Problem 1681

1. Equal Areas of Sub-Triangles:
- It is well known that the medians of a triangle divide it into six smaller triangles of equal area. Therefore, the areas of triangles $AFG$, $BDG$, and $CDG$ are equal.

2. Inradius and Semi-Perimeter:
- For any triangle with area $A$, semi-perimeter $s$, and inradius $r$, the relationship $A = rs$ holds. Since the incircles of triangles $AFG$, $BDG$, and $CDG$ are congruent, they have the same inradius $r$. This implies that the semi-perimeters of these triangles are equal.

3. Medians and Centroid:
- The centroid $G$ of a triangle divides each median into a ratio of 2:1, with the longer segment being closer to the vertex. Let $m_a$, $m_b$, and $m_c$ be the lengths of the medians from vertices $A$, $B$, and $C$ respectively. Let the side lengths of triangle $ABC$ be $2a$, $2b$, and $2c$, where $2a$ is opposite $A$, $2b$ is opposite $B$, and $2c$ is opposite $C$.

4. Equating Perimeters:
- Consider triangles $GCD$ and $GBD$. The semi-perimeters of these triangles can be expressed as:
$$\frac{2m_b}{3} + \frac{m_a}{3} + a = \frac{2m_c}{3} + \frac{m_a}{3} + a$$
Simplifying, we get:
$$m_b = m_c$$

5. Using Stewart's Theorem:
- By Stewart's Theorem, the length of a median can be expressed as:
$$m_a = \sqrt{2b^2 + 2c^2 - a^2}$$
Since $m_b = m_c$, we have:
$$\sqrt{2a^2 + 2c^2 - b^2} = \sqrt{2a^2 + 2b^2 - c^2}$$
Squaring both sides and simplifying, we get:
$$2a^2 + 2c^2 - b^2 = 2a^2 + 2b^2 - c^2$$
$$b^2 = c^2 \implies b = c$$

6. Equating Other Perimeters:
- Now consider triangles $GBD$ and $GAF$. The semi-perimeters of these triangles can be expressed as:
$$\frac{2m_b}{3} + \frac{m_a}{3} + a = \frac{2m_a}{3} + \frac{m_c}{3} + b$$
Using $b = c$ and simplifying, we get:
$$m_b + 3a = m_a + 3b$$
Substituting the median lengths, we get:
$$\sqrt{2a^2 + b^2} + 3a = \sqrt{4b^2 - a^2} + 3b$$
Simplifying further, we get:
$$9(a - b)^2 = 5b^2 + a^2 - 2\sqrt{-2a^4 + 7a^2b^2 + 4b^4}$$
Dividing by $b^2$ and letting $\frac{a}{b} = u$, we get:
$$9(u-1)^2 = 5 + u^2 - 2\sqrt{-2u^4 + 7u^2 + 4}$$
Simplifying further, we get:
$$4u^2 - 9u + 2 = -\sqrt{-2u^4 + 7u^2 + 4}$$
$$16u^4 - 72u^3 + 97u^2 - 36u + 4 = -2u^4 + 7u^2 + 4$$
$$u^4 - 4u^3 + 5u^2 - 2u = 0$$
$$u(u-1)^2(u-2) = 0$$
The solutions are $u = 0$, $u = 1$, and $u = 2$. Clearly, $u = 0$ and $u = 2$ do not work. Checking $u = 1$ shows that it works, implying $a = b = c$.

7. Conclusion:
- Since $a = b = c$, triangle $ABC$ is equilateral.

$\blacksquare$