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?

  • $\begingroup$ Note: This is from APMO 2005. $\endgroup$
    – Calvin Lin
    Jan 17 at 16:41

2 Answers 2


The decomposition is possible because $2005 = 5\cdot 401$ 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 $$AB = p\sqrt{n}, AC = q\sqrt{n}\quad\text{ and }\quad BC = n$$ 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

$$BD = p^2, AD = pq\quad\text{ and }\quad CD = q^2$$ 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.


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 $2005=401\cdot 5$. 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

  • $\begingroup$ 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? $\endgroup$ Nov 26, 2018 at 2:49
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    $\begingroup$ The $30-60-90$ triangle can be divided into three triangles. Start by bisecting the 60 angle. $\endgroup$
    – Empy2
    Nov 26, 2018 at 3:07
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    $\begingroup$ 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? $\endgroup$ Nov 26, 2018 at 3:11
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    $\begingroup$ It seems Michael Beeson has proved that $7$ is impossible recently. this appears on his web site, not yet available on arXiv... $\endgroup$ Nov 26, 2018 at 4:00
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    $\begingroup$ @achillehui: Gosh — a lovely paper, and very hot off the press! $\endgroup$ Nov 26, 2018 at 10:21

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