# word problem with distance

#### eric beans

##### Junior Member
I was stumped. How do you even approach this problem when you don't know what the map looks like?

#### Subhotosh Khan

##### Super Moderator
Staff member
I was stumped. How do you even approach this problem when you don't know what the map looks like?View attachment 28293
Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:

#### lev888

##### Senior Member
I was stumped. How do you even approach this problem when you don't know what the map looks like?View attachment 28293
I think the problem makes sense only if we replace "depending on" with "regardless of".

#### Dr.Peterson

##### Elite Member
I think the problem makes sense only if we replace "depending on" with "regardless of".
I think the wording is valid, though easy to misunderstand.

Which route is shortest depends on the location of the towns; they want to know which route is not the shortest for any layout. Your "regardless of" may be clearer, but I'd also want to put it at the end rather than the start of the question. I'd say this:

Which of the following could not be the shortest journey for any layout of the towns?​

I was stumped. How do you even approach this problem when you don't know what the map looks like?

I think the first thing I'd do is to try making a layout that fits each option, and hope to find one that I struggle with. Then try to convince yourself that your inability to do it is not just your fault, but inherent in the route.

#### lev888

##### Senior Member
I think the wording is valid, though easy to misunderstand.
I can't parse it. It's either depending on or independent of, no?

#### eric beans

##### Junior Member
Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:

there could be any infinite number of possible layout of the 4 towns. how was i to try them all? dejected, and clueless, what I did was i just drew a box with each of the 4 corners x, a, b, c. then i tried each of the answer choices. i didn't see any major insights. A and D traveled the longest distance and had the same distance. (i don't understand how they can predict which would be longest without a specific map of the 4 towns.) then i had the feeling i didn't understand the question and/or missed something.

I get the sense i missed something but i don't know what.

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#### JeffM

##### Elite Member
Consider choice A. Can you create a map where that route is shortest?

AXBC going left to right all on a straight line works. So choice A is wrong; you can draw a map where total distance is shortest with XAXBCBX

You do not have the map as you said. But you do not need the correct map. You need to find a situation where no map is possible.

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#### lev888

##### Senior Member
there could be any infinite number of possible layout of the 4 towns. how was i to try them all? dejected, and clueless, what I did was i just drew a box with each of the 4 corners x, a, b, c. then i tried each of the answer choices. i didn't see any major insights. A and D traveled the longest distance and had the same distance. (i don't understand how they can predict which would be longest without a specific map of the 4 towns.) then i had the feeling i didn't understand the question and/or missed something.

I get the sense i missed something but i don't know what.

View attachment 28294View attachment 28295
For each of the 5 answers you need to try to come up with a map where the answer is the shortest route. You will _not_ find such a map for one of the answers. Then you'll need to formally prove that such map does not exist.

#### eric beans

##### Junior Member
Consider choice A. Can you create a map where that route is shortest?

AXBC going left to right all on a straight line works

You do not have the map as you said. But you do not need the correct map. You need to find a situation where no map is possible.
how did you know to try AXBC?

like this? i get distance of 6. but how does that get me to figuring out the answer?

#### JeffM

##### Elite Member
I did not know in advance. But the shortest distance between two points is a straight line. Therefore, if the shortest distance between A and B runs through X, X must lie on the line AC and lie between A and B. And if the shortest line between X and C runs through B, then B must lie on the line XC and B must lie between B and C.

#### Dr.Peterson

##### Elite Member
how did you know to try AXBC?

View attachment 28296like this? i get distance of 6. but how does that get me to figuring out the answer?
there could be any infinite number of possible layout of the 4 towns. how was i to try them all? dejected, and clueless, what I did was i just drew a box with each of the 4 corners x, a, b, c. then i tried each of the answer choices. i didn't see any major insights. A and D traveled the longest distance and had the same distance. (i don't understand how they can predict which would be longest without a specific map of the 4 towns.) then i had the feeling i didn't understand the question and/or missed something.

I get the sense i missed something but i don't know what.

View attachment 28294View attachment 28295

I think you may be missing what "layout" means. The points could be anywhere on your paper, not necessarily in your little square; the lengths of the edges don't have to be the same. I hope you understand that "AXBC" didn't mean going around in a square as you have drawn it, but literally laid out along a line, A----X----B----C. As you drew it, XAXBCBX is far from the shortest route; the shortest there would be (for example) XACBX. In the straight-line version, XAXBCBX is as short as you can get.

As we have said, try different layouts, hoping to make a given path the shortest. So far, we know that the first choice is not the answer, because there is a layout in which that route is as short as you can get. Yes, it takes some thought to see that. (I will tell you that my first attempts were all along a line, because that seemed like the easiest thing to try. But not all will be.)

#### eric beans

##### Junior Member
I think you may be missing what "layout" means. The points could be anywhere on your paper, not necessarily in your little square; the lengths of the edges don't have to be the same. I hope you understand that "AXBC" didn't mean going around in a square as you have drawn it, but literally laid out along a line, A----X----B----C. As you drew it, XAXBCBX is far from the shortest route; the shortest there would be (for example) XACBX. In the straight-line version, XAXBCBX is as short as you can get.

As we have said, try different layouts, hoping to make a given path the shortest. So far, we know that the first choice is not the answer, because there is a layout in which that route is as short as you can get. Yes, it takes some thought to see that. (I will tell you that my first attempts were all along a line, because that seemed like the easiest thing to try. But not all will be.)
Oh, I didn't understand. I thought the map of the 4 cities could be literally any infinite number of 2D arrangements. But what you're saying is it's confined to 1D arrangement along a linear line? That seems to be poor wording on the problem. I didn't get that "layout" meant automatically linear arrangement of towns.

