Conjecture: If $A^prime$ is outside the circumcircle of $triangle ABC$, then $triangle A^prime BC$ has a...
While solving a few problems, I came across a property of triangle. It looks simple, but I am not able to prove it. Simply stating it:
Conjecture. Given $triangle ABC$ and its circumcircle, consider a point $A^prime$ outside the circumcircle of $triangle ABC$. The circumradius of $triangle A^prime BC$ is greater than the circumradius of $triangle ABC$.
Note: I am not sure whether the statement is completely correct or not. It's just an observation.
geometry
edited Dec 13 '18 at 13:41
Blue
47.7k870151
asked Dec 13 '18 at 10:48
saisanjeevsaisanjeev
947212
add a comment |
While solving a few problems, I came across a property of triangle. It looks simple, but I am not able to prove it. Simply stating it:
Conjecture. Given $triangle ABC$ and its circumcircle, consider a point $A^prime$ outside the circumcircle of $triangle ABC$. The circumradius of $triangle A^prime BC$ is greater than the circumradius of $triangle ABC$.
Note: I am not sure whether the statement is completely correct or not. It's just an observation.
geometry
edited Dec 13 '18 at 13:41
Blue
47.7k870151
asked Dec 13 '18 at 10:48
saisanjeevsaisanjeev
947212
Without additional conditions, the conjecture is false. In particular, it's possible that $A^prime$ is outside $bigcirc ABC$ while also $A$ is outside $bigcirc A^prime BC$, with distinct circumradii, so one radius is greater, but one is smaller.
– Blue
Dec 13 '18 at 14:06
What might be the additional conditions?
– saisanjeev
Dec 15 '18 at 6:02
Just to overcome the problem I noted, you need to require $A$ to be inside $bigcirc A^prime BC$. Whether that's enough to guarantee the conjecture ... I haven't really thought about it.
– Blue
Dec 15 '18 at 8:52
1
If your intent is to guarantee a larger circle, you could use this: On the side of $overleftrightarrow{BC}$ containing the major arc $stackrel{frown}{BC}$, point $A^prime$ must lie outside $bigcirc ABC$; on the side containing the minor arc, $A^prime$ must lie inside. (If the case of semicircles, $A^prime$ can lie inside or outside.) With this formulation, it doesn't matter where $A$ is relative to $bigcirc A^prime BC$. (Note that $A^prime$ on $overleftrightarrow{BC}$ determines a line, which you may want to consider not-a-circle or else a circle of infinite radius.)
– Blue
Dec 15 '18 at 9:24
add a comment |
While solving a few problems, I came across a property of triangle. It looks simple, but I am not able to prove it. Simply stating it:
Conjecture. Given $triangle ABC$ and its circumcircle, consider a point $A^prime$ outside the circumcircle of $triangle ABC$. The circumradius of $triangle A^prime BC$ is greater than the circumradius of $triangle ABC$.
Note: I am not sure whether the statement is completely correct or not. It's just an observation.
geometry
edited Dec 13 '18 at 13:41
Blue
47.7k870151
asked Dec 13 '18 at 10:48
saisanjeevsaisanjeev
947212
While solving a few problems, I came across a property of triangle. It looks simple, but I am not able to prove it. Simply stating it:
Conjecture. Given $triangle ABC$ and its circumcircle, consider a point $A^prime$ outside the circumcircle of $triangle ABC$. The circumradius of $triangle A^prime BC$ is greater than the circumradius of $triangle ABC$.
Note: I am not sure whether the statement is completely correct or not. It's just an observation.
geometry
geometry
edited Dec 13 '18 at 13:41
Blue
47.7k870151
asked Dec 13 '18 at 10:48
saisanjeevsaisanjeev
947212
edited Dec 13 '18 at 13:41
Blue
47.7k870151
asked Dec 13 '18 at 10:48
saisanjeevsaisanjeev
947212
edited Dec 13 '18 at 13:41
Blue
47.7k870151
edited Dec 13 '18 at 13:41
Blue
47.7k870151
edited Dec 13 '18 at 13:41
Blue
47.7k870151
47.7k870151
asked Dec 13 '18 at 10:48
saisanjeevsaisanjeev
947212
asked Dec 13 '18 at 10:48
saisanjeevsaisanjeev
947212
asked Dec 13 '18 at 10:48
saisanjeevsaisanjeev
947212
947212
Without additional conditions, the conjecture is false. In particular, it's possible that $A^prime$ is outside $bigcirc ABC$ while also $A$ is outside $bigcirc A^prime BC$, with distinct circumradii, so one radius is greater, but one is smaller.
