Monday, March 5, 2007

Problems 3-5 thru 3-6-07: Geometry and Reasoning

Note: The problem below was chosen for the March 23rd Carnival of Mathematics. If you'd like to see a different kind of mathematical challenge, I recommend you also visit the posting for 3-30-07 (Challenging Geometry: Circles Inscribed in Quadrilaterals, Right Triangles).



Today's questions involve well-known ideas from geometry. Similar questions have appeared on SATs and math contests.
Some suggestions: Use it for in-class enrichment or assign it for extra credit outside of class. The first part lends itelf to a fairly simply visual approach (cutting up the square and matching the pieces), but the second is more sophisticated. Encourage the visualization but require the analytical approach as well!
Reviews: 30-60-90, equilateral, areas, symmetry, etc.

7 comments:

Dave Marain said...

No answers - just a few thoughts...
1. How many students would assume that the first rectangle is a square? So another way to start the discussion might be to ask the class: "Explain why DEFG CANNOT be a square!" (it's easy but a nice way to start the juices flowing).
2. I know some of you will be wondering why I didn't ask the students if DEFG is the rectangle of maximum area that can be inscribed in the triangle. OR is it the square? So go ahead and ask it! Do they need calculus to prove it or just a knowledge of quadratic functions...
3. How might one generalize these results? Relax the equilateral condition and make it isosceles? A bit harder or much harder? Derive an expression for the side of the square LMNO in terms of the side of the triangle?
4. Do problems like these really have learning value or are they merely 'puzzles'?

Darmok said...

It's pretty obvious that it's half; all you have to do is fold the three triangles onto the rectangle to see that their sum is the same as the area of the rectangle! (It'd be easy to prove since the vertices are at the midpoints, just separate into horizontal and vertical components.

Dave Marain said...

darmok--
folding is a much better way of expressing it than my 'cutting it up' phrase! thanks!

Totally_clueless said...

One way to look at problem 1 is that the area of the square is equal to twice the area of an equilateral triangle of side half as long as the original triangle.

For (2), it was not quite obvious that the folding will give a figure on the square only with overlaps but no gaps. It took some extra reasoning to convince myself.

TC

Dave Marain said...

thanks tc--
the key is your phrase 'extra reasoning'! This is my raison d'etre for writing these questions!How many of our students are exposed daily to thinking at a deeper level? Standard college prep? Fundamentals level? Honors? You can open up any issue of Mathematics Teacher from NCTM and you'll find a month's worth of great problems. What % of classes see these or are provided an incentive to try them? Certainly some do but my guess is that it would be a few honors teachers, that's about it.

jonathan said...

First, establish that DEFG is half of ABC. Then, since ILM and ADE are congruent, we can move the diagrams productively onto each other. Notice that B lies on JL and G on OL. Now we are really comparing the area of trapezoid BJKC with rectangle GONF...

Anonymous said...

Darmok's method of folding the triangle to derive an explanation may not be feasible.

As the diagram did indicate that it is not drawn to scale.

Base on my observation, it is impossible to assume the sum if the the diagrams are not drawn to scale.