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\begin{document}
\title{Theorem on altitudes and the Jacobi identity}
\author{A. Zaslavskiy and M. Skopenkov}
\date{}
\maketitle

{\bf Solutions.}

\smallskip

\noindent First let us give a table containing the answers to all
the problems:

\bigskip
\small

\noindent\begin{tabular}{|l|l|} \hline
{Algebraic object}&{Geometric sense}\\[5pt]\hline
$\vec A$&{a point $A$}\\
$\vec a$&{a spherical line $a$}\\
$[\vec A,\vec B]$&{the line passing through $A$ and $B$}\\
$[\vec a,\vec b]$&{the intersection point of $a$ and $b$}\\
$[\vec A,\vec b]$&{the perpendicular dropped from the point $A$ to the line $b$}\\
$\vec A+\vec B$&{the middle of the segment $AB$}\\
$(\vec a,\vec b)$&{cosine of the angle between $a$ and $b$}\\
$|[\vec a,\vec b]|$&{sine of the angle between $a$ and $b$}\\
$(\vec A,\vec B)$&{cosine of the length of the segment $AB$ (i. e. the arc of a big circle between $A$ and $B$)}\\
$|[\vec A,\vec B]|$&{sine of the length of the segment $AB$}\\
$(\vec A,\vec b)$&{sine of the distance between $A$ and $b$}\\
$|[\vec A,\vec b]|$&{cosine of the distance between $A$ and $b$}\\
$\vec A+\vec B+\vec C=0$&{1) the points $A$, $B$ and $C$ are collinear}\\
&{2) the centers of the circles $A$, $B$ and $C$ are collinear}\\
$\vec a+\vec b+\vec c=0$&{the lines $a$, $b$ and $c$ have a common point}\\
$[\vec A,[\vec B,\vec C]]+[\vec B,[\vec C,\vec A]]+[\vec C ,[\vec
A,\vec B]]=0$&{the altitudes of a triangle $ABC$
have a common point}\\
$[\vec A,\vec B+\vec C]+[\vec B,\vec C+\vec A]+[\vec C,\vec A+\vec B]=0$
&{the medians of a triangle $ABC$ have a common point}\\
$[\vec A,\vec B]+[\vec B,\vec C]+[\vec C,\vec A]$&{the center of
the circumscribed circle of the triangle $ABC$}\\
$(\vec A,\vec B,\vec C)$&{sine of the length of the segment $AB$
$\cdot$ sine of the distance between $C$ and $AB$}\\
$(\vec A,[\vec B,\vec C])=(\vec B,[\vec C,\vec A])=(\vec C,[\vec A,\vec B])$&{The sines theorem for the triangle $ABC$}\\
$(\vec A,\vec B)=1$&{the circles $A$ and $B$ are orthogonal}\\[2pt]
$\frac{[\vec A,\vec B]+[\vec B,\vec C]+[\vec C,\vec A]}{(\vec
A,\vec B,\vec C)}$&{the circle orthogonal to the circles $A$, $B$ and $C$}\\[2pt]
$\vec A-\vec B+\vec C-\vec D=0$ &{there exists a circle orthogonal
to the circles $A$, $B$, $C$ and $D$}\\[2pt]
$\frac{d_B}{d_B-d_A}\vec A-\frac{d_A}{d_B-d_A}\vec B$&{bisector of
the circles $A$ and $B$}\\[4pt]
\hline
\end{tabular}

\bigskip

\noindent{\bf 8.} Let $\vec A$, $\vec B$ and $\vec C$ be the
vectors corresponding to the vertices of the spherical triangle
$ABC$. By Problem 1 the vector $[\vec A,\vec B]$ corresponds to
the spherical line passing through the points $A$ and $B$. Then by
Problem 3 the vector $[\vec A,[\vec B,\vec C]$ corresponds to the
perpendicular dropped from the point $A$ to the line $BC$, that
is, it corresponds to the line containing the altitude $h_A$ of
the triangle $ABC$. Analogously the vectors $[\vec B,[\vec C,\vec
A]$ and $[\vec C,[\vec A,\vec B]$ correspond to the lines
containing two other altitudes of the triangle $ABC$. So by
Problem 5 the Jacobi identity means that the three constructed
lines have a common point, i. e. we get the theorem on altitudes
of a spherical triangle.

