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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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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.

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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.

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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.

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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?

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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 by Kioshanta
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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. )

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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.

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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

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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... ;)

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