Prove that there exists a triangle which can be cut into 2005 congruent triangles.
I thought maybe we can start with congruent triangle and try to cut it similar to how we create a Sierpinski's Triangle? However, the number of smaller triangles we get is a power of $4$ so it does not work. Any ideas?
geometry contest-math
edited Nov 26 at 2:08
Akash Roy
1
asked Nov 26 at 1:59
mathnoob
1,759422
add a comment |
I thought maybe we can start with congruent triangle and try to cut it similar to how we create a Sierpinski's Triangle? However, the number of smaller triangles we get is a power of $4$ so it does not work. Any ideas?
geometry contest-math
edited Nov 26 at 2:08
Akash Roy
1
asked Nov 26 at 1:59
mathnoob
1,759422
add a comment |
I thought maybe we can start with congruent triangle and try to cut it similar to how we create a Sierpinski's Triangle? However, the number of smaller triangles we get is a power of $4$ so it does not work. Any ideas?
geometry contest-math
edited Nov 26 at 2:08
Akash Roy
1
asked Nov 26 at 1:59
mathnoob
1,759422
I thought maybe we can start with congruent triangle and try to cut it similar to how we create a Sierpinski's Triangle? However, the number of smaller triangles we get is a power of $4$ so it does not work. Any ideas?
geometry contest-math
geometry contest-math
edited Nov 26 at 2:08
Akash Roy
1
asked Nov 26 at 1:59
mathnoob
1,759422
edited Nov 26 at 2:08
Akash Roy
1
asked Nov 26 at 1:59
mathnoob
1,759422
edited Nov 26 at 2:08
Akash Roy
1
edited Nov 26 at 2:08
Akash Roy
1
edited Nov 26 at 2:08
Akash Roy
1
1
asked Nov 26 at 1:59
mathnoob
1,759422
asked Nov 26 at 1:59
mathnoob
1,759422
asked Nov 26 at 1:59
mathnoob
1,759422
1,759422
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
The decomposition is possible because $2005 = 5cdot 401$ and both $5$ and $401$ are primes of the form $4k+1$. This allow $2005$ to be written as a sum of squares.
$$2005 = 22^2 + 39^2 = 18^2+41^2$$
For any integer $n = p^2 + q^2$ that can be written as a sum of squares.
Consider a right-angled triangle $ABC$ with
$$AB = psqrt{n}, AC = qsqrt{n}quadtext{ and }quad BC = n$$
Let $D$ on $BC$ be the foot of attitude at $A$. It is easy to
see $triangle DBA$ and $triangle DAC$ are similar to $triangle ABC$ with
$$BD = p^2, AD = pqquadtext{ and }quad CD = q^2$$
One can split $triangle DBA$ into $p^2$ and $triangle DAC$ into $q^2$
triangles with sides $p, q$ and $sqrt{n}$.
As an example, following is a subdivision of a triangle into $13 = 2^2 + 3^2$ congruent triangles.
In the literature, this is known as a biquadratic tiling of a triangle. For more information about subdividing triangles into congruent triangles, look at answers in this MO post. In particular, the list of papers by Michael Beeson there. The construction described here is based on what I have learned from one of Michael's papers.
answered Nov 26 at 2:57
achille hui
95.2k5129256
add a comment |
Since $2005$ is the sum of two squares, there exists this sort of triangle. In a contest setting, if you're trying to show $2005$ is a sum of two squares, realize that $2005=401cdot 5$. Since 401 and 5 are sums of squares, we know that their product is also a sum of squares.
Now in general if we have $n=x^2+y^2$, let our triangle be an $x$ by $y$ right triangle. Then, we can split this triangle along the altitude to the hypotenuse, which gives us two similar triangles with hypotenuses $x$ and $y$ respectively. Finally, we can decompose each of these into $x^2$ and $y^2$ similar right triangles with hypotenuses $1$, via a stretched version of the following picture.
answered Nov 26 at 2:47
Isaac Browne
4,60731132
I know this sum of squares is a sufficient condition, but I'm not sure if it's necessary. Any thoughts on this? As a particular example, would $3$ or $7$ work?
– Isaac Browne
Nov 26 at 2:49
4
The $30-60-90$ triangle can be divided into three triangles. Start by bisecting the 60 angle.
– Empy2
Nov 26 at 3:07
2
Alright, so $3$ works. I also just thought of the construction of connecting vertices of an equilateral triangle to the center, which would create three congruent triangles. Any others?
– Isaac Browne
Nov 26 at 3:11
6
It seems Michael Beeson has proved that $7$ is impossible recently. this appears on his web site, not yet available on arXiv...
