All chords pass through the midpoint of each other












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$begingroup$


Bob draws $n$ chords in a circle, and every chord passes through the midpoint of another chord.



Is it true that all $n$ chords are diameters?



I am not sure how to do these problems.



I tried drawing some circles and lines and it seems always true but I do not know why.










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  • 1




    $begingroup$
    I am not sure if this will help you with tje proof but 2 chords can't bisect each other unless they are both diameters.
    $endgroup$
    – Fareed AF
    Dec 18 '18 at 12:47
















0












$begingroup$


Bob draws $n$ chords in a circle, and every chord passes through the midpoint of another chord.



Is it true that all $n$ chords are diameters?



I am not sure how to do these problems.



I tried drawing some circles and lines and it seems always true but I do not know why.










share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    I am not sure if this will help you with tje proof but 2 chords can't bisect each other unless they are both diameters.
    $endgroup$
    – Fareed AF
    Dec 18 '18 at 12:47














0












0








0





$begingroup$


Bob draws $n$ chords in a circle, and every chord passes through the midpoint of another chord.



Is it true that all $n$ chords are diameters?



I am not sure how to do these problems.



I tried drawing some circles and lines and it seems always true but I do not know why.










share|cite|improve this question









$endgroup$




Bob draws $n$ chords in a circle, and every chord passes through the midpoint of another chord.



Is it true that all $n$ chords are diameters?



I am not sure how to do these problems.



I tried drawing some circles and lines and it seems always true but I do not know why.







combinatorics circles






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share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 18 '18 at 12:32









user627514user627514

393




393








  • 1




    $begingroup$
    I am not sure if this will help you with tje proof but 2 chords can't bisect each other unless they are both diameters.
    $endgroup$
    – Fareed AF
    Dec 18 '18 at 12:47














  • 1




    $begingroup$
    I am not sure if this will help you with tje proof but 2 chords can't bisect each other unless they are both diameters.
    $endgroup$
    – Fareed AF
    Dec 18 '18 at 12:47








1




1




$begingroup$
I am not sure if this will help you with tje proof but 2 chords can't bisect each other unless they are both diameters.
$endgroup$
– Fareed AF
Dec 18 '18 at 12:47




$begingroup$
I am not sure if this will help you with tje proof but 2 chords can't bisect each other unless they are both diameters.
$endgroup$
– Fareed AF
Dec 18 '18 at 12:47










2 Answers
2






active

oldest

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2












$begingroup$

A good start would be this. Take an arbitrary chord. Then consider a second chord that goes through the middle of the first chord.



enter image description here



Here, $AB$ is an arbitrary chord, $C$ is its midpoint, and $DE$ is a chord that goes through $C$. $F$ is the midpoint of $DE$. What can you tell about $OF$ and $OC$?



Unless $O equiv F equiv C$, then $OF < OC$. That's because $OF$ is perpendicular to $FC$.



But that is problematic, because the farthest chord from the center still needs to go through the center of a chord.



Here's a formal proof:



Let $DE$ be the chord furthest away from the center, along the perpendicular line from the center to the chord (if there are several equidistant, it can be any of them). $F$ is the midpoint of $DE$. Assume $DE$ is not a diameter.



There exists a chord $AB$ such that $DE$ goes through the midpoint of $AB$. Call that midpoint $C$. Since DE is furthest away, then $OC leq OF$. $DE$ is not a diameter, so $O$, $F$, and $C$ don't coincide. But $angle OFC$ is 90 degrees, so $OF < OC$, contradicting that $DE$ is furthest away.



So $DE$ must be diameter. Since it's the furthest chord from the center, all chords must be diameters.






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    0












    $begingroup$

    Assume you have one of them, which has a given length and assume this length is smaller than the diameter. Then it is passing through the middle of another chord in the set. How does the length of this one compare to the first? From this you should be able to conclude.






    share|cite|improve this answer









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      2 Answers
      2






      active

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      2 Answers
      2






      active

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      active

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      active

      oldest

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      2












      $begingroup$

      A good start would be this. Take an arbitrary chord. Then consider a second chord that goes through the middle of the first chord.



      enter image description here



      Here, $AB$ is an arbitrary chord, $C$ is its midpoint, and $DE$ is a chord that goes through $C$. $F$ is the midpoint of $DE$. What can you tell about $OF$ and $OC$?



      Unless $O equiv F equiv C$, then $OF < OC$. That's because $OF$ is perpendicular to $FC$.



      But that is problematic, because the farthest chord from the center still needs to go through the center of a chord.



      Here's a formal proof:



      Let $DE$ be the chord furthest away from the center, along the perpendicular line from the center to the chord (if there are several equidistant, it can be any of them). $F$ is the midpoint of $DE$. Assume $DE$ is not a diameter.



      There exists a chord $AB$ such that $DE$ goes through the midpoint of $AB$. Call that midpoint $C$. Since DE is furthest away, then $OC leq OF$. $DE$ is not a diameter, so $O$, $F$, and $C$ don't coincide. But $angle OFC$ is 90 degrees, so $OF < OC$, contradicting that $DE$ is furthest away.



