PUMaC 2012 · 几何（B 组） · 第 7 题
PUMaC 2012 — Geometry (Division B) — Problem 7
题目详情
- [ 7 ] Assume the earth is a perfect sphere with a circumference of 60 units. A great circle is a circle on a sphere whose center is also the center of the sphere. There are three train tracks on three great circles of the earth. One is along the equator and the other two pass through the poles, intersecting at a 90 degree angle. If each track has a train of length L traveling at the same speed, what is the maximum value of L such that the trains can travel without crashing?
解析
- [ 7 ] Assume the earth is a perfect sphere with a circumference of 60 units. A great circle is a circle on a sphere whose center is also the center of the sphere. There are three train tracks on three great circles of the earth. One is along the equator and the other two pass through the poles, intersecting at a 90 degree angle. If each track has a train of length L traveling at the same speed, what is the maximum value of L such that the trains can travel without crashing? Solution: Let E be the circumference of the earth. Note that if no train barely-crashes into another train (“barely-crashing” refers to the case where a train enters an intersection just as the tail of another train barely leaves the intersection), then we can simply extend ev- ery train until one train barely-crashes into another. Therefore we can assume without loss of generality that train A is barely leaving the north pole just as train B is entering the north pole. For the sake of clarity, we will label the intersections 1 through 6, with 1 and 6 being the north and south poles respectively, 4 and 2 on train A ’s track, and 3 and 5 on train B ’s track, as in the diagram on the next page. Consider the possibility that the trains are longer than E/ 4. If we can make this work, then we no longer need to consider scenarios for trains to be shorter. Then at the moment the tail of train A barely leaves the north pole (and train B barely enters the north pole), train 3 C is either heading towards intersection 3 (where train B is) or heading towards intersection 4 (where train A is), on the longer 3 E/ 4 portion of its track that passes through intersections 2 and 5. We will first consider the case for when train C is heading towards intersection 4 (where train A sits): In order for it to not collide into train A , it must be at least a distance of E/ 4 away from intersection 4. In order to not collide into train B when train B comes down to intersection 5, train C must also either be of length at most E/ 4 (if it initially sits with its front at intersection 5), in which case it will have left the intersection by the time train B arrives, or it must be much more than E/ 2 away from intersectin 4, so that it doesn’t arrive at intersection 5 when train B arrives, which leaves much less than E/ 4 room for it to exist (between intersections 2 and 3). Since we are considering the possibility when trains are longer than E/ 4, neither scenario is acceptable, as both cases give it at most E/ 4 length. Therefore if the trains are longer than E/ 4, train C cannot be heading towards train A at intersection 4. Now let’s consider the other case where train C is heading towards intersection 3 (where train B sits) at the moment the tail of train A barely leaves the north pole. Let s denote the offset of train C ’s tail from intersection 4: 4 In order for train C to not collide into train B at intersection 3, the distance between the front of train C and the intersection must be at least the distance between the tail of train B and the intersection: 3 E/ 4 − ( L + s ) ≥ L − E/ 4 In order for train B to not collide into train C , we simply have to make sure s is nonnegative (as in, the distance between the front of train B and intersection 5 (which is E/ 4) must be at least the distance between the tail of train C and intersection 5 (which is E/ 4 − s ), so s simply has to be at least 0). In order for train A to not collide with train C at intersection 2, the distance between the front of train A and the intersection must be at least the distance between the tail of train C and the intersection: 3 E/ 4 − L ≥ E/ 2 − s The first inequality can be rearranged to become E − s L ≤ 2 and the second inequality can be rearranged to become E + 4 s L ≤ 4 As s increases from 0 to E/ 2, the first bound decreases while the second increases, so the maximum occurs when the two are equal: 2 E − 2 s = E + 4 s E s = 6 5 E L ≤ 12 5 5 E Since L = is the maximum length that satisfies both inequalities, this length is the max- 12 5 E imum possible length when the length is greater than E/ 4. Note that when L = , if we 12 advance time such that the trains move by s = E/ 6, we have the same exact situation as before, except with the new north pole at 3, train C becoming the new train B , train B becoming the new train A , and train A becoming the new train C , with the same offset s = E/ 6, and no collisions having occurred. Therefore this length of 5 E/ 12 is possible with no collisions, and we no longer need to consider any cases where the train has a length of less than E/ 4. Plugging in E = 60, we have L = 25 . Problem contributed by John Stogin