Let's calculate the area of a right triangle if the lengths of the segments $m$ and $n$ established by the point of tangency of the inscribed circle on the hypotenuse are known.
I. solution: For the inscribed circle, the equality of the tangent segments drawn from the vertices and the fact that the triangle is right-angled result in the lengths of the legs and the hypotenuse being (Fig. 1):

![](https://cdn.mathpix.com/cropped/2024_05_02_350d994ede0a9e5da257g-1.jpg?height=490&width=634&top_left_y=243&top_left_x=732)

$$
m+\varrho, \quad n+\varrho, \quad m+n
$$

Thus, twice the area we are looking for is the product of the legs:

$$
2 t=(m+\varrho)(n+\varrho)=m n+(m+n+\varrho) \varrho
$$

From here, we need to eliminate $\varrho$. Another general relationship involving the area and $\varrho$ is $t=s \varrho$, but this introduces the semiperimeter $s$ as another unknown into our calculations. However, based on (1), $s$ can also be expressed in terms of $m, n, \varrho$: $s=m+n+\varrho$. Moreover, the second product on the right side of (2) is precisely $s \varrho=t$, which allows us to eliminate $\varrho$ in one step. Substituting this and rearranging, the formula for the area we are looking for is $t=m n$.

Máté Zsolt (Szeged, Radnóti M. Gym. II. Class)

Remark. The Pythagorean theorem also provides a relationship between the given data and the $\varrho$ to be eliminated, using (1): $(m+\varrho)^{2}+(n+\varrho)^{2}=(m+n)^{2}$. Simplifying, halving, and factoring, we get $(m+n+\varrho) \varrho=m n$. Substituting this into (2) (or recognizing $s \varrho=t$ on the left side), we get $t=m n$.

Pósa Lajos (Bp. XIII., Sziget utcai Secondary School V. Class)

II. solution: It occurs to us that among the factors of the product under the square root in Heron's formula, the three differences are the lengths of the tangent segments drawn from the vertices to the inscribed circle, in our case $s-a=m, s-b=n$, $s-c=\varrho$. Substituting these into the formula and recognizing another expression for the area in the product $s \varrho$,

$$
t=\sqrt{s(s-c)(s-a)(s-b)}=\sqrt{s \varrho m n}=\sqrt{t m n}
$$

and from this, by squaring and dividing, we get $t=m n$.

Török László (Ózd, József A. Gym. I. Class)

III. solution: We can also solve the problem using the method of area transformation. Complete the right-angled triangle $ABC$ to a rectangle with vertex $D$, extend the tangent radii $OA'$ and $OB'$ of the inscribed circle to the intersections $E$ and $F$ on sides $AD$ and $BD$, respectively, denote the intersection of $A'E$ and $B'F$ with the diagonal $AB$ by $G$ and $H$, respectively, and the point of tangency of the circle on $AB$ by $I$ (Fig. 2).

![](https://cdn.mathpix.com/cropped/2024_05_02_350d994ede0a9e5da257g-1.jpg?height=431&width=590&top_left_y=1915&top_left_x=754)

Then, using the letters to denote the corresponding areas,

$$
A B C=B A D=B H F+F H G E D+G A E=O H I+F H G E D+G O I=F O E D
$$

indeed, the right-angled triangles $B H F$ and $O H I$, as well as $G A E$ and $G O I$, are congruent because the acute angles at $H$ and $G$ are vertical angles, and the legs opposite them are equal to the radius of the inscribed circle. Therefore,

$$
t=F O \cdot O E=B A^{\prime} \cdot B^{\prime} A=B I \cdot I A=m n
$$

Mezey Ferenc (Bp. II., Rákóczi F. Gym. II. Class)