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Distance between two stations
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A railway line is divided into $10$ sections by the stations $A, B, C, D, E, F, G, H, I, J$ and $K$. The distance between $A$ and $K$ is $56$ km. A trip along two successive sections never exceeds $12$ km. A trip along three successive sections is at least $17$ km. What is the distance between $B$ and $G$?
I have no idea how to solve this question. I thought about taking the distance between each set of successive stations as a variable, but this gets too messy. And taking $56over10=5.6$ doesn't work as well. The inequalities look like they're important, but I can not make use of them anywhere.
Please help.
algebra-precalculus contest-math
edited Aug 17 at 16:03
Stefan4024
29.5k53377
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up vote
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A railway line is divided into $10$ sections by the stations $A, B, C, D, E, F, G, H, I, J$ and $K$. The distance between $A$ and $K$ is $56$ km. A trip along two successive sections never exceeds $12$ km. A trip along three successive sections is at least $17$ km. What is the distance between $B$ and $G$?
I have no idea how to solve this question. I thought about taking the distance between each set of successive stations as a variable, but this gets too messy. And taking $56over10=5.6$ doesn't work as well. The inequalities look like they're important, but I can not make use of them anywhere.
Please help.
algebra-precalculus contest-math
edited Aug 17 at 16:03
Stefan4024
29.5k53377
4
This is quite a nice problem. Where did you get it from?
– TheSimpliFire
Aug 17 at 16:34
1
@TheSimpliFire A mock test for the PRMO.
– user585025
Aug 17 at 16:58
add a comment |Â
up vote
11
down vote
favorite
up vote
11
down vote
favorite
A railway line is divided into $10$ sections by the stations $A, B, C, D, E, F, G, H, I, J$ and $K$. The distance between $A$ and $K$ is $56$ km. A trip along two successive sections never exceeds $12$ km. A trip along three successive sections is at least $17$ km. What is the distance between $B$ and $G$?
I have no idea how to solve this question. I thought about taking the distance between each set of successive stations as a variable, but this gets too messy. And taking $56over10=5.6$ doesn't work as well. The inequalities look like they're important, but I can not make use of them anywhere.
Please help.
algebra-precalculus contest-math
edited Aug 17 at 16:03
Stefan4024
29.5k53377
A railway line is divided into $10$ sections by the stations $A, B, C, D, E, F, G, H, I, J$ and $K$. The distance between $A$ and $K$ is $56$ km. A trip along two successive sections never exceeds $12$ km. A trip along three successive sections is at least $17$ km. What is the distance between $B$ and $G$?
I have no idea how to solve this question. I thought about taking the distance between each set of successive stations as a variable, but this gets too messy. And taking $56over10=5.6$ doesn't work as well. The inequalities look like they're important, but I can not make use of them anywhere.
Please help.
algebra-precalculus contest-math
edited Aug 17 at 16:03
Stefan4024
29.5k53377
edited Aug 17 at 16:03
Stefan4024
29.5k53377
edited Aug 17 at 16:03
Stefan4024
29.5k53377
edited Aug 17 at 16:03
Stefan4024
29.5k53377
29.5k53377
asked Aug 17 at 15:31
user585025
4
This is quite a nice problem. Where did you get it from?
– TheSimpliFire
Aug 17 at 16:34
1
@TheSimpliFire A mock test for the PRMO.
– user585025
Aug 17 at 16:58
add a comment |Â
4
This is quite a nice problem. Where did you get it from?
– TheSimpliFire
Aug 17 at 16:34
1
@TheSimpliFire A mock test for the PRMO.
– user585025
Aug 17 at 16:58
4
4
This is quite a nice problem. Where did you get it from?
– TheSimpliFire
Aug 17 at 16:34
This is quite a nice problem. Where did you get it from?
– TheSimpliFire
Aug 17 at 16:34
1
1
@TheSimpliFire A mock test for the PRMO.
– user585025
Aug 17 at 16:58
@TheSimpliFire A mock test for the PRMO.
– user585025
Aug 17 at 16:58
add a comment |Â
7 Answers
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Note that the the distance between $A$ and $D$ is at least $17$. On the other side the distance between $B$ and $D$ is at most $12$. This the distance between $A$ and $B$ is at least $5$. With similar reasoning we get that the length of each section is at least $5$.
So we get that the distance between $G$ and $K$ is at least $22$ (at least $17$ for three sections and at least $5$ for the final one). Thus the distance from $A$ to $G$ is at most $56-22=34$. On the other side there are $6$ sections between $A$ and $G$, so the distance is at least $34$ (we have two times three sections). Thus we conclude that the distance between $A$ and $G$ is $34$.
On the other side the distance between $B$ and $K$ is at least $51$, as it consists of nine sections. Thus the distance $AB$ is at most $56-51=5$. From this and the first paragraph we conclude that $AB=5$ and finally
$$BG = AG - AB = 34-5=29$$
edited Aug 17 at 16:32
TheSimpliFire
10.5k62053
answered Aug 17 at 16:01
Stefan4024
29.5k53377
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Let $x_1,...,x_10$ represent the distances between each section, i.e., $x_1$ is the distance between $A$ and $B$.
Note that $x_1$ and $x_10$ are likely to be quite small, since they are on the edges, and only have to deal with the trip along three successive sections at least 17 km. As such, $x_3$ and $x_8$ are likely to be large in order to ensure the 17 property of $x_1,x_2,x_3$, and $x_8,x_9,x_10$. We also know by symmetry that $x_n=x_11-n$
Hence, after some guess and check, I came up with
$$x_1,...x_10 = 5, 5, 7, 5, 6, 6, 5, 7, 5, 5$$
So, $overlineBG=5+7+5+6+6=textbf29$
answered Aug 17 at 15:45
Rushabh Mehta
1,548218
2
How do you know that they have to be integers?
– TheSimpliFire
Aug 17 at 15:51
I didn't. I just started there cuz why not
– Rushabh Mehta
Aug 17 at 15:51
3
This suffices to show that if there is a unique solution that it would have to be $29$, however it has yet to be shown that there must be a unique solution. It is possible that there is a whole range of values that $overlineBG$ could take, leaving this answer as being incomplete.
– JMoravitz
Aug 17 at 15:52
Now you have to show that this is the only solution :)
– TheSimpliFire
Aug 17 at 15:52
1
@TheSimpliFire He didn't say prove it, so given that I presume this is competition math, this is sufficient
– Rushabh Mehta
Aug 17 at 15:53
 |Â
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|____|____|____|____|____|____|____|____|____|____|
A B C D E F G H I J K
From the given information we can say that any single section can be taken as the difference of some $3$ successive sections and subset of $2$ successive sections. So, a single section should be atleast $5$ km long.
Also the section JK is the total line minus $3$ sets of three successive sections AD, DG, and GJ. These three successive sections should be at least length of $51$ km. The section JK can be at most $5$ km. By symmetry AB should also be exactly $5$ km. The lay out of the $3$ sets of successive sections so as to isolate the sections DE or GH, then the same argument as above can be used to conclude that each of them is exactly $5$ km. Since the $3$ sets of $2$ successive sections remaining, namely, BD, EG, and HJ they can sum up to at most $3cdot12=36$ km and at the same time they must cover the remaining distance. So, $56-(4cdot5)=36$. So, these three sets of two successive sections must be exactly $12$ km. So, the total length from B to G is exactly $12+5+12=29$ km
answered Aug 17 at 15:53
Key Flex
1
From the given information we can say that any single section can be taken as the difference of some 3 successive sections and subset of 2 successive sections. What does this mean?
