Sergey Markelov’s Best
Nikolay Konstantinov, the creator and the organizer of the Tournament of the Towns, discussed some of his favorite tournament problems in a recent Russian interview. He mentioned two beautiful geometry problems by Sergey Markelov that I particularly loved. The first one is from the 2003 tournament.
An ant is sitting on the corner of a brick. A brick means a solid rectangular parallelepiped. The ant has a math degree and knows the shortest way to crawl to any point on the surface of the brick. Is it true that the farthest point from the ant is the opposite corner?
The other one is from 1995.
Share:There are six pine trees on the shore of a circular lake. A treasure is submerged on the bottom of the lake. The directions to the treasure say that you need to divide the pine trees into two groups of three. Each group forms a triangle, and the treasure is at the midpoint between the two triangles’ orthocenters. Unfortunately, the directions do not explain how exactly to divide the trees into the groups. How many times do you need to dive in order to guarantee finding the treasure?
Bill:
I really like the first question. My answer is a counterintuitive No, not necessarily.
Let’s work with a brick that’s 16cm x 35cm x 68cm.
The ant begins on the point where Side A (16cm x 68cm), Side B (35cm x 68cm), and Side C (16cm x 35cm) meet.
The opposite corner is the point where Side D (16cm x 68cm), Side E (35cm x 68cm), and Side F (16cm x 35cm) meet.
Imagine unfolding the brick so that Side A is adjacent to and on the same plane as Side E. This forms a 51cm x 68cm rectangle, which the Pythagorian theorem tells us we can traverse in 85cm.
But there are parts of Side F that the ant cannot reach in 85cm. If the ant had instead travelled from corner to corner across Side B and Side F, it would have traversed a 35cm x 84cm rectangle, and the trip would have been 91cm. There is an area on Side F close to the DEF corner that is farther away from the ant’s original position than is the DEF corner itself.
6 January 2011, 7:47 pmIlya:
The second problem is very beautiful. I found a two-line analytic solution based on Euler line: Place the circumcenter of a triangle at the origin. Then its orthocenter is given by t(A+B+C) thanks to the Euler line theorem. Hence t=1, and the point we are looking for is the half of the sum of the locations of the six pine trees. Do you know a geometric solution that doesn’t involve computations?
7 January 2011, 2:09 pmTanya Khovanova:
Ilya,
I solved the second problem by asking myself, what would have happened if instead of the orthocenter we used the intersection of medians.
7 January 2011, 2:26 pmIlya:
Well, with medians it is clear, but it is an affine problem in this case: the answer is the average of the points, and, for example, the condition that the lake is a disc is not necessary. However, with orthocenters the problem is not affine, and I don’t see how to pass from the centers of mass to the orthocenters (I mean geometrically)
7 January 2011, 3:25 pmTanya Khovanova:
Ilya,
You go from the centers of mass to the orthocenters by the Euler line.
7 January 2011, 4:12 pmAustin:
I also liked the first problem. There are six reasonable candidates for the shortest path between opposite corners. (By “reasonable,” I mean that the path travels over only two faces, and is a straight line when the brick surface is unfolded appropriately.)
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.” 🙂
7 January 2011, 7:35 pmIlya:
ok 🙂
8 January 2011, 3:54 amcolorblind:
Suppose the brick has sides of length a, b, and c where a>=b>=c. The side touching the ant’s starting point with dimensions a by b is A, with dimensions a by c is B, and dimensions b by c is C. the side opposite A is D, opposite B is E, and opposite C is F. For the sake of clarity XY will mean a path crosses faces X and Y in that order, and XYZ will mean a path crosses faces X and Y and Z in that order. XY’ and XYZ’ will mean that a variation of the full paths XY and XYZ are being considered.
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.
9 January 2011, 4:14 amAustin:
Hi, colorblind. I don’t think you meant “Pick a point along path BF or CE” (in the third paragraph). I think you *did* mean “Pick a point on the b-by-c face incident to the original destination, at distance d from each of the edges of that face that meet at the old terminating point.” Such a point needn’t be on path BF or CE.
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.
17 January 2011, 5:58 pmTanya Khovanova’s Math Blog » Blog Archive » Tetrahedron Problems:
[…] around for nice problems, for my readers often send them to me. In response to my blog about him, Sergey Markelov’s Best, Markelov sent me more of his problems. Here is a cute tetrahedron problem that he designed: Six […]
18 February 2011, 2:25 pmmashplum:
If the ant’s starting corner is placed at the origin and the opposite corner is at point (l,w,h) then any other point on the surface of the brick can be described by an ordered triple (x,y,z) where x≤l, y≤w, and z≤h. If the brick is trimmed so that (x,y,z) is the new opposite corner, the shortest path to (x,y,z) must be less than the shortest path to (l,w,h) since we are taking about a smaller brick. That same shortest path to (x,y,z) can be followed on the original brick, since in both cases he could avoid the trimmed area. So the opposite corner is always the farthest distance.
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.
5 April 2011, 6:46 am