2
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Sorry for the ignorance but how does the direction of DE tell us anything about the ratio αE:βE. Could you explain that part? $\endgroup$ – Dood Dec 3 '18 at 19:49
  • $\begingroup$ 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. $\endgroup$ – Dood Dec 3 '18 at 19:59
  • 1
    $\begingroup$ @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. $\endgroup$ – achille hui Dec 4 '18 at 0:30
  • 1
    $\begingroup$ For the first question, $\vec{DE} \propto \frac{\vec{a}}{a} - \frac{\vec{b}}{b}$ implies existance of $\lambda_E$ such that $$\vec{E}-\vec{D} = \lambda_E\left(\frac{\vec{a}}{a} -\frac{\vec{b}} {b}\right) = \lambda_E\left(\frac{\vec{A}-\vec{D}}{a} - \frac{\vec{B}-\vec{D}}{a}\right)\\ \iff \vec{E} = \frac{\lambda_E}{a}\vec{A} - \frac{\lambda_E}{b}\vec{B} + \left(1 - \frac{\lambda_E}{a} + \frac{\lambda_E}{b}\right)\vec{D} $$ $\endgroup$ – achille hui Dec 4 '18 at 0:38
1
$\begingroup$

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. $\square$

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Dood Dec 3 '18 at 11:35
  • $\begingroup$ Think of the points as position vectors, which add coordinate-wise. The lengths act as scalar multipliers. $\endgroup$ – Blue Dec 3 '18 at 11:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.