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Problem 12015.
设ABC为三角形，G为其重心，D、E、F分别为BC、CA、AB的中点。对于 ABC 平面上的任意点 P，则
\[PA+PB+PC\leq 2(PD+PE+PF)+3PG\]
出处:
Monthly 12015.pdf
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Solution I by Giuseppe Fera and Giorgio Tescaro, Vicenza, Italy. For each of the lines $A B$, $B C$, and $C A$, define its inside to be the half-plane containing the interior of $\triangle A B C$. By symmetry, it suffices to consider three regions for the location of $P: \triangle B C G, \Omega_1$, and $\Omega_2$, where $\Omega_1$ is the intersection of the outsides of $A B$ and $A C$, and $\Omega_2$ is the intersection of the inside of $A B$, the inside of $A C$, and the outside of $B C$. All three regions include their boundaries.
Since $\overrightarrow{P A}=3 \overrightarrow{P G}-2 \overrightarrow{P D}$, we have $P A \leq 2 P D+3 P G$, with equality if and only if $P$ lies on the line segment $G D$. Thus it suffices to prove $P B+P C \leq 2(P E+P F)$ for $P$ in the three regions above.
If $P$ lies in $\triangle B C G$, then $G$ lies in $\triangle E F P$, so
$$
P B+P C \leq G B+G C=2(G E+G F) \leq 2(P E+P F),
$$
where both equalities hold if and only if $P=G$.
Now suppose that $P$ is in $\Omega_1$ or $\Omega_2$. Define point $Q$ such that $\overrightarrow{A Q}=2 \overrightarrow{A P}$. Since $P$ lies in $\triangle B C Q$, we have $P B+P C \leq$ $Q B+Q C=2(P F+P E)$, with equality if and only if $P=Q=A$. This completes the proof and shows that equality holds if and only if $P=G$. |
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