Сфера в конус

[Гост]

Да се намери радиусът на сферата, вписана в конуса с основа единичната окръжност в ху-равнината и връх точката Т(3;4;8).

[Добромир Глухаров]

Осно сечение.png (14.46 KiB) Прегледано 440 пъти

$A(-\frac{3}{5};-\frac{4}{5};0),B(\frac{3}{5};\frac{4}{5};0),C(3;4;8),D(3;4;0)$

$\vec{AB}(\frac{6}{5};\frac{8}{5};0),\vec{AC}(\frac{18}{5};\frac{24}{5};8),\vec{BC}(\frac{12}{5};\frac{16}{5};8)$

$\vec{AB}\times\vec{AC}=\begin{array}{|ccc|}\vec{i}&\vec{j}&\vec{k}\\\frac{6}{5}&\frac{8}{5}&0\\\frac{18}{5}&\frac{24}{5}&8\end{array}=\frac{64}{5}\cdot\vec{i}-\frac{48}{5}\cdot\vec{j}+0.\vec{k}$

$S_{\Delta ABC}=\frac{1}{2}|\vec{AB}\times\vec{AC}|=\frac{1}{2}\sqrt{\left(\frac{64}{5}\right)^2+\left(\frac{48}{5}\right)^2+0^2}=\frac{1}{2}\sqrt{\frac{8^2(8^2+6^2)}{25}}=\frac{8}{2.5}\sqrt{100}=8$

$S_{\Delta ABC}=p.r$

$p=\frac{1}{2}(AB+BC+AC)=\frac{1}{2}\left(\sqrt{\left(\frac{6}{5}\right)^2+\left(\frac{8}{5}\right)^2+0^2}+\sqrt{\left(\frac{12}{5}\right)^2+\left(\frac{16}{5}\right)^2+8^2}+\sqrt{\left(\frac{18}{5}\right)^2+\left(\frac{24}{5}\right)^2+8^2}\right)$

$p=\frac{1}{2}\left(\sqrt{\frac{100}{25}}+\sqrt{\frac{400+64.25}{25}}+\sqrt{\frac{900+64.25}{25}}\right)=\frac{1}{2}\left(2+\sqrt{\frac{2000}{25}}+\sqrt{\frac{2500}{25}}\right)$

$p=\frac{1}{2}(2+\sqrt{80}+10)=\frac{1}{2}(12+4\sqrt{5})=6+2\sqrt{5}$

$\Rightarrow r=\frac{8}{6+2\sqrt{5}}=4\cdot\frac{3-\sqrt{5}}{(3-\sqrt{5})(3+\sqrt{5})}=4\cdot\frac{3-\sqrt{5}}{9-5}=3-\sqrt{5}$