\(% Differentiation % https://tex.stackexchange.com/a/60546/ \newcommand{\Diff}{\mathop{}\!\mathrm{d}} \newcommand{\DiffFrac}[2]{\frac{\Diff #1}{\Diff #2}} \newcommand{\DiffOp}[1]{\frac{\Diff}{\Diff #1}} \newcommand{\Ndiff}[1]{\mathop{}\!\mathrm{d}^{#1}} \newcommand{\NdiffFrac}[3]{\frac{\Ndiff{#1} #2}{\Diff {#3}^{#1}}} \newcommand{\NdiffOp}[2]{\frac{\Ndiff{#1}}{\Diff {#2}^{#1}}} % Evaluation \newcommand{\LEvalAt}[2]{\left.#1\right\vert_{#2}} \newcommand{\SqEvalAt}[2]{\left[#1\right]_{#2}} % Epsilon & Phi \renewcommand{\epsilon}{\varepsilon} \renewcommand{\phi}{\varphi} % Sets \newcommand{\NN}{\mathbb{N}} \newcommand{\ZZ}{\mathbb{Z}} \newcommand{\QQ}{\mathbb{Q}} \newcommand{\RR}{\mathbb{R}} \newcommand{\CC}{\mathbb{C}} \newcommand{\PP}{\mathbb{P}} \renewcommand{\emptyset}{\varnothing} % Probabililty \DeclareMathOperator{\Cov}{Cov} \DeclareMathOperator{\Corr}{Corr} \DeclareMathOperator{\Var}{Var} \DeclareMathOperator{\Expt}{E} \DeclareMathOperator{\Prob}{P} % Distribution \DeclareMathOperator{\Binomial}{B} \DeclareMathOperator{\Poisson}{Po} \DeclareMathOperator{\Normal}{N} \DeclareMathOperator{\Exponential}{Exp} \DeclareMathOperator{\Geometric}{Geo} \DeclareMathOperator{\Uniform}{U} % Complex Numbers \DeclareMathOperator{\im}{Im} \DeclareMathOperator{\re}{Re} % Missing Trigonometric & Hyperbolic functions \DeclareMathOperator{\arccot}{arccot} \DeclareMathOperator{\arcsec}{arcsec} \DeclareMathOperator{\arccsc}{arccsc} \DeclareMathOperator{\sech}{sech} \DeclareMathOperator{\csch}{csch} \DeclareMathOperator{\arsinh}{arsinh} \DeclareMathOperator{\arcosh}{arcosh} \DeclareMathOperator{\artanh}{artanh} \DeclareMathOperator{\arcoth}{arcoth} \DeclareMathOperator{\arsech}{arsech} \DeclareMathOperator{\arcsch}{arcsch} % UK Notation \DeclareMathOperator{\cosec}{cosec} \DeclareMathOperator{\arccosec}{arccosec} \DeclareMathOperator{\cosech}{cosech} \DeclareMathOperator{\arcosech}{arcosech} % Paired Delimiters \DeclarePairedDelimiter{\ceil}{\lceil}{\rceil} \DeclarePairedDelimiter{\floor}{\lfloor}{\rfloor} \DeclarePairedDelimiter{\abs}{\lvert}{\rvert} \DeclarePairedDelimiter{\ang}{\langle}{\rangle} % Vectors \newcommand{\vect}[1]{\mathbf{#1}} \newcommand{\bvect}[1]{\overrightarrow{#1}} % https://tex.stackexchange.com/a/28213 % \DeclareMathSymbol{\ii}{\mathalpha}{letters}{"10} % \DeclareMathSymbol{\jj}{\mathalpha}{letters}{"11} % \newcommand{\ihat}{\vect{\hat{\ii}}} % \newcommand{\jhat}{\vect{\hat{\jj}}} \newcommand{\ihat}{\textbf{\^{ı}}} \newcommand{\jhat}{\textbf{\^{ȷ}}} \newcommand{\khat}{\vect{\hat{k}}} % Other Functions \DeclareMathOperator{\sgn}{sgn} \DeclareMathOperator{\tr}{tr} % Other Math Symbols \DeclareMathOperator{\modulo}{mod} \newcommand{\divides}{\mid} \newcommand{\notdivides}{\nmid} \newcommand{\LHS}{\text{LHS}} \newcommand{\RHS}{\text{RHS}} \newcommand{\degree}{^{\circ}}\)
Let the tetrahedron be \(OABC\), and let \(\abs *{OA} = a, \abs *{OB} = b, \abs *{OC} = c, \abs *{BC} = d, \abs *{AC} = e, \abs *{AB} = f\).
This tetrahedron is isosceles, if and only if \(a = d, b = e,\) and \(c = f\).
The perimeter of the face \(OAB\) is \(a + b + f\), of face \(OBC\) is \(b + c + d\), of face \(OAC\) is \(a + c + e\), and of face \(ABC\) is \(d + e + f\).
If the tetrahedron is isosceles, \(a = d, b = e\) and \(c = f\), then all the faces have perimeter \(a + b + c\) and are equal.
If all faces have equal perimeter, then comparing the perimeters of faces \(OAB\), \(OBC\) and \(OAC\), \(a + f = c + d\), \(b + f = c + e\), \(b + d = a + e\).
Hence, \(a - d = b - e = c - f\). Let the difference be \(t\), and \(a = d + t, b = e + t, c = f + t\).
Comparing the perimeter of face \(OAB\) and face \(ABC\) this time, we have \((d + t) + (e + t) = d + e\), which gives \(t = 0\).
