[bonanova]

An ant stands in a fenced triangular field 1 meter on a side.

The fences are painted, respectively: red, white and blue.

With really nothing else to do, the ant walks to each of the three

fences; he returns to his starting point after each fence is visited.

If his starting point is randomly chosen, how far would we expect

the ant would walk if he takes the shortest possible route?

[Guest]

Not sure if this is right, but isn't the expected distance the same as the distance from the point which is equidistant from all 3 sides (the incenter)? The shortest distance from a point to a side is the perpendicular from the point.

The inradius times the semiperimeter is the area, which is 1/2 x 1/2x sqrt(3) = (sqrt3)/4. The semiperimeter is 3/2, so the inradius is (sqrt3)/6.

The expected distance is just sqrt(3)x6 = sqrt(3)...I think

Edited March 21, 2009 by bonanova
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[bonanova]

Not sure if this is right, but isn't the expected distance the same as the distance from the point which is equidistant from all 3 sides (the incenter)? The shortest distance from a point to a side is the perpendicular from the point.

The inradius times the semiperimeter is the area, which is 1/2 x 1/2x sqrt(3) = (sqrt3)/4. The semiperimeter is 3/2, so the inradius is (sqrt3)/6.

The expected distance is just sqrt(3)x6 = sqrt(3)...I think

Yes.

There is an even stronger argument for the answer to your first question.

Can you find it?

[CaptainEd]

I presume that ANY location inside the triangle would give the same result. If so, then if he starts at one corner and simply goes to the opposite side and back, it would be the same length. That height is Sqrt(3)/2, and going there and back is sqrt(3).

[bonanova]

I presume that ANY location inside the triangle would give the same result. If so, then if he starts at one corner and simply goes to the opposite side and back, it would be the same length. That height is Sqrt(3)/2, and going there and back is sqrt(3).

That's correct, and fairly straightforward to prove.

Once you have that, you can choose any point.

Your choice is the simplest; Next's was nice also.

BTW, any point along the line from the center to a vertex works also.

As one distance increases, the other two decrease at sin[30^o] - 1/2 that rate keeping the total constant.

For points off the lines of symmetry, inspect the areas of the three triangles defined by the starting point and the three pairs of vertices.

Equate their sum to the area of the large triangle and you have it.

[Guest]

If I understand the responses so far, the claim is that the ant travels the same distance no matter where in the triangle he starts.

But

If the ant starts at the center of the triangle (a distance of Sqrt(3)/4 from each side), his total distance travelled is 3 * sqrt(3) /2. But, if the ant starts in a corner, the distance to the far side is almost 1, so that the ant's total distance travelled appears to approach 2 as the ant gets closer to a corner of the triangle.

[bonanova]

If I understand the responses so far, the claim is that the ant travels the same distance no matter where in the triangle he starts.

But

If the ant starts at the center of the triangle (a distance of Sqrt(3)/4 from each side), his total distance travelled is 3 * sqrt(3) /2. But, if the ant starts in a corner, the distance to the far side is almost 1, so that the ant's total distance travelled appears to approach 2 as the ant gets closer to a corner of the triangle.

From center to side is 1/3 of the altitude or Sqrt(3)/6.

Three short trips = 1 long one.