\noindent{\bf 9.} Denote by $\vec a$ {\it the direction vector} of
the line $a$, that is, a vector parallel to the line $a$.
Analogously, let $\vec b$ and $\vec c$ be the direction vectors of
the lines $b$ and $c$. Then obviously the vector $[\vec b,\vec c]$
is parallel to the common perpendicular to the lines $b$ and $c$,
that is, to the line $a'$. Thus the vector $[\vec a,[\vec b,\vec
c]]$ is parallel to the common perpendicular to the lines $a$ and
$a'$, that is, to the line $a''$. Analogously, the vectors $[\vec
b,[\vec c,\vec a]]$ and $[\vec c,[\vec a,\vec b]]$ are parallel to
the lines  $b''$ and $c''$ respectively. By the Jacobi identity
the last three vectors are parallel to one plane. Thus the three
lines $a''$, $b''$ и $c''$ are parallel to one plane.

\noindent{\bf 10.} By symmetry the vector $\vec A+\vec B$
corresponds to the middle of the arc of the big circle passing
through the points $A$ and $B$. It is natural to say that this
point is {\it the middle} of the spherical segment with ends $A$
and $B$.

\noindent{\it Remark.} In fact a pair of points $A$ and $B$ on the
sphere determines two distinct segments. If {\it one} of the two
vectors $\vec A$ and $\vec B$ is replaced by the opposite vector,
then their sum corresponds to the middle of the other segment with
ends $A$ and $B$.

\noindent{\bf 11.} Let $\vec A$, $\vec B$ and $\vec C$ be unit
vectors pointing to $A$, $B$ and $C$ respectively from the center
of the sphere. Then by Problem 10 the vector $\vec B+\vec C$
corresponds to the middle of the segment $BC$. Then by Problem 1
the vector $[\vec A,\vec B+\vec C]$ corresponds to {\it the
median} of the triangle $ABC$. Then by Problem 5 the identity
$[\vec A,\vec B+\vec C]+[\vec B,\vec C+\vec A]+[\vec C,\vec A+\vec
B]=0$ means that the medians of the triangle $ABC$ have a common
point.

\noindent The theorem on medians can also be proved by
consideration of the vector $\vec A+\vec B+\vec C$.

\noindent{\it Remark.} (A. Mafusalov) Unlike Euclidean geometry,
the theorem on medians of a spherical triangle has an external
analog: two 'external' medians $m_A$, $m_B$ and the opposite
'internal' median $m_C$ of a triangle have a common point.

\noindent{\bf 12.} First let us prove that the vectors $\vec
A-\vec B$, $\vec B-\vec C$ and $\vec C-\vec A$ correspond to the
{\it middle perpendiculars} to the sides of a spherical triangle
$ABC$. Indeed, since
$$
(\vec A-\vec B,\vec A+\vec B)=(\vec A,\vec A)-(\vec B,\vec
B)=1-1=0,
$$
it follows that the line corresponding to $\vec A-\vec B$ passes
through the middle of $AB$. And since
$$
(\vec A-\vec B,[\vec A,\vec B])=(\vec A,[\vec A,\vec B])-(\vec
B,[\vec A,\vec B])=0,
$$
it follows that this line is orthogonal to the segment $AB$. Thus
the vector $\vec A-\vec B$ corresponds to the middle perpendicular
to the segment $AB$.

\noindent Denote by $\vec V=[\vec A,\vec B]+[\vec B,\vec C]+[\vec
C,\vec A]$. Let $V$ be the point on the sphere corresponding to
this vector. Let us check that the point $V$ belongs to all the
three constructed middle perpendiculars. Indeed, since
$$
(\vec V,\vec A-\vec B)=([\vec B,\vec C],\vec A)-([\vec C, \vec
A],\vec B)=0,
$$
it follows that $V$ belongs to the line corresponding to the
vector $\vec A-\vec B$. Analogously, $V$ belongs to two other
middle perpendiculars.

 \noindent{\it Remark.} The identity $(\vec A-\vec B)+(\vec
B-\vec C)+(\vec C-\vec A)=0$ and Problem 5 also imply that the
three middle perpendiculars to the sides of a triangle have a
common point.

