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A string is wound symmetrically around a circular rod. The string goes exactly
4 times around the rod. The circumference of the rod is 4 cm and its length is 12 cm.
Find the length of the string.
Show all your work.

(It was presented at a meeting of the European Mathematical Society in 2001,
"Reference levels in mathematics in Europe at age16").

Can you solve it? You may want to try before seing the solution.
[I sometimes train olympiad students at my university, so I like such problems].

restart;
eq:= 2/Pi*cos(t), 2/Pi*sin(t), 3/2/Pi*t; # The equations of the helix, t in 0 .. 8*Pi:
               
p:=plots[spacecurve]([eq, t=0..8*Pi],scaling=constrained,color=red, thickness=5, axes=none):
plots:-display(plottools:-cylinder([0,0,0], 2/Pi, 12, style=surface, color=yellow),
                         p, scaling=constrained,axes=none);
 

VectorCalculus:-ArcLength(<eq>, t=0..8*Pi);

                           20

 

Let's look at the first loop around the rod.
If we develop the corresponding 1/4 of the cylinder, it results a rectangle  whose sides are 4 and 12/4 = 3.
The diagonal is 5 (ask Pythagora why), so the length of the string is 4*5 = 20.

 

A duck, pursued by a fox, escapes to the center of a perfectly circular pond. The fox cannot swim, and the duck cannot take flight from the water. The fox is four times faster than the duck. Assuming the fox and duck pursue optimum strategies, is it possible for the duck to reach the edge of the pond and fly away without being eaten? If so, how?

http://www.crazyforcode.com/fox-duck-puzzle/

there is an animation here

https://www.youtube.com/watch?v=Zw9cHEnhzWo

wonder if the equations of motion can be derived usingg maple and an animaton...?

Whassup homies?

http://www.mathsisfun.com/puzzles/who-lives-in-the-city--solution.html

tried to solve this using C.Loves program, but didn't quite get their solution...

Who_Lives_in_the_Cit.mw

Vars:= [PN,Name, TV, Dest,Ages,Hair,Lives]:
PN:=[$1..5]:
Name:= [Bob, Keeley, Rachael, Eilish, Amy]:
TV:=[Simpsons, Coronation, Eastenders, Desperate, Neighbours]:
Dest:= [Fra, Aus, Eng, Afr,Ita]:
Ages:= [14, 21, 46, 52, 81]:
Hair:=[afro, long, straight, curly , bald]:
Lives:= [town, city, village, farm, youth]:
Con1:= Desperate=3: Con2:= Bob=1: Con3:= NextTo(Simpsons,youth,PN): Con4:= Succ(Afr,Rachael,PN): Con5:= village=52: Con6:= Aus=straight: Con7:= Afr=Desperate: Con8:= 14=5: Con9:= Amy=Eastenders: Con10:= Ita=long: Con11:= Keeley=village: Con12:= bald=46: Con13:= Eng=4: Con14:= NextTo(Desperate,Neighbours,PN): Con15:= NextTo(Coronation,afro,PN): Con16:= NextTo(Rachael,afro,PN): Con17:= 21=youth: Con18:= Coronation=long: Con19:= 81=farm: Con20:= Fra=town: Con21:= Eilish<>straight:

read "LogicProblem.mpl"; City:= LogicProblem(Vars): with(City);

 

I'm trying to write a program that solves sudoku's using a Groebner basis. I introduced 81 variables x1 to x81, this is a linearisation of the sudoku board.

The space of valid sudokus is defined by:


for i=1,…,81 : Fi=(xi−1)(xi−2)⋯(xi−9) This represents the fact that all squares have integer values between 1 and 9.

for all xi and xj which...

The June edition of the IBM Ponder This website poses the following puzzle:

Assume that cars have a length of two units and that they are parked along the circumference of a circle whose length is 100 units, which is marked as 100 segments, each one exactly one unit long.

A car can park on any two adjacent free segments (i.e., it does not need any extra maneuvering space).

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