Prove following points lie on a circle.

Question

I found this in a textbook without a solution and I wasnt able to solve it myself.

Let ABCD be a tetrahedron with all faces acute. Let E be the mid point of the longer arc AB on a circle ABD. Let F be the mid point of the longer arc BC on a circle BCD. Let G be the mid point of the longer arc AC on a circle ACD.

Show that points D,E,F,G lie on a circle.

My approach to that was to try to show these point were co-planear. Since they all lie on one sphere (the one with inscribed tetrahedron ABCD) that would solve the problem. Needles to say I failed at that.

Answer 1

Let $\vec{a} = \vec{DA}, \vec{b} = \vec{DB}, \vec{c} = \vec{DC}$ and $a,b,c$ be corresponding magnitudes.

Let us look at what happens on the plane holding circle $ABD$. Let $X$ be the circle's center and $E'$ be the mid point of the shorter arc $AB$. It is not hard to see

$$\angle E'DB = \frac12 \angle E'XB = \frac12 \angle AXE' = \angle ADE'$$

This implies $DE'$ is the angular bisector of $\angle ADB$ and $\vec{DE'}$ is pointing along the direction $\frac{\vec{a}}{a} + \frac{\vec{b}}{b}$. Since $DE$ is perpendicular to $DE'$, $\vec{DE}$ is pointing along the direction $\frac{\vec{a}}{a} - \frac{\vec{b}}{b}$.

To proceed, we will re-express this fact in terms of barycentric coordinates.

For any $P \in \mathbb{R}^3$, the barycentric coorindates of $P$ with respect to tetrahedron $ABCD$ is a 4-tuple $(\alpha_P, \beta_P, \gamma_P, \delta_P)$ which satisfies: $$\alpha_P + \beta_P + \gamma_P + \delta_P = 1\quad\text{ and }\quad \vec{P} = \alpha_P \vec{A} + \beta_P \vec{B} + \gamma_P \vec{C} + \delta_P\vec{D}$$

In particular, the barycentric coordinates for $D$ is $(0,0,0,1)$.

Let's look at point $E$. Since $E$ lies on the plane holding $ABD$, $\gamma_E = 0$. Since $DE$ is pointing along the direction $\frac{\vec{a}}{a} - \frac{\vec{b}}{b}$, we find $\alpha_E : \beta_E = \frac1a : -\frac1b$. From this, we can deduce there is a $\lambda_E$ such that

$$(\alpha_E, \beta_E, \gamma_E, \delta_E) = \left(\frac{\lambda_E}{a}, -\frac{\lambda_E}{b}, 0, 1 + \lambda_E\frac{a - b}{ab}\right)$$

By a similar argument, we can find $\lambda_F$ and $\lambda_G$ such that

$$(\alpha_F, \beta_F, \gamma_F, \delta_F) = \left(0,\frac{\lambda_F}{b}, -\frac{\lambda_F}{c}, 1 + \lambda_F\frac{b - c}{bc}\right)\\ (\alpha_G, \beta_G, \gamma_G, \delta_G) = \left(-\frac{\lambda_G}{a},0,\frac{\lambda_G}{c}, 1 + \lambda_G\frac{c - a}{ca}\right)$$ In terms of barycentric coordinates, $D,E,F,G$ are coplanar when and only when following determinant evaluates to zero.

$$\mathcal{D}\stackrel{def}{=}\left|\begin{matrix} \alpha_E & \beta_E & \gamma_E & \delta_E\\ \alpha_F & \beta_F & \gamma_F & \delta_F\\ \alpha_G & \beta_G & \gamma_G & \delta_G\\ \alpha_D & \beta_D & \gamma_D & \delta_D \end{matrix}\right| = \left|\begin{matrix} \alpha_E & \beta_E & \gamma_E & \delta_E\\ \alpha_F & \beta_F & \gamma_F & \delta_F\\ \alpha_G & \beta_G & \gamma_G & \delta_G\\ 0 & 0 & 0 & 1 \end{matrix}\right| = \left|\begin{matrix} \alpha_E & \beta_E & \gamma_E\\ \alpha_F & \beta_F & \gamma_F\\ \alpha_G & \beta_G & \gamma_G\\ \end{matrix}\right| $$ Substitute above expression of barycentric coordinates of $E,F,G$ into last determinant, we find $$\mathcal{D} = \lambda_E\lambda_F\lambda_G \left| \begin{matrix} \frac1a & -\frac1b & 0\\ 0 & \frac1b & -\frac1c\\ -\frac1a & 0 &\frac1c \end{matrix} \right| = 0 $$ as the rows of determinant on RHS sum to zero.