\noindent{\bf 13.} The solution follows directly from the formulas
$(\vec {\mathstrut a},\vec {\mathstrut b})=|\vec {\mathstrut
a}|\cdot|\vec {\mathstrut b}|\cdot \cos \gamma$ and $|[\vec
{\mathstrut a},\vec {\mathstrut b}]|=|\vec {\mathstrut a}|\cdot
|\vec {\mathstrut b}|\cdot \sin \gamma$, where $\gamma$ is the
angle between the vectors $\vec {\mathstrut a}$ and $\vec
{\mathstrut b}$.


\noindent{\bf 14.} Assume that the lengths of the vectors $\vec
A$, $\vec B$ and $\vec C$ are equal to 1. Then by Problem 13 we
obtain $|([\vec A,\vec B],\vec C)|=\sin AB \sin h_C$, where $h_C$
is the length of the altitude of the triangle $ABC$ passing
through the vertex $C$. The identity $(\vec A,[\vec B,\vec
C])=(\vec B,[\vec C,\vec A])=(\vec C,[\vec A,\vec B])$ means that
$\sin AB \sin h_C=\sin BC \sin h_A=\sin CA \sin h_B$ (1). Writing
an analogous identity for the unit vectors corresponding to the
lines containing the sides of the triangle $ABC$, we get $\sin
\angle C \sin h_C=\sin \angle A \sin h_A=\sin \angle B \sin h_B$
(2). Dividing (1) by (2), we get spherical {\it sine theorem}:
$$
\frac{\sin AB}{\sin \angle C}=\frac{\sin BC}{\sin \angle
A}=\frac{\sin CA}{\sin \angle  B}.
$$

\noindent{\it Remark.} Mixed product of three vectors has a clear
meaning --- it is the volume of the parallelepiped spanned by
these three vectors, with a sign. We do not know such a natural
interpretation of this number in terms of spherical geometry.

\noindent{\bf 15.} {\it First solution.} It is proved analogously
to the case Euclidean geometry that {\it the sides} of a spherical
triangle uniquely define the angles of the triangle. Now it
remains to consider the triangle dual to the initial one (that is,
a triangle such that the vectors pointing to its vertices are
orthogonal to the planes containing the sides of the initial
triangle).

\noindent{\it Second solution.}Analogously to the solution of
Problem 6 one can prove the identity $([\vec a,\vec c],[\vec
b,\vec c])=(\vec a,\vec b)(\vec c,\vec c)-(\vec a,\vec c)(\vec
b,\vec c)$. Applying the results of Problem 13, we obtain the
formula ({\it the spherical cosine theorem})
$$
\sin \angle A\sin \angle B\cos AB = \cos \angle C-\cos \angle
A\cos \angle B.
$$
This implies directly that the angles of a spherical triangle
uniquely determine the sides of the triangle.

\noindent{\it Remark*.} The obtained results remain true also in
the Lobachevskiy (hyperbolic) geometry, because it can be
transformed to spherical by multiplication of all coordinates by
$i$. Then the words 'the lines have a common point' in all the
formulations above should be replaced by 'the lines belong to one
sheaf'. Formally, our above proofs remain valid in this case, if
one takes the unit sphere in the quasi-euclidean space and puts
$[\vec A,\vec B]:=*(\vec A\wedge \vec B)$ [1].

\smallskip

\noindent{\bf 19.} It is easy to check that the directions of both
vectors $u$ and $v$ do not depend on the choice of points $A$ and
$B$: the vector $u$ is parallel to our plane, and the vector $v$
is orthogonal to the plane, passing through our line and the point
$O$. The length of the vector $u$ is equal to $AB$, and the length
of the vector $v$ is equal to two areas of the triangle $ABO$,
that is, to $AB\cdot h$, where $h$ is the distance between $O$ and
our line. Thus for another choice of the points $A$ and $B$ the
vectors $u$ and $v$ are multiplied by the same number.

\noindent{\bf 20.} Indeed, our line should lie in the plane
passing through the point $O$ and orthogonal to the vector $v$.
Our line should also be parallel to the vector $v$, and the
distance between our line and the point $O$ is equal to $|v|/|u|$.
The vector $[u,v]$ points, in which 'direction' with respect to
the point $O$ our line lies. Thus our line is uniquely defined by
the pair of vectors $u$ and $v$.

