Let $dfrac{AB}{OB}=dfrac{AC}{PO}=2$ and also $AB=AC$ then prove that $dfrac{OP}{MO}=dfrac{1+sqrt{5}}{2}$
$begingroup$
Let $dfrac{AB}{OB}=dfrac{AC}{PO}=2$ and also $AB=AC$
then prove that $dfrac{OP}{MO}=dfrac{1+sqrt{5}}{2}$
My Try :
I know that by Thales's theorem we have $dfrac{AB}{OB}=dfrac{AC}{PO}=dfrac{OP}{BC}=2$
and by $AB=AC$ we have $AB=AC=BC=2$ and $BO=PC=AP=AO =OP=1$
Now what ?
geometry euclidean-geometry
asked Dec 31 '18 at 18:47
Almot1960Almot1960
2,318825
$endgroup$
add a comment |
$begingroup$
Let $dfrac{AB}{OB}=dfrac{AC}{PO}=2$ and also $AB=AC$
then prove that $dfrac{OP}{MO}=dfrac{1+sqrt{5}}{2}$
My Try :
I know that by Thales's theorem we have $dfrac{AB}{OB}=dfrac{AC}{PO}=dfrac{OP}{BC}=2$
and by $AB=AC$ we have $AB=AC=BC=2$ and $BO=PC=AP=AO =OP=1$
Now what ?
geometry euclidean-geometry
asked Dec 31 '18 at 18:47
Almot1960Almot1960
2,318825
$endgroup$
1
$begingroup$
The equation $dfrac{AB}{OB}=dfrac{AC}{PO}=dfrac{1}{2}$ doesn't match with the figure.
$endgroup$
– mouthetics
Dec 31 '18 at 18:57
$begingroup$
The figure looks like you are trying to make a 5--point star which would have the proportions of $phi$ but your original equation does not make sense if it relates to the drawing.
$endgroup$
– poetasis
Dec 31 '18 at 19:24
add a comment |
$begingroup$
Let $dfrac{AB}{OB}=dfrac{AC}{PO}=2$ and also $AB=AC$
then prove that $dfrac{OP}{MO}=dfrac{1+sqrt{5}}{2}$
My Try :
I know that by Thales's theorem we have $dfrac{AB}{OB}=dfrac{AC}{PO}=dfrac{OP}{BC}=2$
and by $AB=AC$ we have $AB=AC=BC=2$ and $BO=PC=AP=AO =OP=1$
Now what ?
geometry euclidean-geometry
asked Dec 31 '18 at 18:47
Almot1960Almot1960
2,318825
$endgroup$
Let $dfrac{AB}{OB}=dfrac{AC}{PO}=2$ and also $AB=AC$
then prove that $dfrac{OP}{MO}=dfrac{1+sqrt{5}}{2}$
My Try :
I know that by Thales's theorem we have $dfrac{AB}{OB}=dfrac{AC}{PO}=dfrac{OP}{BC}=2$
and by $AB=AC$ we have $AB=AC=BC=2$ and $BO=PC=AP=AO =OP=1$
Now what ?
geometry euclidean-geometry
geometry euclidean-geometry
asked Dec 31 '18 at 18:47
Almot1960Almot1960
2,318825
asked Dec 31 '18 at 18:47
Almot1960Almot1960
2,318825
edited Dec 31 '18 at 19:42
Almot1960
asked Dec 31 '18 at 18:47
Almot1960Almot1960
2,318825
asked Dec 31 '18 at 18:47
Almot1960Almot1960
2,318825
asked Dec 31 '18 at 18:47
Almot1960Almot1960
2,318825
2,318825
1
$begingroup$
The equation $dfrac{AB}{OB}=dfrac{AC}{PO}=dfrac{1}{2}$ doesn't match with the figure.
$endgroup$
– mouthetics
Dec 31 '18 at 18:57
$begingroup$
The figure looks like you are trying to make a 5--point star which would have the proportions of $phi$ but your original equation does not make sense if it relates to the drawing.
