is a quadrilateral, then its midpoint quadrilateral MNOP is the quadrilateral with vertices at the midpoints of the sides: M = midpoint AB, N = midpoint BC, etc.
Answers: Here are some student answers, all of them good, all different, but some open to discussion about possible ways to be more concise. In each case, ask
What criterion for parallelogram is used here?
Is it necessary to repeat so many steps? If you state what you prove as something like "any side of the midpoint quadrilateral is parallel to the opposite side", then would this make it possible to make the argument only once?
Answer A: | Answer B | Answer C
Prove, if true. Construct a counterexample if false.
Draw a triangle ABC and construct midpoints D, E, F as in the figure. Let G be the intersection of the two medians CF and BD. Then let H be the midpoint of BG and I be the midpoint of CG.
DFHI is the midpoint quadrilateral of CABG, so this follows from 4.3.
The diagonals of the parallelgram DFHI bisect each other, so DG = HG. Also, since H is midpoint of BG, HG = BH. Since BD = DH + HG + GD, BG/BD = 2GD/3GD = 2/3.
The same argument applies to any median intersecting another median, so CG/CF = 2/3 also.
Repeating the same reasoning BJ/BD = AJ/AE = 2./3.
But this says that BJ = (2/3)BD = BG. But there is only one point in segment BD whose distance from B is (2/3)BD, so the two points G and J must be the same, and all the medians are concurrent at G.
Suppose you place your ruler on paper and without moving it, draw lines along both edges of the ruler. Then turn the ruler (no special angle) and draw lines along both edges again.
More formally, there are given two parallel lines m1 and m2 which are distance d apart. Let n1 and n2 be another pair of parallel lines that are the same distance d apart, but which are not parallel to the first pair. Then the lines will intersect at four points, forming a parallelogram (by definition!). Prove that this parallelogram is a rhombus.
Student Answer (correct). Note that the standard mistake on this problem is to confuse the distance between lines (given as equal) with the distances between vertices (to be proved equal).
Given an isosceles triangle ABC, with |AB| = |AC| = b and |BC| = a, what is the radius R of the circumcircle of ABC? (The answer should be in terms of a and b, if possible.)