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Where's the F@*&$^# is the Road pt. 2


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Question

After parachuting into the woods at night, you land right next to a sign that says : "Road: One Mile." But it has fallen over, and there's no way to know what direction the road is. Describe the shortest path you might take that will guarantee reaching (touching) the road. (You can't see the road until you reach it!) What's the maximum distance you might walk on your path?

The only difference, is that you know for a fact that wherever the road is, it is definitely a straight line.

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You have landed at the center of a 1-mile radius circle, at least one point of which is in common with a road.

Worst case situation is that exactly one point of the circle is shared with the road.

Moreover, the road cannot be seen from any non-zero distance.

That is, the road must be discovered using the Braille method.

Thus every point on that circle must be traveled to ensure finding that worst-case single point.

Walk 1 mile to get to the circle.

Then walk the circumference.

Worst case distance traveled = radius + circumference = 1 + 2pi miles.

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You have landed at the center of a 1-mile radius circle, at least one point of which is in common with a road.

Worst case situation is that exactly one point of the circle is shared with the road.

Moreover, the road cannot be seen from any non-zero distance.

That is, the road must be discovered using the Braille method.

Thus every point on that circle must be traveled to ensure finding that worst-case single point.

Walk 1 mile to get to the circle.

Then walk the circumference.

Worst case distance traveled = radius + circumference = 1 + 2pi miles.

radius is shorter than quarter arc.

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You have landed at the center of a 1-mile radius circle, at least one point of which is in common with a road.

Worst case situation is that exactly one point of the circle is shared with the road.

Moreover, the road cannot be seen from any non-zero distance.

That is, the road must be discovered using the Braille method.

Thus every point on that circle must be traveled to ensure finding that worst-case single point.

Walk 1 mile to get to the circle.

Then walk the circumference.

Worst case distance traveled = radius + circumference = 1 + 2pi miles.

There is a shorter path

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Walk 1 mile in any direction, then walk 3/4 of a circle. If you still haven't hit the road, continue on the straight line tangent to the circle from this point on until you hit the road. The longest path you will have to walk is 2+3/2pi miles.

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Walk 1 mile in any direction, then walk 3/4 of a circle. If you still haven't hit the road, continue on the straight line tangent to the circle from this point on until you hit the road. The longest path you will have to walk is 2+3/2pi miles.

Much closer but there is still an even shorter path.

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I believe that there is no statement in the question that describes the road as infinite; I believe it can't be. It also doesn't say the road is tangent against the circle with a radius of 1 mile. If the road is starting at 1 mile away and continuing away, the minimum path is 1+2pi.

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This op unlike part 1 states in the last line that it is a straight line. Lines go on forever.

I believe that there is no statement in the question that describes the road as infinite; I believe it can't be. It also doesn't say the road is tangent against the circle with a radius of 1 mile. If the road is starting at 1 mile away and continuing away, the minimum path is 1+2pi.

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Very nice problem.

I verified TSLF's solution numerically, but then realized the optimal condition can be calculated directly.

Which is probably what TSLF did.

Since I went through the calculation I thought I'd share it.

Calculation of optimal condition

With respect to the normal passing through the center of the circle.
Let the angle from the center to the 1st road point be a
Let the angle from the center to the 2nd road point be b

Total path = tan(a) + 1/cos(a) + 2(pi-a-b) + tan(b)

Compare distance from center to 1st road point back to 1st point of tangency by two routes:
Difference = tan(a) + 1/cos(a) - (1+2a)
Setting the derivative to zero gives

1/cos2(a) + tan(a)/cos(a) = 2, from which sin(a) = 1/2

a=30o

Comparing the distance from the 2nd point of tangency to the road by two routes:

Difference = tan(b) - 2b

Setting the derivative to zero gives sec2(b) = 2

b = 45o

Total path = (1/cos(30)) + tan(30) +7pi/6 + tan(45) = 6.397242237

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