Another way to think of it is this: If I’m at the South Pole, there is no spot on the surface of the earth that is farther away than the North Pole. I could travel directly to the North Pole and then continue to travel one more mile to destination B. But that doesn’t make destination B farther away from the South Pole, because I could have found a shorter path to it.

]]>Assuming the above correction is, well, correct, I agree with your approach but not with your computations. I think there is a sign error, and your conclusion should be exactly the opposite:

d+a > b+c.

Thus, if a > b+c, then there is a point farther from the ant than the opposite corner.

I came to the same conclusion by the following visualization. Considering the points on the far b-by-c face, those which are closer than the opposite corner must lie inside of one several regions cut from that face by circles. Assuming a>b, a sufficiently small neighborhood of the opposite corner will be intersected by only two of these circles: those whose radii are equal to the length of the paths you call BF and CE. If a=b+c, then these two circles are tangent, and if a>b+c, then the angles subtended by these circles at the opposite corner shrink further. Thus in these cases, there are some points arbitrarily close to the opposite corner which are not covered by these two circles. However, even if a<b+c, there may be an interior region of the b-by-c face which lies farther than the opposite corner. For this to happen, the two aforementioned circles must intersect close enough to the opposite corner, where the criterion for “close enough” depends on the lengths of the paths AF, CD, BD, and AE. I don’t feel like working out the exact criterion algebraically, so I am going to stick with my “perhaps” in this case of the problem.

At any rate, I think it is pretty neat that when there *are* destinations farther than the opposite corner, the locus of all such points is generally cut out by three (or more?) concave circular arcs.

]]>As Austin noted there are six candidate paths. the paths BF and CE will have length sqrt(a^2+b^2+c^2+2ab), two more (AF, CD) will have lengths sqrt(a^2+b^2+c^2+2ac), and the last two (BD, AE)will have length sqrt(a^2+b^2+c^2+2bc). Of course these paths are all of equal length if and only if a=b=c. In that case, the far corner is clearly the farthest point. However, in any other case, a>c so paths BF and CE will be longer than the paths BD and AE.

Now suppose we pick a point along path BF or CE sufficiently close to the opposite corner, say distance d from each of the edges of the second face that meet at the old terminating point. One of the two paths BF or CE will be shortened. suppose BF’ is the shortened path. BF’ is still longer than the optimal paths BD and AE. The question is how will BD’ and AE’, the paths to the point terminating BF’, be different than BD and AE. The lengths of BD’ and AE’ are sqrt((b+c-d)^2+(a+d)^2). Now we just need to determine if sqrt(a^2+B^2+c^2+2bc) is ever less than sqrt((b+c-d)^2+(a+d)^2). That is true if a^2+b^2+c^2+2bc is ever less than (b+c-d)^2+(a+d)^2 or if there are conditions where a^2+b^2+c^2+2bc-(b+c-d)^2+(a+d)^2 < 0.

Simplifying:

a^2+b^2+c^2+2bc-(b+c-d)^2+(a+d)^2 < 0

b^2+2bc-2bd+c^2-2cd+2d^2+2ad+a^2-b^2-2bc-c^2-a^2 <0

2d^2+2ad-2bd-2cd <0

d+a-b-c<0

d+a<b+c

Therefore, whenever a<b+c we can chooses a point such that the path from one corner to the opposite corner is shorter.

]]>Generically, two of these six paths will be (equally) the shortest, and if we perturb the destination by a small enough distance, the shortest path to the new destination will be a small perturbation of one of those two paths. But are there destinations for which both perturbed paths are longer than the original two paths? I think the answer is yes (and thus NO to the original question) if the largest dimension of the brick is at least as great as the sum of the other two dimensions. Otherwise, my answer is “perhaps.” 🙂

]]>You go from the centers of mass to the orthocenters by the Euler line.

]]>I solved the second problem by asking myself, what would have happened if instead of the orthocenter we used the intersection of medians.

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