Guest Posted January 8, 2009 Report Share Posted January 8, 2009 This seems to be a classic but I searched and didn't find a clue that this will be a repost??? There are three horses those are standing at the corners of an equilateral triangular area, having each side as 200 meters. Suddenly, each horse begins to run towards to the horse left of it. It's obvious that their track will be a curved line, ending at the middle of the triangular area. Till to that joining point, how many meters will each horse run? Quote Link to comment Share on other sites More sharing options...
0 Guest Posted January 8, 2009 Report Share Posted January 8, 2009 well!! How are the horses standing? Is it possible that the horses are facing in opposite directions, such that, 2 horses run towards each other.... Quote Link to comment Share on other sites More sharing options...
0 Guest Posted January 8, 2009 Report Share Posted January 8, 2009 well!! How are the horses standing? Is it possible that the horses are facing in opposite directions, such that, 2 horses run towards each other.... As I posted, each horse runs towards the horse left of it. I mean each runs at clockwise, none two of them run face to face. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted January 8, 2009 Report Share Posted January 8, 2009 (edited) 112 Meters approximately. If all three horses are standing facing towards the center of the triangle, and each starts to run towards the other two, or, if all three are standing the way I have shown in the attachment.horse_puzzle.doc Edited January 8, 2009 by vchopra Quote Link to comment Share on other sites More sharing options...
0 Guest Posted January 8, 2009 Report Share Posted January 8, 2009 Ah - that would be the other left then. Right! Quote Link to comment Share on other sites More sharing options...
0 Guest Posted January 8, 2009 Report Share Posted January 8, 2009 112 Meters approximately. If all three horses are standing facing towards the center of the triangle, and each starts to run towards the other two, or, if all three are standing the way I have shown in the attachment. Nope, Each horse departs to its own left, not left according to you. Think it as clockwise. Quote Link to comment Share on other sites More sharing options...
0 Prof. Templeton Posted January 8, 2009 Report Share Posted January 8, 2009 each will have traveled 200 meters when they meet in the middle. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted January 8, 2009 Report Share Posted January 8, 2009 technically speaking, the horses will never run into each other, but in reality taking into consideration the horses actually take up space, they will have run about 145 M before meeting in the middle. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted January 8, 2009 Report Share Posted January 8, 2009 All horses go their respective lefts as shown in the attachment....if they are all going at the same speed, each of them will cover 112 meters before meeting in the center. If you think this is not the correct answer please provide the detail as what direction each horse is facing at the starting position.horse_puzzle.doc Quote Link to comment Share on other sites More sharing options...
0 Prof. Templeton Posted January 8, 2009 Report Share Posted January 8, 2009 All horses go their respective lefts as shown in the attachment....if they are all going at the same speed, each of them will cover 112 meters before meeting in the center. If you think this is not the correct answer please provide the detail as what direction each horse is facing at the starting position. the horses (let's call them A,B, & C)start and remain in an equilateral formation as they spiral toward the center of the triangle, the vectors of B & C never brings them closer to A. It is only A's vector that brings it closer to B (likewise B to C and C to A) so it's as if B stood still and A moved toward B. Therefore the distance traveled must be equal to the original distance of the side of the triangle which is 200m. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted January 8, 2009 Report Share Posted January 8, 2009 Here's a picture of how I see the problem. I wonder if the solution depends on the size of the horses? If they are just points, do they ever meet?Horses.doc Quote Link to comment Share on other sites More sharing options...
0 Joe's Student Posted January 8, 2009 Report Share Posted January 8, 2009 I was wondering PT if you could show me the math that proves what your saying? Or elborate a bit more on why it is the distance you said it is. If you have he time that is. Quote Link to comment Share on other sites More sharing options...
