Given a regular tetrahedron $(G,DEF)$ with lenght of his sides $a$, see picture. $H$ is the centroid of $\triangle DEF, HE = \frac{\sqrt{3}}{3}a, GH = \sqrt{\frac{2}{3}}a , LH = \frac{1}{4}GH = \frac{1}{4}\sqrt{\frac{2}{3}}a$ and area $\triangle DEF = \frac{1}{4}\sqrt{3}a^{2}$. Point $L$ is the centroid of the tetrahedron $G,DEF$. Cutting this tetrahedron by a plane $\rightarrow \triangle ABC$ with lenght of the sides $k \cdot a$. $I$ is the centroid of $\triangle ABC, BI = \frac{\sqrt{3}}{3}ka, GI = \sqrt{\frac{2}{3}}ka , KI = \frac{1}{4}GI = \frac{1}{4}\sqrt{\frac{2}{3}}ka$ and area $\triangle ABC = \frac{1}{4}\sqrt{3}k^{2}a^{2}$. Point $K$ is the centroid of the tetrahedron $G,ABC$. Point $M$ is the centroid of the frustum $(DEF,ABC)$. There are two ways to calculate $HM$. 1) From http://mathworld.wolfram.com/PyramidalFrustum.html The height of the frustum $(DEF,ABC): h=IH = GH - GI = (1-k)\sqrt{\frac{2}{3}}a$. $A_{1}=$ area $\triangle DEF$, $A_{2}=$ area $\triangle ABC$. \[HM =\frac{h(A_{1}+2\sqrt{A_{1}A_{2}}+3A_{2})}{4(A_{1}+\sqrt{A_{1}A_{2}}+A_{2})}\] \[HM = \frac{\sqrt{\frac{2}{3}}a(1+k+k^{2}-3k^{3})}{4(1+k+k^{2})} \quad (1)\] 2) From an old theorem http://adcs.home.xs4all.nl/stevin/weegconst/85.html , number 90 $V_{1}=$ volume frustum $(DEF,ABC)=\frac{1}{12}\sqrt{2}a^{3}(1-k^{3})$, $V_{2}= $ volume $(G,DEF)=\frac{1}{12}\sqrt{2}a^{3}k^{3}$. $KL= KI + IL = KI +IH-LH = \frac{3}{4}\sqrt{\frac{2}{3}}a(1-k)$. \[\frac{KL}{LM}=\frac{V_{1}}{V_{2}}\] \[LM =\frac{\frac{3}{4}\sqrt{\frac{2}{3}}ak^{3}}{1+k+k^{2}}\] \[HM=LH-LM=\frac{1}{4}\sqrt{\frac{2}{3}}a-\frac{\frac{3}{4}\sqrt{\frac{2}{3}}ak^{3}}{1+k+k^{2}}=(1)\] \[GM = GH - HM = \sqrt{\frac{2}{3}}a - \frac{\sqrt{\frac{2}{3}}a(1+k+k^{2}-3k^{3})}{4(1+k+k^{2})}\] \[GM = \frac{3}{4}\sqrt{\frac{2}{3}}a \frac{1+k+k^{2}+k^{3}}{1+k+k^{2}}\] If we can prove that $ \frac{3}{4}GI = \frac{2}{3}GM \quad \quad(2)$, then: \[\frac{3}{4}\sqrt{\frac{2}{3}}ka =\frac{2}{3}\frac{3}{4}\sqrt{\frac{2}{3}}a \frac{1+k+k^{2}+k^{3}}{1+k+k^{2}}\] \[3k(1+k+k^{2})=2(1+k+k^{2}+k^{3})\] \[k^{3}+k^{2}+k-2=0\] To prove (2), we use analytic geometry, maybe there is another way. $D(\frac{a}{2},0,0),E(-\frac{a}{2},0,0),F(0,\frac{\sqrt{3}}{6}a,\sqrt{\frac{2}{3}}a),G(0,\frac{\sqrt{3}}{2}a,0)$. The centroid $H(0,\frac{\sqrt{3}}{18}a,\sqrt{\frac{2}{3}}\frac{a}{3})$. Equation of $HG$: \[\left\{\begin{array}{ll}x=0\\ -\sqrt{2}(y-\frac{\sqrt{3}}{2}a)=4z\end{array}\right.\] Equation of the sphere with midpoint $G$ and radius $GM$: \[x^{2}+(y-\frac{\sqrt{3}}{2}a)^{2}+z^{2}=GM^{2}\] $HG \cap $sphere $=M : y_{M}= \frac{\sqrt{3}}{2}a -\frac{2\sqrt{2}}{3}GM$. Equation of the sfere with midpoint $G$ and radius $GI$: \[x^{2}+(y-\frac{\sqrt{3}}{2}a)^{2}+z^{2}=GI^{2}\] $HG \cap $sfere $=I : I(0,\frac{\sqrt{3}}{2}a -\frac{2\sqrt{2}}{3}GI,\frac{GI}{3})$. Constructing a plane, through $I$ and $\bot HG: 4y-\sqrt{2}z=2\sqrt{3}a-3\sqrt{2}GI$ This plane cuts the $xy-$plane in $y=\frac{1}{4}(2\sqrt{3}a-3\sqrt{2}GI)$. This value must equal $y_{M}$. Therefore: \[\frac{\sqrt{3}}{2}a -\frac{2\sqrt{2}}{3}GM=\frac{1}{4}(2\sqrt{3}a-3\sqrt{2}GI)\] \[\frac{3}{4}GI = \frac{2}{3}GM\]

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Equilateral $\triangle ABC$ with area $S$ is horizontal base of regular tetrahedron $ABCD$. $O$ is center of $\triangle ABC$ and $A'$ midpoint of $BC.$ $h = DO$ and $\varphi = \angle ODA'$ $\Longrightarrow$ $\cos \varphi = \frac{DO}{DA'}= \frac{\sqrt{DA'^2 - OA'^2}}{DA'} = \sqrt{\frac{_8}{^9}}$. Frustrum volume is $V= \tfrac{1}{3} hS(1 - x^3)$. $G$ is frustrum centroid $\Longrightarrow$ $g = DG = \frac{h^2S}{V} \int_x^1 z^3\text dz = \tfrac{3}{4} h \cdot \frac{1-x^4}{1-x^3}$. Tip over equilibrium means that $hx = g \cos^2 \varphi = \tfrac{8}{9} g$ or $x = \tfrac{8}{9} \cdot \tfrac{3}{4} \cdot \frac{1-x^4}{1-x^3} = \tfrac{2}{3} \cdot \frac{(1+x)(1+x^2)}{1 + x + x^2} = \frac{_2}{^3}\left(\frac{1}{1+x+x^2} + x \right)$ or $x^3 + x^2 + x = 2$.