Prove that $Q$, $S$ and $T$ lie on the same line.
$begingroup$
Let $(C)$ be a circle , and $ABCD$ be a quadrilateral inscribed in $(C)$
Let $P$ be the intersection of $(AD)$ and $(BC)$. And $Q$ the intersection of $(AB)$ and $(CD)$
Let $S$ and $T$ be points in $(C)$ such that $(PS)$ and $(PT)$ are tangents to $(C)$
this problem can be done by Lahire theorem or projective geometry or polarisation.
But I wanna if there is a simple solution by angle chasing or radical axis
here is what I think we should do :
$$widehat{OSP}=180-widehat{OTP}=90$$
so OSPT is cyclic
so the problem reduces itself to prove that Q lies on the radical axis of $(C)$ and the circle where $OSPT$ is inscribed.
N.B: This problem is taken from a preparation test for IMO 2020 in Morocco.
geometry contest-math
asked Dec 21 '18 at 18:01
user600785user600785
12011
$endgroup$
add a comment |
$begingroup$
Let $(C)$ be a circle , and $ABCD$ be a quadrilateral inscribed in $(C)$
Let $P$ be the intersection of $(AD)$ and $(BC)$. And $Q$ the intersection of $(AB)$ and $(CD)$
Let $S$ and $T$ be points in $(C)$ such that $(PS)$ and $(PT)$ are tangents to $(C)$
this problem can be done by Lahire theorem or projective geometry or polarisation.
But I wanna if there is a simple solution by angle chasing or radical axis
here is what I think we should do :
$$widehat{OSP}=180-widehat{OTP}=90$$
so OSPT is cyclic
so the problem reduces itself to prove that Q lies on the radical axis of $(C)$ and the circle where $OSPT$ is inscribed.
N.B: This problem is taken from a preparation test for IMO 2020 in Morocco.
geometry contest-math
asked Dec 21 '18 at 18:01
user600785user600785
12011
$endgroup$
$begingroup$
Is the solution based on poles and polar lines interesting. I have such one.
$endgroup$
– Oldboy
Jan 4 at 15:57
$begingroup$
I aleady have a solution based on poles nd polar lines , but I'am looking for a an easy solution based in angle chasing or radical axis
$endgroup$
– user600785
Jan 5 at 7:01
$begingroup$
I don not have a solution based in angle chasing,but it is not based on pole and polar lines as well, you interested?
$endgroup$
– StAKmod
Feb 18 at 0:53
$begingroup$
@StAKmod lemme see and thanks
$endgroup$
– user600785
Feb 23 at 13:14
add a comment |
$begingroup$
Let $(C)$ be a circle , and $ABCD$ be a quadrilateral inscribed in $(C)$
Let $P$ be the intersection of $(AD)$ and $(BC)$. And $Q$ the intersection of $(AB)$ and $(CD)$
Let $S$ and $T$ be points in $(C)$ such that $(PS)$ and $(PT)$ are tangents to $(C)$
this problem can be done by Lahire theorem or projective geometry or polarisation.
But I wanna if there is a simple solution by angle chasing or radical axis
here is what I think we should do :
$$widehat{OSP}=180-widehat{OTP}=90$$
so OSPT is cyclic
so the problem reduces itself to prove that Q lies on the radical axis of $(C)$ and the circle where $OSPT$ is inscribed.
N.B: This problem is taken from a preparation test for IMO 2020 in Morocco.
geometry contest-math
asked Dec 21 '18 at 18:01
user600785user600785
12011
$endgroup$
Let $(C)$ be a circle , and $ABCD$ be a quadrilateral inscribed in $(C)$
Let $P$ be the intersection of $(AD)$ and $(BC)$. And $Q$ the intersection of $(AB)$ and $(CD)$
Let $S$ and $T$ be points in $(C)$ such that $(PS)$ and $(PT)$ are tangents to $(C)$
this problem can be done by Lahire theorem or projective geometry or polarisation.