– user585025
Aug 17 at 16:00
@Buddha I concluded the above statement from the given question "A trip along two successive sections never exceeds 12 km. A trip along three successive sections is at least 17 km"
– Key Flex
Aug 17 at 16:06
1
And why is it that $AB$ is exactly $5$, instead of at most $5$ like $JK$?
– user585025
Aug 17 at 16:10
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Here's a solution that requires the least English :)
Let the distance between $A$ and $B$ be $d_1$, that between $B$ and $C$ be $d_2$, and so on, so that the distance between $J$ and $K$ is $d_10$. By symmetry we need only consider $d_1$ to $d_5$. Note also that each section must be at least $5$ kilometres. We seek $$d_2+d_3+d_4+d_5+d_6=d_2+d_3+d_4+2d_5$$ due to symmetry.
Let's now create a list of inequalities:
$$d_1+d_2le12tag1$$$$d_1+d_2+d_3ge17tag2$$gives$$d_3ge5tag3$$ and $$d_2+d_3le12tag4$$$$d_2+d_3+d_4ge17tag5$$ gives$$d_4ge5tag6$$and put $(3)$ into $(5)$ to get $$d_2le7tag7$$ Now put $(6)$ into $(1)$ to get $$d_1le5implies d_1=5tag8$$ Put $(7)$ into $(5)$ to give $$d_3+d_4ge10tag9$$ and $(3)$ gives$$d_4le5tag10$$ Equating $(6)$ and $(10)$ together gives $$d_4=5tag11$$ Put $(11)$ into $(5)$ to get $$d_2+d_3ge12tag12$$ and equate with $(4)$ to get $$d_2+d_3=12tag13$$ Finally, $$d_1+d_2+d_3+d_4+d_5=28$$ due to symmetry so $$d_5=28-d_1-d_4-(d_2+d_3)=6tag14$$ Hence the required distance is $$(d_2+d_3)+d_4+2d_5=12+5+2cdot6=29$$
answered Aug 17 at 16:28
TheSimpliFire
10.5k62053
There are possibilities where $d_5 neq d_6$. Example: $5, 5, 7, 5, 5, 7, 5, 5, 7, 5$
– tehtmi
Aug 18 at 2:58
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This is a bruteforce solution, but one way to do this is to simply formulate it as a linear programming feasability problem. Let $x=(x_1,ldots,x_10)$ be the length of the sections. Then, $x$ is a point in the polytope $ y mid A y leq b $ where
$$
A=beginpmatrix1 & 1\
& phantom-1 & phantom-1\
& & phantom-1 & phantom-1\
& & & ddots & ddots\
& & & & phantom-1 & phantom-1\
-1 & -1 & -1\
& -1 & -1 & -1\
& & ddots & ddots & ddots\
& & & -1 & -1 & -1\
phantom-1 & phantom-1 & phantom-1 & phantom-1 & cdots & phantom-1\
-1 & -1 & -1 & -1 & cdots & -1
endpmatrix
qquadtextandqquad
b=beginpmatrixphantom-12\
phantom-12\
phantom-12\
phantom-vdots\
phantom-12\
-17\
-17\
phantom-vdots\
-17\
phantom-56\
-56
endpmatrix.
$$
answered Aug 17 at 15:46
parsiad
16.1k32253
1
Umm, this is a contest problem so I don't think this would be very useful?
– TheSimpliFire
Aug 17 at 15:53
@TheSimpliFire Useful to learn something new outside of the contest, which can probably be applied later...
– user202729
Aug 18 at 16:16
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Let the ten distances between adjacent stations be $x_1$ to $x_10$.
$$17 leq x_1 + (x_2 + x_3) leq x_1 + 12$$
So
$$x_1 geq 5$$
And then
$$56 = x_1 + (x_2 + x_3 + x_4) + (x_5 + x_6 + x_7) + (x_8 + x_9 + x_10) geq 5 + 17 + 17 + 17 = 56$$
There is no slack in this inequality, so in fact $x_1 = 5$ and $x_2 + x_3 + x_4 = x_5 + x_6 + x_7 = x_8 + x_9 + x_10 = 17$.
We can similarly get the symmetric equations $x_10 = 5$ and $x_1 + x_2 + x_3 = x_4 + x_5 + x_6 = x_7 + x_8 + x_9 = 17$.
Together, these give $x_1 = x_4 = x_7 = x_10 = 5$ and $x_2 + x_3 = x_5 + x_6 = x_8 + x_9 = 12$.
The solution is then $(x_2 + x_3) + x_4 + (x_5 + x_6) = 12 + 5 + 12 = 29$.
Addendum 1: Multiple solutions to the full system are possible. E.g. $5, 6, 6, 5, 6, 6, 5, 6, 6, 5$ versus $5, 5, 7, 5, 5, 7, 5, 5, 7, 5$.
Addendum 2: As in Key Flex's answer, it is also possible to use splits like $(x_1 + x_2 + x_3) + x_4 + (x_5 + x_6 + x_7) + (x_8 + x_9 + x_10)$.
answered Aug 18 at 2:56
tehtmi
53429
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1
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Using algebraic method.
Label the stations with $A$ through $K$:
$hspace4cm$
The given conditions are:
$$K-A=56 textand begincases
A-Cge -12\
B-Dge -12\
C-Ege -12\
D-Fge -12\
E-Gge -12\
F-Hge -12\
G-Ige -12\
H-Jge -12\
I-Kge -12\
endcases textand
begincases
D-Age 17\
E-Bge 17\
F-Cge 17\
G-Dge 17\
H-Ege 17\
I-Fge 17\
J-Gge 17\
K-Hge 17\
endcases$$
One the one hand:
$$(D-A)+(G-D)=G-Age 34;$$
on the other hand:
$$(K-H)+(H-J)=K-Jge 5;\
(K-J)+(J-G)=K-Gge 22;\
K-G=(56+A)-Gge 22 Rightarrow \
G-Ale 34.$$
Hence: $colorredG-A=34$.
One the one hand:
$$(B-D)+(D-A)=B-Age 5 Rightarrow A-Ble -5;\
$$
on the other hand:
$$(E-B)+(H-E)=H-B=ge 34;\
(H-B)+(K-H)=K-Bge 51;\
K-B=(56+A)-Bge 51 Rightarrow \
A-Bge -5.$$
Hence: $colorblueA-B=-5$.
In conclusion: $(G-A)+(A-B)=G-B=34+(-5)=29$.
answered Aug 18 at 7:41
farruhota
14.7k2633
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7 Answers
7
active
oldest
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7 Answers
7
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
8
down vote
accepted
Note that the the distance between $A$ and $D$ is at least $17$. On the other side the distance between $B$ and $D$ is at most $12$. This the distance between $A$ and $B$ is at least $5$. With similar reasoning we get that the length of each section is at least $5$.
So we get that the distance between $G$ and $K$ is at least $22$ (at least $17$ for three sections and at least $5$ for the final one). Thus the distance from $A$ to $G$ is at most $56-22=34$. On the other side there are $6$ sections between $A$ and $G$, so the distance is at least $34$ (we have two times three sections). Thus we conclude that the distance between $A$ and $G$ is $34$.
On the other side the distance between $B$ and $K$ is at least $51$, as it consists of nine sections. Thus the distance $AB$ is at most $56-51=5$. From this and the first paragraph we conclude that $AB=5$ and finally
$$BG = AG - AB = 34-5=29$$
edited Aug 17 at 16:32
TheSimpliFire
10.5k62053
answered Aug 17 at 16:01
Stefan4024
29.5k53377
add a comment |Â
up vote
8
down vote
accepted
Note that the the distance between $A$ and $D$ is at least $17$. On the other side the distance between $B$ and $D$ is at most $12$. This the distance between $A$ and $B$ is at least $5$. With similar reasoning we get that the length of each section is at least $5$.
