bonanova Posted December 13, 2008 Report Share Posted December 13, 2008 In the remarkable county of Semicircle, two roads connect the towns of A and F. One road is perfectly straight; and it forms the diameter of the other road - a perfect semicircle. Somewhere along the semicircular road are the towns of B, C, D and E. Remarkably, both A and F towns are an exact number of miles from the other four, as are A and F themselves. That is, the distance from town A to the other five towns is an exact number of miles; the same is true for town F. This could not be true if the straight road were any shorter. What is the distance of town A from town F? Quote Link to comment Share on other sites More sharing options...
0 Guest Posted December 13, 2008 Report Share Posted December 13, 2008 In the remarkable county of Semicircle, two roads connect the towns of A and F. One road is perfectly straight; and it forms the diameter of the other road - a perfect semicircle. Somewhere along the semicircular road are the towns of B, C, D and E. Remarkably, both A and F towns are an exact number of miles from the other four, as are A and F themselves. That is, the distance from town A to the other five towns is an exact number of miles; the same is true for town F. This could not be true if the straight road were any shorter. What is the distance of town A from town F? 10 miles, but still trying to prove it. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted December 13, 2008 Report Share Posted December 13, 2008 In the remarkable county of Semicircle, two roads connect the towns of A and F. One road is perfectly straight; and it forms the diameter of the other road - a perfect semicircle. Somewhere along the semicircular road are the towns of B, C, D and E. Remarkably, both A and F towns are an exact number of miles from the other four, as are A and F themselves. That is, the distance from town A to the other five towns is an exact number of miles; the same is true for town F. This could not be true if the straight road were any shorter. What is the distance of town A from town F? If I understand correctly,65 miles. Quote Link to comment Share on other sites More sharing options...
0 bonanova Posted December 13, 2008 Author Report Share Posted December 13, 2008 If I understand correctly,65 miles. Yup. Nice. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted December 13, 2008 Report Share Posted December 13, 2008 (edited) I don't think 65 miles works. The lowest I could get is 325 miles (If I understand the prolem, that is). Edit: Just saw your post, nevermind Edited December 13, 2008 by Kioshanta Quote Link to comment Share on other sites More sharing options...
0 Guest Posted December 13, 2008 Report Share Posted December 13, 2008 (edited) I realize what I did wrong, but clearly that's not all. Now that I correct it, I get 25 miles. Could you provide a solution? Edited December 13, 2008 by Kioshanta Quote Link to comment Share on other sites More sharing options...
0 Guest Posted December 13, 2008 Report Share Posted December 13, 2008 I realize what I did wrong, but clearly that's not all. Now that I correct it, I get 25 miles. Could you provide a solution? Two diametrically opposite, and any third point, on a circle form a right triangle, where the diameter is the hypotenuse. To solve the problem, we need to find the first two Pythagorean triples (from symmetry, each one will give us two additional towns on the ring-road). They are 3,4,5 and 5,12,13. The LCM of 5 and 13 is 65. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted December 13, 2008 Report Share Posted December 13, 2008 Two diametrically opposite, and any third point, on a circle form a right triangle, where the diameter is the hypotenuse. To solve the problem, we need to find the first two Pythagorean triples (from symmetry, each one will give us two additional towns on the ring-road). They are 3,4,5 and 5,12,13. The LCM of 5 and 13 is 65. I did that, but another triple is 7, 24, 25. The LCM of 5 (from 3,4,5) and 25 (7,24,25) is 25. Would that not work? Quote Link to comment Share on other sites More sharing options...
0 Guest Posted December 14, 2008 Report Share Posted December 14, 2008 I did that, but another triple is 7, 24, 25. The LCM of 5 (from 3,4,5) and 25 (7,24,25) is 25. Would that not work? Yes, it would. Nice. Quote Link to comment Share on other sites More sharing options...
0 bonanova Posted December 14, 2008 Author Report Share Posted December 14, 2008 Yes, it would. Nice. Just to tweak things ... What if B, C, D and E were all closer to A then to F? That excludes mirror-image placements. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted December 14, 2008 Report Share Posted December 14, 2008 (edited) Just to tweak things ... What if B, C, D and E were all closer to A then to F? That excludes mirror-image placements. That's how I got 325 earlier. That is the LCM of 65, 25, 13 and 5. Edited December 14, 2008 by Kioshanta Quote Link to comment Share on other sites More sharing options...
0 bonanova Posted December 14, 2008 Author Report Share Posted December 14, 2008 That's how I got 325 earlier. That is the LCM of 65, 25, 13 and 5. It can be shorter than that. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted December 15, 2008 Report Share Posted December 15, 2008 It can be shorter than that. 85 miles? These 4 non-multiple Pythagorean triples have an LCM of 85: [3,4,5] [15,8,17] [13,84,85] [77,36,85] (By multiple I mean triples like this set that are simply multiples of each other and don't provide unique coordinates for a town: [3,4,5], [8,6,10], [9,12,15], etc. ) Quote Link to comment Share on other sites More sharing options...
0 bonanova Posted December 16, 2008 Author Report Share Posted December 16, 2008 85 miles? These 4 non-multiple Pythagorean triples have an LCM of 85: [3,4,5] [15,8,17] [13,84,85] [77,36,85] (By multiple I mean triples like this set that are simply multiples of each other and don't provide unique coordinates for a town: [3,4,5], [8,6,10], [9,12,15], etc. ) Shorter still I think. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted December 16, 2008 Report Share Posted December 16, 2008 Shorter still I think. I'm curious to see, I hope you're right - you did incorrectly proclaim 65 as the correct answer for the first one. Quote Link to comment Share on other sites More sharing options...
0 bonanova Posted December 16, 2008 Author Report Share Posted December 16, 2008 I'm curious to see, I hope you're right - you did incorrectly proclaim 65 as the correct answer for the first one. 652 = 632 + 162 652 = 602 + 252 652 = 562 + 332 652 = 522 + 392 Quote Link to comment Share on other sites More sharing options...
0 Guest Posted December 16, 2008 Report Share Posted December 16, 2008 652 = 632 + 162 652 = 602 + 252 652 = 562 + 332 652 = 522 + 392 Someone answered 65 and you said 'yup'. But the answer was later shown to be 25. I woke up last night, though I had it - the center point I reasoned, could be considered either half of the semicircle. But then I realized I still needed a triple, and I was no closer. But you are right these do indeed work. My table of 'triples' was lacking. Nice Quote Link to comment Share on other sites More sharing options...
0 bonanova Posted December 16, 2008 Author Report Share Posted December 16, 2008 Someone answered 65 and you said 'yup'. But the answer was later shown to be 25. I woke up last night, though I had it - the center point I reasoned, could be considered either half of the semicircle. But then I realized I still needed a triple, and I was no closer. But you are right these do indeed work. My table of 'triples' was lacking. Nice Yes I thought they had it -- but then realized they were re-using symmetric points. Rather than say more, I just excluded symmetry to see if the other points would be found... Quote Link to comment Share on other sites More sharing options...
Question
bonanova
In the remarkable county of Semicircle, two roads connect the towns of A and F.
One road is perfectly straight; and it forms the diameter of the other road - a perfect semicircle.
Somewhere along the semicircular road are the towns of B, C, D and E.
Remarkably, both A and F towns are an exact number of miles from the other four, as are A and F themselves.
That is, the distance from town A to the other five towns is an exact number of miles; the same is true for town F.
This could not be true if the straight road were any shorter.
What is the distance of town A from town F?
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