An ant is trapped on a spider web inside a can with open top. The can has a radius R and height H. A spider sitting on the outside face of the can, sensed the trapped ant and started crawling towards it. The relative positions of the ant and spider are shown in the figure below.
If the spider moves at a constant crawling speed, calculate the total length of the path to reach the ant at the quickest possible time. Please round-off your answer to the nearest four decimal places.
CLARIFICATIONS:
- It is understood there can be 2 possible positions, given a single y-offset. In this exercise please consider only the position as drawn above. That means at the top view, the ant and the spidert shall only be within the first and third quadrant, respectively.
- The dotted lines is not the spider's path. They are used only to project the heights to the side.
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My results are all different, although the formula gets the right result in all the simple cases like hs=H/2; ha=H/2; xs=0; xa=0; etc...
I’m checking my solution…
Hi William,
I updated the test suites. I made a mistake in using full diameter instead of the semicircle's diameter. I also added a clarification to ensure only one option is considered.,
With regards to your solution. I think that your algorithm will work only if both the ant and spider are in the same side. Please take note that the ant is inside and the spider is outside. Please try again. Thanks.
The dotted line doesn't show it, but presumably the spider has to go up, over the top edge of the can, and down again. Also it can presumably choose whether to go clockwise or counterclockwise.
Hi William, I've included one solution into the problem itself. I hope this clarifies everything. Of course the circumference will not change. But if you would not consider inside/outside, the insects will appear to be closer together, and you will be get shorter distance than the correct one.