Mixing
Constructing Escher's Template for the Tetrahedron 5-Compound
Clash Royale CLAN TAG#URR8PPP
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Consider the above figure. The lengths of all the outer line segments are known, and are specified below. We know that angle $A$ is 60 degrees and that line segment $|CD|$ extends $|AD|$.
This information, as far as I can figure, is not enough to actually construct this figure, though it should be enough to derive all required information.
I also have a suspicion that $|BC| = |CD| = |BD|$, which if true would be enough information, but thus far I've been unable to prove it.
The more specific facts can be found at: http://mathworld.wolfram.com/Tetrahedron5-Compound.html
But all we really need to know is this:
The side lengths are:
$|AB| = |AE| = 1$
$|DE| = phi^-1 approx 0.618033988749895$
$|BC| = |CD| = phi^-1/sqrt2 approx 0.437016024448821$
and what I've already mentioned, where $phi$ denotes the golden ratio.
It's all about the line segment $|CD|$; that is the real unknown. I basically just need one single piece of information, other than what I've already got, and then I won't have a problem solving the problem. It doesn't really matter what that information is either, except perhaps angle D. I'm not quite sure that would help.
So that's it, this being my first post, I do hope I've not messed up too badly. Also I'm certain this is not a hard problem, but I've been struggling with it on a off for a couple of years now, believe it or now, so some help would be much appreciated.
Best regards.
geometry
edited Jul 16 at 20:58
MvG
29.7k44597
asked Jul 16 at 20:06
Zacariaz
132
add a comment |Â
up vote
2
down vote
favorite
Consider the above figure. The lengths of all the outer line segments are known, and are specified below. We know that angle $A$ is 60 degrees and that line segment $|CD|$ extends $|AD|$.
This information, as far as I can figure, is not enough to actually construct this figure, though it should be enough to derive all required information.
I also have a suspicion that $|BC| = |CD| = |BD|$, which if true would be enough information, but thus far I've been unable to prove it.
The more specific facts can be found at: http://mathworld.wolfram.com/Tetrahedron5-Compound.html
But all we really need to know is this:
The side lengths are:
$|AB| = |AE| = 1$
$|DE| = phi^-1 approx 0.618033988749895$
$|BC| = |CD| = phi^-1/sqrt2 approx 0.437016024448821$
and what I've already mentioned, where $phi$ denotes the golden ratio.
It's all about the line segment $|CD|$; that is the real unknown. I basically just need one single piece of information, other than what I've already got, and then I won't have a problem solving the problem. It doesn't really matter what that information is either, except perhaps angle D. I'm not quite sure that would help.
So that's it, this being my first post, I do hope I've not messed up too badly. Also I'm certain this is not a hard problem, but I've been struggling with it on a off for a couple of years now, believe it or now, so some help would be much appreciated.
Best regards.
geometry
edited Jul 16 at 20:58
MvG
29.7k44597
asked Jul 16 at 20:06
Zacariaz
132
Could you give the precise values of $|DE|$ and $|BC|$? This could be crucial for solving the problem. Also, clarify, is your question about how to find the remaining information about the figure given what you specified?
– Sambo
Jul 16 at 20:37
Added exact values for the side lengths, scaled up from the cited MathWorld article. I assume there is a typo about $lvert CDrvert$ being the real unknown. Did you mean $lvert BDrvert$?
– MvG
Jul 16 at 21:58
@MvG No, I actually did mean |CD|, but I probably didn't phrase it well. What I really meant I suppose, is simply that trying to construct with compass and ruler, this particular line segment cannot be placed and it implies two solutions. But still, thanks for your assistance, I got my answer, rather more quickly than I expected I might add
– Zacariaz
Jul 16 at 22:19
@Sambo I have in fact given values for those line segments. As for clarification, I just need enough information to be able to find the information I need in order to be able to construct it, but I do believe MgVs answer has provided the complete picture.
