Dacă Q este centrul sferei înscrise în tetraedrul tridreptunghic, atunci [math]Q\in Int_{OABC}[/math] şi d(Q,(OAB))=d(Q,(OBC))=d(Q,(OCA))=r[br][math]V_{OABC}=V_{QOAB}+V_{QOBC}+V_{QOCA}+V_{QABC}[/math] [math]\Rightarrow[/math] [math]\frac{\left(a\cdot b\cdot c\right)}{6}=\frac{r}{3}\cdot\left(\frac{\left(a\cdot b\right)}{2}+\frac{\left(b\cdot c\right)}{2}+\frac{\left(c\cdot a\right)}{2}+\frac{\sqrt{\left(a\cdot b\right)^2+\left(b\cdot c\right)^2+\left(c\cdot a\right)^2}}{2}\right)[/math] [math]\Rightarrow[/math] [math]r=\frac{\left(a\cdot b\cdot c\right)}{a\cdot b+b\cdot c+c\cdot a+\sqrt{\left(a\cdot b\right)^2+\left(b\cdot c\right)^2+\left(c\cdot a\right)^2}}[/math].

Centrul sferei înscrise în tetraedrul tridreptunghic: Q(r,r,r)[br][math]T_1\in\left(xOy\right)\Rightarrow\left(r,r,r\right)-\left(0,0,r\right)=\left(r,r,0\right)[/math][br][math]T_2\in\left(yOz\right)\Rightarrow\left(r,r,r\right)-\left(r,0,0\right)=\left(0,r,r\right)[/math][br][math]T_3\in\left(xOz\right)\Rightarrow\left(r,r,r\right)-\left(0,r,0\right)=\left(r,0,r\right)[/math]

Ecuaţia planului bazei, (ABC), este [math]\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1\Rightarrow\vec{\nu}\left(\frac{1}{a},\frac{1}{b},\frac{1}{c}\right),\nu=\sqrt{\left(\frac{1}{a}\right)^2+\left(\frac{1}{b}\right)^2+\left(\frac{1}{c}\right)^2}=\frac{1}{h}.[/math] [br]Versorul normal la planul bazei [math]\vec{\eta}=\frac{\vec{\nu}}{\nu}=h\cdot\left(\frac{1}{a},\frac{1}{b},\frac{1}{c}\right)[/math] [br]T[sub]4[/sub]: [math]\left(r,r,r\right)+r\cdot\left(\frac{h}{a},\frac{h}{b,},\frac{h}{c}\right)=r\cdot\left(1+\frac{h}{a},1+\frac{h}{b},1+\frac{h}{c}\right)[/math]

Versorul normal la planul bazei: [math]\vec{\eta}\left(\frac{h}{a},\frac{h}{b},\frac{h}{c}\right)_{ }\Rightarrow H:_{ }h\cdot\left(\frac{h}{a},\frac{h}{b},\frac{h}{c}\right)=h^2\cdot\left(\frac{1}{a},\frac{1}{b},\frac{1}{c}\right).[/math]

[list][*][i]o demonstraţie analitică[/i][/*][/list]Dacă J este vârful opus lui O, în cubul înscris în tetraedrul tridreptunghic, de muchie m, atunci [br]J(m,m,m) şi J [math]\in[/math] planului (ABC), de ecuaţie [math]\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1[/math] [math]\Rightarrow[/math] [math]\frac{1}{m}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}[/math] (1) [math]\frac{1}{h}=\sqrt{\left(\frac{1}{a}\right)^2+\left(\frac{1}{b}\right)^2+\left(\frac{1}{c}\right)^2}[/math](2)[br][math]\frac{1}{r}=\frac{a\cdot b+b\cdot c+c\cdot a+\sqrt{\left(a\cdot b\right)^2+\left(b\cdot c\right)^2+\left(c\cdot a\right)^2}}{a\cdot b\cdot c}[/math](3)[br][br][list][*][i]o demonstraţie sintetică[/i][/*][/list]Q[math]\in[/math](OJ), QT[sub]1[math]\parallel[/math] [/sub]JE, T[sub]1[/sub] [math]\in[/math](OE) [math]\Rightarrow[/math] [math]\Delta_{ }QT_1O_{ }\sim\Delta_{ }JEO_{ }\Rightarrow_{ }\frac{r}{m}=\frac{OQ}{OJ}[/math]. (1)[br]QT[sub]4 [math]\perp[/math] [/sub](ABC), OH [math]\bot[/math] (ABC) [math]\Rightarrow[/math] QT[sub]4[/sub] [math]\parallel[/math] OH. Q [math]\in[/math] (OJ), T[sub]4[/sub] [math]\in[/math] (JH) [math]\Rightarrow[/math] [math]\Delta_{ }QT_4J_{ }\sim\Delta_{ }OHJ_{ }\Rightarrow[/math] [math]\frac{r}{h}=\frac{QJ}{OJ}[/math]. (2)[br]Din (1) şi (2) [math]\Rightarrow[/math] [math]\frac{r}{h}+\frac{r}{m}=1[/math] [math]\Leftrightarrow[/math] [math]\frac{1}{h}+\frac{1}{m}=\frac{1}{r}[/math][br]