 Solution

Normalize $v_K$ by $v_K({\pi }_K)=1$ and put $v_K(0)=+\infty$. Write $f(X)={\sum }_{j=0}^d{a}_j{X}^j$, with $a_d=1$ and $a_0\ne 0$. 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 atIt 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 $P=\sum c_j{X}^j$. 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.

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.