– achille hui
Nov 26 at 4:00
1
@achillehui: Gosh — a lovely paper, and very hot off the press!
– Peter LeFanu Lumsdaine
Nov 26 at 10:21
add a comment |
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2 Answers
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2 Answers
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The decomposition is possible because $2005 = 5cdot 401$ and both $5$ and $401$ are primes of the form $4k+1$. This allow $2005$ to be written as a sum of squares.
$$2005 = 22^2 + 39^2 = 18^2+41^2$$
For any integer $n = p^2 + q^2$ that can be written as a sum of squares.
Consider a right-angled triangle $ABC$ with
$$AB = psqrt{n}, AC = qsqrt{n}quadtext{ and }quad BC = n$$
Let $D$ on $BC$ be the foot of attitude at $A$. It is easy to
see $triangle DBA$ and $triangle DAC$ are similar to $triangle ABC$ with
$$BD = p^2, AD = pqquadtext{ and }quad CD = q^2$$
One can split $triangle DBA$ into $p^2$ and $triangle DAC$ into $q^2$
triangles with sides $p, q$ and $sqrt{n}$.
As an example, following is a subdivision of a triangle into $13 = 2^2 + 3^2$ congruent triangles.
In the literature, this is known as a biquadratic tiling of a triangle. For more information about subdividing triangles into congruent triangles, look at answers in this MO post. In particular, the list of papers by Michael Beeson there. The construction described here is based on what I have learned from one of Michael's papers.
answered Nov 26 at 2:57
achille hui
95.2k5129256
add a comment |
The decomposition is possible because $2005 = 5cdot 401$ and both $5$ and $401$ are primes of the form $4k+1$. This allow $2005$ to be written as a sum of squares.
$$2005 = 22^2 + 39^2 = 18^2+41^2$$
For any integer $n = p^2 + q^2$ that can be written as a sum of squares.
Consider a right-angled triangle $ABC$ with
$$AB = psqrt{n}, AC = qsqrt{n}quadtext{ and }quad BC = n$$
Let $D$ on $BC$ be the foot of attitude at $A$. It is easy to
see $triangle DBA$ and $triangle DAC$ are similar to $triangle ABC$ with
$$BD = p^2, AD = pqquadtext{ and }quad CD = q^2$$
One can split $triangle DBA$ into $p^2$ and $triangle DAC$ into $q^2$
triangles with sides $p, q$ and $sqrt{n}$.
As an example, following is a subdivision of a triangle into $13 = 2^2 + 3^2$ congruent triangles.
In the literature, this is known as a biquadratic tiling of a triangle. For more information about subdividing triangles into congruent triangles, look at answers in this MO post. In particular, the list of papers by Michael Beeson there. The construction described here is based on what I have learned from one of Michael's papers.
answered Nov 26 at 2:57
achille hui
95.2k5129256
add a comment |
The decomposition is possible because $2005 = 5cdot 401$ and both $5$ and $401$ are primes of the form $4k+1$. This allow $2005$ to be written as a sum of squares.
$$2005 = 22^2 + 39^2 = 18^2+41^2$$
For any integer $n = p^2 + q^2$ that can be written as a sum of squares.
Consider a right-angled triangle $ABC$ with
$$AB = psqrt{n}, AC = qsqrt{n}quadtext{ and }quad BC = n$$
Let $D$ on $BC$ be the foot of attitude at $A$. It is easy to
see $triangle DBA$ and $triangle DAC$ are similar to $triangle ABC$ with
$$BD = p^2, AD = pqquadtext{ and }quad CD = q^2$$
One can split $triangle DBA$ into $p^2$ and $triangle DAC$ into $q^2$
triangles with sides $p, q$ and $sqrt{n}$.
As an example, following is a subdivision of a triangle into $13 = 2^2 + 3^2$ congruent triangles.
In the literature, this is known as a biquadratic tiling of a triangle. For more information about subdividing triangles into congruent triangles, look at answers in this MO post. In particular, the list of papers by Michael Beeson there. The construction described here is based on what I have learned from one of Michael's papers.
answered Nov 26 at 2:57
achille hui
95.2k5129256
The decomposition is possible because $2005 = 5cdot 401$ and both $5$ and $401$ are primes of the form $4k+1$. This allow $2005$ to be written as a sum of squares.
$$2005 = 22^2 + 39^2 = 18^2+41^2$$
For any integer $n = p^2 + q^2$ that can be written as a sum of squares.