So we get that the distance between $G$ and $K$ is at least $22$ (at least $17$ for three sections and at least $5$ for the final one). Thus the distance from $A$ to $G$ is at most $56-22=34$. On the other side there are $6$ sections between $A$ and $G$, so the distance is at least $34$ (we have two times three sections). Thus we conclude that the distance between $A$ and $G$ is $34$.
On the other side the distance between $B$ and $K$ is at least $51$, as it consists of nine sections. Thus the distance $AB$ is at most $56-51=5$. From this and the first paragraph we conclude that $AB=5$ and finally
$$BG = AG - AB = 34-5=29$$
edited Aug 17 at 16:32
TheSimpliFire
10.5k62053
answered Aug 17 at 16:01
Stefan4024
29.5k53377
add a comment |Â
up vote
8
down vote
accepted
up vote
8
down vote
accepted
Note that the the distance between $A$ and $D$ is at least $17$. On the other side the distance between $B$ and $D$ is at most $12$. This the distance between $A$ and $B$ is at least $5$. With similar reasoning we get that the length of each section is at least $5$.
So we get that the distance between $G$ and $K$ is at least $22$ (at least $17$ for three sections and at least $5$ for the final one). Thus the distance from $A$ to $G$ is at most $56-22=34$. On the other side there are $6$ sections between $A$ and $G$, so the distance is at least $34$ (we have two times three sections). Thus we conclude that the distance between $A$ and $G$ is $34$.
On the other side the distance between $B$ and $K$ is at least $51$, as it consists of nine sections. Thus the distance $AB$ is at most $56-51=5$. From this and the first paragraph we conclude that $AB=5$ and finally
$$BG = AG - AB = 34-5=29$$
edited Aug 17 at 16:32
TheSimpliFire
10.5k62053
answered Aug 17 at 16:01
Stefan4024
29.5k53377
Note that the the distance between $A$ and $D$ is at least $17$. On the other side the distance between $B$ and $D$ is at most $12$. This the distance between $A$ and $B$ is at least $5$. With similar reasoning we get that the length of each section is at least $5$.
So we get that the distance between $G$ and $K$ is at least $22$ (at least $17$ for three sections and at least $5$ for the final one). Thus the distance from $A$ to $G$ is at most $56-22=34$. On the other side there are $6$ sections between $A$ and $G$, so the distance is at least $34$ (we have two times three sections). Thus we conclude that the distance between $A$ and $G$ is $34$.
On the other side the distance between $B$ and $K$ is at least $51$, as it consists of nine sections. Thus the distance $AB$ is at most $56-51=5$. From this and the first paragraph we conclude that $AB=5$ and finally
$$BG = AG - AB = 34-5=29$$
edited Aug 17 at 16:32
TheSimpliFire
10.5k62053
answered Aug 17 at 16:01
Stefan4024
29.5k53377
edited Aug 17 at 16:32
TheSimpliFire
10.5k62053
edited Aug 17 at 16:32
TheSimpliFire
10.5k62053
edited Aug 17 at 16:32
TheSimpliFire
10.5k62053
10.5k62053
answered Aug 17 at 16:01
Stefan4024
29.5k53377
answered Aug 17 at 16:01
Stefan4024
29.5k53377
answered Aug 17 at 16:01
Stefan4024
29.5k53377
29.5k53377
add a comment |Â
add a comment |Â
up vote
4
down vote
Let $x_1,...,x_10$ represent the distances between each section, i.e., $x_1$ is the distance between $A$ and $B$.
Note that $x_1$ and $x_10$ are likely to be quite small, since they are on the edges, and only have to deal with the trip along three successive sections at least 17 km. As such, $x_3$ and $x_8$ are likely to be large in order to ensure the 17 property of $x_1,x_2,x_3$, and $x_8,x_9,x_10$. We also know by symmetry that $x_n=x_11-n$
Hence, after some guess and check, I came up with
$$x_1,...x_10 = 5, 5, 7, 5, 6, 6, 5, 7, 5, 5$$
So, $overlineBG=5+7+5+6+6=textbf29$
answered Aug 17 at 15:45
Rushabh Mehta
1,548218
2
How do you know that they have to be integers?
– TheSimpliFire
Aug 17 at 15:51
I didn't. I just started there cuz why not
– Rushabh Mehta
Aug 17 at 15:51
3
This suffices to show that if there is a unique solution that it would have to be $29$, however it has yet to be shown that there must be a unique solution. It is possible that there is a whole range of values that $overlineBG$ could take, leaving this answer as being incomplete.
– JMoravitz
Aug 17 at 15:52
Now you have to show that this is the only solution :)
– TheSimpliFire
Aug 17 at 15:52
1
@TheSimpliFire He didn't say prove it, so given that I presume this is competition math, this is sufficient
– Rushabh Mehta
Aug 17 at 15:53
 |Â
show 2 more comments
up vote
4
down vote
Let $x_1,...,x_10$ represent the distances between each section, i.e., $x_1$ is the distance between $A$ and $B$.
Note that $x_1$ and $x_10$ are likely to be quite small, since they are on the edges, and only have to deal with the trip along three successive sections at least 17 km. As such, $x_3$ and $x_8$ are likely to be large in order to ensure the 17 property of $x_1,x_2,x_3$, and $x_8,x_9,x_10$. We also know by symmetry that $x_n=x_11-n$
Hence, after some guess and check, I came up with
$$x_1,...x_10 = 5, 5, 7, 5, 6, 6, 5, 7, 5, 5$$
So, $overlineBG=5+7+5+6+6=textbf29$
answered Aug 17 at 15:45
Rushabh Mehta
1,548218
2
How do you know that they have to be integers?
– TheSimpliFire
Aug 17 at 15:51
I didn't. I just started there cuz why not
– Rushabh Mehta
Aug 17 at 15:51
3
This suffices to show that if there is a unique solution that it would have to be $29$, however it has yet to be shown that there must be a unique solution. It is possible that there is a whole range of values that $overlineBG$ could take, leaving this answer as being incomplete.
– JMoravitz
Aug 17 at 15:52
Now you have to show that this is the only solution :)
– TheSimpliFire
Aug 17 at 15:52
1
@TheSimpliFire He didn't say prove it, so given that I presume this is competition math, this is sufficient
– Rushabh Mehta
Aug 17 at 15:53
 |Â
show 2 more comments
up vote
4
down vote
up vote
4
down vote
Let $x_1,...,x_10$ represent the distances between each section, i.e., $x_1$ is the distance between $A$ and $B$.
Note that $x_1$ and $x_10$ are likely to be quite small, since they are on the edges, and only have to deal with the trip along three successive sections at least 17 km. As such, $x_3$ and $x_8$ are likely to be large in order to ensure the 17 property of $x_1,x_2,x_3$, and $x_8,x_9,x_10$. We also know by symmetry that $x_n=x_11-n$
Hence, after some guess and check, I came up with
$$x_1,...x_10 = 5, 5, 7, 5, 6, 6, 5, 7, 5, 5$$
So, $overlineBG=5+7+5+6+6=textbf29$
answered Aug 17 at 15:45
Rushabh Mehta
1,548218
Let $x_1,...,x_10$ represent the distances between each section, i.e., $x_1$ is the distance between $A$ and $B$.
