Find that there exists no relationship between the difference of two angles and a point [on hold] The Next CEO of Stack OverflowReflection of orthocenter about side midpoints lies on circumcircleBisector EqualityGiven the circumcircle, the 9-point circle, and the angular measures for a triangle, construct the triangle?Prove that the intersection of $BM$ and $CN$ is on the circumcircle of triangle $ABC.$show that two angles in a circumcircle are equalGeometry question on triangleProving the points $P,O,N$ are collinearWhat is the missing angle in the isosceles triangle?Some geometry - I have found a paralelogram$AH=AS$ where $H$ is the orthocenter of $triangle ABC$ and $S$ is the midpoint of the arc $BHC$ of the circumcircle of $triangle BHC$
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Find that there exists no relationship between the difference of two angles and a point [on hold]
The Next CEO of Stack OverflowReflection of orthocenter about side midpoints lies on circumcircleBisector EqualityGiven the circumcircle, the 9-point circle, and the angular measures for a triangle, construct the triangle?Prove that the intersection of $BM$ and $CN$ is on the circumcircle of triangle $ABC.$show that two angles in a circumcircle are equalGeometry question on triangleProving the points $P,O,N$ are collinearWhat is the missing angle in the isosceles triangle?Some geometry - I have found a paralelogram$AH=AS$ where $H$ is the orthocenter of $triangle ABC$ and $S$ is the midpoint of the arc $BHC$ of the circumcircle of $triangle BHC$
$begingroup$
If $M$ is the midpoint of $BC$ of the isosceles triangle $Delta ABC$, where $AC=AB$. Let $P$ be a point in the minor arc of $MA$ from the circumcircle of triangle $Delta ABM$. Let $Q$ be a point in the angle $angle BMA$ so $angle QMP=90°$ and $QP=BP$. Demonstrate that $angle MQB - angle CQM$ don't depend on $P$.
geometry contest-math
asked Mar 28 at 9:26
AJMC2002AJMC2002
324
New contributor
AJMC2002 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
put on hold as off-topic by Travis, YiFan, José Carlos Santos, Thomas Shelby, Strants Mar 28 at 18:55
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Travis, YiFan, José Carlos Santos, Thomas Shelby, Strants
add a comment |
$begingroup$
If $M$ is the midpoint of $BC$ of the isosceles triangle $Delta ABC$, where $AC=AB$. Let $P$ be a point in the minor arc of $MA$ from the circumcircle of triangle $Delta ABM$. Let $Q$ be a point in the angle $angle BMA$ so $angle QMP=90°$ and $QP=BP$. Demonstrate that $angle MQB - angle CQM$ don't depend on $P$.
geometry contest-math
asked Mar 28 at 9:26
AJMC2002AJMC2002
324
New contributor
AJMC2002 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
put on hold as off-topic by Travis, YiFan, José Carlos Santos, Thomas Shelby, Strants Mar 28 at 18:55
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Travis, YiFan, José Carlos Santos, Thomas Shelby, Strants
$begingroup$
"Let $Q$ be a point in the angle $angle BMA$...". Could you clarify what this means? I may just be dumb...
$endgroup$
– Milten
Mar 28 at 10:28
$begingroup$
It means that Q belongs in the region of the angle BMA.
$endgroup$
– AJMC2002
Mar 28 at 17:40
$begingroup$
Could you provide a picture? That will make the problem more attractive. At least I would try to solve it now, but without a picture..., it's too much work.
$endgroup$
– Dr. Mathva
Mar 28 at 18:06
add a comment |
$begingroup$
If $M$ is the midpoint of $BC$ of the isosceles triangle $Delta ABC$, where $AC=AB$. Let $P$ be a point in the minor arc of $MA$ from the circumcircle of triangle $Delta ABM$. Let $Q$ be a point in the angle $angle BMA$ so $angle QMP=90°$ and $QP=BP$. Demonstrate that $angle MQB - angle CQM$ don't depend on $P$.
geometry contest-math
asked Mar 28 at 9:26
AJMC2002AJMC2002
324
New contributor
AJMC2002 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
If $M$ is the midpoint of $BC$ of the isosceles triangle $Delta ABC$, where $AC=AB$. Let $P$ be a point in the minor arc of $MA$ from the circumcircle of triangle $Delta ABM$. Let $Q$ be a point in the angle $angle BMA$ so $angle QMP=90°$ and $QP=BP$. Demonstrate that $angle MQB - angle CQM$ don't depend on $P$.
geometry contest-math
geometry contest-math
asked Mar 28 at 9:26
AJMC2002AJMC2002
324
New contributor
AJMC2002 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Mar 28 at 9:26
AJMC2002AJMC2002
324
New contributor
AJMC2002 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Mar 28 at 9:26
AJMC2002AJMC2002
324
New contributor
AJMC2002 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Mar 28 at 9:26
AJMC2002AJMC2002
324
asked Mar 28 at 9:26
AJMC2002AJMC2002
324
324
New contributor
AJMC2002 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
AJMC2002 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
AJMC2002 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
put on hold as off-topic by Travis, YiFan, José Carlos Santos, Thomas Shelby, Strants Mar 28 at 18:55
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Travis, YiFan, José Carlos Santos, Thomas Shelby, Strants
put on hold as off-topic by Travis, YiFan, José Carlos Santos, Thomas Shelby, Strants Mar 28 at 18:55
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Travis, YiFan, José Carlos Santos, Thomas Shelby, Strants
$begingroup$
"Let $Q$ be a point in the angle $angle BMA$...". Could you clarify what this means? I may just be dumb...
$endgroup$
– Milten
Mar 28 at 10:28
$begingroup$
It means that Q belongs in the region of the angle BMA.
$endgroup$
– AJMC2002
Mar 28 at 17:40
$begingroup$
Could you provide a picture? That will make the problem more attractive. At least I would try to solve it now, but without a picture..., it's too much work.
$endgroup$
– Dr. Mathva
Mar 28 at 18:06
add a comment |
$begingroup$
"Let $Q$ be a point in the angle $angle BMA$...". Could you clarify what this means? I may just be dumb...
$endgroup$
– Milten
Mar 28 at 10:28
$begingroup$
It means that Q belongs in the region of the angle BMA.
$endgroup$
– AJMC2002
Mar 28 at 17:40
$begingroup$
Could you provide a picture? That will make the problem more attractive. At least I would try to solve it now, but without a picture..., it's too much work.
$endgroup$
– Dr. Mathva
Mar 28 at 18:06
$begingroup$
"Let $Q$ be a point in the angle $angle BMA$...". Could you clarify what this means? I may just be dumb...
$endgroup$
– Milten
Mar 28 at 10:28
$begingroup$
"Let $Q$ be a point in the angle $angle BMA$...". Could you clarify what this means? I may just be dumb...
$endgroup$
– Milten
Mar 28 at 10:28
$begingroup$
It means that Q belongs in the region of the angle BMA.
$endgroup$
– AJMC2002
Mar 28 at 17:40
$begingroup$
It means that Q belongs in the region of the angle BMA.
$endgroup$
– AJMC2002
Mar 28 at 17:40
$begingroup$
Could you provide a picture? That will make the problem more attractive. At least I would try to solve it now, but without a picture..., it's too much work.
$endgroup$
– Dr. Mathva
Mar 28 at 18:06
$begingroup$
Could you provide a picture? That will make the problem more attractive. At least I would try to solve it now, but without a picture..., it's too much work.
$endgroup$
– Dr. Mathva
Mar 28 at 18:06
add a comment |
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$begingroup$
"Let $Q$ be a point in the angle $angle BMA$...". Could you clarify what this means? I may just be dumb...
$endgroup$
– Milten
Mar 28 at 10:28
$begingroup$
It means that Q belongs in the region of the angle BMA.
$endgroup$
– AJMC2002
Mar 28 at 17:40
$begingroup$
Could you provide a picture? That will make the problem more attractive. At least I would try to solve it now, but without a picture..., it's too much work.
$endgroup$
– Dr. Mathva
Mar 28 at 18:06