Given a triangle on the plane, construct inside the triangle the point $P$ for which the centroid of the triangle formed by the three projections of $P$ onto the sides of the triangle happens to be $P$.
1. **Define the Problem and Setup:**
   Let \( \triangle ABC \) be a given triangle. We need to find a point \( P \) inside the triangle such that the centroid of the triangle formed by the projections of \( P \) onto the sides of \( \triangle ABC \) is \( P \) itself.

2. **Projections and Midpoints:**
   Let \( P_A, P_B, P_C \) be the projections of \( P \) onto the sides \( BC, CA, \) and \( AB \) respectively. Denote \( M_A \) as the midpoint of \( P_B P_C \), \( M_B \) as the midpoint of \( P_A P_C \), and \( M_C \) as the midpoint of \( P_A P_B \).

3. **Locus of Points:**
   Consider the locus of points \( P \) such that \( PM_A \perp BC \). This locus is denoted by \( \ell_A \). Similarly, define \( \ell_B \) and \( \ell_C \) for the other sides.

4. **Linearity and Cevian:**
   If \( U \) and \( V \) lie on \( \ell_A \), then every point on the line \( UV \) also lies on \( \ell_A \). This is because the mapping \( P \mapsto M_A \) via perpendicularity and midpoint properties are linear. Hence, \( \ell_A \) must be a line passing through \( A \), making it an \( A \)-cevian.

5. **Simson Line and Symmedian:**
   To find another point on \( \ell_A \), consider \( P \) on the circumcircle of \( \triangle ABC \). In this case, \( P_A = M_A \), and \( \angle PM_A B = \angle PCA \). By spiral similarity, \( PM_A \) and \( PA \) are isogonal in \( \angle BPC \). Therefore, \( P \) lies on the \( A \)-symmedian, implying \( \ell_A \) is the \( A \)-symmedian.

6. **Intersection of Symmedians:**
   The desired point \( P \) is the intersection of the symmedians \( \ell_A, \ell_B, \ell_C \). This point is known as the Lemoine point (or the symmedian point) of \( \triangle ABC \).

\[
\boxed{\text{The Lemoine point}}
\]