0 EventHorizon Posted January 8, 2009 Report Share Posted January 8, 2009 The point sized horses form an equilateral triangle that spins as it shrinks to a point. The distance from one corner of the triangle to the center initially is 200/sqrt(3) meters. The angle each horse will always be running will be 30 degrees to the left of the direction directly to the center of the triangle. For a total distance moved of epsilon in an infinitesimal time period, epsilon*sqrt(3)/2 will be the amount the distance the center would decrease and epsilon/2 would be the amount traveled that is wasted in the clockwise movement. The total wasted in clockwise movement will then be 200/3 meters. Combining distance to the center and the total wasted in clockwise movement (root sum square...yay for Pythagorus! (you can do this because you are simply combining 30 degree angles together -- it doesn't matter that they are infinitesimal lengths being combined or that the actual distance is being rotated slightly after each epsilon)) results in 400/3 m = about 133.33 meters. Anyone see an error? Quote Link to comment Share on other sites More sharing options...
0 Guest Posted January 8, 2009 Report Share Posted January 8, 2009 Using just geometry and assuming each horse would follow a smooth arc that would lead them to the adjacent vertex (where his target horse would arrive if they didn't collide in the center point) I calculated... 120.5 to nearest tenth. Calculate the segment height of an arc which passes through center enroute to adjacent vertex as 57.14m using trig. Chord length for the arc is 200m (length of each side of triangle). Using formula for arc length given chord and segment height yields 241.03m and horses crash at halfway point = 120.5m Quote Link to comment Share on other sites More sharing options...
0 EventHorizon Posted January 8, 2009 Report Share Posted January 8, 2009 Using just geometry and assuming each horse would follow a smooth arc that would lead them to the adjacent vertex (where his target horse would arrive if they didn't collide in the center point) I calculated... 120.5 to nearest tenth. Calculate the segment height of an arc which passes through center enroute to adjacent vertex as 57.14m using trig. Chord length for the arc is 200m (length of each side of triangle). Using formula for arc length given chord and segment height yields 241.03m and horses crash at halfway point = 120.5m Their paths will not be on circular arcs. Their paths will be smooth and will be arced, just not along the path of a circle. Notice that at any point during the horses run, their remaining path must be a scaled version of their total path from start to finish. This is because they are still forming an equilateral triangle and the horses are still chasing the one to the left, in other words, it is a scaled version of the original problem. When close to the center point Circular arcs will result in the horses running towards the center as opposed to running towards the other horse. Quote Link to comment Share on other sites More sharing options...
0 EventHorizon Posted January 8, 2009 Report Share Posted January 8, 2009 (edited) ...If they are just points, do they ever meet? Either that, or they end up endlessly running clockwise in the smallest circle they can manage (crazy point horses...). But theoretically, yes, they do meet....as long as their speed does not decrease below a given positive threshold (the path itself has finite length). Edited January 8, 2009 by EventHorizon Quote Link to comment Share on other sites More sharing options...
0 Prof. Templeton Posted January 8, 2009 Report Share Posted January 8, 2009 I was wondering PT if you could show me the math that proves what your saying? Or elborate a bit more on why it is the distance you said it is. If you have he time that is. Well, I was wrong about their vectors not bringing them closer, that only works for a square, so it's less then the original 200m. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted January 9, 2009 Report Share Posted January 9, 2009 Although I'm not fully convinced, The original source of this problem gives solution as: At each moment after the horses depart, the positions of horses is a new equilateral triangle. This triangle gets smaller at each second until being a point. During this time, edge of triangle become from 200m to 0. Thus during this time, each horse runs 200m. There is no details in answer. This solution requires that: When a horse runs x meters, the new triangles edge will be x meters smaller than the previous triangle. I'm not sure that this is correct. We need masters comment on this. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted January 9, 2009 Report Share Posted January 9, 2009 Although I'm not fully convinced, The original source of this problem gives solution as: At each moment after the horses depart, the positions of horses is a new equilateral triangle. This triangle gets smaller at each second until being a point. During this time, edge of triangle become from 200m to 0. Thus during this time, each horse runs 200m. There is no details in answer. This solution requires that: When a horse runs x meters, the new triangles edge will be x meters smaller than the previous triangle. I'm not sure that this is correct. We need masters comment on this. That sounds plausible - the horses are always going to be facing along the edge of the triangle. And if you do the sum over smaller and smaller intervals then presumably they don't deviate from the line sufficiently to reduce the distance required to travel. I did have a feeling that solution to the old fours horses problem depended on them being in a square, and hence the vetors were perpendicular, but I'm now about 95% convinced that you could generalise it to any shape. Quote Link to comment Share on other sites More sharing options...
