BrainDen.com - Brain Teasers
• 0

# Waiting spider

## Question

The spider on  its web just in front of  the tower clock's center   plans to strike the bugs on
the tips of the clock's  6" , 8"  &  10" hands.  It must move straight toward the edge to
encounter all its preys in one passing as near as possible.  If the spider speed is one inch
per minute..at what time should it start the attack?

## Recommended Posts

• 0

I don't know whether the longest hand is the second hand or the minute hand, so I'll consider both situations.

Either way, the optimal trajectory must start exactly in the same direction as the second hand. This is because this provides the best trade-offs in errors. Meanwhile, since the hour hand is obviously the 6" hand, we also know that the spider must set off 6 minutes in advance of it, which translates to approximately half a tick counter-clockwise of the spider's trajectory.

If the longest hand is the minute hand, then the spider must set off 10 minutes in advance of it, which translates to 10 full ticks. This means the spider's best opportunity presents itself when the minute hand is around 9.5 ticks behind the hour hand. This obviously occurs about once an hour, so we have eleven candidates. Since we are adjusting the second hand such that it lines up exactly with the spider's starting trajectory, out of the eleven times where the minute hand is exactly 9.5 ticks behind the hour hand, we are interested in the time where the second hand is the least far from where it needs to be.

The position of the second hand satisfies the relation, (-11/720)*s = 9.5 (mod 60).

All solutions are in the form, s = -9.5*(720/11) + (43200/11)*n, for all integers n.

From this, we can glean that the second hand differs by about 27+ ticks between our hourly candidates. If we assume that the hour & minute hands move forward by about 5+ ticks per candidate, we can infer that the second hand error is offset by around 22 ticks between candidates.

In one solution, the second hand is on 38 ticks from 12 o' clock at 10 minutes to midnight/noon. This makes the positive second hand error about 38 ticks. Incidentally, since adding 22 to this gives us a round number, this suggests that the next hour has the spider's optimal time of attack at around 12:55:05.

The same reasoning applies if the longest hand is the second hand, except this time the spider must set off when the minute hand is approximately 7.5 ticks behind the hour hand.

The position of the second hand satisfies the relation, (-11/720)*s = 7.5 (mod 60).

All solutions are in the form, s = -7.5*(720/11) + (43200/11)*n, for all integers n.

Again, we can assume that the second hand error is offset by around 22 ticks per candidate.

In one solution, the second hand is on 49 ticks from 12 o' clock at 8 minutes to midnight/noon. From here, we can make a very rough estimate of the remaining positive errors: 11, 33, 55, 17, 39, 1, 23, 45, 7, 29. Our seventh candidate appears to be the closest. Calculating that time gives us 6:24:34.

Edited by gavinksong
##### Share on other sites

• 0

OP says ... It must move straight toward the edge to encounter all its preys in one passing as near as possible ... at what time should it start the attack?. At first reading I thought it was asking how quickly the spider can get all three bugs eaten. But that would not involve waiting. Then I thought the spider should get to them all in one place, but they will never be in one place. Then I thought it asks that the spider move along a strictly radial path. That would make an interesting calculation indeed. Is that it?

If so, the spider's desired path begins at time t0 minutes along a radial direction of a0 degrees (modulo 360) measured CW from 12:00. In {r, a} coordinates, r measured in inches, the spider's path is {(t-t0), a0}. Let's name the bugs h, m and s, for the hands on which they sit. At midnight their respective positions are {6, 0}, {8, 0} and {10, 0}. At time t thereafter they are {6, ct/12}, {8, ct} and {10, 60ct}, where c = 6 degrees/minute.

Then we desire a t0 such that a0 = c(t0+4)/12 = c(t0+6) = 60c(t0+8) modulo 360.

But that may be overly restrictive. The OP does not require that spider does not stop, once he begins, until poor s is eaten. Yet, without that condition, a0 can be any value, and t0 also can be any value. The spider picks any angle and starts whenever he wants. He goes 6", waits for h to arrive, eats him, goes 2" more, waits for m, eats him, goes 2" more, waits for s and eats him. So there must be another condition. That the spider does not stop seems like a good one, because the spider then eats the last bug as quickly as possible, after launching his attack. In fact, that could simply be the statement of the problem.

##### Share on other sites

• 0

@bonanova,

Yes it asks that the spider move along a strictly radial path.

The attack does not involve stopping..just a quick bite to paralize as it passes to them.
It would devour their juices later.While not exactly under its fangs..the bugs must be near
enough for the strikes.

## Join the conversation

You can post now and register later. If you have an account, sign in now to post with your account.

×   Pasted as rich text.   Paste as plain text instead

Only 75 emoji are allowed.

×   Your previous content has been restored.   Clear editor

×   You cannot paste images directly. Upload or insert images from URL.