Skip to content

Puzzle – half way around a sector

July 12, 2009 pm31 1:18 pm

I was looking at a MathNotations puzzle, really a borrowing from Totally Clueless, over here. And I thought, what else could we do with arcs and radii? And I came up with a series that all require some uglyish algebra, starting with the first at the level of a challenge problem in algebra 1.

  • Puzzle version 1:  Consider a quarter circle AOB with center O. Starting at point A, what is the halfway point around the perimeter?

Now, that looks fairly algebraic to me. Take the perimeter, divide it in half, be careful with a little subtraction. Does the radius fall out?

  • Puzzle version 2: Consider a sector of a circle CND with central  ∠N. Find N if the paths from C to D are equidistant.

Still fairly algebraic. Maybe even a little easier, though we could get some challenge back, just a bit, for asking for the answer in degrees.

  • Puzzle version 3: Consider a sector of a circle EMF, with central ∠M. Find the point on the perimeter at the greatest distance from E in terms of M and the radius.

Of course we expect the radius to drop out, and we’re left with some annoying algebra, and two cases, depending on M. Annoying, but not too tricky. Try it though.

New set.

  • Puzzle B, version 1: Consider a quarter circle GLH with center L. Consider a segment with one endpoint at G that divides the sector into two pieces of equal area. Where is the other endpoint of this segment?

Ooh, something to think about before you start calculating…

  • Puzzle B, version 2: Consider a sector IKJ, with central ∠K. Consider a segment with one endpoint at K that divides the sector into two pieces of equal area. Where is the other endpoint of this segment? Are there conditions under which this puzzle does not have a solution?

That last might be too hard for me! Try to avoid leaving comments on questions that are too easy for you. Thanks.

Final challenge – can you pull out something that would be a fair challenge for an advanced middle school class? That’s the harder stuff to write…

4 Comments leave one →
  1. Clueless (Totally) permalink
    July 12, 2009 pm31 3:23 pm 3:23 pm

    A simpler version of the last two problems may be:

    Find the central angle of a sector AOB such that the area of the triangle with the center O and the two end points A& B as vertices is equal to the area of the figure bounded by the segment AB and the curved part.

    This also gives some insight as to where the third point that bisects the sector is located.


    • July 12, 2009 pm31 9:11 pm 9:11 pm

      Question is easily understandable (quick sketch tells all). But am I looking at the angle in terms of a function of the angle? Yecch. Kate(t) had a hard question like that a while back….

  2. July 16, 2009 pm31 6:04 pm 6:04 pm

    It seems to me that the last question is not as bad as you seem to think it is (I may, of course, be missing something!), and that the solution always exists, split into two cases plus a boundary case in the way that TC suggests. For angles smaller than that boundary, the solution is “nice” (and it’s a three-or-so-liner). For larger angles, the solution is indeed not all that nice, or at least mine isn’t, though it’s quick easy to get as far as you’re going to get.

  3. Anonymous permalink
    October 3, 2009 am31 5:20 am 5:20 am

    give me a puzzle?

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s

%d bloggers like this: