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.
geometry
edited Jan 19 at 8:45
idriskameni
749321
asked Jan 19 at 5:15
Anson ChanAnson Chan
162
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add a comment |
$begingroup$
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.
geometry
edited Jan 19 at 8:45
idriskameni
749321
asked Jan 19 at 5:15
Anson ChanAnson Chan
162
$endgroup$
$begingroup$
Draw a picture and prove it with a pentagon first.
$endgroup$
– fleablood
Jan 19 at 6:19
add a comment |
$begingroup$
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.
geometry
edited Jan 19 at 8:45
idriskameni
749321
asked Jan 19 at 5:15
Anson ChanAnson Chan
162
$endgroup$
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
geometry
edited Jan 19 at 8:45
idriskameni
749321
asked Jan 19 at 5:15
Anson ChanAnson Chan
162
edited Jan 19 at 8:45
idriskameni
749321
asked Jan 19 at 5:15
Anson ChanAnson Chan
162
edited Jan 19 at 8:45
idriskameni
749321
edited Jan 19 at 8:45
idriskameni
749321
edited Jan 19 at 8:45
idriskameni
749321
749321
asked Jan 19 at 5:15
Anson ChanAnson Chan
162
asked Jan 19 at 5:15
Anson ChanAnson Chan
162
asked Jan 19 at 5:15
Anson ChanAnson Chan
162
162
$begingroup$
Draw a picture and prove it with a pentagon first.
$endgroup$
– fleablood
Jan 19 at 6:19
add a comment |
$begingroup$
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
add a comment |
2 Answers
2
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$begingroup$
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.
answered Jan 19 at 9:18
Christian BlatterChristian Blatter
177k9115328
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$begingroup$
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
$$
left{fracpi{n},,,(k+1)fracpi{n},,,(n-2-k)fracpi{n}right}
$$
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:
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.
answered Jan 19 at 9:54
robjohn♦robjohn
271k27316643
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add a comment |
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2 Answers
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active
oldest
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2 Answers
2
active
oldest
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$begingroup$
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.
answered Jan 19 at 9:18
Christian BlatterChristian Blatter
177k9115328
$endgroup$
add a comment |
$begingroup$
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.
answered Jan 19 at 9:18
Christian BlatterChristian Blatter
177k9115328
$endgroup$
add a comment |
$begingroup$
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.
answered Jan 19 at 9:18
Christian BlatterChristian Blatter
177k9115328
$endgroup$
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.
answered Jan 19 at 9:18
Christian BlatterChristian Blatter
177k9115328
answered Jan 19 at 9:18
Christian BlatterChristian Blatter
177k9115328
answered Jan 19 at 9:18
Christian BlatterChristian Blatter
177k9115328
answered Jan 19 at 9:18
Christian BlatterChristian Blatter
177k9115328
177k9115328
add a comment |
add a comment |
$begingroup$
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
$$
left{fracpi{n},,,(k+1)fracpi{n},,,(n-2-k)fracpi{n}right}
$$
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:
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.
answered Jan 19 at 9:54
robjohn♦robjohn
271k27316643
$endgroup$
add a comment |
$begingroup$
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
$$
left{fracpi{n},,,(k+1)fracpi{n},,,(n-2-k)fracpi{n}right}
$$
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:
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.
answered Jan 19 at 9:54
robjohn♦robjohn
271k27316643
$endgroup$
add a comment |
$begingroup$
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
$$
left{fracpi{n},,,(k+1)fracpi{n},,,(n-2-k)fracpi{n}right}
$$
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:
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.
answered Jan 19 at 9:54
robjohn♦robjohn
271k27316643
$endgroup$
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
$$
left{fracpi{n},,,(k+1)fracpi{n},,,(n-2-k)fracpi{n}right}
$$
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:
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.
answered Jan 19 at 9:54
robjohn♦robjohn
271k27316643
edited Jan 20 at 3:03
answered Jan 19 at 9:54
robjohn♦robjohn
271k27316643
answered Jan 19 at 9:54
robjohn♦robjohn
271k27316643
answered Jan 19 at 9:54
robjohn♦robjohn
271k27316643
271k27316643
add a comment |
add a comment |
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$begingroup$
Draw a picture and prove it with a pentagon first.
$endgroup$
– fleablood
Jan 19 at 6:19