Yesterday e (who I just blogrolled – I think we are related) from e’s ponderings posted the following problem:

Prove that the midpoint quadrilateral (a quadrilateral obtained by connecting consecutive midpoints of sides) of an isosceles trapezoid is a rhombus.

and invited readers to offer more than one proof.

I found more than one proof, and then asked: how can I modify this question to be appropriate for younger students? for older students? for weaker students? for stronger students?

Generate good, lousy, in between … toss out the bad, … sometimes stumble on something nice. … Ask yourself, “Let’s change the problem” and then ask “Is this new problem any good for … my students? … For anyone I know? ” And then change it again.

(Lots of variations below the fold. Harder problem for puzzle hounds at the bottom) —->

This was a little puzzle for me, not a big “Problem!” Technically, I might it fell on the line for me between problem and exercise; while I had never seen the question, I knew that it would collapse in the face of coordinates. I also know a euclidean (but high school style) proof that the midpoint quadrilateral of an arbitrary quadrilateral is a parallelogram (including, btw, the degenerate quadrilateral with 3 colinear vertices!), so I knew I could extend that for the second proof.

I distinguish between “problem” and “exercise” in the following way: If the solver has a ready algorithm that will work, if the solver has an approach that will get to the ‘answer,’ — this is an exercise. If the solver does not know how to start, or is unsure if an approach will lead to a solution, that would be a problem. It’s a fluid boundary. With experience, exercises may become problems. A problem for some students will be an exercise for others.

Dave Marain wants to know why I call things puzzles and not problems. I never thought about that. I think that problems (or even challenging exercises) that I work on, I call puzzles, because they are recreational, I enjoy them.

New Problems – just a little easier

• Include a diagram – always makes a difference. This applies to every other variant problem that follows.
• Add information that pushes the student in the ‘right’ direction: include (one? both?) diagonal(s) to prompt the euclidean proof, or instructions to use coordinates for the coordinate proof.
• Name some steps involved: “Using our theorem about the segment connecting midpoints of a triangle…” or “Place an isosceles trapezoid on a coordinate grid, and, using the midpoint formula and the distance formula, prove…” There are other options that accomplish the same steering, but leave plenty of work for the student.

Related problems – easier

• Supply coordinates (2a,0), (2b,2c) etc.
• Supply coordinates with numbers instead of variables: “Show that the midpoint quadrilateral of the isosceles trapezoid with coordinates (8,0),(6,4),(-6,4),(-8,0) is a rhombus”
• For easier theorem-based work, supply lengths: “Show that the midpoint quadrilateral of the isosceles trapezoid with sides of length 8,10,20,and 10 is a rhombus” A diagram with at least one diagonal drawn would help.

Related problems – even easier

• Give the lengths of the sides (as above) and ask for the area of the trapezoid and the area of the midpoint quadrilateral – this can be done without discovering that a rhombus is involved, but it might be noticed along the way. The sides I offered give easy to work with 6/8/10 triangles to make finding altitudes easier.
• Same as above, but with coordinates. As one standard coordinate-based way of finding irregular looking areas is to box them and subtract unwanted triangles, it is quite likely that this question would force the discovery that the midpoint quadrilateral is a rhombus. (I like this one! I could follow up by redoing it with variable coordinates – corresponds roughly to the extra proof I offered on e’s page).
• Find the perimeter of …. – Always easier for kiddies to cope with finding a number than with “showing” or “proving”
• Give them all the coordinates – trapezoid and rhombus, and ask them to verify that each figure is as claimed. Can they generate another example? – Again, numbers are easier. Proof is hard.

More difficult variations

• The question includes a conclusion. We could try something more open: Describe the midpoint quadrilateral (a quadrialteral obtained by connecting consecutive midpoints of sides) of an isosceles trapezoid. Prove that your description is correct
• We could ask for a ratio. Even strong students tremble when they see that word: Given isosceles trapezoid with M and N the midpoints of the bases and X the midpoint of another side, find the ratio MX:XN.

Even more difficult – letting the question range

• Without mentioning rhombus, express the area of the midpoint quadrilateral of an isosceles trapezoid in terms of its bases, B and b, and its other side, a (or, and its height, h. Less challenging, but still hard.) Hmm. Finding the area of any trapezoid in terms of its sides, without the height, might be a challenge.
• Same thing, but look for the perimeter. (In general, perimeters are easier to work with than area, but in this case it looks to be the opposite)
• Given isosceles trapezoid with midpoint of one base M, midpoints of the two congruent sides J and K, and intersection point of the diagonals, L, prove JLMK is a kite. – unfamiliar figure adds to the difficulty. Denying a diagram does as well.

Finally, for the puzzle-seekers who try their skills on harder stuff here

• The midpoints of a quadrilateral form a rhombus. Describe the quadrilateral. (what must it be, what can it not be, what might it be).

Now, I do want to point out, many of these proposed variations are mediocre. But when I work on a problem or puzzle, part of (per Polya) “Looking Back” is to generate lots and lots of new problems. Generate good, lousy, in between, don’t worry about quality at first, toss out the bad, try out the ones that look interesting, and sometimes stumble on something nice. It is a habit. Ask yourself, “Let’s change the problem” and then ask “Is this new problem any good for me? For my students? For other students? For anyone I know? ” And then change it again.

