Prove that there exists a triangle which can be cut into 2005 congruent triangles.












25














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?






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    25














    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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      25












      25








      25


      10





      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







      share|cite|improve this question















      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






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      share|cite|improve this question




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      edited Nov 26 at 2:08









      Akash Roy

      1




      1










      asked Nov 26 at 1:59









      mathnoob

      1,759422




      1,759422






















          2 Answers
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          active

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          46














          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.



          tiling by 13 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.






          share|cite|improve this answer































            12














            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.



            Triangle Tiling






            share|cite|improve this answer





















            • 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











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            2 Answers
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            46














            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.



            tiling by 13 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.






            share|cite|improve this answer




























              46














              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.



              tiling by 13 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.






              share|cite|improve this answer


























                46












                46








                46






                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.



                tiling by 13 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.






                share|cite|improve this answer














                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.



                tiling by 13 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.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Nov 27 at 0:23

























                answered Nov 26 at 2:57









                achille hui

                95.2k5129256




                95.2k5129256























                    12














                    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.



                    Triangle Tiling






                    share|cite|improve this answer





















                    • 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
















                    12














                    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.



                    Triangle Tiling






                    share|cite|improve this answer





















                    • 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














                    12












                    12








                    12






                    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.



                    Triangle Tiling






                    share|cite|improve this answer












                    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.



                    Triangle Tiling







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    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


















                    • 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


















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