A right-angled triangle has side lengths that are integers, which have no common divisor greater than 1. The radius of the circle tangent to the hypotenuse is 420 units. Determine the lengths of the sides.
The side lengths form a Pythagorean triple, specifically a primitive one, so there exist relatively prime, different parity positive integers $m, n$, with $m > n$, such that the sides are

$$
a=m^{2}-n^{2}, \quad b=2 m n, \quad c=m^{2}+n^{2}
$$

where $c$ is the hypotenuse. 1.

![](https://cdn.mathpix.com/cropped/2024_05_02_78823e638c905ecd0fedg-1.jpg?height=486&width=591&top_left_y=421&top_left_x=759)

It is known, on the other hand, that in every right triangle, the radius of the circle tangent to the hypotenuse, $O E_{a}=O E_{b}$, is equal to half the perimeter (see the notation in the figure). Indeed, due to the right angle, $O, E_{a}, C, E_{b}$ are the vertices of a square, so the radius is equal to the distance $C E_{b}=C E_{a}$ from the right angle vertex to the points of tangency on the legs. This distance, however, is equal to half the perimeter in any triangle, because the lengths of the tangents from an external point to a circle are equal, so

$$
\begin{gathered}
C E_{b}+C E_{a}=C A+A E_{b}+C B+B E_{a}=C A+C B+\left(A E_{c}+B E_{c}\right)= \\
=b+a+c=2 s
\end{gathered}
$$

thus $C E_{b}=s$.

Accordingly, in our case, expressed in terms of $m, n$,

$$
\varrho_{c}=m^{2}+m n=m(m+n)=420
$$

Here, $m+n$ is certainly odd, as the sum of an even and an odd number, so $m$ is even, and in fact divisible by 4. Furthermore, $m+n > m$, since $n$ is positive, so $m$ is smaller, and $m+n$ is larger, than the square root of 420, which is $20.4 \ldots$, i.e., $m \leq 20$ and $m+n \geq 21$. On the other hand, $n < 420, \quad m > \sqrt{210}=14.4 \ldots
$$

so $m \geq 16$.

Accordingly, $m$ can only be 16 or 20. However, 420 is not divisible by 16; with the other possibility,

$$
m=20, \quad m+n=21, \quad n=1
$$

and thus from (1) $a=399, b=40, c=401$.

Jenő Lehel (Budapest, Apáczai Csere J. g. II. o. t.)

Remarks. 1. With similar reasoning, we can reach the goal using the formulas for primitive Pythagorean triples

$$
a=u v, \quad b=\frac{u^{2}-v^{2}}{2}, \quad c=\frac{u^{2}+v^{2}}{2}
$$

where $u, v$ are relatively prime, odd, and $u > v$.

2. Most solutions chose the appropriate pair of integers satisfying (2), or the corresponding equality in the indicated second method, with longer reasoning. [^0]

[^0]:    ${ }^{1}$ See K.M.L. (1961) 3. footnote

    ${ }^{2}$ There