$endgroup$
– poetasis
Dec 31 '18 at 19:24
add a comment |
1
$begingroup$
The equation $dfrac{AB}{OB}=dfrac{AC}{PO}=dfrac{1}{2}$ doesn't match with the figure.
$endgroup$
– mouthetics
Dec 31 '18 at 18:57
$begingroup$
The figure looks like you are trying to make a 5--point star which would have the proportions of $phi$ but your original equation does not make sense if it relates to the drawing.
$endgroup$
– poetasis
Dec 31 '18 at 19:24
1
1
$begingroup$
The equation $dfrac{AB}{OB}=dfrac{AC}{PO}=dfrac{1}{2}$ doesn't match with the figure.
$endgroup$
– mouthetics
Dec 31 '18 at 18:57
$begingroup$
The equation $dfrac{AB}{OB}=dfrac{AC}{PO}=dfrac{1}{2}$ doesn't match with the figure.
$endgroup$
– mouthetics
Dec 31 '18 at 18:57
$begingroup$
The figure looks like you are trying to make a 5--point star which would have the proportions of $phi$ but your original equation does not make sense if it relates to the drawing.
$endgroup$
– poetasis
Dec 31 '18 at 19:24
$begingroup$
The figure looks like you are trying to make a 5--point star which would have the proportions of $phi$ but your original equation does not make sense if it relates to the drawing.
$endgroup$
– poetasis
Dec 31 '18 at 19:24
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Let $AB = s$, $PN = a$, $OP = b$.
Use the power of the point $P$ with respect to the circle:
$$ (a+b)cdot a = ({sover 2})^2$$
Note that since $APO$ is equlateral we have $s=2b$. Now write $q= b/a$ and we get$$ q^2-q-1=0implies q={1pm sqrt{5}over2}$$
answered Dec 31 '18 at 18:56
Maria MazurMaria Mazur
50.3k1361126
$endgroup$
$begingroup$
How come you are accepting only few questions. Are they not good?
$endgroup$
– Maria Mazur
Jan 8 at 18:35
add a comment |
$begingroup$
$frac{AB}{OB}geq1$ but you say it is $frac{1}{2}$.
You applied Thales' theorem incorrectly. $dfrac{AB}{OB}=dfrac{AC}{PO}$ does not guarantee that $OPparallel BC$. I think it will be $dfrac{AB}{OB}=dfrac{AC}{PC}$.
I know this is a comment but I do not have the required reputation.
answered Dec 31 '18 at 18:57
Nilkantha GhosalNilkantha Ghosal
416
$endgroup$
$begingroup$
Your answer will probably get deleted. As you noted yourself, this should be a comment. The OP is probably still thankful for your help. Regarding that you are aware of this, the comment by @amWhy is a little bit out of place.
$endgroup$
– klirk
Dec 31 '18 at 19:25
$begingroup$
This does not provide an answer to the question. Once you have sufficient reputation you will be able to comment on any post; instead, provide answers that don't require clarification from the asker. - From Review
$endgroup$
– klirk
Dec 31 '18 at 19:25
$begingroup$
@klirk, I know how to solve the question. As soon as the OP corrects his question, I will edit my "answer" to include the correct answer. I am surprised as to why so many people are concerned about my answer and no one is concerned about the fact that the question in its current form makes no sense.
$endgroup$
– Nilkantha Ghosal
Dec 31 '18 at 19:34
$begingroup$
I edited ......
$endgroup$
– Almot1960
Dec 31 '18 at 19:42
$begingroup$
@Almot1960, still it got problems
$endgroup$
– Nilkantha Ghosal
Dec 31 '18 at 19:53
|
show 2 more comments
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let $AB = s$, $PN = a$, $OP = b$.
Use the power of the point $P$ with respect to the circle:
$$ (a+b)cdot a = ({sover 2})^2$$
Note that since $APO$ is equlateral we have $s=2b$. Now write $q= b/a$ and we get$$ q^2-q-1=0implies q={1pm sqrt{5}over2}$$
answered Dec 31 '18 at 18:56
Maria MazurMaria Mazur
50.3k1361126
$endgroup$
$begingroup$
How come you are accepting only few questions. Are they not good?
