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Total number of right angle triangles whose inradius is $2013$ and sides are integer

Attempt: assuming that $a,b,c>0$ are the sides of a triangle.

so form a right angle triangle with sides $a,b,c$ and right angle at $C$

$\displaystyle r=(s-c)\tan \frac{C}{2}=(s-c)=\frac{a+b-c}{2}$ and $c^2=a^2+b^2$

$\displaystyle 2r = a+b-\sqrt{a^2+b^2}$ so $(\sqrt{a^2+b^2})^2 = (a+b-2r)$

$\displaystyle a^2+b^2 = (a+b)^2+4r^2-4r(a+b) = a^2+b^2+2ab-4r(a+b)$

$\displaystyle ab-2r(a+b)=0$

Could someone help me how to solve it, thanks.

gymbvghjkgkjkhgfkl
  • 8,090
DXT
  • 11,241

3 Answers3

1

$\displaystyle a^2+b^2 = (a+b)^2+4r^2-4r(a+b) = a^2+b^2+2ab-4r(a+b)$

This should be $$a^2+b^2=(a+b)^2+4r^2-4r(a+b)=a^2+b^2+2ab\color{red}{+4r^2}-4r(a+b)$$ from which we have $$(a-2r)(b-2r)=2r^2=2^1\times 3^2\times 11^2\times 61^2\tag1$$ where $r=2013=3\times 11\times 61$.

We may suppose that $$-2r\lt a-2r\le b-2r\tag2$$

Note that a necessary and sufficient condition is $(1)$ and $$(c=)\ a+b-2r\gt 0\iff (a-2r)+(b-2r)\gt -2r\tag3$$

  • If both $a-2r$ and $b-2r$ are negative, then, from $(1)$ and $(2)$, there is only one case $(a-2r,b-2r)=(-3721,-2178)$. However, this is not sufficient since this does not satisfy $(3)$.

  • If both $a-2r$ and $b-2r$ are positive, then, from $(1)$ and $(2)$, the number of cases is the half of the number of the positive divisors of $2^1\times 3^2\times 11^2\times 61^2$, and every case is sufficient since every case satisfies $(3)$.

Therefore, the answer is $$\frac{(1+1)\times (2+1)\times (2+1)\times (2+1)}{2}=\color{red}{27}$$

mathlove
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In this post, I have proven that in Euclid's formula, if you use $m=n+1$, you get infinite number of primitive Pythagorean triples, $(2n+1,2n(n+1),2n^2+2n+1)$. The inradius of these triangles is equal to integer $n$. Thus, if you need to find a Pythagorean triple with inradius $2013$, just use $n=2013$.So: $$a=2n+1= 2\times 2013+1=4027$$ $$b=2n(n+1)= 2\times 2013(2013+1)=8108364$$ $$c=2n^2+2n+1= 2\times 2013^2 + 2\times 2013+1=8108365$$

On the other hand, using Euclid's formula, you can generate infinite number of primitive Pythagorean triples, using different $n$ and $m$ values. Using the similar approach in this post, you can prove that inradius of each triangle is equal to $n(m-n)$. The factors of $2013$ is: $$2013=1,3,11,33,61,183,671,2013$$ Since $r=n(m-n)$, $n$ can be only $1,3, 11,$ and $33$ because $n \lt m$. Thus, You can fond following triangles with all having inradius of $2013$: $$n=1,m=2013+1=2014$$ $$n=3,m=671+3=674$$ $$n=11,m=183+11=194$$ $$n=33,m=61+33=94$$ Thus, total number of right angle triangles whose inradius is $2013$ and all sides are integers is $5$.

Mathew Mahindaratne
  • 817
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Remember the Pythagorian triple $(a,b,c)$ where $$c=m^2+n^2, b=m^2-n^2,a=2mn. $$ We know that such triple defines a right triangle. The additional condition $$ab=2r(a+b)-2r^2 $$ implies that $$2mn(m^2-n^2)=2r(m^2-n^2+2mn)-2r^2 .$$ Now finding $m $ and $n$ answers your question.

  • However this is not discard the solution. –  Mar 01 '17 at 14:47
  • Setting $a=2mn,b=m^2-n^2,c=m^2+n^2$ does not give all solutions. For example, $(a,b,c)=(4032,1354749,1354755)$ is a solution, but there are no integers $m,n$ such that $2mn=4032,m^2-n^2=1354749,m^2+n^2=1354755$. – mathlove Mar 03 '17 at 07:07