\noindent{\bf 21.} The construction of the desired line, starting
from the vectors $\vec u$ and $\vec v$, is in fact given in the
solution of the previous problem.

\noindent{\bf 22.} (This is one of the most hard problems of this
set) Define {\it the inner product} of bivectors $\hat a=(u,v)$
and $\hat b=(u',v')$ by the formula
$$
(\hat a,\hat b)=((u,v);(u,v')+(u',v)).
$$
The inner product of two bivectors is a {\it pair} of numbers. It
is easy to check that $(\hat a,[\hat a,\hat b])=(\hat b,[\hat
a,\hat b])=(0;0)$. So our problem is a direct corollary of the
following lemma:

\smallskip

\noindent{\bf Lemma.} If $(\hat a,\hat b)=(0;0)$, then the lines
$a$ and $b$ intersect, and the angle between them is right.

\smallskip

\noindent{\it Proof of the lemma.} The lines $a$ and $b$ are
orthogonal because the first component of the inner product of
$\hat a$ and $\hat b$ is zero: $(u,u')=0$. It remains to shows
that if $(\hat a,\hat b)=(0;0)$, then the lines $a$ and $b$
intersect.

\noindent First consider the case when $u\perp v$ and $u'\perp
v'$.

\noindent Let us check that the inner product does not depend on
the choice of the point $O$. Assume that in the beginning we fix
another point $O_1$. Let $\hat a_1=(u_1,v_1)$, $\hat
b_1=(u'_1,u'_1)$ be the bivectors obtained from the lines $a$ and
$b$ by our construction, if we assume that the fixed point is
$O_1$. Then one can check directly that $u_1=u$,
$v_1=v+[u,\overrightarrow{OO_1}]$ и $u_1'=u'$,
$v_1'=v'+[u',\overrightarrow{OO_1}]$. So
$$
(\hat a_1,\hat
b_1)=((u,u');(u,v'+[u',\overrightarrow{OO_1}])+(v+[u,\overrightarrow{OO_1}],u'))=
((u,u');(u,v')+(v,u'))=(\hat a,\hat b),
$$
because
$(u,[u',\overrightarrow{OO_1}])+([u,\overrightarrow{OO_1}],u')=0$.
Thus indeed the inner product of bivectors does not depend on the
choice of the point $O$.

\noindent Therefore we may assume without loss of generality that
the point $O$ belongs to the line $a$. Then $v=0$. Thus the
condition $(\hat a,\hat b)=(0;0)$ means that the vector $u$ is
orthogonal to the vector $v'$. If $v'=0$, then we are done,
because the point $O$ is a common point for $a$ and $b$ in this
case. If $v'\ne 0$, then both lines lie in the plane passing
through the point $O$ orthogonal to the vector $v'$, and hence
these two lines intersect each other. Thus the case $u\perp v$,
$u'\perp v'$ has been considered.

\noindent Now consider the case when not necessarily $u\perp v$
and $u'\perp v'$. (This case is important for what follows,
because even if this condition is satisfied for the bivectors
$\hat a$ and $\hat b$, it may not be satisfied for the bivector
$[\hat a,\hat b]$.)

\noindent In this case, by definition, the bivector $\hat a=(u,v)$
corresponds to the same line as the bivector $(u,\mathop{pr} v)$.
Denote the last bivector by $\hat a_\perp$. Notice that
$\mathop{pr} v=v+\alpha u$ for some number $\alpha$. Define
analogously the bivector $b_\perp$ and the number $\beta$. Since
$(\hat a,\hat b)=(0;0)$, it follows that $(\hat a_\perp,\hat
b_\perp)=((u,u');(u,v')+\beta(u,u')+(v,u')+\alpha(u,u'))=(\hat
a,\hat b)=(0;0)$, because $(u,u')=0$. Then the second case of the
lemma is reduced to the first one, and we are done.

\noindent{\it Remark*.} The definition of the product of bivectors
comes from the formula for the commutator in the Lie algebra of
the movements group of the 3-space. {\it Sliding vector} is
another geometric interpretation of our bivectors [3].

\noindent{\bf 23.} Notice that $(\hat a+\hat b,[\hat a,\hat b])=
(\hat a,[\hat a,\hat b])+(\hat b,[\hat a,\hat b])=0$. Then by
lemma from the solution of Problem 22 the line corresponding to
the vector $\hat a+\hat b$ intersects the line corresponding to
the vector $[\hat a,\hat b]$. And by Problem 22 the last line is
the common perpendicular to the lines $a$ and $b$.

