Math Problem Solving Team
Clark University
23 Sep 2004
We looked at a couple of the POWs (problems of the week).
I. From the Macalester College POTW.
21 Sep 2004. Problem of the Week 1013. Digit Rotation.
The number 102564 has the property that when it is multiplied by 4, its rightmost digit
moves to the left and the other digits move one space to the right (4 × 102564 = 410256).
Find a number that has this same property when multiplied by 6. Source: Puzzles 101, by
Nobuyuki Yoshigahara, a wonderful new problem book by famed Japanese puzzleman Nob Y.
Published by AK Peters.
We had a couple of ideas. First, the first digit has to be a 1. If it were 2 or greater, then
when you multiplied it by 6, the product would have too many digits. Second, you could start
with a guess for the last digit and that would determine the next one to the left, and that
digit would determine the digit to its left, and so forth. But we didn't know what should be
the last digit, and the computations were beginning to take time. So we went on to another
problem. If you have time, see if you can carry out the computations, or perhaps find other
ways to do this one.
II. United States Military Academy's
Problem of the Week. 23 Sep 2004,
week 3. To determine whether or not
A few computations showed it was true for a lot of n.
We talked about mathematical induction for a while. Sterling's formula was mentioned. But
we didn't spend much time trying to actually solve the problem. Why don't we try this again?
Try it on your own. You might get an insight and come up with a clever argument.
III. From the Geometry Problem of the Week, Math Forum at Drexel. 13 Sep 2004.
(Note: as of October 2005, this Geometry Problem of the Week has gone commercial. Access
is denied to those who don't pay. What a shame.) Partitioning the plane.
One line can divide a plane into two regions. Two lines can divide a plane into four
regions. Three lines intersecting at one point can divide a plane into six regions, but
you can create more regions than that if all the lines don't intersect at a single point
(try it!).
If you use six lines, what's the maximum number of regions into which you can divide a
plane? Don't forget to explain the strategy you used to be sure you have found the
maximum number of regions.
Extra: What if you have n lines? What's the maximum number of regions into which can
you divide the plane?
We got this one. A little careful experimenting gave us the answer for n = 6,
and we saw the pattern for the general solution.
Furthermore, we had a good informal reason why it was right. We should use this as an example
when we take up construction of proofs so we can learn how to explain our solution and turn the
explanation into an airtight argument.
IV. Perdue University's Problem of
the Week. 9/21/04. Problem 5. Incenters of triangles inscribed in ovals.
Suppose K is a smooth closed curve in the plane with convex interior,
A is a point on K, and O is a point in the interior of K.
Show that there exist points B and C on K such that O is the
center of the inscribed circle of the triangle ABC.
As most of the language was
new, all we got to do was look at what the question meant. It's a fun problem, though, and
you may be able to find the answer even if you haven't studied the formal mathematics of
the problem. Think about the Intermediate Value Theorem from calculus that says if a
continuous function takes on two values, then it takes on all intermediate values.
V. Bradley University's
Problem of the Week. 9/20/04. Problem 198. 60 degree lines.
Three lines intersect at a common point O and form 60 angles with each other. Draw
any line in the plane not containing the point of intersection O. This line intersects
the three lines at the points A, B, C. Given any two of the lengths OA, OB, OC, how do
you find the third?
We didn't have much time to look at it. We examined the special case when OA = OC.
We looked at polar coordinates and saw that if the line ABC is made the vertical line
x=1 while the point O is the origin, then the equation of the line in polar coordinates
is r = sec theta. That's when we ran out of time. I think we're on the right track, or
at least we're on one of the right tracks.
Back to Clark's Math Problem Solving Team page
D. Joyce
Department of Mathematics and Computer Science
Clark University
This file is located at http://aleph0.clarku.edu/~djoyce/mpst/