10 Comments leave one →
1. February 12, 2007 am28 12:41 am 12:41 am

I recognize that none of my proposed problems is suitable for K-5. Any suggestions? We could ask about a rectangle, and ask them to show, rather than prove, that the midpoint quad is a rhombus. What else?

Jonathan

February 12, 2007 am28 1:34 am 1:34 am

Heh, the hardest question was the one I immediately thought of upon reading the first part of this post :-)

Now, how to solve it? I finally got the answer, after many investigations, but I don’t think I would of thought how to prove it without your note that the midpoint quadrilateral of any quadrilateral is a parallelogram.

Here’s a (slight) generalization:
a tetrahedron is formed by 4 points in space. Each tetrahedron has several “midpoint quadrilaterals”: given tetrahedron ABCD, we can prove (using the same proof that Jonathan knows, I imagine) that the midpoints of AB, BC, CD and DA form a parallelogram. (This is slightly remarkable, because it’s not immediately clear that those 4 points should even lie in the same plane.) The method of the proof is as follows: in triangle ABC, the segment joining the midpoints of AB and BC is parallel to AC and half its length, and similarly in triange ACD. So two sides of the quadrilateral are parallel and equal, so it is a parallelogram.
Now, this gives us two of the side lengths in terms of AC. Applying the same argument to the other two sides shows that they have length equal to one half BD. So the figure is a rhombus if and only if AC and BD have the same length. For the case where ABCD is a quadrilateral, not a tetrahedron, this means that it has two equal diagonals. For a tetrahedron, it means that the “midpoint quadrilateral” is a rhombus only if the corresponding pair of opposite sides of the tetrahedron have equal lengths.

A natural variation (actually a little easier, I think): what quadrilaterals (or tetrahedra) have as their midpoint quadrilateral a rectangle? Then, putting the two problems together, a square?

3. February 12, 2007 am28 2:03 am 2:03 am

I am not following the logic for “For a tetrahedron, it means that the “midpoint quadrilateral” is a rhombus only if the corresponding pair of opposite sides of the tetrahedron have equal lengths.” If a quad is a special case of a tetrahedron, wouldn’t what holds for the tetr hold for the quad? And in this case, opposite sides cogruent is neither necessary nor sufficient for the quad.

Have I goofed?

February 12, 2007 am28 5:46 am 5:46 am

Sorry, this is what I get for assuming that everyone else has the same picture in their heads that I do :-)

A tetrahedron has 6 edges. Let’s say the tetrahedron is ABCD, with edges AB, BC, CD, DA, AC and BD. These correspond to the 4 sides and two diagonals of a quadrilateral — if you pick one vertex of the quadrilateral up out of the plane, you can’t tell whether the segments connected to it were sides or diagonals any more.

We only want a quadrilateral, so we pick only 4 edges of the tetrahedron. For instance, I chose AB, BC, CD and DA, leaving out one pair of opposite edges (AC and BD). There were 2 other ways I could have done this (excluding AB and CD or excluding AD and BC). Now, when I take the midpoints of those 4 edges, I’m making a particular choice: I’m identifying those 4 edges with the edges of a quadrilateral, which leaves the other two edges to be the diagonals. Thus, the condition “two opposite edges are equal” for a tetrahedron corresponds to the condition “diagonals are equal” for a quadrilateral.

There are 3 different ways that I can choose a “midpoint quadrilateral” of a tetrahedron; having one pair of opposite sides equal only insures that I get one “midpoint rhombus.” It does show something else, though: suppose I have a quadrilateral. I take the midpoints of the two diagonals and two opposite sides and join them. The resulting quadrilateral is a rhombus if and only if the other two sides are equal.

5. February 12, 2007 am28 6:30 am 6:30 am

OK, I make sense out of this by saying: in the planar (degenerate) tetrahedron, two edges are reduced to diagonals… Maybe we are better off looking for the centroid tetrahedron of a tetrahedron. But this goes beyond what I am comfortable playing with quick answers to.

February 12, 2007 pm28 6:10 pm 6:10 pm

Possible question: I haven’t tried this out myself yet.

Quadrilateral B is formed by joining the midpoints of the sides of Quadrilateral A. If Quadrilateral A is similar to Quadrilateral B, what possible forms can Quadrilateral A take. (Simple case: a Square, but are there other answers?)

February 12, 2007 pm28 9:22 pm 9:22 pm

An excellent question, Clueless — Jonathan’s theorem again gives us that the quadrilateral must be a parallelogram to start. Every parallelogram is defined by two side lengths and one angle measure. I think that if you give me two of these values, there will always be a unique value of the third parameter which makes such a parallelogram. (Actually, with one caveat: the parallelogram with sides a, b and angle t is the same as the parallelogram with sides a, b and angle 180 – t.)

For example, I believe that a parallelogram with sides of length 1 and phi (= (rt(5) – 1)/2) and an angle of 60 degrees is such a figure.