Is there a flaw in this proof?
$begingroup$
A new user posted this question on Nov 16:
"$MNPQ$ is a cyclic quadrilateral (that $M$ and $N$ are fixed points on the circumcircle of the cyclic quadrilateral $MNPQ$. The points $P$ and $Q$ can vary on the circle) with $K$ is the intersection between $MP$ and $NQ$. The circumcircles of triangle $KMQ$ and $KNP$ intersect $J≠K$. The perpendicular bisectors of $NQ$, $MP$ intersect $JN$, $JM$ at $X$,$Y$. Prove that $XY$ goes through a fixed point."
OP conjectured that $XY$ goes through the midpoint of $MN$.
[I posted an answer on Nov 19, but it was deleted that same day on the vote of three moderators. On Nov 20 the posted question was put on hold as off-topic: "missing context or other details" etc, and OP deleted the question on Nov 21.]
Does the following prove the truth of OP's conjecture?
Retaining the conditions and diagram of the posted question, where $A$ is the intersection of $XY$ with $NM$, we can prove that $A$ is the mid-point of $NM$, and therefore fixed, by introducing a fourth circle:
Draw the circle through points $A$, $M$, $O$.
Since $OB$ is the perpendicular bisector of $PM$ at $B$, the $angle OBM$ is right. $B$ therefore lies on circle $AMO$, and $OM$ is the diameter. And since $OM$ is the radius of circle $MNPQ$, the circles are internally tangent at $M$.
Hence since chords $AM$ and $NM$ make equal angles with the shared tangent at $M$, the circular segments on them are similar and their bases are proportional to the circles' diameters:$$frac{AM}{NM}=frac{OM}{2OM}$$
Therefore, $A$ bisects $NM$, and since $NM$ is fixed, $A$ is fixed.
Note: It may seem we have taken no account of the conditions on $X$ and $Y$, which determine point $A$, but the proof rests on point $B$, and $B$ determines $Y$. Again, we have not referred to $X$, or the line $OX$ and its intersection with $NQ$ at $C$, but if we make our fourth circle instead through points $A$, $N$, $O$, the same result follows. For since $angle OCN$ is right, then $C$ lies on circle $ANO$, which has diameter $NO$ and is internally tangent at $N$ to circle $MNPQ$, and it follows again by similarity of segments that $NM=2NA$.
euclidean-geometry
asked Dec 10 '18 at 6:34
Edward PorcellaEdward Porcella
1,4311511
$endgroup$
add a comment |
$begingroup$
A new user posted this question on Nov 16:
"$MNPQ$ is a cyclic quadrilateral (that $M$ and $N$ are fixed points on the circumcircle of the cyclic quadrilateral $MNPQ$. The points $P$ and $Q$ can vary on the circle) with $K$ is the intersection between $MP$ and $NQ$. The circumcircles of triangle $KMQ$ and $KNP$ intersect $J≠K$. The perpendicular bisectors of $NQ$, $MP$ intersect $JN$, $JM$ at $X$,$Y$. Prove that $XY$ goes through a fixed point."
OP conjectured that $XY$ goes through the midpoint of $MN$.
[I posted an answer on Nov 19, but it was deleted that same day on the vote of three moderators. On Nov 20 the posted question was put on hold as off-topic: "missing context or other details" etc, and OP deleted the question on Nov 21.]
Does the following prove the truth of OP's conjecture?
Retaining the conditions and diagram of the posted question, where $A$ is the intersection of $XY$ with $NM$, we can prove that $A$ is the mid-point of $NM$, and therefore fixed, by introducing a fourth circle:
Draw the circle through points $A$, $M$, $O$.
Since $OB$ is the perpendicular bisector of $PM$ at $B$, the $angle OBM$ is right. $B$ therefore lies on circle $AMO$, and $OM$ is the diameter. And since $OM$ is the radius of circle $MNPQ$, the circles are internally tangent at $M$.
Hence since chords $AM$ and $NM$ make equal angles with the shared tangent at $M$, the circular segments on them are similar and their bases are proportional to the circles' diameters:$$frac{AM}{NM}=frac{OM}{2OM}$$
Therefore, $A$ bisects $NM$, and since $NM$ is fixed, $A$ is fixed.