If that's the case, I would need to go through XABC, AXBC, ABXC, ABCX as the four possible choices of "layout"? But still that requires me to try 20 times running through all the answer choices, no?

#### eric beans

##### Junior Member
I did not know in advance. But the shortest distance between two points is a straight line. Therefore, if the shortest distance between A and B runs through X, X must lie on the line AC and lie between A and B. And if the shortest line between X and C runs through B, then B must lie on the line XC and B must lie between B and C.
How did you know it was linear arrangement of towns and not 2D arrangements of town? Layout as I understood seemed ambiguous.

#### eric beans

##### Junior Member
I think you may be missing what "layout" means. The points could be anywhere on your paper, not necessarily in your little square; the lengths of the edges don't have to be the same. I hope you understand that "AXBC" didn't mean going around in a square as you have drawn it, but literally laid out along a line, A----X----B----C. As you drew it, XAXBCBX is far from the shortest route; the shortest there would be (for example) XACBX. In the straight-line version, XAXBCBX is as short as you can get.

As we have said, try different layouts, hoping to make a given path the shortest. So far, we know that the first choice is not the answer, because there is a layout in which that route is as short as you can get. Yes, it takes some thought to see that. (I will tell you that my first attempts were all along a line, because that seemed like the easiest thing to try. But not all will be.)
i guess given 4 elements (XABC) there could be 24 different permutations/layouts. Then I'd have to try all 5 answer choices against the 24 layouts, that's like 124 possibilities. What's the trick to this? I don't think they expected to try all of them because that would take too long. I don't see what the trick is.

#### eric beans

##### Junior Member
what i tried was to look for a layout within the choices and then try to look for the lengths. they all came out the same. so this is not the way to find the answer.

#### lev888

##### Senior Member
Oh, I didn't understand. I thought the map of the 4 cities could be literally any infinite number of 2D arrangements. But what you're saying is it's confined to 1D arrangement along a linear line? That seems to be poor wording on the problem. I didn't get that "layout" meant automatically linear arrangement of towns.

If that's the case, I would need to go through XABC, AXBC, ABXC, ABCX as the four possible choices of "layout"? But still that requires me to try 20 times running through all the answer choices, no?
No, it's not correct to assume that the arrangement is automatically linear for all answers. For some answers it may have to be linear.

#### Dr.Peterson

##### Elite Member
Oh, I didn't understand. I thought the map of the 4 cities could be literally any infinite number of 2D arrangements. But what you're saying is it's confined to 1D arrangement along a linear line? That seems to be poor wording on the problem. I didn't get that "layout" meant automatically linear arrangement of towns.
No, I did not say that. I said the opposite: It is not confined to any one shape, so you can't just draw one kind of picture. What I said about a linear arrangement is that that is a possibility, and often the easiest to imagine, so I started with that option.

If that's the case, I would need to go through XABC, AXBC, ABXC, ABCX as the four possible choices of "layout"? But still that requires me to try 20 times running through all the answer choices, no?
No, you can reason about it as JeffM said in #10.

what i tried was to look for a layout within the choices and then try to look for the lengths. they all came out the same. so this is not the way to find the answer.

View attachment 28311
Don't assume any particular distances (1 unit each, for example). And you aren't comparing between different arrangements, but within one arrangement. You want to find an arrangement for which the given route is the shortest of all routes for that arrangement that start and end at X and go through A, B, and C at least once.

I did not know in advance. But the shortest distance between two points is a straight line. Therefore, if the shortest distance between A and B runs through X, X must lie on the line AC and lie between A and B. And if the shortest line between X and C runs through B, then B must lie on the line XC and B must lie between B and C.
This explains why he chose A----X----B----C. Given that arrangement, for example, the route XAXBXBCBX would be longer than necessary (obviously), and there is no shorter route than XAXBCBX. On the other hand, if C were not quite on the line, then CX would be shorter than CB+BX, so that XAXBCX would be a shorter route.

May I ask what is the context of this problem? Are you studying some topic that is expected to help you solve it, or is this just a puzzle or an intelligence test, or what?

#### eric beans

##### Junior Member
i give up. can someone just plainly show how they got the answer? i don't even know what branch of math this would be under other than just word problems.

#### eric beans

##### Junior Member
Consider choice A. Can you create a map where that route is shortest?

AXBC going left to right all on a straight line works. So choice A is wrong; you can draw a map where total distance is shortest with XAXBCBX

You do not have the map as you said. But you do not need the correct map. You need to find a situation where no map is possible.
"shortest" relative to the other 5 choices? or shortest within the choice A? If it's shortest distance within choice A, there could be infinite number of layouts in 2D space with 4 loci's to consider?

how can i do this without a map? i'm more confused than ever before.

#### JeffM

##### Elite Member
Like Dr. Peterson, I find this problem odd.

You first must try to imagine some arrangement (layout) of points in the plane out of the infinite number of possible arrangements such that a specified route is not longer than any other route. There is no way that you can test every one of an infinite number so you need to find some criteria that restrict what you need to consider.

And you must do this for each route.

I have never seen a problem quite like it. Moreover, there probably is a branch of mathematics that makes it straightforward (topology?), but it is not one commonly studied.

Dr. Peterson said he initially tried a straight line out of simplicity, which just shows that he cleverer than I am.

Here is what I initially thought. Any three points in a plane form a triangle. Then I realized that thought was incorrect. Any three points not on a straight line form a triangle. Thus a special case is three co-linear points. And that divides into two cases, four co-linear points and three co-linear points with a point off that line.

I’ll admit I have not completed the problem. But I suspect that ONE way to solve it is to reduce it to four cases, the co-linear ones already mentioned, and two cases involving triangles, one with the fourth point inside the triangle, the other with the fourth point outside the triangle.