Задание 2976

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     1) Пусть $$ND=MN=2x\Rightarrow$$ $$CM=x$$; $$AB=4$$. Пусть $$CH\left | \right |AB\Rightarrow$$ $$CH=4$$, $$BC=AH=y$$. По т. Пифагора из $$\Delta CDH$$: $$HD=\sqrt{CD^{2}-CH^{2}}=\sqrt{25x^{2}-16}$$

      2) По свойству касательной и секущей : $$\left\{\begin{matrix}BC^{2}=CM*CN\\AD^{2}=DN*DM\end{matrix}\right.\Leftrightarrow$$ $$\left\{\begin{matrix}y^{2}=x*3x\\(y+\sqrt{25x^{2}-16})^{2}=2x*4x\end{matrix}\right.\Leftrightarrow$$ $$\left\{\begin{matrix}y=x\sqrt{3}\\y^{2}+25x^{2}-16+2y\sqrt{25x^{2}-16}=8x^{2}(1)\end{matrix}\right.$$

   Рассмотрим (1): $$25x^{2}+3x^{2}-8x^{2}+2x\sqrt{3}*\sqrt{25x^{2}-16}=16\Leftrightarrow$$$$2 x\sqrt{3}\sqrt{25x^{2}-16}=16-20x^{2}\Leftrightarrow$$$$x\sqrt{3}\sqrt{25x^{2}-16}=8-10x^{2}\Leftrightarrow$$$$\left\{\begin{matrix}3x^{2}(15x^{2}-16)=(8-10x^{2})^{2}(2)\\8-10 x^{2}\geq 0(3)\end{matrix}\right.$$

   Рассмотрим (2): $$75x^{4}-8x^{2}=64-160x^{2}+100x^{4}\Leftrightarrow$$ $$25x^{2}-112x^{2}+64=0\Rightarrow$$ $$D=6144=32^{2}*6$$

   Тогда: $$\left\{\begin{matrix}x_{1}^{2}=\frac{112+32\sqrt{6}}{50} \in (3)\\x_{2}^{2}=\frac{112-32\sqrt{6}}{50}=\frac{56-16\sqrt{6}}{25}=(\frac{4\sqrt{3}-2\sqrt{2}}{5})^{2}\end{matrix}\right.$$

     3) Площадь $$S=\frac{BC+AD}{2}*CH=$$$$\frac{y+y+\sqrt{25x^{2}-16}}{2}*4=$$$$2(2y+\sqrt{25x^{2}-16})$$

   1) $$\sqrt{25x^{2}-16}=\sqrt{25*\frac{56-16\sqrt{6}}{25}-16}=$$$$\sqrt{40-16\sqrt{6}}=\sqrt{(2\sqrt{6}-4)^{2}}=$$$$\left | 2\sqrt{6}-4 \right |=2\sqrt{6}-4$$

   2) $$2y=2*x\sqrt{3}=2\sqrt{3}*\left | \frac{4\sqrt{3}-2\sqrt{2}}{5} \right |=$$$$2\sqrt{3}*\frac{4\sqrt{3}-2\sqrt{2}}{5}=$$$$\frac{24-4\sqrt{6}}{5}$$

   $$S=2(\frac{24-4\sqrt{6}}{5}+2\sqrt{6}-4)=$$$$\frac{2}{5}*(24-4\sqrt{6}+10\sqrt{6}-20)=$$$$\frac{2}{5}(6\sqrt{6}+4)=\frac{4(2+3\sqrt{6})}{5}$$