Let $dfrac{AB}{OB}=dfrac{AC}{PO}=2$ and also $AB=AC$ then prove that $dfrac{OP}{MO}=dfrac{1+sqrt{5}}{2}$












2












$begingroup$


enter image description here




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







share|cite|improve this question











$endgroup$








  • 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
















2












$begingroup$


enter image description here




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







share|cite|improve this question











$endgroup$








  • 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














2












2








2


1



$begingroup$


enter image description here




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







share|cite|improve this question











$endgroup$




enter image description here




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 31 '18 at 19:42







Almot1960

















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














  • 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










2 Answers
2






active

oldest

votes


















1












$begingroup$

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$$ q^2-q-1=0implies q={1pm sqrt{5}over2}$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    How come you are accepting only few questions. Are they not good?
    $endgroup$
    – Maria Mazur
    Jan 8 at 18:35



















1












$begingroup$

$frac{AB}{OB}geq1$ 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.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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












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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

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$$ q^2-q-1=0implies q={1pm sqrt{5}over2}$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    How come you are accepting only few questions. Are they not good?
    $endgroup$
    – Maria Mazur
    Jan 8 at 18:35
















1












$begingroup$

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$$ q^2-q-1=0implies q={1pm sqrt{5}over2}$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    How come you are accepting only few questions. Are they not good?
    $endgroup$
    – Maria Mazur
    Jan 8 at 18:35














1












1








1





$begingroup$

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$$ q^2-q-1=0implies q={1pm sqrt{5}over2}$$






share|cite|improve this answer









$endgroup$



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$$ q^2-q-1=0implies q={1pm sqrt{5}over2}$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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


















  • $begingroup$
    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











1












$begingroup$

$frac{AB}{OB}geq1$ 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.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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
















1












$begingroup$

$frac{AB}{OB}geq1$ 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.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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














1












1








1





$begingroup$

$frac{AB}{OB}geq1$ 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.






share|cite|improve this answer











$endgroup$



$frac{AB}{OB}geq1$ 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.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 31 '18 at 19:52

























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


















  • $begingroup$
    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


















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