Note that $x_1$ and $x_10$ are likely to be quite small, since they are on the edges, and only have to deal with the trip along three successive sections at least 17 km. As such, $x_3$ and $x_8$ are likely to be large in order to ensure the 17 property of $x_1,x_2,x_3$, and $x_8,x_9,x_10$. We also know by symmetry that $x_n=x_11-n$
Hence, after some guess and check, I came up with
$$x_1,...x_10 = 5, 5, 7, 5, 6, 6, 5, 7, 5, 5$$
So, $overlineBG=5+7+5+6+6=textbf29$
answered Aug 17 at 15:45
Rushabh Mehta
1,548218
answered Aug 17 at 15:45
Rushabh Mehta
1,548218
answered Aug 17 at 15:45
Rushabh Mehta
1,548218
answered Aug 17 at 15:45
Rushabh Mehta
1,548218
1,548218
2
How do you know that they have to be integers?
– TheSimpliFire
Aug 17 at 15:51
I didn't. I just started there cuz why not
– Rushabh Mehta
Aug 17 at 15:51
3
This suffices to show that if there is a unique solution that it would have to be $29$, however it has yet to be shown that there must be a unique solution. It is possible that there is a whole range of values that $overlineBG$ could take, leaving this answer as being incomplete.
– JMoravitz
Aug 17 at 15:52
Now you have to show that this is the only solution :)
– TheSimpliFire
Aug 17 at 15:52
1
@TheSimpliFire He didn't say prove it, so given that I presume this is competition math, this is sufficient
– Rushabh Mehta
Aug 17 at 15:53
 |Â
show 2 more comments
2
How do you know that they have to be integers?
– TheSimpliFire
Aug 17 at 15:51
I didn't. I just started there cuz why not
– Rushabh Mehta
Aug 17 at 15:51
3
This suffices to show that if there is a unique solution that it would have to be $29$, however it has yet to be shown that there must be a unique solution. It is possible that there is a whole range of values that $overlineBG$ could take, leaving this answer as being incomplete.
– JMoravitz
Aug 17 at 15:52
Now you have to show that this is the only solution :)
– TheSimpliFire
Aug 17 at 15:52
1
@TheSimpliFire He didn't say prove it, so given that I presume this is competition math, this is sufficient
– Rushabh Mehta
Aug 17 at 15:53
2
2
How do you know that they have to be integers?
– TheSimpliFire
Aug 17 at 15:51
How do you know that they have to be integers?
– TheSimpliFire
Aug 17 at 15:51
I didn't. I just started there cuz why not
– Rushabh Mehta
Aug 17 at 15:51
I didn't. I just started there cuz why not
– Rushabh Mehta
Aug 17 at 15:51
3
3
This suffices to show that if there is a unique solution that it would have to be $29$, however it has yet to be shown that there must be a unique solution. It is possible that there is a whole range of values that $overlineBG$ could take, leaving this answer as being incomplete.
– JMoravitz
Aug 17 at 15:52
This suffices to show that if there is a unique solution that it would have to be $29$, however it has yet to be shown that there must be a unique solution. It is possible that there is a whole range of values that $overlineBG$ could take, leaving this answer as being incomplete.
– JMoravitz
Aug 17 at 15:52
Now you have to show that this is the only solution :)
– TheSimpliFire
Aug 17 at 15:52
Now you have to show that this is the only solution :)
– TheSimpliFire
Aug 17 at 15:52
1
1
@TheSimpliFire He didn't say prove it, so given that I presume this is competition math, this is sufficient
– Rushabh Mehta
Aug 17 at 15:53
@TheSimpliFire He didn't say prove it, so given that I presume this is competition math, this is sufficient
– Rushabh Mehta
Aug 17 at 15:53
 |Â
show 2 more comments
up vote
4
down vote
|____|____|____|____|____|____|____|____|____|____|
A B C D E F G H I J K
From the given information we can say that any single section can be taken as the difference of some $3$ successive sections and subset of $2$ successive sections. So, a single section should be atleast $5$ km long.
Also the section JK is the total line minus $3$ sets of three successive sections AD, DG, and GJ. These three successive sections should be at least length of $51$ km. The section JK can be at most $5$ km. By symmetry AB should also be exactly $5$ km. The lay out of the $3$ sets of successive sections so as to isolate the sections DE or GH, then the same argument as above can be used to conclude that each of them is exactly $5$ km. Since the $3$ sets of $2$ successive sections remaining, namely, BD, EG, and HJ they can sum up to at most $3cdot12=36$ km and at the same time they must cover the remaining distance. So, $56-(4cdot5)=36$. So, these three sets of two successive sections must be exactly $12$ km. So, the total length from B to G is exactly $12+5+12=29$ km
answered Aug 17 at 15:53
Key Flex
1
From the given information we can say that any single section can be taken as the difference of some 3 successive sections and subset of 2 successive sections. What does this mean?
– user585025
Aug 17 at 16:00
@Buddha I concluded the above statement from the given question "A trip along two successive sections never exceeds 12 km. A trip along three successive sections is at least 17 km"
– Key Flex
Aug 17 at 16:06
1
And why is it that $AB$ is exactly $5$, instead of at most $5$ like $JK$?
– user585025
Aug 17 at 16:10
add a comment |Â
up vote
4
down vote
|____|____|____|____|____|____|____|____|____|____|
A B C D E F G H I J K
From the given information we can say that any single section can be taken as the difference of some $3$ successive sections and subset of $2$ successive sections. So, a single section should be atleast $5$ km long.
Also the section JK is the total line minus $3$ sets of three successive sections AD, DG, and GJ. These three successive sections should be at least length of $51$ km. The section JK can be at most $5$ km. By symmetry AB should also be exactly $5$ km. The lay out of the $3$ sets of successive sections so as to isolate the sections DE or GH, then the same argument as above can be used to conclude that each of them is exactly $5$ km. Since the $3$ sets of $2$ successive sections remaining, namely, BD, EG, and HJ they can sum up to at most $3cdot12=36$ km and at the same time they must cover the remaining distance. So, $56-(4cdot5)=36$. So, these three sets of two successive sections must be exactly $12$ km. So, the total length from B to G is exactly $12+5+12=29$ km
answered Aug 17 at 15:53
Key Flex
1
From the given information we can say that any single section can be taken as the difference of some 3 successive sections and subset of 2 successive sections. What does this mean?
– user585025
Aug 17 at 16:00
@Buddha I concluded the above statement from the given question "A trip along two successive sections never exceeds 12 km. A trip along three successive sections is at least 17 km"
– Key Flex
Aug 17 at 16:06
1
And why is it that $AB$ is exactly $5$, instead of at most $5$ like $JK$?
– user585025
Aug 17 at 16:10
add a comment |Â
up vote
4
down vote
up vote
4
down vote
|____|____|____|____|____|____|____|____|____|____|
A B C D E F G H I J K
From the given information we can say that any single section can be taken as the difference of some $3$ successive sections and subset of $2$ successive sections. So, a single section should be atleast $5$ km long.
Also the section JK is the total line minus $3$ sets of three successive sections AD, DG, and GJ. These three successive sections should be at least length of $51$ km. The section JK can be at most $5$ km. By symmetry AB should also be exactly $5$ km. The lay out of the $3$ sets of successive sections so as to isolate the sections DE or GH, then the same argument as above can be used to conclude that each of them is exactly $5$ km. Since the $3$ sets of $2$ successive sections remaining, namely, BD, EG, and HJ they can sum up to at most $3cdot12=36$ km and at the same time they must cover the remaining distance. So, $56-(4cdot5)=36$. So, these three sets of two successive sections must be exactly $12$ km. So, the total length from B to G is exactly $12+5+12=29$ km
answered Aug 17 at 15:53
Key Flex
1
|____|____|____|____|____|____|____|____|____|____|
A B C D E F G H I J K
From the given information we can say that any single section can be taken as the difference of some $3$ successive sections and subset of $2$ successive sections. So, a single section should be atleast $5$ km long.
