# Prove following points lie on a circle.

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.

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.

• Sorry for the ignorance but how does the direction of DE tell us anything about the ratio αE:βE. Could you explain that part? – Dood Dec 3 '18 at 19:49
• Also does the fact that D,E,F,G are coplanar if and only if the determinant D=0 come from the formula for volume of tetrahedron using its verticies in barycentric coordinates? If so could you show the whole formula or at least link to it? I cant find it anywhere but i assume it's simmilar to the formula for a triangle which I did find. Many thanks for your contribution anyway. – Dood Dec 3 '18 at 19:59
• @Dood, yes, it is related to the formula of volume of tetrahedron. No. I don't have a link. This is a well known result and generalise to any finite dimensional simplex. – achille hui Dec 4 '18 at 0:30
• For the first question, $\vec{DE} \propto \frac{\vec{a}}{a} - \frac{\vec{b}}{b}$ implies existance of $\lambda_E$ such that $$\vec{E}-\vec{D} = \lambda_E\left(\frac{\vec{a}}{a} -\frac{\vec{b}} {b}\right) = \lambda_E\left(\frac{\vec{A}-\vec{D}}{a} - \frac{\vec{B}-\vec{D}}{a}\right)\\ \iff \vec{E} = \frac{\lambda_E}{a}\vec{A} - \frac{\lambda_E}{b}\vec{B} + \left(1 - \frac{\lambda_E}{a} + \frac{\lambda_E}{b}\right)\vec{D}$$ – achille hui Dec 4 '18 at 0:38

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, $$B\cdot C = b c \cos\alpha \qquad |B\times C| = b c \sin\alpha$$

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. $$\square$$

• Pardon my ignorance but I'm confused as to what is being written as soon as equality (1). How can a point be a sum of other points and lenghts. – Dood Dec 3 '18 at 11:35
• Think of the points as position vectors, which add coordinate-wise. The lengths act as scalar multipliers. – Blue Dec 3 '18 at 11:38