發新話題
打印

圓的問題

回復 1# tsyr 的帖子

\(\begin{align}
  & {{a}^{2}}+{{b}^{2}}=1 \\
& {{\left( c-2 \right)}^{2}}+{{\left( d-2 \right)}^{2}}=4 \\
&  \\
& {{\left( 2a-2b \right)}^{2}}\le \left( {{a}^{2}}+{{b}^{2}} \right)\left[ {{2}^{2}}+{{\left( -2 \right)}^{2}} \right]=8 \\
& 2a-2b\le 2\sqrt{2} \\
&  \\
& {{\left[ a\left( d-2 \right)-b\left( c-2 \right) \right]}^{2}}\le \left[ {{a}^{2}}+{{\left( -b \right)}^{2}} \right]\left[ {{\left( d-2 \right)}^{2}}+{{\left( c-2 \right)}^{2}} \right]=4 \\
& ad-bc-\left( 2a-2b \right)\le 2 \\
& ad-bc\le 2+2\sqrt{2} \\
& {{\left( ad-bc \right)}^{2}}\le 12+8\sqrt{2} \\
\end{align}\)

等號成立於\(a=-b=\frac{\sqrt{2}}{2},c=d=2+\sqrt{2}\)

TOP

回復 3# tsyr 的帖子

也可以這樣做
\(\begin{align}
  & a=\cos \alpha ,b=\sin \alpha ,c=2+2\cos \beta ,d=2+2\sin \beta  \\
&  \\
& ad-bc=2\cos \alpha -2\sin \alpha +2\left( \sin \beta \cos \alpha -\cos \beta \sin \alpha  \right) \\
& =2\sqrt{2}\sin \left( \frac{\pi }{4}-\alpha  \right)+2\sin \left( \beta -\alpha  \right) \\
& \le 2\sqrt{2}+2 \\
&  \\
& {{\left( ad-bc \right)}^{2}}\le 12+8\sqrt{2} \\
\end{align}\)

等號成立的條件同上

TOP

發新話題