Problem 21. In a triangle with a perimeter of $2 \sqrt{3}$, the product of its three angle bisectors is 1, and the radius of the inscribed circle is $\frac{1}{3}$. Find the angles of the triangle.
Solution. We will use the formula for the angle bisector of a triangle:

20

Example 3. Application of Heron's formula

$$
I_{a}=\frac{2 \sqrt{b c p(p-a)}}{b+c} \text { (prove it!). }
$$

Similarly,

$$
I_{b}=\frac{2 \sqrt{a c p(p-b)}}{a+c} ; I_{c}=\frac{2 \sqrt{a b p(p-c)}}{a+b}
$$

Then, taking into account the condition, we have:

$$
I_{a} \cdot l_{b} \cdot l_{c}=\frac{8 a b c p \sqrt{p(p-a)(p-b)(p-c)}}{(a+b)(b+c)(a+c)}=1
$$

or

$$
\frac{8 a b c \cdot \sqrt{3} \cdot S}{(a+b)(b+c)(a+c)}=1
$$

But $S=p r=\sqrt{3} \cdot \frac{1}{3}$. Then $\frac{8 a b c \cdot \sqrt{3} \cdot \frac{1}{3} \cdot \sqrt{3}}{(a+b)(b+c)(a+c)}=1$, or:

$$
8 a b c=(a+b)(b+c)(a+c)
$$

It is known that for non-negative numbers $a, b, c$:

$$
(a+b)(b+c)(a+c) \geq 8 a b c \text { (prove it!). }
$$

Since in our case equality holds, then $a=b=c$.

Answer. $60^{\circ} ; 60^{\circ} ; 60^{\circ}$.