– Zacariaz
Jul 16 at 22:24
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Consider the above figure. The lengths of all the outer line segments are known, and are specified below. We know that angle $A$ is 60 degrees and that line segment $|CD|$ extends $|AD|$.
This information, as far as I can figure, is not enough to actually construct this figure, though it should be enough to derive all required information.
I also have a suspicion that $|BC| = |CD| = |BD|$, which if true would be enough information, but thus far I've been unable to prove it.
The more specific facts can be found at: http://mathworld.wolfram.com/Tetrahedron5-Compound.html
But all we really need to know is this:
The side lengths are:
$|AB| = |AE| = 1$
$|DE| = phi^-1 approx 0.618033988749895$
$|BC| = |CD| = phi^-1/sqrt2 approx 0.437016024448821$
and what I've already mentioned, where $phi$ denotes the golden ratio.
It's all about the line segment $|CD|$; that is the real unknown. I basically just need one single piece of information, other than what I've already got, and then I won't have a problem solving the problem. It doesn't really matter what that information is either, except perhaps angle D. I'm not quite sure that would help.
So that's it, this being my first post, I do hope I've not messed up too badly. Also I'm certain this is not a hard problem, but I've been struggling with it on a off for a couple of years now, believe it or now, so some help would be much appreciated.
Best regards.
geometry
edited Jul 16 at 20:58
MvG
29.7k44597
asked Jul 16 at 20:06
Zacariaz
132
Consider the above figure. The lengths of all the outer line segments are known, and are specified below. We know that angle $A$ is 60 degrees and that line segment $|CD|$ extends $|AD|$.
This information, as far as I can figure, is not enough to actually construct this figure, though it should be enough to derive all required information.
I also have a suspicion that $|BC| = |CD| = |BD|$, which if true would be enough information, but thus far I've been unable to prove it.
The more specific facts can be found at: http://mathworld.wolfram.com/Tetrahedron5-Compound.html
But all we really need to know is this:
The side lengths are:
$|AB| = |AE| = 1$
$|DE| = phi^-1 approx 0.618033988749895$
$|BC| = |CD| = phi^-1/sqrt2 approx 0.437016024448821$
and what I've already mentioned, where $phi$ denotes the golden ratio.
It's all about the line segment $|CD|$; that is the real unknown. I basically just need one single piece of information, other than what I've already got, and then I won't have a problem solving the problem. It doesn't really matter what that information is either, except perhaps angle D. I'm not quite sure that would help.
So that's it, this being my first post, I do hope I've not messed up too badly. Also I'm certain this is not a hard problem, but I've been struggling with it on a off for a couple of years now, believe it or now, so some help would be much appreciated.
Best regards.
geometry
edited Jul 16 at 20:58
MvG
29.7k44597
asked Jul 16 at 20:06
Zacariaz
132
edited Jul 16 at 20:58
MvG
29.7k44597
edited Jul 16 at 20:58
MvG
29.7k44597
edited Jul 16 at 20:58
MvG
29.7k44597
29.7k44597
asked Jul 16 at 20:06
Zacariaz
132
asked Jul 16 at 20:06
Zacariaz
132
asked Jul 16 at 20:06
Zacariaz
132
132
Could you give the precise values of $|DE|$ and $|BC|$? This could be crucial for solving the problem. Also, clarify, is your question about how to find the remaining information about the figure given what you specified?
– Sambo
Jul 16 at 20:37
Added exact values for the side lengths, scaled up from the cited MathWorld article. I assume there is a typo about $lvert CDrvert$ being the real unknown. Did you mean $lvert BDrvert$?
– MvG
Jul 16 at 21:58
@MvG No, I actually did mean |CD|, but I probably didn't phrase it well. What I really meant I suppose, is simply that trying to construct with compass and ruler, this particular line segment cannot be placed and it implies two solutions. But still, thanks for your assistance, I got my answer, rather more quickly than I expected I might add
– Zacariaz
Jul 16 at 22:19
@Sambo I have in fact given values for those line segments. As for clarification, I just need enough information to be able to find the information I need in order to be able to construct it, but I do believe MgVs answer has provided the complete picture.