\noindent{\bf 24.} The Jacobi identity for the product of
bivectors follows directly from the Jacobi identity for the
product of vectors.

\noindent{\bf 25.} Applying Problem 22 several times, we obtain
that the bivectors $[\hat a,[\hat b,\hat c]]$, $[\hat b,[\hat
c,\hat a]]$ and $[\hat c,[\hat b,\hat a]]$ correspond to the lines
$a''$, $b''$ and $c''$ respectively. By the Jacobi identity it
follows that $[\hat c,[\hat b,\hat a]]=[\hat a,[\hat b,\hat
c]]+[\hat b,[\hat c,\hat a]]$. Then by Problem 23 the line $c''$
intersects the common perpendicular to the lines $a''$ and $b''$.
Denote this common perpendicular by $h$. By Problem 9 the angle
between $c''$ and $h$ is right. So $h$ is a common perpendicular
for $a''$, $b''$ and $c''$, and the Theorem on altitudes of a
triangle follows.

\smallskip

\noindent{\bf 27.} In this solution we will use some facts which
are proved in the further solutions.

\noindent It is sufficient to prove the statement of the problem
for circles on a sphere (since a plane can be mapped into a sphere
by a stereographic projection). For some circle, we will denote
the corresponding vector by the same letter with the arrow over it
(e.g. the vector corresponding to a circle $a$ is denoted by $\vec
a$). By problem~34, $\vec c'=\frac{d_B}{d_B-d_A}\vec
a-\frac{d_A}{d_B-d_A}\vec b$; one can write down the analogous
formulae for $\vec a'$ and $\vec b'$. Then we have
$$
\frac{d_C-d_B}{d_C d_B}\vec a'+\frac{d_A-d_C}{d_A d_C}\vec
b'+\frac{d_B-d_A}{d_B d_A}\vec c'=0.
$$
Applying the lemma from the solution of problem~34, we obtain that
the circles $a$, $b$, and $c$ have a common point.

\noindent{\it Note$^*$.} Three circles in this statement can form
a ``convex'' or an ``concave'' triangle. According to this, our
theorem is equivalent to the theorem on bisectors in spherical
geometry or in Lobachevskiy geometry.

\noindent{\bf 28$^*$.} (Solution by A. Mafusalov) We will present
a plan of the solution, omitting some technical details. It is
based on the following elegant theorem.

\smallskip
\noindent{\bf Theorem on the altitudes of a curved triangle.} Let
$A$, $B$, $C$ be pairwise intersecting circles. Let $C'$ be a
circle perpendicular to $C$ and passing through two points of
intersection of $A$ and $B$. The circles $A'$ and $B'$ are defined
analogously. Then three circles $A'$, $B'$ and $C'$ belong to one
sheaf. (In particular, if two of them intersect, then the third
passes through their common point(s)).
\smallskip

\noindent One can check that the statement of problem~28 is
essentially the theorem on altitudes for the curved triangle
$a'b'c'$.

\smallskip
\noindent{\it Proof of the theorem.} Let $D$ be the radical center
of  $A$, $B$, and $C$. Denote by $d$ the degree of the point  $D$
with respect to our circles. Consider the sphere of radius
$\sqrt{|d|}/2$ touching the initial plane in $D$. Let us perform
the stereographic projection of the plane into this sphere. Three
cases are possible.

\noindent (i) $d<0$. One can check that $A$, $B$ and $C$ are
mapped into three big circles on the sphere. In this case our
theorem follows from the theorem on altitudes of a spherical
triangle (problem~8).

\noindent (ii) $d=0$. Let us perform an inversion by a point $D$.
Then our theorem is essentially the theorem on altitudes of
euclidean triangle.


