Lời giải
Chọn hệ trục tọa độ $Oxyz$ sao cho $O(0;0;0), A(2a;0;0), B(0;a;0), S(0;0;h)\Rightarrow C(-2a;0;0), D(0;-a;0)$
a) Ta có: $\left\{ \begin{array}{l} SC\bot (\alpha )\\ SC\bot BD(BD\bot SAC) \end{array} \right. \Rightarrow BD//(\alpha )\Rightarrow B’D’//BD$
$\overrightarrow{SC}=(-2;0;-h) $
$\Rightarrow $ phương trình tham số $SC:\left\{ \begin{array}{l} x=-2a+2at\\ y=0\\z=ht \end{array} \right. $
Phương trình $(\alpha ):2a(x-2a)+hz=0\Leftrightarrow 2ax+hz-4a^2=0$
$\Rightarrow C'(\frac{8a^3-2ah^2}{4a^2+h^2};0;\frac{8a^2h}{4a^2+h^2} )$
$\overrightarrow{SB}=(0;a;-h) \Rightarrow $ Phương trình tham số $SB:\left\{ \begin{array}{l} x=0\\ y=a+at\\z=-ht \end{array} \right. $
$\Rightarrow B'(0;\frac{ah^2-4a^3}{h^2};\frac{4a^2}{h} )$
$D’B’//DB\Rightarrow I$ là trung điểm $D’B’$
Mặt khác, $D’B’\bot (SAC)(do DB\bot (SAC))$ nên $D’B’\bot AC’\Rightarrow \Delta B’C’D’$ cân tại $C’\Rightarrow \Delta IB’C’$ là nửa tam giác đều
Để $\Delta B’C’D’$ đều thì $IC’=IB’\sqrt{3} $
$\begin{array}{l}
\Leftrightarrow {\left( {\frac{{8{a^3} – 2a{h^2}}}{{4{a^2} + {h^2}}}} \right)^2} + \left( {\frac{{4{a^2}{h^2} – 16{a^4}}}{{{h^2}(4{a^2} + {h^2})}}} \right) = 3{\left( {\frac{{4{a^3} – a{h^2}}}{{{h^2}}}} \right)^2}\\
\Leftrightarrow {\left( {\frac{{4{a^3} – a{h^2}}}{{4{a^2} + {h^2}}}} \right)^2} + \left( {4 + \frac{{16{a^2}}}{{{h^2}}}} \right) = 3{\left( {\frac{{4{a^3} – a{h^2}}}{{{h^2}}}} \right)^2} \Leftrightarrow \frac{4}{{4{a^2} + {h^2}}} = \frac{3}{{{h^2}}}\\
\Leftrightarrow h = 2a\sqrt 3
\end{array}$
b) Ta có:
$\begin{array}{l}
{V_{S.ABCD}} = \frac{1}{3}r.{S_{tp}}\\
{S_{\Delta SAB}} = \frac{1}{2}|{\rm{[}}\overrightarrow {SA} {\rm{,}}\overrightarrow {SB} {\rm{]}}| = \frac{1}{2}|{\rm{[(2a;0; – h),(0;a; – h)]}}|\\
= \frac{1}{2}|(ah;2ah;2{a^2})| = \frac{a}{2}\sqrt {4{a^2} + 5{h^2}}
\end{array}$
Mà $\Delta SAB=\Delta SBC=\Delta SCD=\Delta SDA$, nên:
$\begin{array}{l}
{S_{tp}} = 4{S_{\Delta SAB}} = 2a\sqrt {4{a^2} + 5{h^2}} \,\,;\,\,\,{S_{ABCD}} = \frac{1}{2}AC.BD = 4{a^2}\\
\Rightarrow {S_{tp}} = 4{a^2} + 2a\sqrt {4{a^2} + 5{h^2}} \\
{V_{S.ABCD}} = \frac{1}{3}h.{S_{ABCD}} = \frac{{4{a^2}h}}{3} \Rightarrow r = \frac{{3V}}{{{S_{tp}}}} = \frac{{2ah}}{{2a + \sqrt {4{a^2} + 5{h^2}} }}
\end{array}$
Trả lời