– Zacariaz
Jul 16 at 22:24
add a comment |Â
Could you give the precise values of $|DE|$ and $|BC|$? This could be crucial for solving the problem. Also, clarify, is your question about how to find the remaining information about the figure given what you specified?
– Sambo
Jul 16 at 20:37
Added exact values for the side lengths, scaled up from the cited MathWorld article. I assume there is a typo about $lvert CDrvert$ being the real unknown. Did you mean $lvert BDrvert$?
– MvG
Jul 16 at 21:58
@MvG No, I actually did mean |CD|, but I probably didn't phrase it well. What I really meant I suppose, is simply that trying to construct with compass and ruler, this particular line segment cannot be placed and it implies two solutions. But still, thanks for your assistance, I got my answer, rather more quickly than I expected I might add
– Zacariaz
Jul 16 at 22:19
@Sambo I have in fact given values for those line segments. As for clarification, I just need enough information to be able to find the information I need in order to be able to construct it, but I do believe MgVs answer has provided the complete picture.
– Zacariaz
Jul 16 at 22:24
Could you give the precise values of $|DE|$ and $|BC|$? This could be crucial for solving the problem. Also, clarify, is your question about how to find the remaining information about the figure given what you specified?
– Sambo
Jul 16 at 20:37
Could you give the precise values of $|DE|$ and $|BC|$? This could be crucial for solving the problem. Also, clarify, is your question about how to find the remaining information about the figure given what you specified?
– Sambo
Jul 16 at 20:37
Added exact values for the side lengths, scaled up from the cited MathWorld article. I assume there is a typo about $lvert CDrvert$ being the real unknown. Did you mean $lvert BDrvert$?
– MvG
Jul 16 at 21:58
Added exact values for the side lengths, scaled up from the cited MathWorld article. I assume there is a typo about $lvert CDrvert$ being the real unknown. Did you mean $lvert BDrvert$?
– MvG
Jul 16 at 21:58
@MvG No, I actually did mean |CD|, but I probably didn't phrase it well. What I really meant I suppose, is simply that trying to construct with compass and ruler, this particular line segment cannot be placed and it implies two solutions. But still, thanks for your assistance, I got my answer, rather more quickly than I expected I might add
– Zacariaz
Jul 16 at 22:19
@MvG No, I actually did mean |CD|, but I probably didn't phrase it well. What I really meant I suppose, is simply that trying to construct with compass and ruler, this particular line segment cannot be placed and it implies two solutions. But still, thanks for your assistance, I got my answer, rather more quickly than I expected I might add
– Zacariaz
Jul 16 at 22:19
@Sambo I have in fact given values for those line segments. As for clarification, I just need enough information to be able to find the information I need in order to be able to construct it, but I do believe MgVs answer has provided the complete picture.
– Zacariaz
Jul 16 at 22:24
@Sambo I have in fact given values for those line segments. As for clarification, I just need enough information to be able to find the information I need in order to be able to construct it, but I do believe MgVs answer has provided the complete picture.
– Zacariaz
Jul 16 at 22:24
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
Let's first check whether you have enough information. You need 5 points in the plane, each with 2 real degrees of freedom. But you don't care about position or rotation of the figure as a whole, which accounts for 3 real degrees of freedom. Which means you need 7 measurements to hope for full specification. You have 5 lengths, 1 angle and 1 collinearity. Could work.
