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kuing
Post time 2024-1-13 22:32
用向量证比较简单。
我们有
\[PA^2=\bigl(\vv{PG}+\vv{GA}\bigr)^2=PG^2+2\vv{PG}\cdot\vv{GA}+GA^2,\]
`PB^2` 等类似，一共四式，由重心有 `\vv{GA}+\vv{GB}+\vv{GC}+\vv{GD}=\bm0`，所以四式相加得
\[\sum PA^2=4PG^2+\sum GA^2,\]
又由 `\vv{AB}+\vv{AC}+\vv{AD}=4\vv{AG}` 得
\begin{align*}
16GA^2&=\bigl(\vv{AB}+\vv{AC}+\vv{AD}\bigr)^2\\
&=3(AB^2+AC^2+AD^2)-\bigl(\vv{AB}-\vv{AC}\bigr)^2-\bigl(\vv{AB}-\vv{AD}\bigr)^2-\bigl(\vv{AC}-\vv{AD}\bigr)^2\\
&=3(AB^2+AC^2+AD^2)-BC^2-BD^2-CD^2,
\end{align*}
`GB^2` 等类似，也是一共四式，求和即得
\[16\sum GA^2=4(a^2+b^2+c^2+m^2+n^2+l^2),\]
代入上面即
\[\sum PA^2=4PG^2+\frac{a^2+b^2+c^2+m^2+n^2+l^2}4.\] |
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