= Solution
Normalize $v_K$ by $v_K(\pi_K)=1$ and put $v_K(0)=+\infty$. Write $f(X)=\sum_{j=0}^d a_jX^j$, with $a_d=1$ and $a_0\ne0$. In the usual coefficient-exponent convention, the <Newton polygon> $N_K(f)$ is the lower boundary of the <convex hull> of the upward vertical rays starting at
$$
\boxed{(j,v_K(a_j))\qquad(0\leq j\leq d,\ a_j\ne0).}
$$
It runs from $(0,v_K(a_0))$ to $(d,0)$, with nondecreasing slopes from left to right. Zero coefficients contribute no finite point. Equivalently it is the largest convex piecewise-linear function lying below all these coefficient points.
The <Newton polygon root valuation theorem> says that a segment of slope $s$ and horizontal length $r$ accounts for exactly $r$ roots of <valuation> $-s$, counted with multiplicity, using the extended <valuation> on an <algebraic closure>. The nonzero constant coefficient excludes a zero root.
For a short justification, use the <weighted Gauss valuation> $w_t(P)=\min_j(v(c_j)+jt)$ for $P=\sum c_jX^j$. It is multiplicative: after scaling $X$ by an element of <valuation> $t$ in a valued extension, the initial nonzero residue polynomials multiply without vanishing. Rational $t$ suffice. Factoring a monic <polynomial> into linear factors gives $w_t(f)=\sum_r\min(t,v(r))$. This supporting-line function changes derivative at precisely the root <valuations>; its derivative drops count their multiplicities. Supporting lines to the lower polygon have slope $-t$, proving the sign and horizontal-length assertion.
\b[Slope convention for part (b).] Reflecting the horizontal axis, by plotting $(d-j,v_K(a_j))$, produces the <reflected Newton polygon convention>; its slopes are the root <valuations> themselves. Part (b)'s wording with $m$ instead of $-m$ is correct under this positive-root-<valuation> convention. The proof below gives both forms explicitly, so the result does not depend on silently changing the sign.
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