Show that every point in the interior of one circle is the orthocentre of another triangle inscribed in...












0












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Let $C_1$ and $C_2$ be two circles in the plane with radius R and 3R respectively. Show that every point in the interior of $C_2$ is the orthocentre of some triangle inscribed in $C_1$.
enter image description here
I gave a construction as follows. Take any point, call it H in the interior of $C_2$. Join OH, it intersects the circle $C_1$ at two points say $A$ and $X$ with $A$ being nearer to $H$. Construct perpendicular bisector of $AX$. Let it intersect $C_1$ at $B$ and $C$. I tell that $ABC$ is the required triangle.



If I assume $H$ to be the orthocentre then all the properties are matching. However, I am unable to prove that the above construction will guarantee that H will be the orthocentre of triangle ABC.



Any help will be appreciated. Thanks in advance










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$endgroup$








  • 1




    $begingroup$
    This sound highly unlikely to be true.
    $endgroup$
    – Maria Mazur
    Dec 24 '18 at 8:55










  • $begingroup$
    any better ideas for such a construction. I would also like to know how to prove it wrong
    $endgroup$
    – saisanjeev
    Dec 24 '18 at 8:56










  • $begingroup$
    I was talking about the problem.
    $endgroup$
    – Maria Mazur
    Dec 24 '18 at 8:56










  • $begingroup$
    oh. any way we can find such a point.
    $endgroup$
    – saisanjeev
    Dec 24 '18 at 12:03










  • $begingroup$
    This can't be true, please read carefully your post again.
    $endgroup$
    – Maria Mazur
    Dec 24 '18 at 12:11
















0












$begingroup$


Let $C_1$ and $C_2$ be two circles in the plane with radius R and 3R respectively. Show that every point in the interior of $C_2$ is the orthocentre of some triangle inscribed in $C_1$.
enter image description here
I gave a construction as follows. Take any point, call it H in the interior of $C_2$. Join OH, it intersects the circle $C_1$ at two points say $A$ and $X$ with $A$ being nearer to $H$. Construct perpendicular bisector of $AX$. Let it intersect $C_1$ at $B$ and $C$. I tell that $ABC$ is the required triangle.



If I assume $H$ to be the orthocentre then all the properties are matching. However, I am unable to prove that the above construction will guarantee that H will be the orthocentre of triangle ABC.



Any help will be appreciated. Thanks in advance










share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    This sound highly unlikely to be true.
    $endgroup$
    – Maria Mazur
    Dec 24 '18 at 8:55










  • $begingroup$
    any better ideas for such a construction. I would also like to know how to prove it wrong
    $endgroup$
    – saisanjeev
    Dec 24 '18 at 8:56










  • $begingroup$
    I was talking about the problem.
    $endgroup$
    – Maria Mazur
    Dec 24 '18 at 8:56










  • $begingroup$
    oh. any way we can find such a point.
    $endgroup$
    – saisanjeev
    Dec 24 '18 at 12:03










  • $begingroup$
    This can't be true, please read carefully your post again.
    $endgroup$
    – Maria Mazur
    Dec 24 '18 at 12:11














0












0








0


1



$begingroup$


Let $C_1$ and $C_2$ be two circles in the plane with radius R and 3R respectively. Show that every point in the interior of $C_2$ is the orthocentre of some triangle inscribed in $C_1$.
enter image description here
I gave a construction as follows. Take any point, call it H in the interior of $C_2$. Join OH, it intersects the circle $C_1$ at two points say $A$ and $X$ with $A$ being nearer to $H$. Construct perpendicular bisector of $AX$. Let it intersect $C_1$ at $B$ and $C$. I tell that $ABC$ is the required triangle.



If I assume $H$ to be the orthocentre then all the properties are matching. However, I am unable to prove that the above construction will guarantee that H will be the orthocentre of triangle ABC.