Consider a right-angled triangle $ABC$ with
$$AB = psqrt{n}, AC = qsqrt{n}quadtext{ and }quad BC = n$$
Let $D$ on $BC$ be the foot of attitude at $A$. It is easy to
see $triangle DBA$ and $triangle DAC$ are similar to $triangle ABC$ with
$$BD = p^2, AD = pqquadtext{ and }quad CD = q^2$$
One can split $triangle DBA$ into $p^2$ and $triangle DAC$ into $q^2$
triangles with sides $p, q$ and $sqrt{n}$.
As an example, following is a subdivision of a triangle into $13 = 2^2 + 3^2$ congruent triangles.
In the literature, this is known as a biquadratic tiling of a triangle. For more information about subdividing triangles into congruent triangles, look at answers in this MO post. In particular, the list of papers by Michael Beeson there. The construction described here is based on what I have learned from one of Michael's papers.
answered Nov 26 at 2:57
achille hui
95.2k5129256
edited Nov 27 at 0:23
answered Nov 26 at 2:57
achille hui
95.2k5129256
answered Nov 26 at 2:57
achille hui
95.2k5129256
answered Nov 26 at 2:57
achille hui
95.2k5129256
95.2k5129256
add a comment |
add a comment |
Since $2005$ is the sum of two squares, there exists this sort of triangle. In a contest setting, if you're trying to show $2005$ is a sum of two squares, realize that $2005=401cdot 5$. Since 401 and 5 are sums of squares, we know that their product is also a sum of squares.
Now in general if we have $n=x^2+y^2$, let our triangle be an $x$ by $y$ right triangle. Then, we can split this triangle along the altitude to the hypotenuse, which gives us two similar triangles with hypotenuses $x$ and $y$ respectively. Finally, we can decompose each of these into $x^2$ and $y^2$ similar right triangles with hypotenuses $1$, via a stretched version of the following picture.
answered Nov 26 at 2:47
Isaac Browne
4,60731132
I know this sum of squares is a sufficient condition, but I'm not sure if it's necessary. Any thoughts on this? As a particular example, would $3$ or $7$ work?
– Isaac Browne
Nov 26 at 2:49
4
The $30-60-90$ triangle can be divided into three triangles. Start by bisecting the 60 angle.
– Empy2
Nov 26 at 3:07
2
Alright, so $3$ works. I also just thought of the construction of connecting vertices of an equilateral triangle to the center, which would create three congruent triangles. Any others?
– Isaac Browne
Nov 26 at 3:11
6
It seems Michael Beeson has proved that $7$ is impossible recently. this appears on his web site, not yet available on arXiv...
– achille hui
Nov 26 at 4:00
1
@achillehui: Gosh — a lovely paper, and very hot off the press!
– Peter LeFanu Lumsdaine
Nov 26 at 10:21
add a comment |
Since $2005$ is the sum of two squares, there exists this sort of triangle. In a contest setting, if you're trying to show $2005$ is a sum of two squares, realize that $2005=401cdot 5$. Since 401 and 5 are sums of squares, we know that their product is also a sum of squares.
Now in general if we have $n=x^2+y^2$, let our triangle be an $x$ by $y$ right triangle. Then, we can split this triangle along the altitude to the hypotenuse, which gives us two similar triangles with hypotenuses $x$ and $y$ respectively. Finally, we can decompose each of these into $x^2$ and $y^2$ similar right triangles with hypotenuses $1$, via a stretched version of the following picture.
answered Nov 26 at 2:47
Isaac Browne
4,60731132
I know this sum of squares is a sufficient condition, but I'm not sure if it's necessary. Any thoughts on this? As a particular example, would $3$ or $7$ work?
– Isaac Browne
Nov 26 at 2:49
4
The $30-60-90$ triangle can be divided into three triangles. Start by bisecting the 60 angle.
– Empy2
Nov 26 at 3:07
2
Alright, so $3$ works. I also just thought of the construction of connecting vertices of an equilateral triangle to the center, which would create three congruent triangles. Any others?
– Isaac Browne
Nov 26 at 3:11
6
It seems Michael Beeson has proved that $7$ is impossible recently. this appears on his web site, not yet available on arXiv...
– achille hui
Nov 26 at 4:00
1
@achillehui: Gosh — a lovely paper, and very hot off the press!
– Peter LeFanu Lumsdaine
Nov 26 at 10:21
add a comment |
Since $2005$ is the sum of two squares, there exists this sort of triangle. In a contest setting, if you're trying to show $2005$ is a sum of two squares, realize that $2005=401cdot 5$. Since 401 and 5 are sums of squares, we know that their product is also a sum of squares.