Also the section JK is the total line minus $3$ sets of three successive sections AD, DG, and GJ. These three successive sections should be at least length of $51$ km. The section JK can be at most $5$ km. By symmetry AB should also be exactly $5$ km. The lay out of the $3$ sets of successive sections so as to isolate the sections DE or GH, then the same argument as above can be used to conclude that each of them is exactly $5$ km. Since the $3$ sets of $2$ successive sections remaining, namely, BD, EG, and HJ they can sum up to at most $3cdot12=36$ km and at the same time they must cover the remaining distance. So, $56-(4cdot5)=36$. So, these three sets of two successive sections must be exactly $12$ km. So, the total length from B to G is exactly $12+5+12=29$ km
answered Aug 17 at 15:53
Key Flex
1
edited Aug 17 at 16:23
user585025
answered Aug 17 at 15:53
Key Flex
1
answered Aug 17 at 15:53
Key Flex
1
answered Aug 17 at 15:53
Key Flex
1
1
From the given information we can say that any single section can be taken as the difference of some 3 successive sections and subset of 2 successive sections. What does this mean?
– user585025
Aug 17 at 16:00
@Buddha I concluded the above statement from the given question "A trip along two successive sections never exceeds 12 km. A trip along three successive sections is at least 17 km"
– Key Flex
Aug 17 at 16:06
1
And why is it that $AB$ is exactly $5$, instead of at most $5$ like $JK$?
– user585025
Aug 17 at 16:10
add a comment |Â
From the given information we can say that any single section can be taken as the difference of some 3 successive sections and subset of 2 successive sections. What does this mean?
– user585025
Aug 17 at 16:00
@Buddha I concluded the above statement from the given question "A trip along two successive sections never exceeds 12 km. A trip along three successive sections is at least 17 km"
– Key Flex
Aug 17 at 16:06
1
And why is it that $AB$ is exactly $5$, instead of at most $5$ like $JK$?
– user585025
Aug 17 at 16:10
From the given information we can say that any single section can be taken as the difference of some 3 successive sections and subset of 2 successive sections. What does this mean?– user585025
Aug 17 at 16:00
From the given information we can say that any single section can be taken as the difference of some 3 successive sections and subset of 2 successive sections. What does this mean?– user585025
Aug 17 at 16:00
@Buddha I concluded the above statement from the given question "A trip along two successive sections never exceeds 12 km. A trip along three successive sections is at least 17 km"
– Key Flex
Aug 17 at 16:06
@Buddha I concluded the above statement from the given question "A trip along two successive sections never exceeds 12 km. A trip along three successive sections is at least 17 km"
– Key Flex
Aug 17 at 16:06
1
1
And why is it that $AB$ is exactly $5$, instead of at most $5$ like $JK$?
– user585025
Aug 17 at 16:10
And why is it that $AB$ is exactly $5$, instead of at most $5$ like $JK$?
– user585025
Aug 17 at 16:10
add a comment |Â
up vote
3
down vote
Here's a solution that requires the least English :)
Let the distance between $A$ and $B$ be $d_1$, that between $B$ and $C$ be $d_2$, and so on, so that the distance between $J$ and $K$ is $d_10$. By symmetry we need only consider $d_1$ to $d_5$. Note also that each section must be at least $5$ kilometres. We seek $$d_2+d_3+d_4+d_5+d_6=d_2+d_3+d_4+2d_5$$ due to symmetry.
Let's now create a list of inequalities:
$$d_1+d_2le12tag1$$$$d_1+d_2+d_3ge17tag2$$gives$$d_3ge5tag3$$ and $$d_2+d_3le12tag4$$$$d_2+d_3+d_4ge17tag5$$ gives$$d_4ge5tag6$$and put $(3)$ into $(5)$ to get $$d_2le7tag7$$ Now put $(6)$ into $(1)$ to get $$d_1le5implies d_1=5tag8$$ Put $(7)$ into $(5)$ to give $$d_3+d_4ge10tag9$$ and $(3)$ gives$$d_4le5tag10$$ Equating $(6)$ and $(10)$ together gives $$d_4=5tag11$$ Put $(11)$ into $(5)$ to get $$d_2+d_3ge12tag12$$ and equate with $(4)$ to get $$d_2+d_3=12tag13$$ Finally, $$d_1+d_2+d_3+d_4+d_5=28$$ due to symmetry so $$d_5=28-d_1-d_4-(d_2+d_3)=6tag14$$ Hence the required distance is $$(d_2+d_3)+d_4+2d_5=12+5+2cdot6=29$$
answered Aug 17 at 16:28
TheSimpliFire
10.5k62053
There are possibilities where $d_5 neq d_6$. Example: $5, 5, 7, 5, 5, 7, 5, 5, 7, 5$
– tehtmi
Aug 18 at 2:58
add a comment |Â
up vote
3
down vote
Here's a solution that requires the least English :)
Let the distance between $A$ and $B$ be $d_1$, that between $B$ and $C$ be $d_2$, and so on, so that the distance between $J$ and $K$ is $d_10$. By symmetry we need only consider $d_1$ to $d_5$. Note also that each section must be at least $5$ kilometres. We seek $$d_2+d_3+d_4+d_5+d_6=d_2+d_3+d_4+2d_5$$ due to symmetry.
Let's now create a list of inequalities:
$$d_1+d_2le12tag1$$$$d_1+d_2+d_3ge17tag2$$gives$$d_3ge5tag3$$ and $$d_2+d_3le12tag4$$$$d_2+d_3+d_4ge17tag5$$ gives$$d_4ge5tag6$$and put $(3)$ into $(5)$ to get $$d_2le7tag7$$ Now put $(6)$ into $(1)$ to get $$d_1le5implies d_1=5tag8$$ Put $(7)$ into $(5)$ to give $$d_3+d_4ge10tag9$$ and $(3)$ gives$$d_4le5tag10$$ Equating $(6)$ and $(10)$ together gives $$d_4=5tag11$$ Put $(11)$ into $(5)$ to get $$d_2+d_3ge12tag12$$ and equate with $(4)$ to get $$d_2+d_3=12tag13$$ Finally, $$d_1+d_2+d_3+d_4+d_5=28$$ due to symmetry so $$d_5=28-d_1-d_4-(d_2+d_3)=6tag14$$ Hence the required distance is $$(d_2+d_3)+d_4+2d_5=12+5+2cdot6=29$$
answered Aug 17 at 16:28
TheSimpliFire
10.5k62053
There are possibilities where $d_5 neq d_6$. Example: $5, 5, 7, 5, 5, 7, 5, 5, 7, 5$
– tehtmi
Aug 18 at 2:58
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Here's a solution that requires the least English :)
Let the distance between $A$ and $B$ be $d_1$, that between $B$ and $C$ be $d_2$, and so on, so that the distance between $J$ and $K$ is $d_10$. By symmetry we need only consider $d_1$ to $d_5$. Note also that each section must be at least $5$ kilometres. We seek $$d_2+d_3+d_4+d_5+d_6=d_2+d_3+d_4+2d_5$$ due to symmetry.
