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Prove $fracBCCA cdot fracAEEF cdot fracFDDB = 1$ in Convex hexagon
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Let $ABCDEF$ be a convex hexagon such that $angle B+angle D+angle F=360^circ $and
$fracABBC cdot fracCDDE cdot fracEFFA = 1$ .
Prove that $fracBCCA cdot fracAEEF cdot fracFDDB = 1$.
I know it is an old but nice question it belongs to IMO shortlist in 1998.you can find out some proof in below link but unfortunately there is no synthetic solution in the link
https://artofproblemsolving.com/community/c6h1117p3488
I think we should consider this fact that the circumcircles of triangles $ABC, CDE, EFA$ have a common point $P$.
Because, if $P$ is the intersection of circumcircles $CDE, EFA$ , then:
$angle EPC=180-angle D$
$angle APE=180-angle F$
then: $angle APC=360-(angle D+angle F)=angle B$
and it means that $P$ is on circumcircle of $ABC$ ,too.
Please post synthetic answers. Thanks!
geometry differential-geometry analytic-geometry plane-geometry
asked Jul 24 at 14:30
math enthusiastic
479112
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up vote
4
down vote
favorite
Let $ABCDEF$ be a convex hexagon such that $angle B+angle D+angle F=360^circ $and
$fracABBC cdot fracCDDE cdot fracEFFA = 1$ .
Prove that $fracBCCA cdot fracAEEF cdot fracFDDB = 1$.
I know it is an old but nice question it belongs to IMO shortlist in 1998.you can find out some proof in below link but unfortunately there is no synthetic solution in the link
https://artofproblemsolving.com/community/c6h1117p3488
I think we should consider this fact that the circumcircles of triangles $ABC, CDE, EFA$ have a common point $P$.
Because, if $P$ is the intersection of circumcircles $CDE, EFA$ , then:
$angle EPC=180-angle D$
$angle APE=180-angle F$
then: $angle APC=360-(angle D+angle F)=angle B$
and it means that $P$ is on circumcircle of $ABC$ ,too.
Please post synthetic answers. Thanks!
geometry differential-geometry analytic-geometry plane-geometry
asked Jul 24 at 14:30
math enthusiastic
479112
2
What's wrong with answer #8? It looks fairly complete to me, assuming you understand basic principles of inversion. I mean I can copy it here and get some reputation for free: artofproblemsolving.com/community/c6h1117p533560
– Oldboy
Jul 24 at 14:55
1
@Oldboy it is a nice problem I think it should have some more beautiful solutions I didn't post it just to have a solution there are a lot of solutions but not synthetic. I want math StackExchange users to think about it and discuss it. I am sure it worth. thanks!
– math enthusiastic
Jul 26 at 7:59
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
Let $ABCDEF$ be a convex hexagon such that $angle B+angle D+angle F=360^circ $and
$fracABBC cdot fracCDDE cdot fracEFFA = 1$ .
Prove that $fracBCCA cdot fracAEEF cdot fracFDDB = 1$.
I know it is an old but nice question it belongs to IMO shortlist in 1998.you can find out some proof in below link but unfortunately there is no synthetic solution in the link
https://artofproblemsolving.com/community/c6h1117p3488
I think we should consider this fact that the circumcircles of triangles $ABC, CDE, EFA$ have a common point $P$.
Because, if $P$ is the intersection of circumcircles $CDE, EFA$ , then:
$angle EPC=180-angle D$
$angle APE=180-angle F$
then: $angle APC=360-(angle D+angle F)=angle B$
and it means that $P$ is on circumcircle of $ABC$ ,too.
Please post synthetic answers. Thanks!
geometry differential-geometry analytic-geometry plane-geometry
asked Jul 24 at 14:30
math enthusiastic
479112
Let $ABCDEF$ be a convex hexagon such that $angle B+angle D+angle F=360^circ $and
$fracABBC cdot fracCDDE cdot fracEFFA = 1$ .
Prove that $fracBCCA cdot fracAEEF cdot fracFDDB = 1$.
I know it is an old but nice question it belongs to IMO shortlist in 1998.you can find out some proof in below link but unfortunately there is no synthetic solution in the link
https://artofproblemsolving.com/community/c6h1117p3488
I think we should consider this fact that the circumcircles of triangles $ABC, CDE, EFA$ have a common point $P$.
Because, if $P$ is the intersection of circumcircles $CDE, EFA$ , then:
$angle EPC=180-angle D$
$angle APE=180-angle F$
then: $angle APC=360-(angle D+angle F)=angle B$
and it means that $P$ is on circumcircle of $ABC$ ,too.
