The sides of a triangle are given. A circle is drawn inside the triangle that touches the sides. Calculate the sides, angles, and area of the triangle determined by the points of tangency.

Let's denote the sides of the original triangle as \(a\), \(b\), and \(c\). The points where the incircle touches the sides of the triangle divide each side into two segments. Let's denote these segments as follows:
- The segment from vertex \(A\) to the point of tangency on side \(BC\) is \(s - a\),
- The segment from vertex \(B\) to the point of tangency on side \(CA\) is \(s - b\),
- The segment from vertex \(C\) to the point of tangency on side \(AB\) is \(s - c\),

where \(s\) is the semiperimeter of the triangle, given by \(s = \frac{a + b + c}{2}\).

The sides of the triangle formed by the points of tangency are:
- \(a' = s - a\),
- \(b' = s - b\),
- \(c' = s - c\).

The angles of the triangle formed by the points of tangency are the same as the angles of the original triangle, because the points of tangency form a smaller triangle that is similar to the original triangle.

The area of the triangle formed by the points of tangency can be calculated using the formula for the area of a triangle in terms of its sides and the semiperimeter. The area \(A'\) of the triangle formed by the points of tangency is given by:
\[ A' = \frac{r \cdot s'}{2} \]
where \(r\) is the inradius of the original triangle, and \(s'\) is the semiperimeter of the triangle formed by the points of tangency, given by:
\[ s' = \frac{a' + b' + c'}{2} = \frac{(s - a) + (s - b) + (s - c)}{2} = \frac{3s - (a + b + c)}{2} = \frac{3s - 2s}{2} = \frac{s}{2} \]

Thus, the area \(A'\) is:
\[ A' = \frac{r \cdot \frac{s}{2}}{2} = \frac{r \cdot s}{4} \]

The inradius \(r\) of the original triangle can be calculated using the formula:
\[ r = \frac{A}{s} \]
where \(A\) is the area of the original triangle.

Therefore, the area \(A'\) of the triangle formed by the points of tangency is:
\[ A' = \frac{\frac{A}{s} \cdot s}{4} = \frac{A}{4} \]

So, the sides, angles, and area of the triangle determined by the points of tangency are:
- Sides: \(a' = s - a\), \(b' = s - b\), \(c' = s - c\),
- Angles: Same as the angles of the original triangle,
- Area: \(A' = \frac{A}{4}\).
Let the points of tangency on the sides of triangle $ABC$ be $A_1, B_1, C_1$. Since

$$
AB_1 = AC_1, \quad BC_1 = BA_1 \quad \text{and} \quad CA_1 = CB_1,
$$

it follows that

$$
s = \frac{a + b + c}{2} = A_1C + BC_1 + C_1A = A_1C + c,
$$

and thus

$$
CA_1 = CB_1 = s - c.
$$

Similarly,

$$
AB_1 = AC_1 = s - a \quad \text{and} \quad BC_1 = BA_1 = s - b.
$$

Therefore, the side $A_1B_1$ of triangle $A_1B_1C_1$ is the base of an isosceles triangle whose legs are $CA_1 = CB_1 = s - c$, and the angle at vertex $C$ is $\gamma$; thus,

$$
A_1B_1 = 2(s - c) \sin \frac{\gamma}{2} = s(s - c) \sqrt{\frac{(s - a)(s - b)}{ab}}.
$$

Similarly,

$$
\begin{aligned}
& A_1C_1 = 2(s - b) \sin \frac{\beta}{2} = 2(s - b) \sqrt{\frac{(s - a)(s - c)}{ac}}, \\
& B_1C_1 = 2(s - a) \sin \frac{\alpha}{2} = 2(s - a) \sqrt{\frac{(s - b)(s - c)}{bc}}.
\end{aligned}
$$

If the points $A_1, B_1, C_1$ are connected to the center $O$ of the inscribed circle of the triangle, then

$$
B_1C_1 \perp AO \quad \text{and} \quad OC_1 \perp AB,
$$

so

$$
\angle OC_1B_1 = \angle BAO = \frac{\alpha}{2},
$$

and

$$
A_1C_1 \perp BO \quad \text{and} \quad OC_1 \perp AB,
$$

so

$$
\angle OC_1A_1 = \angle ABO = \frac{\beta}{2},
$$

and thus

$$
\angle A_1C_1B_1 = \angle BAO + \angle ABO = \frac{\alpha + \beta}{2},
$$

which implies

$$
\sin \angle A_1C_1B_1 = \sin \frac{\alpha + \beta}{2} = \cos \frac{\gamma}{2} = \sqrt{\frac{s(s - c)}{ab}},
$$

and similarly,

$$
\begin{aligned}
& \sin \angle A_1B_1C_1 = \sin \frac{\alpha + \gamma}{2} = \cos \frac{\beta}{2} = \sqrt{\frac{s(s - b)}{ac}}, \\
& \sin \angle B_1A_1C_1 = \sin \frac{\beta + \gamma}{2} = \cos \frac{\alpha}{2} = \sqrt{\frac{s(s - a)}{bc}}.
\end{aligned}
$$

The area of triangle $A_1B_1C_1$ is

$$
\begin{gathered}
T = \frac{A_1B_1 \cdot A_1C_1}{2} \sin \frac{\beta + \gamma}{2} = \\
= \frac{2(s - a)(s - b)(s - c)}{abc} \sqrt{s(s - a)(s - b)(s - c)}.
\end{gathered}
$$

(Aurél Erdős.)

The problem was also solved by: Barna D., Bojedain F., Dénes A., Détshy K., Devecis M., Fekete J., Freibauer E., Goldziher K., Kármán T., Laczkó E., Lukhaub Gy., Orlowszky F., Spitzer Ö., Szabó I., Szabó K., Weisz J.