<s>[INST] Determining the average area as the point $X$ varies We consider a fixed triangle $ABC$ with side lengths $a = BC$, $b = CA$, $c = AB$, and
a variable point $X$ in the interior. The lines through $X$ parallel to $AB$ and $AC$,
together with line $BC$, determine a triangle $T_a$. The triangles $T_b$ and $T_c$ are defined
in a similarly way .
Let $S$ and $p$ denote the average area and perimeter, respectively, of the three triangles $T_a, T_b, T_c$.
(a) Determine all possible values of $S$ as $X$ varies, in terms of $a, b, c.$
(b) Determine all possible values of $p$ as $X$ varies, in terms of $a, b, c$.

Can someone give me hints for this problem ? I don't know how to start (a).
I got that (b) is $\frac {1}{3}(a+b+c)$, because of the parallel property , sum of perimeters of $T_a, T_b, T_c$  is $a+b+c$.
 [/INST] Let $[ABC]=S_0$ , its perimeter $= P_0$
Let bases (parallel to $a$) of ${T_a, T_b, T_c}$ be ${x,y,z}$ respectively.
Note  each $T_i \sim \triangle ABC$ with similarity ratios  ${x/a, y/a, z/a}$ respectively.
These three triangles leave three parallelograms out of $\triangle ABC$. Equating base lengths, we get $x+y+z=a$
Thus $$P=\frac{1}{3} \cdot \big( \dfrac{x}{a}+\dfrac{y}{a}+\dfrac{z}{a} \big)P_0$$
$$S=\frac{1}{3} \cdot \big( \dfrac{x^2}{a^2}+\dfrac{y^2}{a^2}+\dfrac{z^2}{a^2} \big)S_0$$
Now find the range of $$\dfrac{x^2+y^2+z^2}{a^2}$$
</s>