Any help will be appreciated. Thanks in advance










share|cite|improve this question











$endgroup$




Let $C_1$ and $C_2$ be two circles in the plane with radius R and 3R respectively. Show that every point in the interior of $C_2$ is the orthocentre of some triangle inscribed in $C_1$.
enter image description here
I gave a construction as follows. Take any point, call it H in the interior of $C_2$. Join OH, it intersects the circle $C_1$ at two points say $A$ and $X$ with $A$ being nearer to $H$. Construct perpendicular bisector of $AX$. Let it intersect $C_1$ at $B$ and $C$. I tell that $ABC$ is the required triangle.



If I assume $H$ to be the orthocentre then all the properties are matching. However, I am unable to prove that the above construction will guarantee that H will be the orthocentre of triangle ABC.



Any help will be appreciated. Thanks in advance







geometric-construction






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share|cite|improve this question













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edited Feb 20 at 19:36









MarianD

2,1711618




2,1711618










asked Dec 24 '18 at 8:52









saisanjeevsaisanjeev

1,073312




1,073312








  • 1




    $begingroup$
    This sound highly unlikely to be true.
    $endgroup$
    – Maria Mazur
    Dec 24 '18 at 8:55










  • $begingroup$
    any better ideas for such a construction. I would also like to know how to prove it wrong
    $endgroup$
    – saisanjeev
    Dec 24 '18 at 8:56










  • $begingroup$
    I was talking about the problem.
    $endgroup$
    – Maria Mazur
    Dec 24 '18 at 8:56










  • $begingroup$
    oh. any way we can find such a point.
    $endgroup$
    – saisanjeev
    Dec 24 '18 at 12:03










  • $begingroup$
    This can't be true, please read carefully your post again.
    $endgroup$
    – Maria Mazur
    Dec 24 '18 at 12:11














  • 1




    $begingroup$
    This sound highly unlikely to be true.
    $endgroup$
    – Maria Mazur
    Dec 24 '18 at 8:55










  • $begingroup$
    any better ideas for such a construction. I would also like to know how to prove it wrong
    $endgroup$
    – saisanjeev
    Dec 24 '18 at 8:56










  • $begingroup$
    I was talking about the problem.
    $endgroup$
    – Maria Mazur
    Dec 24 '18 at 8:56










  • $begingroup$
    oh. any way we can find such a point.
    $endgroup$
    – saisanjeev
    Dec 24 '18 at 12:03










  • $begingroup$
    This can't be true, please read carefully your post again.
    $endgroup$
    – Maria Mazur
    Dec 24 '18 at 12:11








1




1




$begingroup$
This sound highly unlikely to be true.
$endgroup$
– Maria Mazur
Dec 24 '18 at 8:55




$begingroup$
This sound highly unlikely to be true.
$endgroup$
– Maria Mazur
Dec 24 '18 at 8:55












$begingroup$
any better ideas for such a construction. I would also like to know how to prove it wrong
$endgroup$
– saisanjeev
Dec 24 '18 at 8:56




$begingroup$
any better ideas for such a construction. I would also like to know how to prove it wrong
$endgroup$
– saisanjeev
Dec 24 '18 at 8:56












$begingroup$
I was talking about the problem.
$endgroup$
– Maria Mazur
Dec 24 '18 at 8:56




$begingroup$
I was talking about the problem.
$endgroup$
– Maria Mazur
Dec 24 '18 at 8:56












$begingroup$
oh. any way we can find such a point.
$endgroup$
– saisanjeev
Dec 24 '18 at 12:03




$begingroup$
oh. any way we can find such a point.
$endgroup$
– saisanjeev
Dec 24 '18 at 12:03












$begingroup$
This can't be true, please read carefully your post again.
$endgroup$
– Maria Mazur
Dec 24 '18 at 12:11




$begingroup$
This can't be true, please read carefully your post again.
$endgroup$
– Maria Mazur
Dec 24 '18 at 12:11










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