I. solution. Let the midpoint of the hypotenuse $AB = c$ be $F$, the radius of the inscribed circle be $\varrho$, and the points of tangency on $AB$ and $AC$ be $D$ and $E$, respectively. Let $AF$ be the segment that, according to the condition, is seen at a right angle. Then, in the right triangle $AOF$, by the altitude theorem (1st figure),

$$
DO^2 = AD \cdot DF = \varrho^2
$$

# 1984-11-391-1.eps 

1st figure

We also know that

$$
\varrho = OE = CE = s - c = \frac{a + b - c}{2}
$$

and

$$
DA = s - a = \frac{b + c - a}{2}
$$

and

$$
DF = \frac{c}{2} - AD = \frac{a - b}{2}
$$

Substituting the expressions for $AD$, $DF$, and $\varrho$ into (1),

$$
\left(\frac{b + c - a}{2}\right)\left(\frac{a - b}{2}\right) = \left(\frac{a + b - c}{2}\right)^2
$$

which simplifies to $3ac + bc = 2a^2 + 2b^2 + c^2$. From this and the relation $a^2 + b^2 = c^2$,

$$
b = 3c - 3a
$$

Substituting this into $b^2 = c^2 - a^2$ and simplifying,

$$
5\left(\frac{a}{c}\right)^2 - 9\left(\frac{a}{c}\right) + 4 = 0
$$

From this equation, $\frac{a}{c} = 1$ or $\frac{a}{c} = \frac{4}{5}$. The first value clearly does not satisfy the condition, since $c > a$. If $\frac{a}{c} = \frac{4}{5}$, then $\frac{b}{c} = \frac{3c - 3a}{c} = \frac{3}{5}$, so the ratio of the sides of the right triangle is $3:4:5$. With these values, (2) is satisfied, and thus (1) is also satisfied, meaning that the segment $AF$ is indeed seen at a right angle from $O$.

II. solution. Let the midpoint of the hypotenuse $AB$ be $F$, and assume that $O$ is closer to $A$ than to $B$ (2nd figure).

## 1984-11-391-2.eps

2nd figure

The segments $OA$, $OB$, and $OC$ are the angle bisectors of the triangle, so

$$
\angle AOB = 180^\circ - (\angle OAB + \angle OBA) = 180^\circ - \left(\frac{\alpha}{2} + \frac{\beta}{2}\right) = 180^\circ - 45^\circ = 135^\circ,
$$

since the triangle is right-angled at $C$ and thus $\alpha + \beta = 90^\circ$. According to the condition, $\angle AOF = 90^\circ$, so $\angle BOF = 135^\circ - 90^\circ = 45^\circ$. Enlarging the triangle $BOF$ by a factor of 2 from $B$, $F$ moves to $A$, and $O$ moves to, say, $P$. Due to the enlargement,

$$
AP = 2 \cdot FO \quad \text{and} \quad \angle BPA = \angle OPA = \angle BOF = 45^\circ.
$$

In the triangle $OAP$, the angle at $P$ is $45^\circ$, and the angle at $O$ is $180^\circ - \angle BOA = 45^\circ$. Therefore, $OAP$ is isosceles: $AO = AP$, so

$$
AO : OF = AP : OF = 2 : 1
$$

Let the points of tangency of the inscribed circle on the sides $AB$, $BC$, and $CA$ be $R$, $S$, and $T$, respectively. The right triangles $AOF$ and $ARO$ are similar, as they share the angle at $A$, so $AR : RO = AO : OF = 2 : 1$, hence

$$
AR = AT = 2OR = 2\varrho
$$

where $\varrho$ is the radius of the inscribed circle. The quadrilateral $OSCT$ has right angles at $S$, $C$, and $T$, and $OS = OT = \varrho$, so it is a square, and thus

$$
CS = CT = \varrho
$$

Finally, we calculate the unknown segment $BR = BS = x$ using the Pythagorean theorem:

$$
(AR + x)^2 = (CS + x)^2 + CA^2
$$

or

$$
(2\varrho + x)^2 = (\varrho + x)^2 + 9\varrho^2
$$

from which $x = 3\varrho$. Therefore, the ratio of the sides of the triangle is

$$
AB : BC : CA = (AR + RB) : (BS + SC) : (CT + TA) = 5\varrho : 4\varrho : 3\varrho
$$

in agreement with the values obtained in the first solution.
In a right-angled triangle, the center $O$ of the inscribed circle is such that the half of the hypotenuse closer to $O$ is seen at a right angle from $O$. What is the ratio of the sides of the triangle?