Hence, \(a = d\), \(b = e\), \(c = f\), and the tetrahedron is isosceles.
Applying the cosine rule in triangle \(OBC\), we have \[ \abs *{\vect {a}}^2 = \abs *{\vect {b}}^2 + \abs *{\vect {c}}^2 - 2 \abs *{\vect {b}} \abs *{\vect {c}} \cos \angle COB \] and using the dot-product formula \[ \vect {b} \cdot \vect {c} = \abs *{\vect {b}} \abs *{\vect {c}} \cos \angle COB, \] rearranging gives us \[ 2 \vect {b} \cdot \vect {c} = \abs *{\vect {b}}^2 + \abs *{\vect {c}}^2 - \abs *{\vect {a}}^2. \]
Similarly, we have \begin {align*} 2 \vect {a} \cdot \vect {b} & = \abs *{\vect {a}}^2 + \abs *{\vect {b}}^2 - \abs *{\vect {c}}^2, \\ 2 \vect {a} \cdot \vect {c} & = \abs *{\vect {a}}^2 + \abs *{\vect {c}}^2 - \abs *{\vect {b}}^2. \end {align*}
Summing these two, we get \begin {align*} 2 \vect {a} \cdot \vect {b} + 2 \vect {a} \cdot \vect {c} & = 2 \abs *{\vect {a}}^2 \\ \vect {a} \cdot \vect {b} + \vect {a} \cdot \vect {c} & = \abs *{\vect {a}}^2 \\ \vect {a} \cdot \left (\vect {b} + \vect {c}\right ) & = \abs *{\vect {a}}^2. \end {align*}
Let \(\vect {g}\) be the position vector for \(G\). \(\abs *{OG} = \abs *{\vect {g}} = \frac {1}{4} \abs *{\vect {a} + \vect {b} + \vect {c}}\).
Consider the distance between \(A\) and \(G\). \begin {align*} \abs *{AG} & = \abs *{\bvect {AG}} \\ & = \abs *{\vect {g} - \vect {a}} \\ & = \frac {1}{4} \abs *{- 3 \vect {a} + \vect {b} + \vect {c}}. \end {align*}
We want to show that \(\abs *{\vect {a} + \vect {b} + \vect {c}} = \abs *{- 3 \vect {a} + \vect {b} + \vect {c}}\). The following are equivalent
\begin {align*} \abs *{\vect {a} + \vect {b} + \vect {c}} & = \abs *{- 3 \vect {a} + \vect {b} + \vect {c}} \\ \abs *{\vect {a} + \vect {b} + \vect {c}}^2 & = \abs *{- 3 \vect {a} + \vect {b} + \vect {c}}^2 \\ \abs *{\vect {a}}^2 + \abs *{\vect {b}}^2 + \abs *{\vect {c}}^2 + 2 \vect {a} \cdot \vect {b} + 2 \vect {a} \cdot \vect {c} + 2 \vect {b} \cdot \vect {c} & = 9 \abs *{\vect {a}}^2 + \abs *{\vect {b}}^2 + \abs *{\vect {c}}^2 - 6 \vect {a} \cdot \vect {b} - 6 \vect {a} \cdot \vect {c} + 2 \vect {b} \cdot \vect {c} \\ 8 \vect {a} \cdot \vect {b} + 8 \vect {a} \cdot \vect {c} & = 8 \abs *{\vect {a}}^2 \\ \vect {a} \cdot \left (\vect {b} + \vect {c}\right ) & = \abs *{\vect {a}}^2 \end {align*}
and this is true from the previous part.
Hence, \(\abs *{OG} = \abs *{AG}\). By symmetry, \(\abs *{OG} = \abs *{AG} = \abs *{BG} = \abs *{CG}\) and hence \(G\) is equidistant from all four vertices of the tetrahedron.
Notice that \begin {align*} \abs *{\vect {a} - \vect {b} - \vect {c}}^2 & = \abs *{\vect {a}}^2 + \abs *{\vect {b}}^2 + \abs *{\vect {c}}^2 - 2 \vect {a} \cdot \vect {b} - 2 \vect {a} \cdot \vect {c} + 2 \vect {b} \cdot \vect {c} \\ & = \abs *{\vect {a}}^2 + \abs *{\vect {b}}^2 + \abs *{\vect {c}}^2 - 2 \abs *{\vect {a}}^2 + \left (\abs *{\vect {b}}^2 + \abs *{\vect {c}}^2 - \abs *{\vect {a}}^2\right ) \\ & = -2 \abs *{\vect {a}}^2 + 2 \abs *{\vect {b}}^2 + 2 \abs *{\vect {c}}^2 \\ & = 2 \left (\abs *{\vect {b}}^2 + \abs *{\vect {c}}^2 - \abs *{\vect {a}}^2\right ) \\ & = 4 \vect {b} \cdot \vect {c}, \end {align*}
and since the left-hand side is a square, it is non-negative, which means the dot product is non-negative.
Hence, \(\cos \angle BOC \geq 0\), which means it must not be obtuse. By symmetry, this means none of the angles are obtuse.
If one of them is a right angle, say \(\angle BOC\), then the dot product evaluates to \(0\), which must mean \(\abs *{\vect {a} - \vect {b} - \vect {c}} = 0\).
Hence, \(\vect {a} = \vect {b} + \vect {c}\), which means \(A\) lies in the plane containing \(O, B, C\). This will not be a tetrahedron, and hence no angles can be right angles.