Let's now create a list of inequalities:
$$d_1+d_2le12tag1$$$$d_1+d_2+d_3ge17tag2$$gives$$d_3ge5tag3$$ and $$d_2+d_3le12tag4$$$$d_2+d_3+d_4ge17tag5$$ gives$$d_4ge5tag6$$and put $(3)$ into $(5)$ to get $$d_2le7tag7$$ Now put $(6)$ into $(1)$ to get $$d_1le5implies d_1=5tag8$$ Put $(7)$ into $(5)$ to give $$d_3+d_4ge10tag9$$ and $(3)$ gives$$d_4le5tag10$$ Equating $(6)$ and $(10)$ together gives $$d_4=5tag11$$ Put $(11)$ into $(5)$ to get $$d_2+d_3ge12tag12$$ and equate with $(4)$ to get $$d_2+d_3=12tag13$$ Finally, $$d_1+d_2+d_3+d_4+d_5=28$$ due to symmetry so $$d_5=28-d_1-d_4-(d_2+d_3)=6tag14$$ Hence the required distance is $$(d_2+d_3)+d_4+2d_5=12+5+2cdot6=29$$
answered Aug 17 at 16:28
TheSimpliFire
10.5k62053
Here's a solution that requires the least English :)
Let the distance between $A$ and $B$ be $d_1$, that between $B$ and $C$ be $d_2$, and so on, so that the distance between $J$ and $K$ is $d_10$. By symmetry we need only consider $d_1$ to $d_5$. Note also that each section must be at least $5$ kilometres. We seek $$d_2+d_3+d_4+d_5+d_6=d_2+d_3+d_4+2d_5$$ due to symmetry.
Let's now create a list of inequalities:
$$d_1+d_2le12tag1$$$$d_1+d_2+d_3ge17tag2$$gives$$d_3ge5tag3$$ and $$d_2+d_3le12tag4$$$$d_2+d_3+d_4ge17tag5$$ gives$$d_4ge5tag6$$and put $(3)$ into $(5)$ to get $$d_2le7tag7$$ Now put $(6)$ into $(1)$ to get $$d_1le5implies d_1=5tag8$$ Put $(7)$ into $(5)$ to give $$d_3+d_4ge10tag9$$ and $(3)$ gives$$d_4le5tag10$$ Equating $(6)$ and $(10)$ together gives $$d_4=5tag11$$ Put $(11)$ into $(5)$ to get $$d_2+d_3ge12tag12$$ and equate with $(4)$ to get $$d_2+d_3=12tag13$$ Finally, $$d_1+d_2+d_3+d_4+d_5=28$$ due to symmetry so $$d_5=28-d_1-d_4-(d_2+d_3)=6tag14$$ Hence the required distance is $$(d_2+d_3)+d_4+2d_5=12+5+2cdot6=29$$
answered Aug 17 at 16:28
TheSimpliFire
10.5k62053
answered Aug 17 at 16:28
TheSimpliFire
10.5k62053
answered Aug 17 at 16:28
TheSimpliFire
10.5k62053
answered Aug 17 at 16:28
TheSimpliFire
10.5k62053
10.5k62053
There are possibilities where $d_5 neq d_6$. Example: $5, 5, 7, 5, 5, 7, 5, 5, 7, 5$
– tehtmi
Aug 18 at 2:58
add a comment |Â
There are possibilities where $d_5 neq d_6$. Example: $5, 5, 7, 5, 5, 7, 5, 5, 7, 5$
– tehtmi
Aug 18 at 2:58
There are possibilities where $d_5 neq d_6$. Example: $5, 5, 7, 5, 5, 7, 5, 5, 7, 5$
– tehtmi
Aug 18 at 2:58
There are possibilities where $d_5 neq d_6$. Example: $5, 5, 7, 5, 5, 7, 5, 5, 7, 5$
– tehtmi
Aug 18 at 2:58
add a comment |Â
up vote
2
down vote
This is a bruteforce solution, but one way to do this is to simply formulate it as a linear programming feasability problem. Let $x=(x_1,ldots,x_10)$ be the length of the sections. Then, $x$ is a point in the polytope $ y mid A y leq b $ where
$$
A=beginpmatrix1 & 1\
& phantom-1 & phantom-1\
& & phantom-1 & phantom-1\
& & & ddots & ddots\
& & & & phantom-1 & phantom-1\
-1 & -1 & -1\
& -1 & -1 & -1\
& & ddots & ddots & ddots\
& & & -1 & -1 & -1\
phantom-1 & phantom-1 & phantom-1 & phantom-1 & cdots & phantom-1\
-1 & -1 & -1 & -1 & cdots & -1
endpmatrix
qquadtextandqquad
b=beginpmatrixphantom-12\
phantom-12\
phantom-12\
phantom-vdots\
phantom-12\
-17\
-17\
phantom-vdots\
-17\
phantom-56\
-56
endpmatrix.
$$
answered Aug 17 at 15:46
parsiad
16.1k32253
1
Umm, this is a contest problem so I don't think this would be very useful?
– TheSimpliFire
Aug 17 at 15:53
@TheSimpliFire Useful to learn something new outside of the contest, which can probably be applied later...
– user202729
Aug 18 at 16:16
add a comment |Â
up vote
2
down vote
This is a bruteforce solution, but one way to do this is to simply formulate it as a linear programming feasability problem. Let $x=(x_1,ldots,x_10)$ be the length of the sections. Then, $x$ is a point in the polytope $ y mid A y leq b $ where
$$
A=beginpmatrix1 & 1\
& phantom-1 & phantom-1\
& & phantom-1 & phantom-1\
& & & ddots & ddots\
& & & & phantom-1 & phantom-1\
-1 & -1 & -1\
& -1 & -1 & -1\
& & ddots & ddots & ddots\
& & & -1 & -1 & -1\
phantom-1 & phantom-1 & phantom-1 & phantom-1 & cdots & phantom-1\
-1 & -1 & -1 & -1 & cdots & -1
endpmatrix
qquadtextandqquad
b=beginpmatrixphantom-12\
phantom-12\
phantom-12\
phantom-vdots\
phantom-12\
-17\
-17\
phantom-vdots\
-17\
phantom-56\
-56
endpmatrix.
$$
answered Aug 17 at 15:46
parsiad
16.1k32253
1
Umm, this is a contest problem so I don't think this would be very useful?
– TheSimpliFire
Aug 17 at 15:53
@TheSimpliFire Useful to learn something new outside of the contest, which can probably be applied later...
