A regular $2017$-gon is partitioned into triangles by a set of non-intersecting diagonals. Prove that only...












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A regular $2017$-gon is partitioned into triangles by a set of non-intersecting diagonals. Prove that among those triangles only one is a acute angled.










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    Draw a picture and prove it with a pentagon first.
    $endgroup$
    – fleablood
    Jan 19 at 6:19
















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$begingroup$


A regular $2017$-gon is partitioned into triangles by a set of non-intersecting diagonals. Prove that among those triangles only one is a acute angled.










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  • $begingroup$
    Draw a picture and prove it with a pentagon first.
    $endgroup$
    – fleablood
    Jan 19 at 6:19














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$begingroup$


A regular $2017$-gon is partitioned into triangles by a set of non-intersecting diagonals. Prove that among those triangles only one is a acute angled.










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A regular $2017$-gon is partitioned into triangles by a set of non-intersecting diagonals. Prove that among those triangles only one is a acute angled.







geometry






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edited Jan 19 at 8:45









idriskameni

749321




749321










asked Jan 19 at 5:15









Anson ChanAnson Chan

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  • $begingroup$
    Draw a picture and prove it with a pentagon first.
    $endgroup$
    – fleablood
    Jan 19 at 6:19


















  • $begingroup$
    Draw a picture and prove it with a pentagon first.
    $endgroup$
    – fleablood
    Jan 19 at 6:19
















$begingroup$
Draw a picture and prove it with a pentagon first.
$endgroup$
– fleablood
Jan 19 at 6:19




$begingroup$
Draw a picture and prove it with a pentagon first.
$endgroup$
– fleablood
Jan 19 at 6:19










2 Answers
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Since $2017$ is odd there is no diagonal going through the centre of the circle. Therefore there is exactly one triangle containing $M$ in its interior. This triangle is acute, all others are obtuse.






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    A triangle will have $k$ vertices to one side and $n-3-k$ to the other, for $0le kle n-3$. The inscribed angle theorem says that the angles of this triangle are
    $$
    left{fracpi{n},,,(k+1)fracpi{n},,,(n-2-k)fracpi{n}right}
    $$



    enter image description here



    This means that every such triangulation of a regular $n$-gon will have an identical set if triangles, just oriented and positioned differently. Each triangle is labeled with its $k$ value:



    enter image description here



    If $n$ is odd, only when $k=frac{n-3}2$ will the triangle be acute, with angles
    $$
    left{fracpi{n},,,frac{n-1}2fracpi{n},,,frac{n-1}2fracpi{n}right}
    $$

    If $n$ is even, when $k=frac{n-2}2$ or $k=frac{n-4}2$, the triangle will be right, with angles
    $$
    left{fracpi{n},,,frac{n-2}2fracpi{n},,,fracpi2right}
    $$

    and there will be no acute triangles.






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      2 Answers
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      2 Answers
      2






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      active

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      active

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      5












      $begingroup$

      Since $2017$ is odd there is no diagonal going through the centre of the circle. Therefore there is exactly one triangle containing $M$ in its interior. This triangle is acute, all others are obtuse.






      share|cite|improve this answer









      $endgroup$


















        5












        $begingroup$

        Since $2017$ is odd there is no diagonal going through the centre of the circle. Therefore there is exactly one triangle containing $M$ in its interior. This triangle is acute, all others are obtuse.






        share|cite|improve this answer









        $endgroup$
















          5












          5








          5





          $begingroup$

          Since $2017$ is odd there is no diagonal going through the centre of the circle. Therefore there is exactly one triangle containing $M$ in its interior. This triangle is acute, all others are obtuse.






          share|cite|improve this answer









          $endgroup$



          Since $2017$ is odd there is no diagonal going through the centre of the circle. Therefore there is exactly one triangle containing $M$ in its interior. This triangle is acute, all others are obtuse.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 19 at 9:18









