1695

在 ${\displaystyle \mathrm{\Delta }ABC}$ 中, 角 ${\displaystyle A,B,C}$ 的对边分别为 ${\displaystyle a,b,c}$, 面积为 ${\displaystyle S}$, 当 ${\displaystyle a^2-2b^2=\lambda c^2}$ (${\displaystyle \lambda }$ 为常数) 时, ${\displaystyle \frac{S}{c^2}}$ 的最大值为 ${\displaystyle \frac{1}{4}}$, 则 ${\displaystyle \lambda =}$ _____.

我们用边来表达 ${\displaystyle S}$: $$\begin{array}{rl}& 16S^2=4a^2{b}^2{\sin }^{2}C=4a^2{b}^2(1-{\cos }^{2}C)\\ =& 4a^2{b}^2(1-\frac{(a^2+b^2-c^2{)}^2}{4a^2{b}^2})=4a^2{b}^2-(a^2+b^2-c^2{)}^2.\end{array}$$ 代入 ${\displaystyle a^2=2b^2+\lambda c^2}$, 有$$\begin{array}{rl}16S^2& =4b^2(2b^2+\lambda c^2)-(3b^2+(\lambda -1)c^2{)}^2\\ & =-b^4+(6-2\lambda )b^2{c}^2-(\lambda -1)c^4,\end{array}$$ 所以两侧同除 ${\displaystyle c^4}$: $$16{\left(\frac{S}{c^2}\right)}^2=-{\left(\frac{b}{c}\right)}^4+(6-2\lambda ){\left(\frac{b}{c}\right)}^2-(\lambda -1).$$ 记 ${\displaystyle f(x)=-x^2+(6-2\lambda )x-(\lambda -1)}$, 它的最大值为 ${\displaystyle f(3-\lambda )=8-4\lambda }$. 所以 $$16{\left(\frac{S}{c^2}\right)}^2\le 8-4\lambda =16\cdot {\left(\frac{1}{4}\right)}^2,$$ 解得 ${\displaystyle \lambda =\frac{7}{4}}$.