Saudi Arabian Mathematical Competitions

Let $AB = c$, $BC = a$, $CA = b$, $BA' = x$, $CB' = y$, $AC' = z$. The condition in the problem $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = \frac{3}{2}$ is equivalent to
$$\begin{equation*} bcx + cay + abz = \frac{3}{2}abc \tag{1} \end{equation*}$$
From Ceva's Theorem,
$$\frac{x}{a-x} \cdot \frac{y}{b-y} \cdot \frac{z}{c-z} = 1,$$
so
$$\begin{gathered} xyz = (a-x)(b-y)(c-z) = abc - (bcx + cay + abz) \\ + (axy + byz + czx) - xyz \end{gathered}$$
or
$$2xyz = \frac{1}{4}abc + \frac{3}{4}abc - (bcx + cay + abz) + (axy + byz + czx).$$
The (1) is equivalent to
$$2xyz = \frac{1}{4}abc - \frac{1}{2}(bcx + cay + abz) + (axy + byz + czx),$$
which can be written as
$$8xyz - 4(axy + byz + czx) + 2(bcx + cay + abz) - abc = 0$$
so $(2x - a)(2y - b)(2z - c) = 0$.

This means that at least one of the segments $AA'$, $BB'$, $CC'$ is a median in triangle $ABC$ and the conclusion follows.

Solution 2:

Denote
$$x = \frac{A'C}{A'B}, \quad y = \frac{B'A}{B'C}, \quad z = \frac{C'B}{C'A}.$$
From Ceva's Theorem, we have $xyz = 1$.
The condition in the problem is equivalent to
$$\frac{1}{1+x} + \frac{1}{1+y} + \frac{1}{1+z} = \frac{3}{2}$$
That is
$$\begin{gathered} 2[(x+1)(y+1) + (y+1)(z+1) + (z+1)(x+1)] \\ = 3(x+1)(y+1)(z+1) \end{gathered}$$
We obtain
$$\begin{gathered} 2(xy + yz + zx) + 4(x + y + z) + 6 = 3xyz + 3(xy + yz + zx) \\ + 3(x + y + z) + 3 \end{gathered}$$
hence
$$3xyz - (x + y + z) + (xy + yz + zx) - 3 = 0$$
that is
$$1 - (x + y + z) + (xy + yz + zx) - 1 = 0.$$
It follows
$$xyz - (x + y + z) + (xy + yz + zx) - 1 = 0$$
hence
$$(x - 1)(y - 1)(z - 1) = 0$$
and the conclusion follows.