$endgroup$
– Maria Mazur
Jan 8 at 18:35
add a comment |
$begingroup$
Let $AB = s$, $PN = a$, $OP = b$.
Use the power of the point $P$ with respect to the circle:
$$ (a+b)cdot a = ({sover 2})^2$$
Note that since $APO$ is equlateral we have $s=2b$. Now write $q= b/a$ and we get$$ q^2-q-1=0implies q={1pm sqrt{5}over2}$$
answered Dec 31 '18 at 18:56
Maria MazurMaria Mazur
50.3k1361126
$endgroup$
$begingroup$
How come you are accepting only few questions. Are they not good?
$endgroup$
– Maria Mazur
Jan 8 at 18:35
add a comment |
$begingroup$
Let $AB = s$, $PN = a$, $OP = b$.
Use the power of the point $P$ with respect to the circle:
$$ (a+b)cdot a = ({sover 2})^2$$
Note that since $APO$ is equlateral we have $s=2b$. Now write $q= b/a$ and we get$$ q^2-q-1=0implies q={1pm sqrt{5}over2}$$
answered Dec 31 '18 at 18:56
Maria MazurMaria Mazur
50.3k1361126
$endgroup$
Let $AB = s$, $PN = a$, $OP = b$.
Use the power of the point $P$ with respect to the circle:
$$ (a+b)cdot a = ({sover 2})^2$$
Note that since $APO$ is equlateral we have $s=2b$. Now write $q= b/a$ and we get$$ q^2-q-1=0implies q={1pm sqrt{5}over2}$$
answered Dec 31 '18 at 18:56
Maria MazurMaria Mazur
50.3k1361126
answered Dec 31 '18 at 18:56
Maria MazurMaria Mazur
50.3k1361126
answered Dec 31 '18 at 18:56
Maria MazurMaria Mazur
50.3k1361126
answered Dec 31 '18 at 18:56
Maria MazurMaria Mazur
50.3k1361126
50.3k1361126
$begingroup$
How come you are accepting only few questions. Are they not good?
$endgroup$
– Maria Mazur
Jan 8 at 18:35
add a comment |
$begingroup$
How come you are accepting only few questions. Are they not good?
$endgroup$
– Maria Mazur
Jan 8 at 18:35
$begingroup$
How come you are accepting only few questions. Are they not good?
$endgroup$
– Maria Mazur
Jan 8 at 18:35
$begingroup$
How come you are accepting only few questions. Are they not good?
$endgroup$
– Maria Mazur
Jan 8 at 18:35
add a comment |
$begingroup$
$frac{AB}{OB}geq1$ but you say it is $frac{1}{2}$.
You applied Thales' theorem incorrectly. $dfrac{AB}{OB}=dfrac{AC}{PO}$ does not guarantee that $OPparallel BC$. I think it will be $dfrac{AB}{OB}=dfrac{AC}{PC}$.
I know this is a comment but I do not have the required reputation.
answered Dec 31 '18 at 18:57
Nilkantha GhosalNilkantha Ghosal
416
$endgroup$
$begingroup$
Your answer will probably get deleted. As you noted yourself, this should be a comment. The OP is probably still thankful for your help. Regarding that you are aware of this, the comment by @amWhy is a little bit out of place.
$endgroup$
– klirk
Dec 31 '18 at 19:25
$begingroup$
This does not provide an answer to the question. Once you have sufficient reputation you will be able to comment on any post; instead, provide answers that don't require clarification from the asker. - From Review
$endgroup$
– klirk
Dec 31 '18 at 19:25
$begingroup$
@klirk, I know how to solve the question. As soon as the OP corrects his question, I will edit my "answer" to include the correct answer. I am surprised as to why so many people are concerned about my answer and no one is concerned about the fact that the question in its current form makes no sense.
$endgroup$
– Nilkantha Ghosal
Dec 31 '18 at 19:34
$begingroup$
I edited ......
