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Puzzle: Area = 1

January 24, 2007 am31 8:10 am

Dave Marain has been posting engaging problems for use with (mostly) middle school students (some are ok for upper elementary, others for algebra or geometry students). Today he put up a coordinate geometry problem in this group of questions,which made me think.

The problem was along the lines of: Find x such that the area of the polygon formed by (6,7), (12,-2), (15,-2), (x,7) is 117. (He had different numbers, but same idea). Decide what figure is formed, and use a well-known formula to decide what x is. When you click the link and look at Dave’s problem, please make sure to read the comments. He has a talent for generating rich discussions.

I do a lot of “what if” as I solve problems, to see if I can come up with generalizations, or interesting alternates. Try this one, but be warned, I think it is hard:

Find all possible values of (x,y) such that the area of the polygon formed by (0,1), (0,0), (1,0), (x,y) is 1.


7 Comments leave one →
  1. Clueless permalink
    January 24, 2007 pm31 7:46 pm 7:46 pm

    Wow, that’s a great generalization!

    The solution does not turn out to be so hard. There seem to be three possible answer classes (each having infinite possibilites) depending on where you choose (x,y) to be, and this also changes the nomenclature of the quadrilateral (whether it is ABCD, ABDC, or ACBD).


  2. January 24, 2007 pm31 9:24 pm 9:24 pm

    I have two bounded sets of solutions, but am still playing. I know where to search for the third, but have not done so yet. I also need to recheck if the bounds I set really exist, and also am suspicious that there are two non-obvious cases that bear further examination. I will wait for more discussion here before posting details, though.

  3. rdt permalink
    January 24, 2007 pm31 10:26 pm 10:26 pm

    My first thought (on about 5 minutes reflection) is that there are four different cases to consider, depending on which quadrant the point (x,y) is in.

  4. Clueless permalink
    January 24, 2007 pm31 11:08 pm 11:08 pm

    I spoke too soon. There appears to be one more class of solutions for the polygon, and this is qualitatively different from the three solutions I mentioned before.

    If you open up the problem to produce two triangles which do not make a single polygon, then there appear to be two extra classes of solutions, which have a completely different character.

    This is getting curiouser and curiouser.

  5. rdt permalink
    January 25, 2007 am31 12:27 am 12:27 am

    For (x,y) in the first quadrant, I come up with the condition x+y=2.

  6. rdt permalink
    January 25, 2007 am31 12:51 am 12:51 am

    I’m doing this without as much thought-checking as I probably should, but…

    For (x,y) in the 3rd quadrant: x+y = -2.

    For (x,y) in the 2nd quadrant: y-x-(xy)/(1+x) = 2.

    For (x,y) in the 4th quadrant: x-y-(xy)/(1+y) = 2.

    I think the 2nd and 4th quadrant cases are the ones that correspond to the triangles not making a single polygon, and the approach I took was to find the area of the two triangles (which is still [x+y]/2) and subtract the intersection. I the 2nd quardrant case the “height” of the intersection is -x and the “base” is y/(1+x).

  7. January 25, 2007 am31 6:35 am 6:35 am

    The puzzle statement needs refinement:

    Do the vertices need to occur in the order provided? (let’s assume no)

    May the resulting figure be concave? (let’s assume yes)

    May the resulting figure cross itself? Let’s consider this a second puzzle.

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