Problem 1690

1. Define the problem and given conditions:
- Let $\triangle ABC$ have orthocenter $H$.
- The lengths of the altitudes from $A, B,$ and $C$ are $h_a, h_b,$ and $h_c$ respectively.
- The semi-perimeter of $\triangle ABC$ is $p$.
- It is given that $AH \cdot h_a + BH \cdot h_b + CH \cdot h_c = \frac{2}{3} \cdot p^2$.

2. Introduce the feet of the altitudes:
- Let $A', B', C'$ be the feet of the altitudes from $A, B,$ and $C$ respectively.

3. Use the Power of a Point (PoP) and the Law of Cosines:
- By the Power of a Point and the Law of Cosines, we have:
$$AH \cdot h_a = AB' \cdot b = bc \cos A = \frac{b^2 + c^2 - a^2}{2}$$
- Similarly,
$$BH \cdot h_b = \frac{c^2 + a^2 - b^2}{2}$$
$$CH \cdot h_c = \frac{a^2 + b^2 - c^2}{2}$$

4. Rewrite the given equality using these expressions:
- The given equality is:
$$AH \cdot h_a + BH \cdot h_b + CH \cdot h_c = \frac{2}{3} \cdot p^2$$
- Substituting the expressions from step 3, we get:
$$\frac{b^2 + c^2 - a^2}{2} + \frac{c^2 + a^2 - b^2}{2} + \frac{a^2 + b^2 - c^2}{2} = \frac{2}{3} \cdot p^2$$

5. Simplify the left-hand side:
- Combine the terms:
$$\frac{(b^2 + c^2 - a^2) + (c^2 + a^2 - b^2) + (a^2 + b^2 - c^2)}{2} = \frac{2}{3} \cdot p^2$$
- This simplifies to:
$$\frac{2(a^2 + b^2 + c^2)}{2} = \frac{2}{3} \cdot p^2$$
- Therefore:
$$a^2 + b^2 + c^2 = \frac{2}{3} \cdot p^2$$

6. **Express the semi-perimeter $p$:**
- The semi-perimeter $p$ is:
$$p = \frac{a + b + c}{2}$$
- Therefore:
$$p^2 = \left( \frac{a + b + c}{2} \right)^2 = \frac{(a + b + c)^2}{4}$$

7. **Substitute $p^2$ into the equation:**
- Substitute $p^2$ into the equation $a^2 + b^2 + c^2 = \frac{2}{3} \cdot p^2$:
$$a^2 + b^2 + c^2 = \frac{2}{3} \cdot \frac{(a + b + c)^2}{4}$$
- Simplify:
$$a^2 + b^2 + c^2 = \frac{(a + b + c)^2}{6}$$

8. Expand and simplify:
- Expand $(a + b + c)^2$:
$$a^2 + b^2 + c^2 = \frac{a^2 + b^2 + c^2 + 2ab + 2bc + 2ca}{6}$$
- Multiply both sides by 6:
$$6(a^2 + b^2 + c^2) = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca$$
- Rearrange:
$$5(a^2 + b^2 + c^2) = 2(ab + bc + ca)$$

9. Use the identity for squares:
- Rearrange to:
$$5(a^2 + b^2 + c^2) - 2(ab + bc + ca) = 0$$
- This can be written as:
$$(a - b)^2 + (b - c)^2 + (c - a)^2 = 0$$

10. Conclude that the triangle is equilateral:
- Since the sum of squares is zero, each term must be zero:
$$(a - b)^2 = 0, \quad (b - c)^2 = 0, \quad (c - a)^2 = 0$$
- Therefore:
$$a = b = c$$
- Hence, $\triangle ABC$ is equilateral.

$\blacksquare$