– user202729
Aug 18 at 16:16
add a comment |Â
up vote
2
down vote
up vote
2
down vote
This is a bruteforce solution, but one way to do this is to simply formulate it as a linear programming feasability problem. Let $x=(x_1,ldots,x_10)$ be the length of the sections. Then, $x$ is a point in the polytope $ y mid A y leq b $ where
$$
A=beginpmatrix1 & 1\
& phantom-1 & phantom-1\
& & phantom-1 & phantom-1\
& & & ddots & ddots\
& & & & phantom-1 & phantom-1\
-1 & -1 & -1\
& -1 & -1 & -1\
& & ddots & ddots & ddots\
& & & -1 & -1 & -1\
phantom-1 & phantom-1 & phantom-1 & phantom-1 & cdots & phantom-1\
-1 & -1 & -1 & -1 & cdots & -1
endpmatrix
qquadtextandqquad
b=beginpmatrixphantom-12\
phantom-12\
phantom-12\
phantom-vdots\
phantom-12\
-17\
-17\
phantom-vdots\
-17\
phantom-56\
-56
endpmatrix.
$$
answered Aug 17 at 15:46
parsiad
16.1k32253
This is a bruteforce solution, but one way to do this is to simply formulate it as a linear programming feasability problem. Let $x=(x_1,ldots,x_10)$ be the length of the sections. Then, $x$ is a point in the polytope $ y mid A y leq b $ where
$$
A=beginpmatrix1 & 1\
& phantom-1 & phantom-1\
& & phantom-1 & phantom-1\
& & & ddots & ddots\
& & & & phantom-1 & phantom-1\
-1 & -1 & -1\
& -1 & -1 & -1\
& & ddots & ddots & ddots\
& & & -1 & -1 & -1\
phantom-1 & phantom-1 & phantom-1 & phantom-1 & cdots & phantom-1\
-1 & -1 & -1 & -1 & cdots & -1
endpmatrix
qquadtextandqquad
b=beginpmatrixphantom-12\
phantom-12\
phantom-12\
phantom-vdots\
phantom-12\
-17\
-17\
phantom-vdots\
-17\
phantom-56\
-56
endpmatrix.
$$
answered Aug 17 at 15:46
parsiad
16.1k32253
answered Aug 17 at 15:46
parsiad
16.1k32253
answered Aug 17 at 15:46
parsiad
16.1k32253
answered Aug 17 at 15:46
parsiad
16.1k32253
16.1k32253
1
Umm, this is a contest problem so I don't think this would be very useful?
– TheSimpliFire
Aug 17 at 15:53
@TheSimpliFire Useful to learn something new outside of the contest, which can probably be applied later...
– user202729
Aug 18 at 16:16
add a comment |Â
1
Umm, this is a contest problem so I don't think this would be very useful?
– TheSimpliFire
Aug 17 at 15:53
@TheSimpliFire Useful to learn something new outside of the contest, which can probably be applied later...
– user202729
Aug 18 at 16:16
1
1
Umm, this is a contest problem so I don't think this would be very useful?
– TheSimpliFire
Aug 17 at 15:53
Umm, this is a contest problem so I don't think this would be very useful?
– TheSimpliFire
Aug 17 at 15:53
@TheSimpliFire Useful to learn something new outside of the contest, which can probably be applied later...
– user202729
Aug 18 at 16:16
@TheSimpliFire Useful to learn something new outside of the contest, which can probably be applied later...
– user202729
Aug 18 at 16:16
add a comment |Â
up vote
1
down vote
Let the ten distances between adjacent stations be $x_1$ to $x_10$.
$$17 leq x_1 + (x_2 + x_3) leq x_1 + 12$$
So
$$x_1 geq 5$$
And then
$$56 = x_1 + (x_2 + x_3 + x_4) + (x_5 + x_6 + x_7) + (x_8 + x_9 + x_10) geq 5 + 17 + 17 + 17 = 56$$
There is no slack in this inequality, so in fact $x_1 = 5$ and $x_2 + x_3 + x_4 = x_5 + x_6 + x_7 = x_8 + x_9 + x_10 = 17$.
We can similarly get the symmetric equations $x_10 = 5$ and $x_1 + x_2 + x_3 = x_4 + x_5 + x_6 = x_7 + x_8 + x_9 = 17$.
Together, these give $x_1 = x_4 = x_7 = x_10 = 5$ and $x_2 + x_3 = x_5 + x_6 = x_8 + x_9 = 12$.
The solution is then $(x_2 + x_3) + x_4 + (x_5 + x_6) = 12 + 5 + 12 = 29$.
Addendum 1: Multiple solutions to the full system are possible. E.g. $5, 6, 6, 5, 6, 6, 5, 6, 6, 5$ versus $5, 5, 7, 5, 5, 7, 5, 5, 7, 5$.
Addendum 2: As in Key Flex's answer, it is also possible to use splits like $(x_1 + x_2 + x_3) + x_4 + (x_5 + x_6 + x_7) + (x_8 + x_9 + x_10)$.
answered Aug 18 at 2:56
tehtmi
53429
add a comment |Â
up vote
1
down vote
Let the ten distances between adjacent stations be $x_1$ to $x_10$.
$$17 leq x_1 + (x_2 + x_3) leq x_1 + 12$$
So
$$x_1 geq 5$$
And then
$$56 = x_1 + (x_2 + x_3 + x_4) + (x_5 + x_6 + x_7) + (x_8 + x_9 + x_10) geq 5 + 17 + 17 + 17 = 56$$
There is no slack in this inequality, so in fact $x_1 = 5$ and $x_2 + x_3 + x_4 = x_5 + x_6 + x_7 = x_8 + x_9 + x_10 = 17$.
We can similarly get the symmetric equations $x_10 = 5$ and $x_1 + x_2 + x_3 = x_4 + x_5 + x_6 = x_7 + x_8 + x_9 = 17$.
Together, these give $x_1 = x_4 = x_7 = x_10 = 5$ and $x_2 + x_3 = x_5 + x_6 = x_8 + x_9 = 12$.
The solution is then $(x_2 + x_3) + x_4 + (x_5 + x_6) = 12 + 5 + 12 = 29$.
Addendum 1: Multiple solutions to the full system are possible. E.g. $5, 6, 6, 5, 6, 6, 5, 6, 6, 5$ versus $5, 5, 7, 5, 5, 7, 5, 5, 7, 5$.
Addendum 2: As in Key Flex's answer, it is also possible to use splits like $(x_1 + x_2 + x_3) + x_4 + (x_5 + x_6 + x_7) + (x_8 + x_9 + x_10)$.
answered Aug 18 at 2:56
tehtmi
53429
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Let the ten distances between adjacent stations be $x_1$ to $x_10$.
$$17 leq x_1 + (x_2 + x_3) leq x_1 + 12$$
So
$$x_1 geq 5$$
And then
$$56 = x_1 + (x_2 + x_3 + x_4) + (x_5 + x_6 + x_7) + (x_8 + x_9 + x_10) geq 5 + 17 + 17 + 17 = 56$$
There is no slack in this inequality, so in fact $x_1 = 5$ and $x_2 + x_3 + x_4 = x_5 + x_6 + x_7 = x_8 + x_9 + x_10 = 17$.
We can similarly get the symmetric equations $x_10 = 5$ and $x_1 + x_2 + x_3 = x_4 + x_5 + x_6 = x_7 + x_8 + x_9 = 17$.
Together, these give $x_1 = x_4 = x_7 = x_10 = 5$ and $x_2 + x_3 = x_5 + x_6 = x_8 + x_9 = 12$.
The solution is then $(x_2 + x_3) + x_4 + (x_5 + x_6) = 12 + 5 + 12 = 29$.
Addendum 1: Multiple solutions to the full system are possible. E.g. $5, 6, 6, 5, 6, 6, 5, 6, 6, 5$ versus $5, 5, 7, 5, 5, 7, 5, 5, 7, 5$.
Addendum 2: As in Key Flex's answer, it is also possible to use splits like $(x_1 + x_2 + x_3) + x_4 + (x_5 + x_6 + x_7) + (x_8 + x_9 + x_10)$.
answered Aug 18 at 2:56
tehtmi
53429
Let the ten distances between adjacent stations be $x_1$ to $x_10$.
$$17 leq x_1 + (x_2 + x_3) leq x_1 + 12$$
So
$$x_1 geq 5$$
And then
$$56 = x_1 + (x_2 + x_3 + x_4) + (x_5 + x_6 + x_7) + (x_8 + x_9 + x_10) geq 5 + 17 + 17 + 17 = 56$$
There is no slack in this inequality, so in fact $x_1 = 5$ and $x_2 + x_3 + x_4 = x_5 + x_6 + x_7 = x_8 + x_9 + x_10 = 17$.
