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Post Closed as "Not suitable for this site" by abx, user44191, Timothy Chow, ofer zeitouni, Ben McKay
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Đào Thanh Oai
Đào Thanh Oai
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Please don't close this question. Because there is simple configuration with 57 vote up, and don't close. Why you vote up that question and You vote to close this question?

A problem I posed at here since 2014 but no solution:

Let $ABCD$ be a bicentric quadrilateral, $O$ is center of circle $(ABCD)$. Then Incenter of four triangles $OAB,OBC,OCD,ODA$ lie on a circle.

enter image description here

My question: Could You give a your solution for problem above.

The problem like Bradley’s conjecture. You can see Bradley’s conjecture at here and page 73, here

A problem I posed at here since 2014 but no solution:

Let $ABCD$ be a bicentric quadrilateral, $O$ is center of circle $(ABCD)$. Then Incenter of four triangles $OAB,OBC,OCD,ODA$ lie on a circle.

enter image description here

My question: Could You give a your solution for problem above.

The problem like Bradley’s conjecture. You can see Bradley’s conjecture at here and page 73, here

Please don't close this question. Because there is simple configuration with 57 vote up, and don't close. Why you vote up that question and You vote to close this question?

A problem I posed at here since 2014 but no solution:

Let $ABCD$ be a bicentric quadrilateral, $O$ is center of circle $(ABCD)$. Then Incenter of four triangles $OAB,OBC,OCD,ODA$ lie on a circle.

enter image description here

My question: Could You give a your solution for problem above.

The problem like Bradley’s conjecture. You can see Bradley’s conjecture at here and page 73, here

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Đào Thanh Oai
Đào Thanh Oai
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Four Like Bradley’s conjecture (Four incenters lie on a circle)

A problem I posed at here since 2014 but no solution:

Let $ABCD$ be a bicentric quadrilateral, $O$ is center of circle $(ABCD)$. Then Incenter of four triangles $OAB,OBC,OCD,ODA$ lie on a circle.

enter image description here

My question: Could You give a your solution for problem above.

The problem like Bradley’s conjecture. You can see Bradley’s conjecture at here and page 73, here

Four incenters lie on a circle

A problem I posed at here since 2014 but no solution:

Let $ABCD$ be a bicentric quadrilateral, $O$ is center of circle $(ABCD)$. Then Incenter of four triangles $OAB,OBC,OCD,ODA$ lie on a circle.

enter image description here

My question: Could You give a your solution for problem above.

Like Bradley’s conjecture (Four incenters lie on a circle)

A problem I posed at here since 2014 but no solution:

Let $ABCD$ be a bicentric quadrilateral, $O$ is center of circle $(ABCD)$. Then Incenter of four triangles $OAB,OBC,OCD,ODA$ lie on a circle.

enter image description here

My question: Could You give a your solution for problem above.

The problem like Bradley’s conjecture. You can see Bradley’s conjecture at here and page 73, here

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Martin Sleziak
Martin Sleziak
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Đào Thanh Oai
Đào Thanh Oai
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