1705

${\displaystyle \mathrm{\Delta }ABC}$ 中角 ${\displaystyle A,B,C}$ 的对边分别为 ${\displaystyle a,b,c}$ , 满足 ${\displaystyle a^2+b=c}$ , ${\displaystyle \tan C=\frac{4}{3}}$ .

1. 证明: ${\displaystyle a+b+c<2}$ ;
2. 求 ${\displaystyle \mathrm{\Delta }ABC}$ 的内切圆半径 ${\displaystyle r}$ 的取值范围;
3. 若 ${\displaystyle a=\frac{1}{5}}$ , ${\displaystyle \mathrm{\Delta }ABC}$ 的内切圆上有一点 ${\displaystyle P}$, 求点 ${\displaystyle P}$ 到 ${\displaystyle A,B,C}$ 三点的距离的平方和的最大值.

1. 证明 由题意 ${\displaystyle \cos C=\frac{3}{5}>0}$ (以及 ${\displaystyle \sin C=\frac{4}{5}}$), 所以 $$a^2+b^2-c^2=a^2+(b-c)(b+c)=a^2(1-b-c)>0,$$ 也即 ${\displaystyle b+c<1}$. 所以 ${\displaystyle a+b+c<(b+c)+b+c<2}$.
2. 解 一方面我们有面积关系 $$\begin{array}{}\text{(1)}& \frac{1}{2}(a+b+c)r=\frac{1}{2}ab\sin C\Rightarrow r=\frac{ab\sin C}{a+b+c}.\end{array}$$ 另一方面根据余弦定理 $$\begin{array}{}\text{(2)}& 1+\cos C=1+\frac{a^2+b^2-c^2}{2ab}=\frac{(a+b{)}^2-c^2}{2ab}=\frac{(a+b+c)(a+b-c)}{2ab},\end{array}$$ 两式消去 ${\displaystyle a+b+c}$, 有 $$r=\frac{\sin C(a+b-c)}{2(1+\cos C)}=\frac{a-a^2}{4}=\frac{1}{16}-\frac{1}{4}{(a-\frac{1}{2})}^2.$$
   下面来确定 ${\displaystyle a}$ 的范围. 必要性方面 ${\displaystyle a^2=c-b<a\Rightarrow a<1}$; 充分性方面联立 $$\begin{array}{}\text{(3)}& \{\begin{array}{rl}& c=b+a^2,\\ & c^2=a^2+b^2-2ab\cos C,\end{array}\Rightarrow b=\frac{a-a^3}{2(a+\cos C)}.\end{array}$$ 这说明每一个 ${\displaystyle a\in (0,1)}$ 都可以得到一个正值 ${\displaystyle b}$ (从而 ${\displaystyle c}$ 也是正值), 且满足三角形边长要求. 所以 ${\displaystyle r\in (0,\frac{1}{16}]}$.
3. 解 在 (3) 中代入 ${\displaystyle a=\frac{1}{5}}$ 得 ${\displaystyle b=\frac{3}{25}}$, 这样 ${\displaystyle c=\frac{4}{25}}$, 于是 ${\displaystyle \mathrm{\Delta }ABC}$ 是以 ${\displaystyle A}$ 为直角的直角三角形. 由 (1) 得 ${\displaystyle r=\frac{1}{25}}$. 记内心为 ${\displaystyle I}$. 则 $$\begin{array}{rl}\sum _{\mathrm{cyc}}P{A}^2& =\sum _{\mathrm{cyc}}(\overrightarrow{PI}+\overrightarrow{IA}{)}^2=3P{I}^2+\sum _{\mathrm{cyc}}I{A}^2+2\overrightarrow{PI}\cdot \sum _{\mathrm{cyc}}\overrightarrow{IA}.\\ \text{(4)}& & \le \frac{3}{625}+\sum _{\mathrm{cyc}}I{A}^2+\frac{2}{25}\left|\sum _{\mathrm{cyc}}\overrightarrow{IA}\right|.\end{array}$$
   我们以 ${\displaystyle A}$ 为原点建系, 设 ${\displaystyle B(0,\frac{4}{25})}$, ${\displaystyle C(\frac{3}{25},0)}$. 则 ${\displaystyle I(\frac{1}{25},\frac{1}{25})}$, 于是 $$\begin{array}{rl}& \sum _{\mathrm{cyc}}I{A}^2=(\frac{1}{{25}^2}+\frac{1}{{25}^2})+\frac{1}{{25}^2}+{(\frac{4}{25}-\frac{1}{25})}^2+{(\frac{3}{25}-\frac{1}{25})}^2+\frac{1}{{25}^2}=\frac{17}{625},\\ & \left|\sum _{\mathrm{cyc}}\overrightarrow{IA}\right|=\left|(\frac{3}{25}-\frac{3}{25},\frac{4}{25}-\frac{3}{25})\right|=\frac{1}{25},\end{array}$$ 于是代入 (4) 得 ${\displaystyle \sum _{\mathrm{cyc}}P{A}^2\le \frac{22}{625}}$.

延续我们设的坐标, 设 ${\displaystyle P(\frac{1}{25}+\frac{1}{25}\cos \theta ,\frac{1}{25}+\frac{1}{25}\sin \theta )}$, 则$$\begin{array}{rl}& P{A}^2+P{B}^2+P{C}^2\\ =& {(\frac{1}{25}+\frac{1}{25}\cos \theta )}^2+{(\frac{1}{25}+\frac{1}{25}\sin \theta )}^2\\ & +{(\frac{1}{25}+\frac{1}{25}\cos \theta )}^2+{(\frac{1}{25}+\frac{1}{25}\sin \theta -\frac{4}{25})}^2\\ & +{(\frac{1}{25}+\frac{1}{25}\cos \theta -\frac{3}{25})}^2+{(\frac{1}{25}+\frac{1}{25}\sin \theta )}^2\\ =& \frac{20+2\sin \theta }{625}\le \frac{22}{625}.\end{array}$$