We can similarly get the symmetric equations $x_10 = 5$ and $x_1 + x_2 + x_3 = x_4 + x_5 + x_6 = x_7 + x_8 + x_9 = 17$.
Together, these give $x_1 = x_4 = x_7 = x_10 = 5$ and $x_2 + x_3 = x_5 + x_6 = x_8 + x_9 = 12$.
The solution is then $(x_2 + x_3) + x_4 + (x_5 + x_6) = 12 + 5 + 12 = 29$.
Addendum 1: Multiple solutions to the full system are possible. E.g. $5, 6, 6, 5, 6, 6, 5, 6, 6, 5$ versus $5, 5, 7, 5, 5, 7, 5, 5, 7, 5$.
Addendum 2: As in Key Flex's answer, it is also possible to use splits like $(x_1 + x_2 + x_3) + x_4 + (x_5 + x_6 + x_7) + (x_8 + x_9 + x_10)$.
answered Aug 18 at 2:56
tehtmi
53429
answered Aug 18 at 2:56
tehtmi
53429
answered Aug 18 at 2:56
tehtmi
53429
answered Aug 18 at 2:56
tehtmi
53429
53429
add a comment |Â
add a comment |Â
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Using algebraic method.
Label the stations with $A$ through $K$:
$hspace4cm$
The given conditions are:
$$K-A=56 textand begincases
A-Cge -12\
B-Dge -12\
C-Ege -12\
D-Fge -12\
E-Gge -12\
F-Hge -12\
G-Ige -12\
H-Jge -12\
I-Kge -12\
endcases textand
begincases
D-Age 17\
E-Bge 17\
F-Cge 17\
G-Dge 17\
H-Ege 17\
I-Fge 17\
J-Gge 17\
K-Hge 17\
endcases$$
One the one hand:
$$(D-A)+(G-D)=G-Age 34;$$
on the other hand:
$$(K-H)+(H-J)=K-Jge 5;\
(K-J)+(J-G)=K-Gge 22;\
K-G=(56+A)-Gge 22 Rightarrow \
G-Ale 34.$$
Hence: $colorredG-A=34$.
One the one hand:
$$(B-D)+(D-A)=B-Age 5 Rightarrow A-Ble -5;\
$$
on the other hand:
$$(E-B)+(H-E)=H-B=ge 34;\
(H-B)+(K-H)=K-Bge 51;\
K-B=(56+A)-Bge 51 Rightarrow \
A-Bge -5.$$
Hence: $colorblueA-B=-5$.
In conclusion: $(G-A)+(A-B)=G-B=34+(-5)=29$.
answered Aug 18 at 7:41
farruhota
14.7k2633
add a comment |Â
up vote
1
down vote
Using algebraic method.
Label the stations with $A$ through $K$:
$hspace4cm$
The given conditions are:
$$K-A=56 textand begincases
A-Cge -12\
B-Dge -12\
C-Ege -12\
D-Fge -12\
E-Gge -12\
F-Hge -12\
G-Ige -12\
H-Jge -12\
I-Kge -12\
endcases textand
begincases
D-Age 17\
E-Bge 17\
F-Cge 17\
G-Dge 17\
H-Ege 17\
I-Fge 17\
J-Gge 17\
K-Hge 17\
endcases$$
One the one hand:
$$(D-A)+(G-D)=G-Age 34;$$
on the other hand:
$$(K-H)+(H-J)=K-Jge 5;\
(K-J)+(J-G)=K-Gge 22;\
K-G=(56+A)-Gge 22 Rightarrow \
G-Ale 34.$$
Hence: $colorredG-A=34$.
One the one hand:
$$(B-D)+(D-A)=B-Age 5 Rightarrow A-Ble -5;\
$$
on the other hand:
$$(E-B)+(H-E)=H-B=ge 34;\
(H-B)+(K-H)=K-Bge 51;\
K-B=(56+A)-Bge 51 Rightarrow \
A-Bge -5.$$
Hence: $colorblueA-B=-5$.
In conclusion: $(G-A)+(A-B)=G-B=34+(-5)=29$.
answered Aug 18 at 7:41
farruhota
14.7k2633
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Using algebraic method.
Label the stations with $A$ through $K$:
$hspace4cm$
The given conditions are:
$$K-A=56 textand begincases
A-Cge -12\
B-Dge -12\
C-Ege -12\
D-Fge -12\
E-Gge -12\
F-Hge -12\
G-Ige -12\
H-Jge -12\
I-Kge -12\
endcases textand
begincases
D-Age 17\
E-Bge 17\
F-Cge 17\
G-Dge 17\
H-Ege 17\
I-Fge 17\
J-Gge 17\
K-Hge 17\
endcases$$
One the one hand:
$$(D-A)+(G-D)=G-Age 34;$$
on the other hand:
$$(K-H)+(H-J)=K-Jge 5;\
(K-J)+(J-G)=K-Gge 22;\
K-G=(56+A)-Gge 22 Rightarrow \
G-Ale 34.$$
Hence: $colorredG-A=34$.
One the one hand:
$$(B-D)+(D-A)=B-Age 5 Rightarrow A-Ble -5;\
$$
on the other hand:
$$(E-B)+(H-E)=H-B=ge 34;\
(H-B)+(K-H)=K-Bge 51;\
K-B=(56+A)-Bge 51 Rightarrow \
A-Bge -5.$$
Hence: $colorblueA-B=-5$.
In conclusion: $(G-A)+(A-B)=G-B=34+(-5)=29$.
answered Aug 18 at 7:41
farruhota
14.7k2633
Using algebraic method.
Label the stations with $A$ through $K$:
$hspace4cm$
The given conditions are:
$$K-A=56 textand begincases
A-Cge -12\
B-Dge -12\
C-Ege -12\
D-Fge -12\
E-Gge -12\
F-Hge -12\
G-Ige -12\
H-Jge -12\
I-Kge -12\
endcases textand
begincases
D-Age 17\
E-Bge 17\
F-Cge 17\
G-Dge 17\
H-Ege 17\
I-Fge 17\
J-Gge 17\
K-Hge 17\
endcases$$
One the one hand:
$$(D-A)+(G-D)=G-Age 34;$$
on the other hand:
$$(K-H)+(H-J)=K-Jge 5;\
(K-J)+(J-G)=K-Gge 22;\
K-G=(56+A)-Gge 22 Rightarrow \
G-Ale 34.$$
Hence: $colorredG-A=34$.
One the one hand:
$$(B-D)+(D-A)=B-Age 5 Rightarrow A-Ble -5;\
$$
on the other hand:
$$(E-B)+(H-E)=H-B=ge 34;\
(H-B)+(K-H)=K-Bge 51;\
K-B=(56+A)-Bge 51 Rightarrow \
A-Bge -5.$$
Hence: $colorblueA-B=-5$.
In conclusion: $(G-A)+(A-B)=G-B=34+(-5)=29$.
answered Aug 18 at 7:41
farruhota
14.7k2633
answered Aug 18 at 7:41
farruhota
14.7k2633
answered Aug 18 at 7:41
farruhota
14.7k2633
answered Aug 18 at 7:41
farruhota
14.7k2633
14.7k2633
add a comment |Â
add a comment |Â
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4
This is quite a nice problem. Where did you get it from?
– TheSimpliFire
Aug 17 at 16:34
1
@TheSimpliFire A mock test for the PRMO.
– user585025
Aug 17 at 16:58