      So $DE$ must be diameter. Since it's the furthest chord from the center, all chords must be diameters.






      share|cite|improve this answer









      $endgroup$


















        2












        $begingroup$

        A good start would be this. Take an arbitrary chord. Then consider a second chord that goes through the middle of the first chord.



        enter image description here



        Here, $AB$ is an arbitrary chord, $C$ is its midpoint, and $DE$ is a chord that goes through $C$. $F$ is the midpoint of $DE$. What can you tell about $OF$ and $OC$?



        Unless $O equiv F equiv C$, then $OF < OC$. That's because $OF$ is perpendicular to $FC$.



        But that is problematic, because the farthest chord from the center still needs to go through the center of a chord.



        Here's a formal proof:



        Let $DE$ be the chord furthest away from the center, along the perpendicular line from the center to the chord (if there are several equidistant, it can be any of them). $F$ is the midpoint of $DE$. Assume $DE$ is not a diameter.



        There exists a chord $AB$ such that $DE$ goes through the midpoint of $AB$. Call that midpoint $C$. Since DE is furthest away, then $OC leq OF$. $DE$ is not a diameter, so $O$, $F$, and $C$ don't coincide. But $angle OFC$ is 90 degrees, so $OF < OC$, contradicting that $DE$ is furthest away.



        So $DE$ must be diameter. Since it's the furthest chord from the center, all chords must be diameters.






        share|cite|improve this answer









        $endgroup$
















          2












          2








          2





          $begingroup$

          A good start would be this. Take an arbitrary chord. Then consider a second chord that goes through the middle of the first chord.



          enter image description here



          Here, $AB$ is an arbitrary chord, $C$ is its midpoint, and $DE$ is a chord that goes through $C$. $F$ is the midpoint of $DE$. What can you tell about $OF$ and $OC$?



          Unless $O equiv F equiv C$, then $OF < OC$. That's because $OF$ is perpendicular to $FC$.



          But that is problematic, because the farthest chord from the center still needs to go through the center of a chord.



          Here's a formal proof:



          Let $DE$ be the chord furthest away from the center, along the perpendicular line from the center to the chord (if there are several equidistant, it can be any of them). $F$ is the midpoint of $DE$. Assume $DE$ is not a diameter.



          There exists a chord $AB$ such that $DE$ goes through the midpoint of $AB$. Call that midpoint $C$. Since DE is furthest away, then $OC leq OF$. $DE$ is not a diameter, so $O$, $F$, and $C$ don't coincide. But $angle OFC$ is 90 degrees, so $OF < OC$, contradicting that $DE$ is furthest away.



          So $DE$ must be diameter. Since it's the furthest chord from the center, all chords must be diameters.






          share|cite|improve this answer









          $endgroup$



          A good start would be this. Take an arbitrary chord. Then consider a second chord that goes through the middle of the first chord.



          enter image description here



          Here, $AB$ is an arbitrary chord, $C$ is its midpoint, and $DE$ is a chord that goes through $C$. $F$ is the midpoint of $DE$. What can you tell about $OF$ and $OC$?



          Unless $O equiv F equiv C$, then $OF < OC$. That's because $OF$ is perpendicular to $FC$.



          But that is problematic, because the farthest chord from the center still needs to go through the center of a chord.



          Here's a formal proof:



          Let $DE$ be the chord furthest away from the center, along the perpendicular line from the center to the chord (if there are several equidistant, it can be any of them). $F$ is the midpoint of $DE$. Assume $DE$ is not a diameter.



          There exists a chord $AB$ such that $DE$ goes through the midpoint of $AB$. Call that midpoint $C$. Since DE is furthest away, then $OC leq OF$. $DE$ is not a diameter, so $O$, $F$, and $C$ don't coincide. But $angle OFC$ is 90 degrees, so $OF < OC$, contradicting that $DE$ is furthest away.



          So $DE$ must be diameter. Since it's the furthest chord from the center, all chords must be diameters.







          share|cite|improve this answer












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          share|cite|improve this answer










          answered Dec 18 '18 at 13:03









          Todor MarkovTodor Markov

          2,420412




          2,420412























              0












              $begingroup$

              Assume you have one of them, which has a given length and assume this length is smaller than the diameter. Then it is passing through the middle of another chord in the set. How does the length of this one compare to the first? From this you should be able to conclude.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                Assume you have one of them, which has a given length and assume this length is smaller than the diameter. Then it is passing through the middle of another chord in the set. How does the length of this one compare to the first? From this you should be able to conclude.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  Assume you have one of them, which has a given length and assume this length is smaller than the diameter. Then it is passing through the middle of another chord in the set. How does the length of this one compare to the first? From this you should be able to conclude.






                  share|cite|improve this answer









                  $endgroup$



                  Assume you have one of them, which has a given length and assume this length is smaller than the diameter. Then it is passing through the middle of another chord in the set. How does the length of this one compare to the first? From this you should be able to conclude.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 18 '18 at 13:01









                  JoceJoce

                  834411




                  834411






























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