The separation theorem

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Anindya Biswas
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The separation theorem

Unread post by Anindya Biswas » Fri Aug 13, 2021 10:52 am

Let $A,B, C, D$ be $4$ distinct points on the plane. Every circle going through $A,C$ intersects with every circle going through $B,D$. Prove that $A,B, C, D$ are either concyclic or collinear.
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Mehrab4226
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Re: The separation theorem

Unread post by Mehrab4226 » Mon Aug 16, 2021 9:52 pm

Anindya Biswas wrote:
Fri Aug 13, 2021 10:52 am
Let $A,B, C, D$ be $4$ distinct points on the plane. Every circle going through $A,C$ intersects with every circle going through $B,D$. Prove that $A,B, C, D$ are either concyclic or collinear.
Let us assume that there is 4 points that are neither on a line nor on a circle, but has that property.

Let the points are $A,B,C,D$ as above. Clearly the perpendicular bisector of $A,C$ and $B,D$ are not the same as $A,B,C,D$ are not colinear.

If they intersect, we can take the intersection of the perpendicular bisectors as the centre and draw 2 concentric circles not having the property. The two circles are different because they cannot be the same by our assumption.

If they are parallel, the lines $AC$ and $BD$ are parallel as well. We can take the points $X,Y$ on the perpendicular bisector on the mid region of the two parallel lines $AC$ and $BD$. Circles $ACX$ and $BDY$ have no common point. A contradiction. $\square$
The Mathematician does not study math because it is useful; he studies it because he delights in it, and he delights in it because it is beautiful.
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