yptsoi has an ingenious design for this problem, so I'll leave this for him

mecrazywong wrote: yptsoi has an ingenious design for this problem, so I'll leave this for him It is not ingenious at all... but anyway here is my solution: $2005=41^2+18^2$, so a right-angled triangle with adjacent sides lengths $1681$ and $738$ can be cut into $41^2$ congruent right-angled triangle with adjacent sides lengths $41$ and $18$. Similarly a right-angled triangle with adjacent sides lengths $738$ and $324$ can be cut into $18^2$ congruent right-angled triangle with adjacent sides lengths $41$ and $18$. Combining these two triangles gives another big right-angled triangle which can be cut into $2005$ congruent triangles. Therefore such triangle exists.

If I only realized that 2005=41 ² +18 ² ..... Nice solution, nevertheless, though I believe this is about the only way to solve the problem...

Also $2005=39^2+22^2$. To find these you can use Gaussian integers (although it is something I realized the next day...): $2005=5 \times 401=(2+i)(2-i)(20+i)(20-i)$ Now multiplying out in this fashion: 1st-3rd, 2nd-4th you get the pair $39, 22$, while doing 1st-4th, 2nd-3rd gives $41, 18$.

Severius wrote: Also $2005=39^2+22^2$. To find these you can use Gaussian integers (although it is something I realized the next day...): $2005=5 \times 401=(2+i)(2-i)(20+i)(20-i)$ Now multiplying out in this fashion: 1st-3rd, 2nd-4th you get the pair $39, 22$, while doing 1st-4th, 2nd-3rd gives $41, 18$. Severius: this is a nice method to decompose an integer as the sum of two squares, but it works only for very special cases: for example you cannot apply it to prove that $73=8^2+3^2$ and $97=9^2+4^2$. Do you know a general algorithm which allows us to decompose an integer as the sum of two squares.

A number can be expressed as the sum of two squares if and only if it does not contain prime factor of the form 4k+3 Still it is clearly possible for 3 or 6. I don't know whether that is possible for 7 or not.

Darkseer wrote: A number can be expressed as the sum of two squares if and only if it does not contain prime factor of the form 4k+3 Still it is clearly possible for 3 or 6. I don't know whether that is possible for 7 or not. Not quite. A number can be expressed as the sum of two squares iff every prime factor of the form 4k+3 is raised to an even power.

yes, that's right... and to al.M.V., i don't think it was necesary to realize that $2005=41^2+18^2$.. Just make the construction in general (i.e., for $n=a^2+b^2$) and then since $2005=4k+1$ it can be written in the form $a^2+b^2$ for some $a,b$, so the triangle exists (it didn't say you had to put an example of such triangle.. just show its existance) ...

That's right, billzhao. Thanks for pointing out my mistake. By the way, 21=4*5+1 can not be expressed as the sum of two squares.

yes sorry... i wanted to say that the prime divisors were of that form (2005 is not a prime!!! ahh!! ), $2005=5.401$ (and hope 401 is prime!) ...

$ 2005=5*401=(2^2+1)(20^2+1) =40^2+20^2+2^2+1 =(40-1)^2*40+20^2+2^2 =39^2+22^2 $

2005 is 5*401, and both 5 and 401 are 1 mod 4, so 2005 is a sum of two squares. Consider an acute triangle $\triangle ABC$. Let $D$ be the foot from $A$ to $BC$. Split right triangles $\triangle ABD$ and $\triangle ACD$ into $b^2$ and $c^2$ smaller similar triangles respectively such that $b^2+c^2=2005.$

john0512 wrote: 2005 is 5*401, and both 5 and 401 are 1 mod 4, so 2005 is a sum of two squares. Consider an acute triangle $\triangle ABC$. Let $D$ be the foot from $A$ to $BC$. Split right triangles $\triangle ABD$ and $\triangle ACD$ into $b^2$ and $c^2$ smaller similar triangles respectively such that $b^2+c^2=2005.$ how do you know those triangles are congruent?

David_Kim_0202 wrote: john0512 wrote: 2005 is 5*401, and both 5 and 401 are 1 mod 4, so 2005 is a sum of two squares. Consider an acute triangle $\triangle ABC$. Let $D$ be the foot from $A$ to $BC$. Split right triangles $\triangle ABD$ and $\triangle ACD$ into $b^2$ and $c^2$ smaller similar triangles respectively such that $b^2+c^2=2005.$ how do you know those triangles are congruent? oh nvm I misread the problem