Prove that $Q$, $S$ and $T$ lie on the same line.












3












$begingroup$


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 $(C)$



enter image description here



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







share|cite|improve this question









$endgroup$












  • $begingroup$
    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
















3












$begingroup$


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 $(C)$



enter image description here



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







share|cite|improve this question









$endgroup$












  • $begingroup$
    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














3












3








3





$begingroup$


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 $(C)$



enter image description here



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







share|cite|improve this question









$endgroup$




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 $(C)$



enter image description here



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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










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


















  • $begingroup$
    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










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