Circle inscribed in Pythagorean triangle

Question

Given the question (from Burton):

"For an arbitrary positive integer $n$, show that there exists a Pythagorean triangle the radius of whose inscribed circle is $n$."

My solution is $3n$,$4n$,$5n$ while textbook hints at a solution of $2n+1$,$2n^2+2n$,$2n^2+2n+1$

The latter seems to describe a Pythagorean triangle with side lengths that form a primitive Pythagorean triple while mine obviously does not except for the case $n=1$. Is that what is expected if the question does not specify? How does one approach generating such a primitive solution?

The text only says " let us define a Pythagorean triangle to be a right triangle whose sides are of integral length"

Answer 1

Your example fully answers the question as put.

For a primitive triple, suppose it has the standard representation $s^2-t^2,2st,s^2+t^2$ where $s$ and $t$ obey the usual conditions. Then the perimeter $p$ is $2st+2s^2$, and the area $A$ is $st(s^2-t^2)$.

If $r$ is the radius of the incircle, then $rp/2=A$, so $rp=2A$. For an incircle of radius $n$, we want $n(2st+2s^2)=2st(s^2-t^2)$. Cancellation simplifies this to $n=t(s-t)$.

The simplest way to achieve this is to let $s=n+1$ and $t=n$. That ensures $s$ and $t$ are relatively prime and of opposite parity. For $n$ odd we can let $t=1$ and $s=n+1$.

These are the only possibilities for $n=1$ or $n$ an odd prime. Other alternatives are available for $n$ with a more complicated multiplicative structure.

Answer 2

For a triangle with sides $a,b,c$ and area $E,$ and inradius $r,$ let $s =(a+b+c)/2.$ We have $E=rs.$ Now if $a^2+b^2=c^2$ then $E=a b/2$ and $r=E/s=a b/(a+b+c).$ When also $a,b,c$ are positive integers, there are positive integers $k,d,e$ with $\{a,b\}=\{2 k d e, k(d^2-e^2)\}$ and $c=k(d^2+e^2).$ $$\text {This gives}\quad r=\frac {a b}{a+b+c}=\frac {2 k^2 d e (d^2-e^2)}{2 k(d e +d^2)}=k e (d-e).$$ Now given positive integer $n$ let $n=2^p q$ where $p$ is a non-negative integer and $q$ is an odd integer.$$\text {Let }\; k=1,\;e=2^p\text { and}\; d=e+q=2^p+q.$$ Then $r=n$ and $\gcd (d,e)=1, $ while $d,e$ are not both odd, so $\{a,b,c\}=\{2 d e, d^2-e^2,d^2+e^2\}$ is a primitive Pythagorean triplet..... Remark: To show that in any triangle ABC with area E, we have $rs=E \;$: Let $O$ be the center of the incircle $\Sigma$ and let $\Sigma$ be tangent to $BC$ at $A^*$. Since $A^*O$ is perpendicular to $BC$, the area of triangle $BOC$ is $(1/2)(BC)(A^*O)=(1/2)r a.$ Similarly the areas of COA and AOC are $(1/2)r b$ and $(1/2)r c.$ Adding these we have area $E=(1/2)(r a+r b+ rc)=rs.$

Answer 3

The given the question from Burton is:
"For an arbitrary positive integer $n$, show that there exists a Pythagorean triangle the radius of whose inscribed circle is $n$."

That means there is a series of Pythagorean triangles whose inradii is $n$, means either $1,2,3,....,n$.

From Euclid's formula, if $m$ and $n$ $(m \gt n)$are positive integer co-primes with different parity, you can achieve all primitive Pythagorean triangles $(a,b,c)$ where $a = m^2-n^2$, $b=2mn$, and $c= m^2+n^2$. Suppose $m=n+1$ where $n$ is all positive integers $(n=1,2,3,..., \infty)$. Thus, $$a = m^2-n^2 = (n+1)^2 - n^2 =(2n+1)(1) = 2n+1$$ $$b = 2mn = 2(n+1)n = 2n(n+1)$$ $$c= m^2+n^2=(n+1)^2+n^2= n^2 + 2n +1 +n^2=2n^2+2n+1$$ Thus, suppose following right triangle has following measurements:

$BC = a$, $AC = b$, and $AB = c$. The perimeter $(p)$ of the triangle is: $$p=a+b+c=(2n+1)+(2n^2+2n)+(2n^2+2n+1)=4n^2+6n+2$$ The area $(q)$ of the triangle is: $$q= \frac12 \times a \times b = \frac12 \times (2n+1) \times (2n^2+2n) = n(2n+1)(n+1) $$ Also, you find: $OF = OD =OE = r$ where $r=$ the radius of inscribed circle, and $$q = BOC + AOC + AOB = \frac12 (a \times r + b \times r +c \times r) = \frac12 \times r(a + b +c)\\ = \frac12 \times rp = \frac12 \times r(4n^2+6n+2)= r(2n^2+2n+1)$$ $$\therefore \ \ r(2n^2+3n+1) = n(2n+1)(n+1)$$ $$r(2n+1)(n+1) = n(2n+1)(n+1)$$ $$\therefore \ \ r = \frac{n(2n+1)(n+1)}{(2n+1)(n+1)} = n$$

Thereore, for an arbitrary positive integer $n$, there exists a Pythagorean triangle $(a,b,c)$, the radius of whose inscribed circle is $n$. The $(a,b,c)$ is describe with $n$ as $(2n+1,2n(n+1),2n^2+2n+1)$, that is what your text book provided.