= Solution
The stated condition is the definition of an <extreme point> in the convex set of <density matrices>. Suppose first that $\rho=|\psi\rangle\langle\psi|$ with $\|\psi\|=1$, and consider any <convex combination> $\rho=\sum_i a_i\rho_i$ with all $a_i>0$. For every $|w\rangle$ <orthogonal> to $|\psi\rangle$,
$$
0=\langle w|\rho|w\rangle=\sum_i a_i\langle w|\rho_i|w\rangle.
$$
Every term is nonnegative, so $\langle w|\rho_i|w\rangle=0$ for each $i$. To see that this implies $\rho_iw=0$, expand $\rho_i=\sum_j\mu_j|u_j\rangle\langle u_j|$, $\mu_j\geq0$. The equality $\sum_j\mu_j|\langle u_j|w\rangle|^2=0$ implies $\mu_j\langle u_j|w\rangle=0$ for every $j$, which kills $\rho_iw$. Thus each <density matrix> $\rho_i$ vanishes on $\psi^\perp$. By self-adjointness its range lies in the line $\mathbb C\psi$, so $\rho_i=c_i|\psi\rangle\langle\psi|$. Its unit <trace> forces $c_i=1$, establishing the required purity.
Conversely, suppose $\rho$ is not rank one. In its <spectral decomposition>, at least two <eigenvalues> are positive. Select a normalized eigenvector $|e_1\rangle$ with positive <eigenvalue> $\lambda_1$; then $0<\lambda_1<1$. Define
$$
\sigma=\frac{\rho-\lambda_1|e_1\rangle\langle e_1|}{1-\lambda_1}.
$$
The remaining <eigenvalues> show that $\sigma$ is a <density matrix>. The decomposition $\rho=\lambda_1|e_1\rangle\langle e_1|+(1-\lambda_1)\sigma$ is a <convex combination> of two different <density matrices>, neither equal to $\rho$. This violates the given purity condition. Hence the <extreme points of the density-operator state space> obey
$$
\boxed{\rho\text{ is pure}\iff\rho=|\psi\rangle\langle\psi|\text{ for a normalized }|\psi\rangle.}
$$
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