$endgroup$
– Almot1960
Dec 31 '18 at 19:42
$begingroup$
@Almot1960, still it got problems
$endgroup$
– Nilkantha Ghosal
Dec 31 '18 at 19:53
|
show 2 more comments
$begingroup$
$frac{AB}{OB}geq1$ but you say it is $frac{1}{2}$.
You applied Thales' theorem incorrectly. $dfrac{AB}{OB}=dfrac{AC}{PO}$ does not guarantee that $OPparallel BC$. I think it will be $dfrac{AB}{OB}=dfrac{AC}{PC}$.
I know this is a comment but I do not have the required reputation.
answered Dec 31 '18 at 18:57
Nilkantha GhosalNilkantha Ghosal
416
$endgroup$
$begingroup$
Your answer will probably get deleted. As you noted yourself, this should be a comment. The OP is probably still thankful for your help. Regarding that you are aware of this, the comment by @amWhy is a little bit out of place.
$endgroup$
– klirk
Dec 31 '18 at 19:25
$begingroup$
This does not provide an answer to the question. Once you have sufficient reputation you will be able to comment on any post; instead, provide answers that don't require clarification from the asker. - From Review
$endgroup$
– klirk
Dec 31 '18 at 19:25
$begingroup$
@klirk, I know how to solve the question. As soon as the OP corrects his question, I will edit my "answer" to include the correct answer. I am surprised as to why so many people are concerned about my answer and no one is concerned about the fact that the question in its current form makes no sense.
$endgroup$
– Nilkantha Ghosal
Dec 31 '18 at 19:34
$begingroup$
I edited ......
$endgroup$
– Almot1960
Dec 31 '18 at 19:42
$begingroup$
@Almot1960, still it got problems
$endgroup$
– Nilkantha Ghosal
Dec 31 '18 at 19:53
|
show 2 more comments
$begingroup$
$frac{AB}{OB}geq1$ but you say it is $frac{1}{2}$.
You applied Thales' theorem incorrectly. $dfrac{AB}{OB}=dfrac{AC}{PO}$ does not guarantee that $OPparallel BC$. I think it will be $dfrac{AB}{OB}=dfrac{AC}{PC}$.
I know this is a comment but I do not have the required reputation.
answered Dec 31 '18 at 18:57
Nilkantha GhosalNilkantha Ghosal
416
$endgroup$
$frac{AB}{OB}geq1$ but you say it is $frac{1}{2}$.
You applied Thales' theorem incorrectly. $dfrac{AB}{OB}=dfrac{AC}{PO}$ does not guarantee that $OPparallel BC$. I think it will be $dfrac{AB}{OB}=dfrac{AC}{PC}$.
I know this is a comment but I do not have the required reputation.
answered Dec 31 '18 at 18:57
Nilkantha GhosalNilkantha Ghosal
416
edited Dec 31 '18 at 19:52
answered Dec 31 '18 at 18:57
Nilkantha GhosalNilkantha Ghosal
416
answered Dec 31 '18 at 18:57
Nilkantha GhosalNilkantha Ghosal
416
answered Dec 31 '18 at 18:57
Nilkantha GhosalNilkantha Ghosal
416
416
$begingroup$
Your answer will probably get deleted. As you noted yourself, this should be a comment. The OP is probably still thankful for your help. Regarding that you are aware of this, the comment by @amWhy is a little bit out of place.
$endgroup$
– klirk
Dec 31 '18 at 19:25
$begingroup$
This does not provide an answer to the question. Once you have sufficient reputation you will be able to comment on any post; instead, provide answers that don't require clarification from the asker. - From Review
$endgroup$
– klirk
Dec 31 '18 at 19:25
$begingroup$
@klirk, I know how to solve the question. As soon as the OP corrects his question, I will edit my "answer" to include the correct answer. I am surprised as to why so many people are concerned about my answer and no one is concerned about the fact that the question in its current form makes no sense.
