Geometry and Elementary Number Theory

$\textbf{Solution}:$ (for junior form students)

Both $m$ and $n$ must be positive, for $(3,3)$ lies inside $\triangle OAB$ . Since both $y$ axis and $x$ axis are tangent to the inscribed circle of $\triangle OAB$ , the in-radius is $r = 3$ .

As shown in the figure above, we have $OM = ON = 3$ . So $AM = n - 3$ and $BN = m - 3$ . By tangent properties, $AP = AM = n -3$ and $BP = BN = m - 3$ respectively, so

$AB = AP + PB = (n-3) + (m-3) = m+n-6$ .

Since $\angle AOB = 90^{\circ}$ , we have $AB = \sqrt{m^2 + n^2}$ . Therefore

$m+n- 6 = \sqrt{m^2 + n^2}$ .

Squaring this on both sides and then cancelling the common terms on both sides yield

$2mn - 12m - 12n + 36 = 0 \Longrightarrow mn - 6m - 6n + 18 = 0 \Longrightarrow (m-6)(n-6) = 18$ .

From the last equation, both $m$ and $n$ must be greater than $6$ geometrically. Now, as both $m-6$ and $n-6$ are positive and $(m-6)(n-6) = 18 = 2^1 \times 3^2$ , unique prime factorization tells us that there are $(1+1)(1+2) = 6$ pairs of $(m-6, n-6)$ for that. Translating $(m-6,n-6)$ to $(m,n)$ is a one-one correspondence. Therefore, there are precisely $6$ solutions, as in terms of the number of pairs of integers $(m,n)$ , to the problem. $\Box$

Side-note: these $6$ triangles are classified as Pythagorean triangles with three distinct Pythagorean triples $(9,12,15), (8,15,17)$ and $(7,24,25)$ respectively.