Please post synthetic answers. Thanks!
geometry differential-geometry analytic-geometry plane-geometry
asked Jul 24 at 14:30
math enthusiastic
479112
edited Jul 26 at 8:08
asked Jul 24 at 14:30
math enthusiastic
479112
asked Jul 24 at 14:30
math enthusiastic
479112
asked Jul 24 at 14:30
math enthusiastic
479112
479112
2
What's wrong with answer #8? It looks fairly complete to me, assuming you understand basic principles of inversion. I mean I can copy it here and get some reputation for free: artofproblemsolving.com/community/c6h1117p533560
– Oldboy
Jul 24 at 14:55
1
@Oldboy it is a nice problem I think it should have some more beautiful solutions I didn't post it just to have a solution there are a lot of solutions but not synthetic. I want math StackExchange users to think about it and discuss it. I am sure it worth. thanks!
– math enthusiastic
Jul 26 at 7:59
add a comment |Â
2
What's wrong with answer #8? It looks fairly complete to me, assuming you understand basic principles of inversion. I mean I can copy it here and get some reputation for free: artofproblemsolving.com/community/c6h1117p533560
– Oldboy
Jul 24 at 14:55
1
@Oldboy it is a nice problem I think it should have some more beautiful solutions I didn't post it just to have a solution there are a lot of solutions but not synthetic. I want math StackExchange users to think about it and discuss it. I am sure it worth. thanks!
– math enthusiastic
Jul 26 at 7:59
2
2
What's wrong with answer #8? It looks fairly complete to me, assuming you understand basic principles of inversion. I mean I can copy it here and get some reputation for free: artofproblemsolving.com/community/c6h1117p533560
– Oldboy
Jul 24 at 14:55
What's wrong with answer #8? It looks fairly complete to me, assuming you understand basic principles of inversion. I mean I can copy it here and get some reputation for free: artofproblemsolving.com/community/c6h1117p533560
– Oldboy
Jul 24 at 14:55
1
1
@Oldboy it is a nice problem I think it should have some more beautiful solutions I didn't post it just to have a solution there are a lot of solutions but not synthetic. I want math StackExchange users to think about it and discuss it. I am sure it worth. thanks!
– math enthusiastic
Jul 26 at 7:59
@Oldboy it is a nice problem I think it should have some more beautiful solutions I didn't post it just to have a solution there are a lot of solutions but not synthetic. I want math StackExchange users to think about it and discuss it. I am sure it worth. thanks!
– math enthusiastic
Jul 26 at 7:59
add a comment |Â
1 Answer
1
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1
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accepted
OFFICIAL SOLUTION:
Take P so that AFE and PDE are similar. [So take ∠PDE = ∠AFE and ∠PED = ∠AEF. Note that you need to take the right point - P' fails below.] So EF/ED = EA/EP. Also ∠DEF = ∠DEA + ∠AEF = ∠DEA + ∠PED = ∠PEA (this is where we need P on the opposite side of DE to A). Hence triangles DEF and PEA are similar. So FD/EF = AP/EA (*).
Triangles AFE, PDE similar also gives FA/EF = DP/DE. So AB/BC = (DE/CD) (FA/EF) (given) = (DE/CD) (DP/DE) = PD/DC. But ∠CDE + ∠EDP + ∠PDC = 360o, or ∠D + ∠F + ∠PDC = 360o. We are given that ∠B + ∠D + ∠F = 360o, so ∠PDC = ∠ABC. Hence triangles ABC and PDC are similar. So CB/CD = CA/CP. ∠BCD = ∠ACD + ∠BCA = ∠ACD + ∠DCP = ∠ACP. So triangles BCD and ACP are similar. Hence BC/BD = AC/AP. Multiplying by (*) gives (BC/BD) (FD/EF) = AC/EA, which is the required equality
answered Jul 26 at 7:35
math enthusiastic
479112
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
OFFICIAL SOLUTION:
Take P so that AFE and PDE are similar. [So take ∠PDE = ∠AFE and ∠PED = ∠AEF. Note that you need to take the right point - P' fails below.] So EF/ED = EA/EP. Also ∠DEF = ∠DEA + ∠AEF = ∠DEA + ∠PED = ∠PEA (this is where we need P on the opposite side of DE to A). Hence triangles DEF and PEA are similar. So FD/EF = AP/EA (*).
Triangles AFE, PDE similar also gives FA/EF = DP/DE. So AB/BC = (DE/CD) (FA/EF) (given) = (DE/CD) (DP/DE) = PD/DC. But ∠CDE + ∠EDP + ∠PDC = 360o, or ∠D + ∠F + ∠PDC = 360o. We are given that ∠B + ∠D + ∠F = 360o, so ∠PDC = ∠ABC. Hence triangles ABC and PDC are similar. So CB/CD = CA/CP. ∠BCD = ∠ACD + ∠BCA = ∠ACD + ∠DCP = ∠ACP. So triangles BCD and ACP are similar. Hence BC/BD = AC/AP. Multiplying by (*) gives (BC/BD) (FD/EF) = AC/EA, which is the required equality
answered Jul 26 at 7:35
math enthusiastic
479112
add a comment |Â
up vote
1
down vote
accepted
OFFICIAL SOLUTION:
Take P so that AFE and PDE are similar. [So take ∠PDE = ∠AFE and ∠PED = ∠AEF. Note that you need to take the right point - P' fails below.] So EF/ED = EA/EP. Also ∠DEF = ∠DEA + ∠AEF = ∠DEA + ∠PED = ∠PEA (this is where we need P on the opposite side of DE to A). Hence triangles DEF and PEA are similar. So FD/EF = AP/EA (*).
