Prove that there exists a triangle which can be cut into 2005 congruent triangles.
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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
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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
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up vote
2
down vote
favorite
up vote
2
down vote
favorite
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
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 2 hours ago
Akash Roy
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asked 2 hours ago
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The decomposition is possible because $2005 = 5cdot 401$ and both $5$ and $401$ are prime of the form $4k+1$. This allow $2005$ can 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 passing through $A$. It is easy to
check $triangle DBA$ and $triangle DAC$ are similar to $triangle ABC$ with
$$AB = p^2, AD = pqquadtext{ and }quad CD = q^2$$
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.
add a comment |
up vote
2
down vote
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.
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
1 hour ago
1
The $30-60-90$ triangle can be divided into three triangles. Start by bisecting the 60 angle.
– Empy2
1 hour ago
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
1 hour ago
It seems Michael Beeson has proved that $7$ is impossible recently. this appears on his web site, not yet available on arXiv...
– achille hui
20 mins ago
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
The decomposition is possible because $2005 = 5cdot 401$ and both $5$ and $401$ are prime of the form $4k+1$. This allow $2005$ can 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 passing through $A$. It is easy to
check $triangle DBA$ and $triangle DAC$ are similar to $triangle ABC$ with
$$AB = p^2, AD = pqquadtext{ and }quad CD = q^2$$
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.
add a comment |
up vote
3
down vote
The decomposition is possible because $2005 = 5cdot 401$ and both $5$ and $401$ are prime of the form $4k+1$. This allow $2005$ can 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 passing through $A$. It is easy to
check $triangle DBA$ and $triangle DAC$ are similar to $triangle ABC$ with
$$AB = p^2, AD = pqquadtext{ and }quad CD = q^2$$
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.
add a comment |
up vote
3
down vote
up vote
3
down vote
The decomposition is possible because $2005 = 5cdot 401$ and both $5$ and $401$ are prime of the form $4k+1$. This allow $2005$ can 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 passing through $A$. It is easy to
check $triangle DBA$ and $triangle DAC$ are similar to $triangle ABC$ with
$$AB = p^2, AD = pqquadtext{ and }quad CD = q^2$$
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.
The decomposition is possible because $2005 = 5cdot 401$ and both $5$ and $401$ are prime of the form $4k+1$. This allow $2005$ can 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 passing through $A$. It is easy to
check $triangle DBA$ and $triangle DAC$ are similar to $triangle ABC$ with
$$AB = p^2, AD = pqquadtext{ and }quad CD = q^2$$
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.
edited 35 mins ago
answered 1 hour ago
achille hui
93.7k5127251
93.7k5127251
add a comment |
add a comment |
up vote
2
down vote
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.
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
1 hour ago
1
The $30-60-90$ triangle can be divided into three triangles. Start by bisecting the 60 angle.
– Empy2
1 hour ago
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
1 hour ago
It seems Michael Beeson has proved that $7$ is impossible recently. this appears on his web site, not yet available on arXiv...
– achille hui
20 mins ago
add a comment |
up vote
2
down vote
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.
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
1 hour ago
1
The $30-60-90$ triangle can be divided into three triangles. Start by bisecting the 60 angle.
– Empy2
1 hour ago
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
1 hour ago
It seems Michael Beeson has proved that $7$ is impossible recently. this appears on his web site, not yet available on arXiv...
– achille hui
20 mins ago
add a comment |
up vote
2
down vote
up vote
2
down vote
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.
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 1 hour ago
Isaac Browne
4,47231031
4,47231031
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
1 hour ago
1
The $30-60-90$ triangle can be divided into three triangles. Start by bisecting the 60 angle.
– Empy2
1 hour ago
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
1 hour ago
It seems Michael Beeson has proved that $7$ is impossible recently. this appears on his web site, not yet available on arXiv...
– achille hui
20 mins ago
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
1 hour ago
1
The $30-60-90$ triangle can be divided into three triangles. Start by bisecting the 60 angle.
– Empy2
1 hour ago
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
1 hour ago
It seems Michael Beeson has proved that $7$ is impossible recently. this appears on his web site, not yet available on arXiv...
– achille hui
20 mins ago
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
1 hour ago
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
1 hour ago
1
1
The $30-60-90$ triangle can be divided into three triangles. Start by bisecting the 60 angle.
– Empy2
1 hour ago
The $30-60-90$ triangle can be divided into three triangles. Start by bisecting the 60 angle.
– Empy2
1 hour ago
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
1 hour ago
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
1 hour ago
It seems Michael Beeson has proved that $7$ is impossible recently. this appears on his web site, not yet available on arXiv...
– achille hui
20 mins ago
It seems Michael Beeson has proved that $7$ is impossible recently. this appears on his web site, not yet available on arXiv...
– achille hui
20 mins ago
add a comment |
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