$endgroup$
– Nilkantha Ghosal
Dec 31 '18 at 19:34
$begingroup$
I edited ......
$endgroup$
– Almot1960
Dec 31 '18 at 19:42
$begingroup$
@Almot1960, still it got problems
$endgroup$
– Nilkantha Ghosal
Dec 31 '18 at 19:53
|
show 2 more comments
$begingroup$
Your answer will probably get deleted. As you noted yourself, this should be a comment. The OP is probably still thankful for your help. Regarding that you are aware of this, the comment by @amWhy is a little bit out of place.
$endgroup$
– klirk
Dec 31 '18 at 19:25
$begingroup$
This does not provide an answer to the question. Once you have sufficient reputation you will be able to comment on any post; instead, provide answers that don't require clarification from the asker. - From Review
$endgroup$
– klirk
Dec 31 '18 at 19:25
$begingroup$
@klirk, I know how to solve the question. As soon as the OP corrects his question, I will edit my "answer" to include the correct answer. I am surprised as to why so many people are concerned about my answer and no one is concerned about the fact that the question in its current form makes no sense.
$endgroup$
– Nilkantha Ghosal
Dec 31 '18 at 19:34
$begingroup$
I edited ......
$endgroup$
– Almot1960
Dec 31 '18 at 19:42
$begingroup$
@Almot1960, still it got problems
$endgroup$
– Nilkantha Ghosal
Dec 31 '18 at 19:53
$begingroup$
Your answer will probably get deleted. As you noted yourself, this should be a comment. The OP is probably still thankful for your help. Regarding that you are aware of this, the comment by @amWhy is a little bit out of place.
$endgroup$
– klirk
Dec 31 '18 at 19:25
$begingroup$
Your answer will probably get deleted. As you noted yourself, this should be a comment. The OP is probably still thankful for your help. Regarding that you are aware of this, the comment by @amWhy is a little bit out of place.
$endgroup$
– klirk
Dec 31 '18 at 19:25
$begingroup$
This does not provide an answer to the question. Once you have sufficient reputation you will be able to comment on any post; instead, provide answers that don't require clarification from the asker. - From Review
$endgroup$
– klirk
Dec 31 '18 at 19:25
$begingroup$
This does not provide an answer to the question. Once you have sufficient reputation you will be able to comment on any post; instead, provide answers that don't require clarification from the asker. - From Review
$endgroup$
– klirk
Dec 31 '18 at 19:25
$begingroup$
@klirk, I know how to solve the question. As soon as the OP corrects his question, I will edit my "answer" to include the correct answer. I am surprised as to why so many people are concerned about my answer and no one is concerned about the fact that the question in its current form makes no sense.
$endgroup$
– Nilkantha Ghosal
Dec 31 '18 at 19:34
$begingroup$
@klirk, I know how to solve the question. As soon as the OP corrects his question, I will edit my "answer" to include the correct answer. I am surprised as to why so many people are concerned about my answer and no one is concerned about the fact that the question in its current form makes no sense.
$endgroup$
– Nilkantha Ghosal
Dec 31 '18 at 19:34
$begingroup$
I edited ......
$endgroup$
– Almot1960
Dec 31 '18 at 19:42
$begingroup$
I edited ......
$endgroup$
– Almot1960
Dec 31 '18 at 19:42
$begingroup$
@Almot1960, still it got problems
$endgroup$
– Nilkantha Ghosal
Dec 31 '18 at 19:53
$begingroup$
@Almot1960, still it got problems
$endgroup$
– Nilkantha Ghosal
Dec 31 '18 at 19:53
|
show 2 more comments
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$begingroup$
The equation $dfrac{AB}{OB}=dfrac{AC}{PO}=dfrac{1}{2}$ doesn't match with the figure.
$endgroup$
– mouthetics
Dec 31 '18 at 18:57
$begingroup$
The figure looks like you are trying to make a 5--point star which would have the proportions of $phi$ but your original equation does not make sense if it relates to the drawing.
$endgroup$
– poetasis
Dec 31 '18 at 19:24