Now in general if we have $n=x^2+y^2$, let our triangle be an $x$ by $y$ right triangle. Then, we can split this triangle along the altitude to the hypotenuse, which gives us two similar triangles with hypotenuses $x$ and $y$ respectively. Finally, we can decompose each of these into $x^2$ and $y^2$ similar right triangles with hypotenuses $1$, via a stretched version of the following picture.
answered Nov 26 at 2:47
Isaac Browne
4,60731132
Since $2005$ is the sum of two squares, there exists this sort of triangle. In a contest setting, if you're trying to show $2005$ is a sum of two squares, realize that $2005=401cdot 5$. Since 401 and 5 are sums of squares, we know that their product is also a sum of squares.
Now in general if we have $n=x^2+y^2$, let our triangle be an $x$ by $y$ right triangle. Then, we can split this triangle along the altitude to the hypotenuse, which gives us two similar triangles with hypotenuses $x$ and $y$ respectively. Finally, we can decompose each of these into $x^2$ and $y^2$ similar right triangles with hypotenuses $1$, via a stretched version of the following picture.
answered Nov 26 at 2:47
Isaac Browne
4,60731132
answered Nov 26 at 2:47
Isaac Browne
4,60731132
answered Nov 26 at 2:47
Isaac Browne
4,60731132
answered Nov 26 at 2:47
Isaac Browne
4,60731132
4,60731132
I know this sum of squares is a sufficient condition, but I'm not sure if it's necessary. Any thoughts on this? As a particular example, would $3$ or $7$ work?
– Isaac Browne
Nov 26 at 2:49
4
The $30-60-90$ triangle can be divided into three triangles. Start by bisecting the 60 angle.
– Empy2
Nov 26 at 3:07
2
Alright, so $3$ works. I also just thought of the construction of connecting vertices of an equilateral triangle to the center, which would create three congruent triangles. Any others?
– Isaac Browne
Nov 26 at 3:11
6
It seems Michael Beeson has proved that $7$ is impossible recently. this appears on his web site, not yet available on arXiv...
– achille hui
Nov 26 at 4:00
1
@achillehui: Gosh — a lovely paper, and very hot off the press!
– Peter LeFanu Lumsdaine
Nov 26 at 10:21
add a comment |
I know this sum of squares is a sufficient condition, but I'm not sure if it's necessary. Any thoughts on this? As a particular example, would $3$ or $7$ work?
– Isaac Browne
Nov 26 at 2:49
4
The $30-60-90$ triangle can be divided into three triangles. Start by bisecting the 60 angle.
– Empy2
Nov 26 at 3:07
2
Alright, so $3$ works. I also just thought of the construction of connecting vertices of an equilateral triangle to the center, which would create three congruent triangles. Any others?
– Isaac Browne
Nov 26 at 3:11
6
It seems Michael Beeson has proved that $7$ is impossible recently. this appears on his web site, not yet available on arXiv...
– achille hui
Nov 26 at 4:00
1
@achillehui: Gosh — a lovely paper, and very hot off the press!
– Peter LeFanu Lumsdaine
Nov 26 at 10:21
I know this sum of squares is a sufficient condition, but I'm not sure if it's necessary. Any thoughts on this? As a particular example, would $3$ or $7$ work?
– Isaac Browne
Nov 26 at 2:49
I know this sum of squares is a sufficient condition, but I'm not sure if it's necessary. Any thoughts on this? As a particular example, would $3$ or $7$ work?
– Isaac Browne
Nov 26 at 2:49
4
4
The $30-60-90$ triangle can be divided into three triangles. Start by bisecting the 60 angle.
– Empy2
Nov 26 at 3:07
The $30-60-90$ triangle can be divided into three triangles. Start by bisecting the 60 angle.
– Empy2
Nov 26 at 3:07
2
2
Alright, so $3$ works. I also just thought of the construction of connecting vertices of an equilateral triangle to the center, which would create three congruent triangles. Any others?
– Isaac Browne
Nov 26 at 3:11
Alright, so $3$ works. I also just thought of the construction of connecting vertices of an equilateral triangle to the center, which would create three congruent triangles. Any others?
– Isaac Browne
Nov 26 at 3:11
6
6
It seems Michael Beeson has proved that $7$ is impossible recently. this appears on his web site, not yet available on arXiv...
– achille hui
Nov 26 at 4:00
It seems Michael Beeson has proved that $7$ is impossible recently. this appears on his web site, not yet available on arXiv...
– achille hui
Nov 26 at 4:00
1
1
@achillehui: Gosh — a lovely paper, and very hot off the press!
– Peter LeFanu Lumsdaine
Nov 26 at 10:21
@achillehui: Gosh — a lovely paper, and very hot off the press!
– Peter LeFanu Lumsdaine
Nov 26 at 10:21
add a comment |
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