You can pick coordinates and use them to express your distance and collinearity constraints.
beginalign*
A&=(0,0) & B&=(1,0) & C&=(C_x,C_y) & D&=(D_x,D_y) & E&=(tfrac12,tfrac12sqrt3)
endalign*
beginalign*
(C_x-1)^2+C_y^2 &= frac12phi^2 = left(fracphi^-1sqrt2right)^2\
(C_x-D_x)^2+(C_y-D_y)^2 &= frac12phi^2 \
(D_x-tfrac12)^2+(D_y-tfrac12sqrt3)^2 &= frac1phi^2 \
fracC_xC_y &= fracD_xD_y
endalign*
This can be rewritten as a set of polynomial equations. Then typical elimination strategies like Gröbner bases or Resultants can be used to find solutions. But I prefer to leave these things to a computer algebra system, which will tell me that there exist two possible solutions:
beginalignat*3
C_x &= tfrac14sqrt5 + tfrac12 &&approx 1.0590169943749474 &
C_x &approx 0.9369017680762171 \
C_y &= tfrac14sqrt3 &&approx 0.4330127018922193 &
C_y &approx 0.4324368378770998 \
D_x &= tfrac18sqrt5 + tfrac38 &&approx 0.6545084971874737 &
D_x &approx 0.5401122930754872 \
D_y &= tfrac14sqrt-tfrac32sqrt5 + tfrac92
&&approx 0.2676165673298174qquad &
D_y &approx 0.2492944939101796
endalignat*
Your picture shows $C$ slightly right of $B$, so presumably the solution with $C_x>1$ is the one you want. Which is good, because the other solutions are roots of a polynomial of degree $10$ and more, with no radical expression available. This also demonstrates that a ruler and compass solution for that second part is impossible. Since going from the problem statement both solutions would be permissible, reaching either using compass and straightedge is going to be tricky.
You can use these coordinates to confirm that yes, indeed $lvert BCrvert=lvert CDrvert=lvert BDrvert$. So another approach would have been to assume this as the additional piece of information, then compute positions for everything else from that and check that the resulting coordinates satisfy all requirements. More specifically, draw $A,B,E$ then construct $D$ as intersection of two circles cooresponding to the distances to $B$ and $E$. Then draw the triangle $BCD$ and check that line $CD$ passes through $A$.
answered Jul 16 at 21:34
MvG
29.7k44597
Absolutely perfect. I will not pretend that I fully comprehend at this point, but I'm sure I will. At any rate, presuming that you are indeed correct, and this seems a reasonable assumption, I am very grateful. Thanks and best regards.
– Zacariaz
Jul 16 at 22:12
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Let's first check whether you have enough information. You need 5 points in the plane, each with 2 real degrees of freedom. But you don't care about position or rotation of the figure as a whole, which accounts for 3 real degrees of freedom. Which means you need 7 measurements to hope for full specification. You have 5 lengths, 1 angle and 1 collinearity. Could work.
You can pick coordinates and use them to express your distance and collinearity constraints.
beginalign*
A&=(0,0) & B&=(1,0) & C&=(C_x,C_y) & D&=(D_x,D_y) & E&=(tfrac12,tfrac12sqrt3)
endalign*
beginalign*
(C_x-1)^2+C_y^2 &= frac12phi^2 = left(fracphi^-1sqrt2right)^2\
(C_x-D_x)^2+(C_y-D_y)^2 &= frac12phi^2 \
(D_x-tfrac12)^2+(D_y-tfrac12sqrt3)^2 &= frac1phi^2 \
fracC_xC_y &= fracD_xD_y
endalign*
This can be rewritten as a set of polynomial equations. Then typical elimination strategies like Gröbner bases or Resultants can be used to find solutions. But I prefer to leave these things to a computer algebra system, which will tell me that there exist two possible solutions:
beginalignat*3
C_x &= tfrac14sqrt5 + tfrac12 &&approx 1.0590169943749474 &
C_x &approx 0.9369017680762171 \
C_y &= tfrac14sqrt3 &&approx 0.4330127018922193 &
C_y &approx 0.4324368378770998 \
D_x &= tfrac18sqrt5 + tfrac38 &&approx 0.6545084971874737 &
D_x &approx 0.5401122930754872 \
D_y &= tfrac14sqrt-tfrac32sqrt5 + tfrac92
&&approx 0.2676165673298174qquad &
D_y &approx 0.2492944939101796
endalignat*
Your picture shows $C$ slightly right of $B$, so presumably the solution with $C_x>1$ is the one you want. Which is good, because the other solutions are roots of a polynomial of degree $10$ and more, with no radical expression available. This also demonstrates that a ruler and compass solution for that second part is impossible. Since going from the problem statement both solutions would be permissible, reaching either using compass and straightedge is going to be tricky.