Note: It may seem we have taken no account of the conditions on $X$ and $Y$, which determine point $A$, but the proof rests on point $B$, and $B$ determines $Y$. Again, we have not referred to $X$, or the line $OX$ and its intersection with $NQ$ at $C$, but if we make our fourth circle instead through points $A$, $N$, $O$, the same result follows. For since $angle OCN$ is right, then $C$ lies on circle $ANO$, which has diameter $NO$ and is internally tangent at $N$ to circle $MNPQ$, and it follows again by similarity of segments that $NM=2NA$.
euclidean-geometry
asked Dec 10 '18 at 6:34
Edward PorcellaEdward Porcella
1,4311511
$endgroup$
$begingroup$
In your diagram, neither of $P,Q$ are on the (shorter) arc $MN.$ Dos that matter, i.e. does the proof use that assumption? [I'm not saying it does, just didn't want to read it through for other case as I can't imagine the complicated array of lines which would result.]
$endgroup$
– coffeemath
Dec 10 '18 at 7:15
1
$begingroup$
@coffeemath--I don't believe it matters. Only that $MN$ is fixed, while $P$ and $Q$ can move on the circle. I just tried to make the figure as distinct as possible, so as to avoid misleading appearances.
$endgroup$
– Edward Porcella
Dec 10 '18 at 14:44
add a comment |
$begingroup$
A new user posted this question on Nov 16:
"$MNPQ$ is a cyclic quadrilateral (that $M$ and $N$ are fixed points on the circumcircle of the cyclic quadrilateral $MNPQ$. The points $P$ and $Q$ can vary on the circle) with $K$ is the intersection between $MP$ and $NQ$. The circumcircles of triangle $KMQ$ and $KNP$ intersect $J≠K$. The perpendicular bisectors of $NQ$, $MP$ intersect $JN$, $JM$ at $X$,$Y$. Prove that $XY$ goes through a fixed point."
OP conjectured that $XY$ goes through the midpoint of $MN$.
[I posted an answer on Nov 19, but it was deleted that same day on the vote of three moderators. On Nov 20 the posted question was put on hold as off-topic: "missing context or other details" etc, and OP deleted the question on Nov 21.]
Does the following prove the truth of OP's conjecture?
Retaining the conditions and diagram of the posted question, where $A$ is the intersection of $XY$ with $NM$, we can prove that $A$ is the mid-point of $NM$, and therefore fixed, by introducing a fourth circle:
Draw the circle through points $A$, $M$, $O$.
Since $OB$ is the perpendicular bisector of $PM$ at $B$, the $angle OBM$ is right. $B$ therefore lies on circle $AMO$, and $OM$ is the diameter. And since $OM$ is the radius of circle $MNPQ$, the circles are internally tangent at $M$.
Hence since chords $AM$ and $NM$ make equal angles with the shared tangent at $M$, the circular segments on them are similar and their bases are proportional to the circles' diameters:$$frac{AM}{NM}=frac{OM}{2OM}$$
Therefore, $A$ bisects $NM$, and since $NM$ is fixed, $A$ is fixed.
Note: It may seem we have taken no account of the conditions on $X$ and $Y$, which determine point $A$, but the proof rests on point $B$, and $B$ determines $Y$. Again, we have not referred to $X$, or the line $OX$ and its intersection with $NQ$ at $C$, but if we make our fourth circle instead through points $A$, $N$, $O$, the same result follows. For since $angle OCN$ is right, then $C$ lies on circle $ANO$, which has diameter $NO$ and is internally tangent at $N$ to circle $MNPQ$, and it follows again by similarity of segments that $NM=2NA$.
euclidean-geometry
asked Dec 10 '18 at 6:34
Edward PorcellaEdward Porcella
1,4311511
$endgroup$
A new user posted this question on Nov 16:
"$MNPQ$ is a cyclic quadrilateral (that $M$ and $N$ are fixed points on the circumcircle of the cyclic quadrilateral $MNPQ$. The points $P$ and $Q$ can vary on the circle) with $K$ is the intersection between $MP$ and $NQ$. The circumcircles of triangle $KMQ$ and $KNP$ intersect $J≠K$. The perpendicular bisectors of $NQ$, $MP$ intersect $JN$, $JM$ at $X$,$Y$. Prove that $XY$ goes through a fixed point."
OP conjectured that $XY$ goes through the midpoint of $MN$.
[I posted an answer on Nov 19, but it was deleted that same day on the vote of three moderators. On Nov 20 the posted question was put on hold as off-topic: "missing context or other details" etc, and OP deleted the question on Nov 21.]
Does the following prove the truth of OP's conjecture?
Retaining the conditions and diagram of the posted question, where $A$ is the intersection of $XY$ with $NM$, we can prove that $A$ is the mid-point of $NM$, and therefore fixed, by introducing a fourth circle:
Draw the circle through points $A$, $M$, $O$.
Since $OB$ is the perpendicular bisector of $PM$ at $B$, the $angle OBM$ is right. $B$ therefore lies on circle $AMO$, and $OM$ is the diameter. And since $OM$ is the radius of circle $MNPQ$, the circles are internally tangent at $M$.