0 bonanova Posted January 9, 2009 Report Share Posted January 9, 2009 the horses would be running on perpendicular paths, and each would travel the length of the square's side. For an equilateral triangle, the horse being pursued is moving toward the pursuing horse at an angle of 30 degrees from perpendicular. That is, the pursued horse is closing with the pursuing horse at 1/2 [sin 30 degrees] of his speed. When they meet, the pursuing horse will have traveled 2/3 of the original separation, or 133.3 ... meters. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted January 9, 2009 Report Share Posted January 9, 2009 I agree entirely with Bonanova. I was just thinking exactly the same and saw he'd already posted it. Prof T. was just about there with his initial reasoning, but had forgotten about the 30 degree angle meaning that the pursued horse is also closing the gap. It's interesting (or actually quite shocking) that the source of the original puzzle didn't notice this though!! You were right not to be convinced nobody! Quote Link to comment Share on other sites More sharing options...
0 bonanova Posted January 9, 2009 Report Share Posted January 9, 2009 Take it to the next step where there are two horses on a two-sided polygon; that is, a line. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted January 9, 2009 Report Share Posted January 9, 2009 Take it to the next step where there are two horses on a two-sided polygon; that is, a line. In this case I assume you would mean the horses would be running directly towards each other, so the distance travelled would be half the length of the line. Extend this out to a regular pentagon. Now the pursued horse is actually travelling away from the pursuer (although not as fast as the pursuer is travelling) and so each horse would travel further than the length of one side. As you approach an infinite number of sides (i.e. a circle) the distance travelled by each horse approaches infinite, as each horse is effectively running away from it's pursuer at the same speed. I don't have time to come up with the exact formula as I've already spent far too much time on this forum so far today! Quote Link to comment Share on other sites More sharing options...
0 Guest Posted January 9, 2009 Report Share Posted January 9, 2009 I agree entirely with Bonanova. I was just thinking exactly the same and saw he'd already posted it. Prof T. was just about there with his initial reasoning, but had forgotten about the 30 degree angle meaning that the pursued horse is also closing the gap. It's interesting (or actually quite shocking) that the source of the original puzzle didn't notice this though!! You were right not to be convinced nobody! I did a quick sketch (attached.) The size of the triangle is shrinking faster than the horses are moving: The horse has moved from A to B, and if the answer was 200m then he would still have the distance B to C left to travel (albeit in a different direction.) But you can see that the distance to the next horse B to D is actually less than BC. I'm glad I left myself 5% uncertainty I had considered the circle case (but not the 2 horse case) but decided that an inifinite number of horses 200m apart would form an inifinite sized circle, and that that would break the maths in some way. untitled.bmp Quote Link to comment Share on other sites More sharing options...
0 bonanova Posted January 9, 2009 Report Share Posted January 9, 2009 I don't have time to come up with the exact formula as I've already spent far too much time on this forum so far today! the sine of the angle between the perpendicular to a horse's direction, and the "next" horse's direction. You subtract it for the triangle; it's zero for the square, and you add it for polygons with greater than four sides. In the case of the circle it's 90 degrees, added, so the horses never close; staying an unaltered course. An eternal merry-go-round. Whee! Quote Link to comment Share on other sites More sharing options...
0 Guest Posted January 9, 2009 Report Share Posted January 9, 2009 agree with bonanova on the solution approach and answer. My smooth arc approach was oversimplified and as noted, at point of collision the direction would be wrong (aimed at vertex instead of target horse). While I'm too lazy to check the calculation, the slightly longer distance makes sense as the arc would be asymetric and wider than a circular arc. Quote Link to comment Share on other sites More sharing options...
Question
Guest
This seems to be a classic but I searched and didn't find a clue that this will be a repost???
There are three horses those are standing at the corners of an equilateral triangular area, having each side as 200 meters.
Suddenly, each horse begins to run towards to the horse left of it.
It's obvious that their track will be a curved line, ending at the middle of the triangular area.
Till to that joining point, how many meters will each horse run?
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