You can use these coordinates to confirm that yes, indeed $lvert BCrvert=lvert CDrvert=lvert BDrvert$. So another approach would have been to assume this as the additional piece of information, then compute positions for everything else from that and check that the resulting coordinates satisfy all requirements. More specifically, draw $A,B,E$ then construct $D$ as intersection of two circles cooresponding to the distances to $B$ and $E$. Then draw the triangle $BCD$ and check that line $CD$ passes through $A$.
answered Jul 16 at 21:34
MvG
29.7k44597
Absolutely perfect. I will not pretend that I fully comprehend at this point, but I'm sure I will. At any rate, presuming that you are indeed correct, and this seems a reasonable assumption, I am very grateful. Thanks and best regards.
– Zacariaz
Jul 16 at 22:12
add a comment |Â
up vote
2
down vote
accepted
Let's first check whether you have enough information. You need 5 points in the plane, each with 2 real degrees of freedom. But you don't care about position or rotation of the figure as a whole, which accounts for 3 real degrees of freedom. Which means you need 7 measurements to hope for full specification. You have 5 lengths, 1 angle and 1 collinearity. Could work.
You can pick coordinates and use them to express your distance and collinearity constraints.
beginalign*
A&=(0,0) & B&=(1,0) & C&=(C_x,C_y) & D&=(D_x,D_y) & E&=(tfrac12,tfrac12sqrt3)
endalign*
beginalign*
(C_x-1)^2+C_y^2 &= frac12phi^2 = left(fracphi^-1sqrt2right)^2\
(C_x-D_x)^2+(C_y-D_y)^2 &= frac12phi^2 \
(D_x-tfrac12)^2+(D_y-tfrac12sqrt3)^2 &= frac1phi^2 \
fracC_xC_y &= fracD_xD_y
endalign*
This can be rewritten as a set of polynomial equations. Then typical elimination strategies like Gröbner bases or Resultants can be used to find solutions. But I prefer to leave these things to a computer algebra system, which will tell me that there exist two possible solutions:
beginalignat*3
C_x &= tfrac14sqrt5 + tfrac12 &&approx 1.0590169943749474 &
C_x &approx 0.9369017680762171 \
C_y &= tfrac14sqrt3 &&approx 0.4330127018922193 &
C_y &approx 0.4324368378770998 \
D_x &= tfrac18sqrt5 + tfrac38 &&approx 0.6545084971874737 &
D_x &approx 0.5401122930754872 \
D_y &= tfrac14sqrt-tfrac32sqrt5 + tfrac92
&&approx 0.2676165673298174qquad &
D_y &approx 0.2492944939101796
endalignat*
Your picture shows $C$ slightly right of $B$, so presumably the solution with $C_x>1$ is the one you want. Which is good, because the other solutions are roots of a polynomial of degree $10$ and more, with no radical expression available. This also demonstrates that a ruler and compass solution for that second part is impossible. Since going from the problem statement both solutions would be permissible, reaching either using compass and straightedge is going to be tricky.
You can use these coordinates to confirm that yes, indeed $lvert BCrvert=lvert CDrvert=lvert BDrvert$. So another approach would have been to assume this as the additional piece of information, then compute positions for everything else from that and check that the resulting coordinates satisfy all requirements. More specifically, draw $A,B,E$ then construct $D$ as intersection of two circles cooresponding to the distances to $B$ and $E$. Then draw the triangle $BCD$ and check that line $CD$ passes through $A$.
answered Jul 16 at 21:34
MvG
29.7k44597
Absolutely perfect. I will not pretend that I fully comprehend at this point, but I'm sure I will. At any rate, presuming that you are indeed correct, and this seems a reasonable assumption, I am very grateful. Thanks and best regards.
