res@cbosgd.UUCP (06/07/83)
The following was going around the office recently: A principal and a teacher are talking as several people walk by. The principal mentions to the teacher: "I know the three people who just past us. The product of their ages is 2450. Can you tell me their individual ages?" The teacher replies: "I do not have enough information to uniquely answer the question." The principal then says: "OK, the sum of their ages is exactly twice your age. Now can you tell me their ages?" The teacher thinks for awhile and replies: "I still don't have enough information to uniquely identify their ages." Then the principal states: "I will tell you this, and this will be conclusive: Each of the three people we are discussing is younger than myself." The teacher says: "Now I can tell you their ages." Can you figure out the ages of *all* the people involved, including the principal and the teacher?
wjk@ihuxf.UUCP (Bill Krauss) (02/10/84)
I've been trying unsuccessfully to solve the following problem, and I wonder if anyone out there can help. Given the square as shown in the picture. | Consider all continuous, non-self-inter- (0,1)|_____________ (1,1) secting curves that lie completely on or | | in the square, such that one endpoint is on |________ | the x axis and the other is on the y axis, | \L | and neither is the point (0,0) (thus divid- | \ | ing the square into two parts). Find the | A \ | curve for which the ratio R = A / L is (0,0)|___________|_|______ maximized, where A is the area of the part (1,0) of the square that contains (0,0) and L is the length of the curve. I found that for the curve consisting of the segments (0,1) to (1,1) and (1,1) to (1,0), R = A / L = 1/2. For the curve x**2 + y**2 = 1, R = A / L = (pi/4) / (pi/2) = 1/2. It seems the maximum lies somewhere between these two examples, possibly something of the form x**p + y**p = 1, where p > 2.
apdoo@alice.UUCP (Alan Weiss) (02/15/84)
I made an error in posting the solution. The correct answer, of course, is the circle of radius (1-sqr(pi/4))/(1-pi/4) in the upper right hand corner, along with segments of the lines y=1 and x=1. The ratio of area to length was correctly stated.
leimkuhl@uiuccsb.UUCP (02/17/84)
#R:ihuxf:-201300:uiuccsb:9700018:000:79 uiuccsb!leimkuhl Feb 16 12:08:00 1984 Love that ironclad reasoning. Anyone have a more convincing argument?
pizer@ecsvax.UUCP (07/04/84)
A friend of mine introduced me to the following puzzle, I think he called it
fractal geometry. Whatever it is, it makes an interesting computer problem.
I have found what I believe to be the most efficient method of solving it, but
I like to see what other people come up with, here goes
The problem is to take a number, and determine whether the bit at that
particular position is either 1 or 0. The position refers to a position in
a string created by taking the string "1001" and then using that as the string,
repeat it using the same pattern, i.e., "1" becomes "1001", "1001" becomes
"1001011001101001", which becomes "10010110011010010110100110010110...". Given
number 3, for example, the bit would be 0, given 5, it would also be 0, given
7, it would be 1.
If you don't quite understand (I'm not terribly good at explaining things
on paper, or for that matter CRT) or if you have a good solution (mine can
almost be put into one BASIC line) please tell me.
Billy Pizer
(pizer@ecsvax)
{happy 4th o'July}jmrobson@watdaisy.UUCP (Mike Robson) (07/04/84)
It looks to me as if the bit in position i in these strings is the bit required to be added to the binary representation of i-1 to give it odd parity. An interesting point is that the whole process could have been started from the string 10 instead of 1001. Mike Robson, University of Waterloo (etc.)