Triangles AFE, PDE similar also gives FA/EF = DP/DE. So AB/BC = (DE/CD) (FA/EF) (given) = (DE/CD) (DP/DE) = PD/DC. But ∠CDE + ∠EDP + ∠PDC = 360o, or ∠D + ∠F + ∠PDC = 360o. We are given that ∠B + ∠D + ∠F = 360o, so ∠PDC = ∠ABC. Hence triangles ABC and PDC are similar. So CB/CD = CA/CP. ∠BCD = ∠ACD + ∠BCA = ∠ACD + ∠DCP = ∠ACP. So triangles BCD and ACP are similar. Hence BC/BD = AC/AP. Multiplying by (*) gives (BC/BD) (FD/EF) = AC/EA, which is the required equality
answered Jul 26 at 7:35
math enthusiastic
479112
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
OFFICIAL SOLUTION:
Take P so that AFE and PDE are similar. [So take ∠PDE = ∠AFE and ∠PED = ∠AEF. Note that you need to take the right point - P' fails below.] So EF/ED = EA/EP. Also ∠DEF = ∠DEA + ∠AEF = ∠DEA + ∠PED = ∠PEA (this is where we need P on the opposite side of DE to A). Hence triangles DEF and PEA are similar. So FD/EF = AP/EA (*).
Triangles AFE, PDE similar also gives FA/EF = DP/DE. So AB/BC = (DE/CD) (FA/EF) (given) = (DE/CD) (DP/DE) = PD/DC. But ∠CDE + ∠EDP + ∠PDC = 360o, or ∠D + ∠F + ∠PDC = 360o. We are given that ∠B + ∠D + ∠F = 360o, so ∠PDC = ∠ABC. Hence triangles ABC and PDC are similar. So CB/CD = CA/CP. ∠BCD = ∠ACD + ∠BCA = ∠ACD + ∠DCP = ∠ACP. So triangles BCD and ACP are similar. Hence BC/BD = AC/AP. Multiplying by (*) gives (BC/BD) (FD/EF) = AC/EA, which is the required equality
answered Jul 26 at 7:35
math enthusiastic
479112
OFFICIAL SOLUTION:
Take P so that AFE and PDE are similar. [So take ∠PDE = ∠AFE and ∠PED = ∠AEF. Note that you need to take the right point - P' fails below.] So EF/ED = EA/EP. Also ∠DEF = ∠DEA + ∠AEF = ∠DEA + ∠PED = ∠PEA (this is where we need P on the opposite side of DE to A). Hence triangles DEF and PEA are similar. So FD/EF = AP/EA (*).
Triangles AFE, PDE similar also gives FA/EF = DP/DE. So AB/BC = (DE/CD) (FA/EF) (given) = (DE/CD) (DP/DE) = PD/DC. But ∠CDE + ∠EDP + ∠PDC = 360o, or ∠D + ∠F + ∠PDC = 360o. We are given that ∠B + ∠D + ∠F = 360o, so ∠PDC = ∠ABC. Hence triangles ABC and PDC are similar. So CB/CD = CA/CP. ∠BCD = ∠ACD + ∠BCA = ∠ACD + ∠DCP = ∠ACP. So triangles BCD and ACP are similar. Hence BC/BD = AC/AP. Multiplying by (*) gives (BC/BD) (FD/EF) = AC/EA, which is the required equality
answered Jul 26 at 7:35
math enthusiastic
479112
answered Jul 26 at 7:35
math enthusiastic
479112
answered Jul 26 at 7:35
math enthusiastic
479112
answered Jul 26 at 7:35
math enthusiastic
479112
479112
add a comment |Â
add a comment |Â
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2
What's wrong with answer #8? It looks fairly complete to me, assuming you understand basic principles of inversion. I mean I can copy it here and get some reputation for free: artofproblemsolving.com/community/c6h1117p533560
– Oldboy
Jul 24 at 14:55
1
@Oldboy it is a nice problem I think it should have some more beautiful solutions I didn't post it just to have a solution there are a lot of solutions but not synthetic. I want math StackExchange users to think about it and discuss it. I am sure it worth. thanks!
– math enthusiastic
Jul 26 at 7:59