– Blue
Dec 13 '18 at 14:06
What might be the additional conditions?
– saisanjeev
Dec 15 '18 at 6:02
Just to overcome the problem I noted, you need to require $A$ to be inside $bigcirc A^prime BC$. Whether that's enough to guarantee the conjecture ... I haven't really thought about it.
– Blue
Dec 15 '18 at 8:52
1
If your intent is to guarantee a larger circle, you could use this: On the side of $overleftrightarrow{BC}$ containing the major arc $stackrel{frown}{BC}$, point $A^prime$ must lie outside $bigcirc ABC$; on the side containing the minor arc, $A^prime$ must lie inside. (If the case of semicircles, $A^prime$ can lie inside or outside.) With this formulation, it doesn't matter where $A$ is relative to $bigcirc A^prime BC$. (Note that $A^prime$ on $overleftrightarrow{BC}$ determines a line, which you may want to consider not-a-circle or else a circle of infinite radius.)
– Blue
Dec 15 '18 at 9:24
add a comment |
Without additional conditions, the conjecture is false. In particular, it's possible that $A^prime$ is outside $bigcirc ABC$ while also $A$ is outside $bigcirc A^prime BC$, with distinct circumradii, so one radius is greater, but one is smaller.
– Blue
Dec 13 '18 at 14:06
What might be the additional conditions?
– saisanjeev
Dec 15 '18 at 6:02
Just to overcome the problem I noted, you need to require $A$ to be inside $bigcirc A^prime BC$. Whether that's enough to guarantee the conjecture ... I haven't really thought about it.
– Blue
Dec 15 '18 at 8:52
1
If your intent is to guarantee a larger circle, you could use this: On the side of $overleftrightarrow{BC}$ containing the major arc $stackrel{frown}{BC}$, point $A^prime$ must lie outside $bigcirc ABC$; on the side containing the minor arc, $A^prime$ must lie inside. (If the case of semicircles, $A^prime$ can lie inside or outside.) With this formulation, it doesn't matter where $A$ is relative to $bigcirc A^prime BC$. (Note that $A^prime$ on $overleftrightarrow{BC}$ determines a line, which you may want to consider not-a-circle or else a circle of infinite radius.)
– Blue
Dec 15 '18 at 9:24
Without additional conditions, the conjecture is false. In particular, it's possible that $A^prime$ is outside $bigcirc ABC$ while also $A$ is outside $bigcirc A^prime BC$, with distinct circumradii, so one radius is greater, but one is smaller.
– Blue
Dec 13 '18 at 14:06
Without additional conditions, the conjecture is false. In particular, it's possible that $A^prime$ is outside $bigcirc ABC$ while also $A$ is outside $bigcirc A^prime BC$, with distinct circumradii, so one radius is greater, but one is smaller.
– Blue
Dec 13 '18 at 14:06
What might be the additional conditions?
– saisanjeev
Dec 15 '18 at 6:02
What might be the additional conditions?
– saisanjeev
Dec 15 '18 at 6:02
Just to overcome the problem I noted, you need to require $A$ to be inside $bigcirc A^prime BC$. Whether that's enough to guarantee the conjecture ... I haven't really thought about it.
– Blue
Dec 15 '18 at 8:52
Just to overcome the problem I noted, you need to require $A$ to be inside $bigcirc A^prime BC$. Whether that's enough to guarantee the conjecture ... I haven't really thought about it.
– Blue
Dec 15 '18 at 8:52
1
1
If your intent is to guarantee a larger circle, you could use this: On the side of $overleftrightarrow{BC}$ containing the major arc $stackrel{frown}{BC}$, point $A^prime$ must lie outside $bigcirc ABC$; on the side containing the minor arc, $A^prime$ must lie inside. (If the case of semicircles, $A^prime$ can lie inside or outside.) With this formulation, it doesn't matter where $A$ is relative to $bigcirc A^prime BC$. (Note that $A^prime$ on $overleftrightarrow{BC}$ determines a line, which you may want to consider not-a-circle or else a circle of infinite radius.)