– Zacariaz
Jul 16 at 22:12
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Let's first check whether you have enough information. You need 5 points in the plane, each with 2 real degrees of freedom. But you don't care about position or rotation of the figure as a whole, which accounts for 3 real degrees of freedom. Which means you need 7 measurements to hope for full specification. You have 5 lengths, 1 angle and 1 collinearity. Could work.
You can pick coordinates and use them to express your distance and collinearity constraints.
beginalign*
A&=(0,0) & B&=(1,0) & C&=(C_x,C_y) & D&=(D_x,D_y) & E&=(tfrac12,tfrac12sqrt3)
endalign*
beginalign*
(C_x-1)^2+C_y^2 &= frac12phi^2 = left(fracphi^-1sqrt2right)^2\
(C_x-D_x)^2+(C_y-D_y)^2 &= frac12phi^2 \
(D_x-tfrac12)^2+(D_y-tfrac12sqrt3)^2 &= frac1phi^2 \
fracC_xC_y &= fracD_xD_y
endalign*
This can be rewritten as a set of polynomial equations. Then typical elimination strategies like Gröbner bases or Resultants can be used to find solutions. But I prefer to leave these things to a computer algebra system, which will tell me that there exist two possible solutions:
beginalignat*3
C_x &= tfrac14sqrt5 + tfrac12 &&approx 1.0590169943749474 &
C_x &approx 0.9369017680762171 \
C_y &= tfrac14sqrt3 &&approx 0.4330127018922193 &
C_y &approx 0.4324368378770998 \
D_x &= tfrac18sqrt5 + tfrac38 &&approx 0.6545084971874737 &
D_x &approx 0.5401122930754872 \
D_y &= tfrac14sqrt-tfrac32sqrt5 + tfrac92
&&approx 0.2676165673298174qquad &
D_y &approx 0.2492944939101796
endalignat*
Your picture shows $C$ slightly right of $B$, so presumably the solution with $C_x>1$ is the one you want. Which is good, because the other solutions are roots of a polynomial of degree $10$ and more, with no radical expression available. This also demonstrates that a ruler and compass solution for that second part is impossible. Since going from the problem statement both solutions would be permissible, reaching either using compass and straightedge is going to be tricky.
You can use these coordinates to confirm that yes, indeed $lvert BCrvert=lvert CDrvert=lvert BDrvert$. So another approach would have been to assume this as the additional piece of information, then compute positions for everything else from that and check that the resulting coordinates satisfy all requirements. More specifically, draw $A,B,E$ then construct $D$ as intersection of two circles cooresponding to the distances to $B$ and $E$. Then draw the triangle $BCD$ and check that line $CD$ passes through $A$.
answered Jul 16 at 21:34
MvG
29.7k44597
Let's first check whether you have enough information. You need 5 points in the plane, each with 2 real degrees of freedom. But you don't care about position or rotation of the figure as a whole, which accounts for 3 real degrees of freedom. Which means you need 7 measurements to hope for full specification. You have 5 lengths, 1 angle and 1 collinearity. Could work.
You can pick coordinates and use them to express your distance and collinearity constraints.
