Determining the value of $frac{BC}{CE}$ in a cyclic pentagon $ABCDE$.
1
$begingroup$
Let $ABCDE$ be a cyclic pentagon, where $AC=2, AD=3, BD=5, BE=1, frac{CD}{DE} = frac{10}{3}$. What is the value of $frac{BC}{CE}$?
I worked with the area of specific triangle with trigonometry. But I thought that it could arise a complex calculation. For my inability to draw the figure of the contextual problem, I was not sure whether it can give birth of paradox or not. But I am pretty sure there can be another way for solving this problem. In which way, should I go?
Please, pardon my error.
geometry contest-math circle polygons
asked Jan 21 at 7:34
Anirban NiloyAnirban Niloy
483115
$endgroup$
add a comment |
1
$begingroup$
Let $ABCDE$ be a cyclic pentagon, where $AC=2, AD=3, BD=5, BE=1, frac{CD}{DE} = frac{10}{3}$. What is the value of $frac{BC}{CE}$?
I worked with the area of specific triangle with trigonometry. But I thought that it could arise a complex calculation. For my inability to draw the figure of the contextual problem, I was not sure whether it can give birth of paradox or not. But I am pretty sure there can be another way for solving this problem. In which way, should I go?
Please, pardon my error.
geometry contest-math circle polygons
asked Jan 21 at 7:34
Anirban NiloyAnirban Niloy
483115
$endgroup$
$begingroup$
Would you please post me a relevant figure of this problem? Than I might be able to realize the fact of BC:CE being not determined since it is variable. And how I can measure the length of CE instead of its ratio?
$endgroup$
– Anirban Niloy
Jan 21 at 11:08
4
$begingroup$
I've deleted my previous comment because it was wrong. I don't think the conditions in the question can be satisfied at all. BDE is a triangle with the sides 1, 5 and DE. Therefore the length of DE lies between 5-1=4 and 5+1=6. Similarly, ACD is a triangle with sides 2, 3, and CD, and therefore CD lies between 3-2=1 and 3+2=5. These ranges do not allow CD:DE to be 10:3.
$endgroup$
– Jaap Scherphuis
Jan 21 at 12:36
$begingroup$
I agree with @JaapScherphuis. a pentagon satisfying all the conditions you gave cannot exist. Could you please verify if you made a typo while writing the data of the problem?
$endgroup$
– simonet
Feb 16 at 16:23
$begingroup$
@simonet No. The question may be wrong because that problem which I suggested appeared in Bangladesh Math olympiad. Its source is sometimes deceitful and wrong and never is corrected again. I didn't even understand that fact. Later I came to know about it.
$endgroup$
– Anirban Niloy
Feb 16 at 16:48
add a comment |
1
1
1
0
$begingroup$
Let $ABCDE$ be a cyclic pentagon, where $AC=2, AD=3, BD=5, BE=1, frac{CD}{DE} = frac{10}{3}$. What is the value of $frac{BC}{CE}$?
I worked with the area of specific triangle with trigonometry. But I thought that it could arise a complex calculation. For my inability to draw the figure of the contextual problem, I was not sure whether it can give birth of paradox or not. But I am pretty sure there can be another way for solving this problem. In which way, should I go?
Please, pardon my error.
geometry contest-math circle polygons
asked Jan 21 at 7:34
Anirban NiloyAnirban Niloy
483115
$endgroup$
Let $ABCDE$ be a cyclic pentagon, where $AC=2, AD=3, BD=5, BE=1, frac{CD}{DE} = frac{10}{3}$. What is the value of $frac{BC}{CE}$?
I worked with the area of specific triangle with trigonometry. But I thought that it could arise a complex calculation. For my inability to draw the figure of the contextual problem, I was not sure whether it can give birth of paradox or not. But I am pretty sure there can be another way for solving this problem. In which way, should I go?
Please, pardon my error.
geometry contest-math circle polygons
geometry contest-math circle polygons
asked Jan 21 at 7:34
Anirban NiloyAnirban Niloy
483115
asked Jan 21 at 7:34
Anirban NiloyAnirban Niloy
483115
edited Feb 16 at 4:33
Anirban Niloy
asked Jan 21 at 7:34
Anirban NiloyAnirban Niloy
483115
asked Jan 21 at 7:34
Anirban NiloyAnirban Niloy
483115
asked Jan 21 at 7:34
Anirban NiloyAnirban Niloy
483115
483115
$begingroup$
Would you please post me a relevant figure of this problem? Than I might be able to realize the fact of BC:CE being not determined since it is variable. And how I can measure the length of CE instead of its ratio?
$endgroup$
– Anirban Niloy
Jan 21 at 11:08
4
$begingroup$
I've deleted my previous comment because it was wrong. I don't think the conditions in the question can be satisfied at all. BDE is a triangle with the sides 1, 5 and DE. Therefore the length of DE lies between 5-1=4 and 5+1=6. Similarly, ACD is a triangle with sides 2, 3, and CD, and therefore CD lies between 3-2=1 and 3+2=5. These ranges do not allow CD:DE to be 10:3.
$endgroup$
– Jaap Scherphuis
Jan 21 at 12:36
$begingroup$
I agree with @JaapScherphuis. a pentagon satisfying all the conditions you gave cannot exist. Could you please verify if you made a typo while writing the data of the problem?