– Blue
Dec 15 '18 at 9:24
If your intent is to guarantee a larger circle, you could use this: On the side of $overleftrightarrow{BC}$ containing the major arc $stackrel{frown}{BC}$, point $A^prime$ must lie outside $bigcirc ABC$; on the side containing the minor arc, $A^prime$ must lie inside. (If the case of semicircles, $A^prime$ can lie inside or outside.) With this formulation, it doesn't matter where $A$ is relative to $bigcirc A^prime BC$. (Note that $A^prime$ on $overleftrightarrow{BC}$ determines a line, which you may want to consider not-a-circle or else a circle of infinite radius.)
– Blue
Dec 15 '18 at 9:24
add a comment |
1 Answer
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It's not true. Consider a triangle $ABC$ with an obtuse angle at $A$, and let $D$ be a point further back from $A$ so that $angle CDB$ is right. Since $angle CDB < angle CAB$, $D$ is outside the circle that contains $A,B,C$. Then the circumcircle of triangle $DBC$ is smaller than that of $ABC$.
Why? Because $BC$ is a diameter of the circumcircle of $DBC$, but not of the circumcircle of $ABC$.
Now, if that were an acute triangle we started with...
answered Dec 13 '18 at 11:35
jmerryjmerry
2,894312
What can happen if it an acute angled triangle?
– saisanjeev
Dec 15 '18 at 6:02
Add the condition that it's an acute triangle, and the modified conjecture is true. Hint: Law of Sines.
– jmerry
Dec 15 '18 at 6:06
yes i got it thanks. Thanks
– saisanjeev
Dec 15 '18 at 6:09
Any idea how to extend it to obtuse angled triangles?
– saisanjeev
Dec 15 '18 at 7:54
add a comment |
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It's not true. Consider a triangle $ABC$ with an obtuse angle at $A$, and let $D$ be a point further back from $A$ so that $angle CDB$ is right. Since $angle CDB < angle CAB$, $D$ is outside the circle that contains $A,B,C$. Then the circumcircle of triangle $DBC$ is smaller than that of $ABC$.
Why? Because $BC$ is a diameter of the circumcircle of $DBC$, but not of the circumcircle of $ABC$.
Now, if that were an acute triangle we started with...
answered Dec 13 '18 at 11:35
jmerryjmerry
2,894312
What can happen if it an acute angled triangle?
– saisanjeev
Dec 15 '18 at 6:02
Add the condition that it's an acute triangle, and the modified conjecture is true. Hint: Law of Sines.
– jmerry
Dec 15 '18 at 6:06
yes i got it thanks. Thanks
– saisanjeev
Dec 15 '18 at 6:09
Any idea how to extend it to obtuse angled triangles?
– saisanjeev
Dec 15 '18 at 7:54
add a comment |
It's not true. Consider a triangle $ABC$ with an obtuse angle at $A$, and let $D$ be a point further back from $A$ so that $angle CDB$ is right. Since $angle CDB < angle CAB$, $D$ is outside the circle that contains $A,B,C$. Then the circumcircle of triangle $DBC$ is smaller than that of $ABC$.
Why? Because $BC$ is a diameter of the circumcircle of $DBC$, but not of the circumcircle of $ABC$.
Now, if that were an acute triangle we started with...
answered Dec 13 '18 at 11:35
jmerryjmerry
2,894312
What can happen if it an acute angled triangle?
– saisanjeev
Dec 15 '18 at 6:02
Add the condition that it's an acute triangle, and the modified conjecture is true. Hint: Law of Sines.
– jmerry
Dec 15 '18 at 6:06
yes i got it thanks. Thanks
– saisanjeev
Dec 15 '18 at 6:09
Any idea how to extend it to obtuse angled triangles?
– saisanjeev
Dec 15 '18 at 7:54
add a comment |
It's not true. Consider a triangle $ABC$ with an obtuse angle at $A$, and let $D$ be a point further back from $A$ so that $angle CDB$ is right. Since $angle CDB < angle CAB$, $D$ is outside the circle that contains $A,B,C$. Then the circumcircle of triangle $DBC$ is smaller than that of $ABC$.