From this, we can conclude $D, E, F, G$ are coplanar. Since $D, E, F, G$ lie on the intersection of a sphere and a plane, they lie on a circle.

Answer 2

I'll write the tetrahedron as $OABC$, with $O$ at the origin, and I'll let $D$, $E$, $F$ be the new points associated with faces $\triangle OBC$, $\triangle OCA$, $\triangle OAB$. Define $$a := |OA| \qquad b := |OB| \qquad c := |OC| \qquad \alpha := \angle BOC \qquad \beta := \angle COA \qquad \gamma = \angle AOB$$ and recall that, for instance, $$B\cdot C = b c \cos\alpha \qquad |B\times C| = b c \sin\alpha$$

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Consider the situation with $\triangle OBC$. The defining arc property for $D$ indicates that this point lies on the perpendicular bisector of $\overline{BC}$ within the plane of $\triangle OBC$. Thus, $\overrightarrow{DD^\prime}$ is perpendicular to both $\overrightarrow{BC}$ and the normal to the plane (that is, $B\times C$). we can write

$$D=D^\prime + |DD^\prime| \frac{( C-B )\times ( B \times C )}{|BC|\,|B\times C|} \tag{1}$$ where $D^\prime := \frac12(B+C)$ is the midpoint of $\overline{BC}$.

Further, by the Inscribed Angle Theorem, $\angle BOC\cong\angle BDC$, and we conclude that $\overline{DD^\prime}$ is the altitude of an isosceles triangle with vertex angle $\alpha$ and base $|BC|$. Therefore, $|DD^\prime| = \frac12|BC|\,\cot\frac12\alpha$, so we have $$\frac{|DD^\prime|}{|BC|\,|B\times C|} = \frac{\cot\frac12\alpha}{2bc\sin\alpha} = \frac{\cos\frac12\alpha\,/\,\sin\frac12\alpha}{4bc\sin\frac12\alpha\cos\frac12\alpha}=\frac{1}{4bc\sin^2\frac12\alpha} = \frac{1}{2bc(1-\cos\alpha)} \tag{2}$$ Further, via a cross product identity, $$\begin{align} (C-B)\times(B\times C) &= \phantom{-}B\,((C-B)\cdot B) - C\,((C-B)\cdot C) \\ &=\phantom{-}B\left(B\cdot C - |B|^2\right)-C\left(|C|^2 - B\cdot C\right) \\ &=-B\,b(b-c\cos\alpha) - C\,c(c-b\cos\alpha) \end{align}\tag{3}$$ Altogether, this gives us $$\begin{align}D\;2bc(1-\cos\alpha) &= (B+C)bc(1-\cos\alpha)-B\,b(b-c\cos\alpha)-C\,c(c-b\cos\alpha) \\ &=Bb(c-b)+Cc(b-c) \\ &=(c-b)(B b-C c) \tag{4a} \end{align}$$ (Note that, if $b=c$, then $D=0$, as we would expect. This confirms that we got our cross-product vector directions correct in $(1)$.) Likewise, $$\begin{align} E\,2ca(1-\cos\beta) &= (a-c)(C c-A a) \tag{4b} \\ F\,2ab(1-\cos\gamma) &= (b-a)(A a-B b) \tag{4c} \end{align}$$

Finally, observe that, for $a$, $b$, $c$ not all equal (the only case with which we need concern ourselves), $$(\text{eq }4a)\;(a-c)(b-a) + (\text{eq }4b)\;(c-b)(b-a)+(\text{eq }4c)\;(a-c)(c-b) \tag{5}$$ gives a non-trivial linear combination of $D$, $E$, $F$ that vanishes. Consequently, $D$, $E$, $F$ are linearly dependent; that is, they lie on a common plane through $O$, and the result follows. $\square$