          Christian BlatterChristian Blatter

          177k9115328




          177k9115328























              0












              $begingroup$

              A triangle will have $k$ vertices to one side and $n-3-k$ to the other, for $0le kle n-3$. The inscribed angle theorem says that the angles of this triangle are
              $$
              left{fracpi{n},,,(k+1)fracpi{n},,,(n-2-k)fracpi{n}right}
              $$



              enter image description here



              This means that every such triangulation of a regular $n$-gon will have an identical set if triangles, just oriented and positioned differently. Each triangle is labeled with its $k$ value:



              enter image description here



              If $n$ is odd, only when $k=frac{n-3}2$ will the triangle be acute, with angles
              $$
              left{fracpi{n},,,frac{n-1}2fracpi{n},,,frac{n-1}2fracpi{n}right}
              $$

              If $n$ is even, when $k=frac{n-2}2$ or $k=frac{n-4}2$, the triangle will be right, with angles
              $$
              left{fracpi{n},,,frac{n-2}2fracpi{n},,,fracpi2right}
              $$

              and there will be no acute triangles.






              share|cite|improve this answer











              $endgroup$


















                0












                $begingroup$

                A triangle will have $k$ vertices to one side and $n-3-k$ to the other, for $0le kle n-3$. The inscribed angle theorem says that the angles of this triangle are
                $$
                left{fracpi{n},,,(k+1)fracpi{n},,,(n-2-k)fracpi{n}right}
                $$



                enter image description here



                This means that every such triangulation of a regular $n$-gon will have an identical set if triangles, just oriented and positioned differently. Each triangle is labeled with its $k$ value:



                enter image description here



                If $n$ is odd, only when $k=frac{n-3}2$ will the triangle be acute, with angles
                $$
                left{fracpi{n},,,frac{n-1}2fracpi{n},,,frac{n-1}2fracpi{n}right}
                $$

                If $n$ is even, when $k=frac{n-2}2$ or $k=frac{n-4}2$, the triangle will be right, with angles
                $$
                left{fracpi{n},,,frac{n-2}2fracpi{n},,,fracpi2right}
                $$

                and there will be no acute triangles.






                share|cite|improve this answer











                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  A triangle will have $k$ vertices to one side and $n-3-k$ to the other, for $0le kle n-3$. The inscribed angle theorem says that the angles of this triangle are
                  $$
                  left{fracpi{n},,,(k+1)fracpi{n},,,(n-2-k)fracpi{n}right}
                  $$



                  enter image description here



                  This means that every such triangulation of a regular $n$-gon will have an identical set if triangles, just oriented and positioned differently. Each triangle is labeled with its $k$ value:



                  enter image description here



                  If $n$ is odd, only when $k=frac{n-3}2$ will the triangle be acute, with angles
                  $$
                  left{fracpi{n},,,frac{n-1}2fracpi{n},,,frac{n-1}2fracpi{n}right}
                  $$

                  If $n$ is even, when $k=frac{n-2}2$ or $k=frac{n-4}2$, the triangle will be right, with angles
                  $$
                  left{fracpi{n},,,frac{n-2}2fracpi{n},,,fracpi2right}
                  $$

                  and there will be no acute triangles.






                  share|cite|improve this answer











                  $endgroup$



                  A triangle will have $k$ vertices to one side and $n-3-k$ to the other, for $0le kle n-3$. The inscribed angle theorem says that the angles of this triangle are
                  $$
                  left{fracpi{n},,,(k+1)fracpi{n},,,(n-2-k)fracpi{n}right}
                  $$



                  enter image description here



                  This means that every such triangulation of a regular $n$-gon will have an identical set if triangles, just oriented and positioned differently. Each triangle is labeled with its $k$ value:



                  enter image description here



                  If $n$ is odd, only when $k=frac{n-3}2$ will the triangle be acute, with angles
                  $$
                  left{fracpi{n},,,frac{n-1}2fracpi{n},,,frac{n-1}2fracpi{n}right}
                  $$

                  If $n$ is even, when $k=frac{n-2}2$ or $k=frac{n-4}2$, the triangle will be right, with angles
                  $$
                  left{fracpi{n},,,frac{n-2}2fracpi{n},,,fracpi2right}
                  $$

                  and there will be no acute triangles.







                  share|cite|improve this answer














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                  edited Jan 20 at 3:03

























                  answered Jan 19 at 9:54









                  robjohnrobjohn

                  271k27316643




                  271k27316643






























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