Why? Because $BC$ is a diameter of the circumcircle of $DBC$, but not of the circumcircle of $ABC$.
Now, if that were an acute triangle we started with...
answered Dec 13 '18 at 11:35
jmerryjmerry
2,894312
It's not true. Consider a triangle $ABC$ with an obtuse angle at $A$, and let $D$ be a point further back from $A$ so that $angle CDB$ is right. Since $angle CDB < angle CAB$, $D$ is outside the circle that contains $A,B,C$. Then the circumcircle of triangle $DBC$ is smaller than that of $ABC$.
Why? Because $BC$ is a diameter of the circumcircle of $DBC$, but not of the circumcircle of $ABC$.
Now, if that were an acute triangle we started with...
answered Dec 13 '18 at 11:35
jmerryjmerry
2,894312
answered Dec 13 '18 at 11:35
jmerryjmerry
2,894312
answered Dec 13 '18 at 11:35
jmerryjmerry
2,894312
answered Dec 13 '18 at 11:35
jmerryjmerry
2,894312
2,894312
What can happen if it an acute angled triangle?
– saisanjeev
Dec 15 '18 at 6:02
Add the condition that it's an acute triangle, and the modified conjecture is true. Hint: Law of Sines.
– jmerry
Dec 15 '18 at 6:06
yes i got it thanks. Thanks
– saisanjeev
Dec 15 '18 at 6:09
Any idea how to extend it to obtuse angled triangles?
– saisanjeev
Dec 15 '18 at 7:54
add a comment |
What can happen if it an acute angled triangle?
– saisanjeev
Dec 15 '18 at 6:02
Add the condition that it's an acute triangle, and the modified conjecture is true. Hint: Law of Sines.
– jmerry
Dec 15 '18 at 6:06
yes i got it thanks. Thanks
– saisanjeev
Dec 15 '18 at 6:09
Any idea how to extend it to obtuse angled triangles?
– saisanjeev
Dec 15 '18 at 7:54
What can happen if it an acute angled triangle?
– saisanjeev
Dec 15 '18 at 6:02
What can happen if it an acute angled triangle?
– saisanjeev
Dec 15 '18 at 6:02
Add the condition that it's an acute triangle, and the modified conjecture is true. Hint: Law of Sines.
– jmerry
Dec 15 '18 at 6:06
Add the condition that it's an acute triangle, and the modified conjecture is true. Hint: Law of Sines.
– jmerry
Dec 15 '18 at 6:06
yes i got it thanks. Thanks
– saisanjeev
Dec 15 '18 at 6:09
yes i got it thanks. Thanks
– saisanjeev
Dec 15 '18 at 6:09
Any idea how to extend it to obtuse angled triangles?
– saisanjeev
Dec 15 '18 at 7:54
Any idea how to extend it to obtuse angled triangles?
– saisanjeev
Dec 15 '18 at 7:54
add a comment |
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Without additional conditions, the conjecture is false. In particular, it's possible that $A^prime$ is outside $bigcirc ABC$ while also $A$ is outside $bigcirc A^prime BC$, with distinct circumradii, so one radius is greater, but one is smaller.
– Blue
Dec 13 '18 at 14:06
What might be the additional conditions?
– saisanjeev
Dec 15 '18 at 6:02
Just to overcome the problem I noted, you need to require $A$ to be inside $bigcirc A^prime BC$. Whether that's enough to guarantee the conjecture ... I haven't really thought about it.
– Blue
Dec 15 '18 at 8:52
1
If your intent is to guarantee a larger circle, you could use this: On the side of $overleftrightarrow{BC}$ containing the major arc $stackrel{frown}{BC}$, point $A^prime$ must lie outside $bigcirc ABC$; on the side containing the minor arc, $A^prime$ must lie inside. (If the case of semicircles, $A^prime$ can lie inside or outside.) With this formulation, it doesn't matter where $A$ is relative to $bigcirc A^prime BC$. (Note that $A^prime$ on $overleftrightarrow{BC}$ determines a line, which you may want to consider not-a-circle or else a circle of infinite radius.)
– Blue
Dec 15 '18 at 9:24