Hence since chords $AM$ and $NM$ make equal angles with the shared tangent at $M$, the circular segments on them are similar and their bases are proportional to the circles' diameters:$$frac{AM}{NM}=frac{OM}{2OM}$$
Therefore, $A$ bisects $NM$, and since $NM$ is fixed, $A$ is fixed.
Note: It may seem we have taken no account of the conditions on $X$ and $Y$, which determine point $A$, but the proof rests on point $B$, and $B$ determines $Y$. Again, we have not referred to $X$, or the line $OX$ and its intersection with $NQ$ at $C$, but if we make our fourth circle instead through points $A$, $N$, $O$, the same result follows. For since $angle OCN$ is right, then $C$ lies on circle $ANO$, which has diameter $NO$ and is internally tangent at $N$ to circle $MNPQ$, and it follows again by similarity of segments that $NM=2NA$.
euclidean-geometry
euclidean-geometry
asked Dec 10 '18 at 6:34
Edward PorcellaEdward Porcella
1,4311511
asked Dec 10 '18 at 6:34
Edward PorcellaEdward Porcella
1,4311511
asked Dec 10 '18 at 6:34
Edward PorcellaEdward Porcella
1,4311511
asked Dec 10 '18 at 6:34
Edward PorcellaEdward Porcella
1,4311511
asked Dec 10 '18 at 6:34
Edward PorcellaEdward Porcella
1,4311511
1,4311511
$begingroup$
In your diagram, neither of $P,Q$ are on the (shorter) arc $MN.$ Dos that matter, i.e. does the proof use that assumption? [I'm not saying it does, just didn't want to read it through for other case as I can't imagine the complicated array of lines which would result.]
$endgroup$
– coffeemath
Dec 10 '18 at 7:15
1
$begingroup$
@coffeemath--I don't believe it matters. Only that $MN$ is fixed, while $P$ and $Q$ can move on the circle. I just tried to make the figure as distinct as possible, so as to avoid misleading appearances.
$endgroup$
– Edward Porcella
Dec 10 '18 at 14:44
add a comment |
$begingroup$
In your diagram, neither of $P,Q$ are on the (shorter) arc $MN.$ Dos that matter, i.e. does the proof use that assumption? [I'm not saying it does, just didn't want to read it through for other case as I can't imagine the complicated array of lines which would result.]
$endgroup$
– coffeemath
Dec 10 '18 at 7:15
1
$begingroup$
@coffeemath--I don't believe it matters. Only that $MN$ is fixed, while $P$ and $Q$ can move on the circle. I just tried to make the figure as distinct as possible, so as to avoid misleading appearances.
$endgroup$
– Edward Porcella
Dec 10 '18 at 14:44
$begingroup$
In your diagram, neither of $P,Q$ are on the (shorter) arc $MN.$ Dos that matter, i.e. does the proof use that assumption? [I'm not saying it does, just didn't want to read it through for other case as I can't imagine the complicated array of lines which would result.]
$endgroup$
– coffeemath
Dec 10 '18 at 7:15
$begingroup$
In your diagram, neither of $P,Q$ are on the (shorter) arc $MN.$ Dos that matter, i.e. does the proof use that assumption? [I'm not saying it does, just didn't want to read it through for other case as I can't imagine the complicated array of lines which would result.]
$endgroup$
– coffeemath
Dec 10 '18 at 7:15
1
1
$begingroup$
@coffeemath--I don't believe it matters. Only that $MN$ is fixed, while $P$ and $Q$ can move on the circle. I just tried to make the figure as distinct as possible, so as to avoid misleading appearances.
$endgroup$
– Edward Porcella
Dec 10 '18 at 14:44
$begingroup$
@coffeemath--I don't believe it matters. Only that $MN$ is fixed, while $P$ and $Q$ can move on the circle. I just tried to make the figure as distinct as possible, so as to avoid misleading appearances.
$endgroup$
– Edward Porcella
Dec 10 '18 at 14:44
add a comment |
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$begingroup$
In your diagram, neither of $P,Q$ are on the (shorter) arc $MN.$ Dos that matter, i.e. does the proof use that assumption? [I'm not saying it does, just didn't want to read it through for other case as I can't imagine the complicated array of lines which would result.]
$endgroup$
– coffeemath
Dec 10 '18 at 7:15
1
$begingroup$
@coffeemath--I don't believe it matters. Only that $MN$ is fixed, while $P$ and $Q$ can move on the circle. I just tried to make the figure as distinct as possible, so as to avoid misleading appearances.
$endgroup$
– Edward Porcella
Dec 10 '18 at 14:44