4. A right-angled triangle $ABC$ with hypotenuse $AB$ and area $S$ has a circle circumscribed around it. The tangent to this circle at point $C$ intersects the tangents drawn from points $A$ and $B$ at points $D$ and $E$. Express the length of segment $DE$ in terms of the length $c$ of the hypotenuse and the area $S$.

The regional round of category C takes place

## on Tuesday, April 7, 2009

so that it starts in the morning and the participants have 4 hours of pure time to solve the problems. For each problem, a participant can earn 6 points, and a successful solver is a student who scores 10 points or more. Allowed aids are writing and drawing tools, school MF tables, and calculators without a graphical display. These details will be communicated to the students before the start of the competition.
4. Let $O$ be the center of the circumscribed circle, i.e., the midpoint of the hypotenuse $AB$ of the given right triangle $ABC$, and $v$ the length of its altitude to the hypotenuse (Fig. 3). The triangle $EDO$ is also clearly a right triangle because its sides $DO$ and $EO$ are perpendicular to the legs of triangle $ABC$; the altitude to the hypotenuse is the segment $OC$ (of length $\frac{1}{2} c$). Due to the symmetry of segment $AC$ with respect to the axis $OD$, the angle at vertex $D$ satisfies $|\angle CDO| = 90^\circ - |\angle COD| = 90^\circ - |\angle AOD| = \alpha$. Therefore, triangles $EDO$ and $ABC$ are similar $(AA)$. The similarity coefficient $k$ is given by the ratio of the lengths of the corresponding altitudes to the hypotenuses, so $k = |OC| / v = \frac{1}{2} c / v$, and since $vc = 2S$, we have

$$
k = \frac{c^2}{4S}
$$

In this similarity, the hypotenuse $AB$ corresponds to the hypotenuse $DE$, so for its length we have

$$
|DE| = k c = \frac{c^3}{4S}
$$

Another solution. From the symmetry of tangents from a point to a circle, it follows that both triangles $ACD$ and $BCE$ are isosceles, $|AD| = |DC|$, $|BE| = |CE|$. The triangles $ACO$ and $BCO$ are also isosceles, where $O$ is the midpoint of the hypotenuse $AB$ (the legs of both triangles have the length of the radius of the circumscribed circle of the right triangle $ABC$, which is $\frac{1}{2} c$). We will show that these are two pairs of similar triangles $ACD \sim BCO$ and $ACO \sim BCE$. To do this, it suffices to notice that in the quadrilateral $AOC D$, which is composed of two congruent right triangles, $|\angle CDA| = 180^\circ - |\angle AOC| = |\angle COB|$. The isosceles triangles $ACD$ and $BCO$ are thus similar by the theorem $AA$. From this similarity, it follows that $|CD| : |CA| = |CO| : |CB|$, so with the usual notation for the legs, we get $|CD| = \frac{1}{2} c b / a$, and from the similarity of triangles $ACO$ and $BCE$ we get $|CE| = \frac{1}{2} c a / b$. In total, we have

$$
|DE| = |DC| + |CE| = \frac{cb}{2a} + \frac{ca}{2b} = \frac{cb^2 + ca^2}{2ab} = \frac{c(a^2 + b^2)}{2 \cdot 2S} = \frac{c^3}{4S}.
$$

Notes. The similarity of the mentioned isosceles triangles can also be derived by noticing the equality of the corresponding angles $\angle ACO$ and $\angle BCE$ at the bases: both supplement the angle $\angle OCB$ to a right angle ($\angle ACB$, resp. $\angle OCE$). Therefore, $ACO \sim BCE$.

Another possibility is to discover the equality $|\angle ADO| = |\angle BAC| = \alpha$ (the legs of one angle are perpendicular to the legs of the other). From the right triangle $ODA$, we have $|AO| : |AD| = \tan |\angle ADO| = \tan \alpha = a : b$, so $|CD| = |AD| = \frac{1}{2} cb / a$, and analogously for the right triangle $OEB$.

For a complete solution, award 6 points. For identifying the appropriate angle equality, award 3 points, and for calculating the length of segment $DE$, award another 3 points.