But I wanna if there is a simple solution by angle chasing or radical axis
here is what I think we should do :
$$widehat{OSP}=180-widehat{OTP}=90$$
so OSPT is cyclic
so the problem reduces itself to prove that Q lies on the radical axis of $(C)$ and the circle where $OSPT$ is inscribed.
N.B: This problem is taken from a preparation test for IMO 2020 in Morocco.
geometry contest-math
geometry contest-math
asked Dec 21 '18 at 18:01
user600785user600785
12011
asked Dec 21 '18 at 18:01
user600785user600785
12011
asked Dec 21 '18 at 18:01
user600785user600785
12011
asked Dec 21 '18 at 18:01
user600785user600785
12011
asked Dec 21 '18 at 18:01
user600785user600785
12011
12011
$begingroup$
Is the solution based on poles and polar lines interesting. I have such one.
$endgroup$
– Oldboy
Jan 4 at 15:57
$begingroup$
I aleady have a solution based on poles nd polar lines , but I'am looking for a an easy solution based in angle chasing or radical axis
$endgroup$
– user600785
Jan 5 at 7:01
$begingroup$
I don not have a solution based in angle chasing,but it is not based on pole and polar lines as well, you interested?
$endgroup$
– StAKmod
Feb 18 at 0:53
$begingroup$
@StAKmod lemme see and thanks
$endgroup$
– user600785
Feb 23 at 13:14
add a comment |
$begingroup$
Is the solution based on poles and polar lines interesting. I have such one.
$endgroup$
– Oldboy
Jan 4 at 15:57
$begingroup$
I aleady have a solution based on poles nd polar lines , but I'am looking for a an easy solution based in angle chasing or radical axis
$endgroup$
– user600785
Jan 5 at 7:01
$begingroup$
I don not have a solution based in angle chasing,but it is not based on pole and polar lines as well, you interested?
$endgroup$
– StAKmod
Feb 18 at 0:53
$begingroup$
@StAKmod lemme see and thanks
$endgroup$
– user600785
Feb 23 at 13:14
$begingroup$
Is the solution based on poles and polar lines interesting. I have such one.
$endgroup$
– Oldboy
Jan 4 at 15:57
$begingroup$
Is the solution based on poles and polar lines interesting. I have such one.
$endgroup$
– Oldboy
Jan 4 at 15:57
$begingroup$
I aleady have a solution based on poles nd polar lines , but I'am looking for a an easy solution based in angle chasing or radical axis
$endgroup$
– user600785
Jan 5 at 7:01
$begingroup$
I aleady have a solution based on poles nd polar lines , but I'am looking for a an easy solution based in angle chasing or radical axis
$endgroup$
– user600785
Jan 5 at 7:01
$begingroup$
I don not have a solution based in angle chasing,but it is not based on pole and polar lines as well, you interested?
$endgroup$
– StAKmod
Feb 18 at 0:53
$begingroup$
I don not have a solution based in angle chasing,but it is not based on pole and polar lines as well, you interested?
$endgroup$
– StAKmod
Feb 18 at 0:53
$begingroup$
@StAKmod lemme see and thanks
$endgroup$
– user600785
Feb 23 at 13:14
$begingroup$
@StAKmod lemme see and thanks
$endgroup$
– user600785
Feb 23 at 13:14
add a comment |
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$begingroup$
Is the solution based on poles and polar lines interesting. I have such one.
$endgroup$
– Oldboy
Jan 4 at 15:57
$begingroup$
I aleady have a solution based on poles nd polar lines , but I'am looking for a an easy solution based in angle chasing or radical axis
$endgroup$
– user600785
Jan 5 at 7:01
$begingroup$
I don not have a solution based in angle chasing,but it is not based on pole and polar lines as well, you interested?
$endgroup$
– StAKmod
Feb 18 at 0:53
$begingroup$
@StAKmod lemme see and thanks
$endgroup$
– user600785
Feb 23 at 13:14