There’s Only One Inradius For Sixteen Triangles? (Take All)

A primitive Pythagorean triple is formed by sides $p^2-q^2$ , $2pq$ , $p^2+q^2$ where $p,q$ are coprime positive integers of opposite parities. The area of the associated right-angled triangle is $pq(p^2-q^2)$ , and the semiperimeter is $p(p+q)$ , so that the inradius is $r = q(p-q)$ . Note that $p-q$ has to be odd, although $q$ can be even. Thus $r$ must have exactly $16$ odd factors that can be equal to $q$ . Since $p-q$ and $q$ are coprime, it follows that $r$ has exactly $4$ odd prime factors.

Now add the requirement that $1 \le r \le 3000$ . Since $3^2 \times 5 \times 7 \times 11 > 3000$ , it follows that $r$ has no repeated odd prime factors. Since $4 \times 3 \times 5 \times 7 \times 11 > 3000$ , $r$ can contain at most one factor of $2$ . Thus the nine possible values of $r$ are $\begin{array}{rclcrclcrcl} 1155 & = & 3 \times 5 \times 7 \times 11 & \hspace{1cm} & 2310 & = & 2 \times 3 \times 5 \times 7 \times 11 & \hspace{1cm} & 1365 & = & 3 \times 5 \times 7 \times 13 \\ 2730 & = & 2 \times3 \times 5 \times 7 \times 13 & & 1785 & = & 3 \times 5 \times 7 \times 17 & & 1995 & = & 3 \times 5 \times 7 \times 19 \\ 2415 & = & 3 \times 5 \times 7 \times 23 & & 2145 & = & 3 \times 5\times 11 \times 13 & & 2805 & = & 3 \times 5 \times 11 \times 17 \end{array}$ and the sum of these numbers is $\boxed{18705}$ .

Draw a rectangular triangle $ABC$ with its right angle at $C$ and opposite sides $a,b,c$ . Draw its incircle with centre $M$ and radius $r$ , Let it touch the sides $BC, CA$ and $AB$ in $D, E$ and $F$ respectively. Notice that $AME \cong AMF$ and $BMD \cong BMF$ . Therefore $BF=BD=a-r, AF=AE=b-r, AB=c=a+b-2r$ .

So
$a^2+b^2=c^2=(a+b-2r)^2$ $0=2ab-4r(a+b)+4r^2$ $b=\frac{2r(a-r)}{a-2r}$

Fom $c=a+b-2r$ it was already clear that 2r had to be an integer, from $b(a-2r)=2r(a-r)$ we now see that r itself has to be an integer.

If we want $a \le b$ , we can set bounds to a: $2r \lt a \le (2+\sqrt{2})r$

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import math

def PrimitiveTriplesForRadius(r):
    count=0
    for a in range(2*r+1, 1+math.floor((math.sqrt(2)+2.0)*r)):
        if 2*r*(a-r) % (a-2*r) == 0:
            b = 2*r*(a-r) // (a-2*r)
            c = a+b-2*r
            if math.gcd(a, math.gcd(b,c)) == 1:
                count += 1
    return count

sum = 0; r_16 = []
for r in range(1, 3001):
    if PrimitiveTriplesForRadius(r) == 16:
        sum += r
        r_16.append(r)
print (sum, r_16)

output:

1

18705 [1155, 1365, 1785, 1995, 2145, 2310, 2415, 2730, 2805]