## Thursday, March 7, 2013

### The spider and the fly (Greek Version)

In the picture we see that in the point (A) is a spider and in (B) a fly. What is the shortest path that can make the spider along the surface of the adjacent cylindrical shape to catch the fly.

This puzzle is a variation on the puzzle "the spider and the fly", but much more difficult.

1. I don't see two cylinders.

2. Maybe its close to 10.95? i mean not exactly 10.95 but around there like 10.94965..... etc. I would like a reply thanks for ur time.Because my maths arent so good please be kind!I "solved" it with painting mostly and some Pithagoras maths and imagination.

3. you have to say more about your solution

if you want you can send the solution at hliaskal000@gmail.com

4. 2（3-√5）+√（100+9π²）

5. how you did that???

can you explain it?

6. Approximately: Arctg0.5=26*34;
then central angle between radii through A and B =90*+2(26*34')=143*8'~143.13*
The shortest way from A and B to cylindrical surface is by radius =3 - sqrt 5 ~ 0.8
The shortest path on the cylinder is by helical line between bases on the arc = 143.13*
Its length = [3п(143.13)]/180 ~ 7.493
The total path 2(0.8)+sqrt [(7.493)^2 + 10^2 =1.6 + 12.5 =14.1

7. 'Not sure what the exact answer is, but I'd start by converting this to a problem in 2D geometry. This is a cylinder with two closed ends, so imagine laying it on a table on it's side. Than let the endcaps fall on to the table. Finally, unwrap the cylinder by slicing it lengthwise and letting it fall to the table. Now roll the endcaps along the edge of the "cylinder" (now just a rectangular sheet) to minimize the distance between A and B. Measure. :)

8. I think, if we spread the cylinder .... If the spider does the easy for us way , it would be 2*(3-sqrt5) for the distance of the bases of cylinder and for the body of cylinder ,if we spread it it would be a rectangular with side a=10 and side b=2*π*r=2*π*3=6π. So the diagonal is d=sqrt(10^2+(6π)^2)=21.32. The total distance is 22,848, so if I have not made some great @#%\$& the reduction to 14,1 seems too much.
Now for the shortest distance if we set a new cycle with r=sqrt5 and remove it from the cycle of the base of cylinder (r=3)we would have a ring of 3-sqrt5 and outer perimeter 2*π*r=2*π*3=18,84 and inner 2*π*sqrt5=14,04. now if we can imagine the whole spread of the cylider it would be a rectangular with 2 trapeziums over and under each one.Thus , the distance will be from the corner of one trapezium to the opposite corner of the other : d=sqrt((10+3-sqrt5)^2+14,04^2)=18,166
The whole thought is based on the fact that if the cylinder is on its side,as it seems, the distance from the floor of B is 3-sqrt5 and A is 3+sqrt5...
I bet I've made at least one mistake....every critisism is acceptable.