$endgroup$
– simonet
Feb 16 at 16:23
$begingroup$
@simonet No. The question may be wrong because that problem which I suggested appeared in Bangladesh Math olympiad. Its source is sometimes deceitful and wrong and never is corrected again. I didn't even understand that fact. Later I came to know about it.
$endgroup$
– Anirban Niloy
Feb 16 at 16:48
add a comment |
$begingroup$
Would you please post me a relevant figure of this problem? Than I might be able to realize the fact of BC:CE being not determined since it is variable. And how I can measure the length of CE instead of its ratio?
$endgroup$
– Anirban Niloy
Jan 21 at 11:08
4
$begingroup$
I've deleted my previous comment because it was wrong. I don't think the conditions in the question can be satisfied at all. BDE is a triangle with the sides 1, 5 and DE. Therefore the length of DE lies between 5-1=4 and 5+1=6. Similarly, ACD is a triangle with sides 2, 3, and CD, and therefore CD lies between 3-2=1 and 3+2=5. These ranges do not allow CD:DE to be 10:3.
$endgroup$
– Jaap Scherphuis
Jan 21 at 12:36
$begingroup$
I agree with @JaapScherphuis. a pentagon satisfying all the conditions you gave cannot exist. Could you please verify if you made a typo while writing the data of the problem?
$endgroup$
– simonet
Feb 16 at 16:23
$begingroup$
@simonet No. The question may be wrong because that problem which I suggested appeared in Bangladesh Math olympiad. Its source is sometimes deceitful and wrong and never is corrected again. I didn't even understand that fact. Later I came to know about it.
$endgroup$
– Anirban Niloy
Feb 16 at 16:48
$begingroup$
Would you please post me a relevant figure of this problem? Than I might be able to realize the fact of BC:CE being not determined since it is variable. And how I can measure the length of CE instead of its ratio?
$endgroup$
– Anirban Niloy
Jan 21 at 11:08
$begingroup$
Would you please post me a relevant figure of this problem? Than I might be able to realize the fact of BC:CE being not determined since it is variable. And how I can measure the length of CE instead of its ratio?
$endgroup$
– Anirban Niloy
Jan 21 at 11:08
4
4
$begingroup$
I've deleted my previous comment because it was wrong. I don't think the conditions in the question can be satisfied at all. BDE is a triangle with the sides 1, 5 and DE. Therefore the length of DE lies between 5-1=4 and 5+1=6. Similarly, ACD is a triangle with sides 2, 3, and CD, and therefore CD lies between 3-2=1 and 3+2=5. These ranges do not allow CD:DE to be 10:3.
$endgroup$
– Jaap Scherphuis
Jan 21 at 12:36
$begingroup$
I've deleted my previous comment because it was wrong. I don't think the conditions in the question can be satisfied at all. BDE is a triangle with the sides 1, 5 and DE. Therefore the length of DE lies between 5-1=4 and 5+1=6. Similarly, ACD is a triangle with sides 2, 3, and CD, and therefore CD lies between 3-2=1 and 3+2=5. These ranges do not allow CD:DE to be 10:3.
$endgroup$
– Jaap Scherphuis
Jan 21 at 12:36
$begingroup$
I agree with @JaapScherphuis. a pentagon satisfying all the conditions you gave cannot exist. Could you please verify if you made a typo while writing the data of the problem?
$endgroup$
– simonet
Feb 16 at 16:23
$begingroup$
I agree with @JaapScherphuis. a pentagon satisfying all the conditions you gave cannot exist. Could you please verify if you made a typo while writing the data of the problem?
$endgroup$
– simonet
Feb 16 at 16:23
$begingroup$
@simonet No. The question may be wrong because that problem which I suggested appeared in Bangladesh Math olympiad. Its source is sometimes deceitful and wrong and never is corrected again. I didn't even understand that fact. Later I came to know about it.
$endgroup$
– Anirban Niloy
Feb 16 at 16:48
$begingroup$
@simonet No. The question may be wrong because that problem which I suggested appeared in Bangladesh Math olympiad. Its source is sometimes deceitful and wrong and never is corrected again. I didn't even understand that fact. Later I came to know about it.
$endgroup$
– Anirban Niloy
Feb 16 at 16:48
add a comment |
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$begingroup$
Would you please post me a relevant figure of this problem? Than I might be able to realize the fact of BC:CE being not determined since it is variable. And how I can measure the length of CE instead of its ratio?
$endgroup$
– Anirban Niloy
Jan 21 at 11:08
4
$begingroup$
I've deleted my previous comment because it was wrong. I don't think the conditions in the question can be satisfied at all. BDE is a triangle with the sides 1, 5 and DE. Therefore the length of DE lies between 5-1=4 and 5+1=6. Similarly, ACD is a triangle with sides 2, 3, and CD, and therefore CD lies between 3-2=1 and 3+2=5. These ranges do not allow CD:DE to be 10:3.
$endgroup$
– Jaap Scherphuis
Jan 21 at 12:36
$begingroup$
I agree with @JaapScherphuis. a pentagon satisfying all the conditions you gave cannot exist. Could you please verify if you made a typo while writing the data of the problem?
$endgroup$
– simonet
Feb 16 at 16:23
$begingroup$
@simonet No. The question may be wrong because that problem which I suggested appeared in Bangladesh Math olympiad. Its source is sometimes deceitful and wrong and never is corrected again. I didn't even understand that fact. Later I came to know about it.
$endgroup$
– Anirban Niloy
Feb 16 at 16:48