beginalign*
A&=(0,0) & B&=(1,0) & C&=(C_x,C_y) & D&=(D_x,D_y) & E&=(tfrac12,tfrac12sqrt3)
endalign*
beginalign*
(C_x-1)^2+C_y^2 &= frac12phi^2 = left(fracphi^-1sqrt2right)^2\
(C_x-D_x)^2+(C_y-D_y)^2 &= frac12phi^2 \
(D_x-tfrac12)^2+(D_y-tfrac12sqrt3)^2 &= frac1phi^2 \
fracC_xC_y &= fracD_xD_y
endalign*
This can be rewritten as a set of polynomial equations. Then typical elimination strategies like Gröbner bases or Resultants can be used to find solutions. But I prefer to leave these things to a computer algebra system, which will tell me that there exist two possible solutions:
beginalignat*3
C_x &= tfrac14sqrt5 + tfrac12 &&approx 1.0590169943749474 &
C_x &approx 0.9369017680762171 \
C_y &= tfrac14sqrt3 &&approx 0.4330127018922193 &
C_y &approx 0.4324368378770998 \
D_x &= tfrac18sqrt5 + tfrac38 &&approx 0.6545084971874737 &
D_x &approx 0.5401122930754872 \
D_y &= tfrac14sqrt-tfrac32sqrt5 + tfrac92
&&approx 0.2676165673298174qquad &
D_y &approx 0.2492944939101796
endalignat*
Your picture shows $C$ slightly right of $B$, so presumably the solution with $C_x>1$ is the one you want. Which is good, because the other solutions are roots of a polynomial of degree $10$ and more, with no radical expression available. This also demonstrates that a ruler and compass solution for that second part is impossible. Since going from the problem statement both solutions would be permissible, reaching either using compass and straightedge is going to be tricky.
You can use these coordinates to confirm that yes, indeed $lvert BCrvert=lvert CDrvert=lvert BDrvert$. So another approach would have been to assume this as the additional piece of information, then compute positions for everything else from that and check that the resulting coordinates satisfy all requirements. More specifically, draw $A,B,E$ then construct $D$ as intersection of two circles cooresponding to the distances to $B$ and $E$. Then draw the triangle $BCD$ and check that line $CD$ passes through $A$.
answered Jul 16 at 21:34
MvG
29.7k44597
edited Jul 16 at 22:04
answered Jul 16 at 21:34
MvG
29.7k44597
answered Jul 16 at 21:34
MvG
29.7k44597
answered Jul 16 at 21:34
MvG
29.7k44597
29.7k44597
Absolutely perfect. I will not pretend that I fully comprehend at this point, but I'm sure I will. At any rate, presuming that you are indeed correct, and this seems a reasonable assumption, I am very grateful. Thanks and best regards.
– Zacariaz
Jul 16 at 22:12
add a comment |Â
Absolutely perfect. I will not pretend that I fully comprehend at this point, but I'm sure I will. At any rate, presuming that you are indeed correct, and this seems a reasonable assumption, I am very grateful. Thanks and best regards.
– Zacariaz
Jul 16 at 22:12
Absolutely perfect. I will not pretend that I fully comprehend at this point, but I'm sure I will. At any rate, presuming that you are indeed correct, and this seems a reasonable assumption, I am very grateful. Thanks and best regards.
– Zacariaz
Jul 16 at 22:12
Absolutely perfect. I will not pretend that I fully comprehend at this point, but I'm sure I will. At any rate, presuming that you are indeed correct, and this seems a reasonable assumption, I am very grateful. Thanks and best regards.
– Zacariaz
Jul 16 at 22:12
add a comment |Â
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Could you give the precise values of $|DE|$ and $|BC|$? This could be crucial for solving the problem. Also, clarify, is your question about how to find the remaining information about the figure given what you specified?
– Sambo
Jul 16 at 20:37
Added exact values for the side lengths, scaled up from the cited MathWorld article. I assume there is a typo about $lvert CDrvert$ being the real unknown. Did you mean $lvert BDrvert$?
– MvG
Jul 16 at 21:58
@MvG No, I actually did mean |CD|, but I probably didn't phrase it well. What I really meant I suppose, is simply that trying to construct with compass and ruler, this particular line segment cannot be placed and it implies two solutions. But still, thanks for your assistance, I got my answer, rather more quickly than I expected I might add
– Zacariaz
Jul 16 at 22:19
@Sambo I have in fact given values for those line segments. As for clarification, I just need enough information to be able to find the information I need in order to be able to construct it, but I do believe MgVs answer has provided the complete picture.
– Zacariaz
Jul 16 at 22:24