\noindent (iii) $d>0$. One may check that the circles $A$, $B$ and
$C$ map into three circles on the sphere, their centers lying on
one spherical line. Denote this line by $p$. Consider a section of
the sphere by the plane containing $p$, and perform an orthogonal
projection from the sphere into this plane. Three circles
constructed will map into three chords of $p$; denote these chords
by $KL$, $MN$, $PQ$. Applying consequently our stereographic
projection and orthogonal projection to $C'$, we obtain some chord
$P'Q'$. One may check that the line $XY$ should pass through the
intersection point of  $KL$ and $MN$, as well as through that of
the lines tangent to $p$ at $P$ and $Q$. The chords $K'L'$ and
$M'N'$ are defined similarly. We need to prove that the lines
$K'L'$, $M'N'$, and $P'Q'$ are concurrent. One can prove this
statement using the Ceva theorem in a trigopometric form. (This
statement is in fact the theorem on altitudes in Lobachevskiy
geometry, being considered in the Klein model.)

\noindent{\bf 29.} Let $A_1$, $B_1$, and $C_1$ be the {\it
'centers'} of circles, corresponding to the vectors $\vec A$,
$\vec B$, and $\vec C$ (i.e. $A_1$ is the point of the sphere such
that the plane that touches the sphere in this point is parallel
to the circle $A$, and so on). Then the vectors
$\overrightarrow{OA_1}$, $\overrightarrow{OB_1}$, and
$\overrightarrow{OC_1}$ are parallel to $\vec A$, $\vec B$, and
$\vec C$ respectively. Arguments similar to those in the solution
of problem~4 show that the points $A_1$, $B_1$, $C_1$, and $O$ lie
in one plane, i.e. the 'centers' of the circles $A$, $B$, and $C$
lie on one spherical line.

\noindent{\bf 30.} Let $A_1$ be the point such that
$\overrightarrow{OA_1}=\vec A$. Denote by $d_A$ the length of a
tangent from $A_1$ to our sphere. Consider an arbitrary point $C$
on the circle $A$. By the Pythagorean theorem for the triangle
$OA_1C$, we obtain $OA_1^2=OC^2+A_1C^2$, i.e. $(\vec A,\vec
A)=1+d_A^2$, QED.

\noindent{\bf 31.} This solution is obtained from the following
(more general) formula for the angle $\gamma$ between $A$ and $B$:
$$
\cos\gamma=\frac{(A,B)-1}{d_A d_B}.
$$

\noindent Let us prove this formula. Let $C$ be one of the
intersection points of $A$ and $B$. Let $\vec C=\overrightarrow
{OC}$. Note that the vectors $[\vec A,\vec C]$ and $[\vec B,\vec
C]$ are parallel to the lines passing through $C$ and tangent to
the circles $A$ and $B$ respectively. Hence
$$
\cos\gamma=\frac{([\vec A,\vec C],[\vec B,\vec C])}{|[\vec A,\vec
C]|\cdot|\vec B,\vec C|}.
$$

Let us show that the right hand part coincides with the required
expression. It is easy to check that $|[\vec A,\vec C]|=|\vec
A|\cdot|\vec C|\cdot \sin \angle (\vec A,\vec C)=d_A.$ Hence the
denominator of the right hand part is equal to $d_A d_B$. Now we
will rearrange the numerator using the identity from the second
solution of problem~15. We have
$$
([\vec A,\vec C],[\vec B,\vec C])=(\vec A,\vec B)(\vec C,\vec
C)-(\vec A,\vec C)(\vec B,\vec C)=(\vec A,\vec B)-1.
$$
Substituting the expressions for the numerator and the
denominator, we obtain the desired formula.

\noindent{\bf 32.} Assume that the vector $ \vec P=\frac{[\vec
A,\vec B]+[\vec B,\vec C]+[\vec C,\vec A]}{(\vec A,\vec B,\vec
C)}$ corresponds to a circle $P$ (such a circle may not exist, in
which case the given vector has no geometric interpretation).
Notice that
$$
(\vec P,\vec A)=\frac{0+(\vec A,[\vec B,\vec C])+0}{(\vec A,\vec
B,\vec C)}=1.
$$
Then by Problem 31, the circle $P$ is orthogonal to the circle
$A$. Analogously, the circle $P$ is orthogonal to the circles $B$
and $C$. In other words, $P$ is {\it the common perpendicular} to
the circles $A$, $B$ and $C$.

\noindent{\it Remark*.} It is more natural to each circle to
assign a pair $(\vec A,h)$, where $\vec A$ is the unit vector
orthogonal to the plane containing the circle, and $h$ is the
distance between the plane and the point $O$. In other words, to
each circle we assign a quaternion. Denote this quaternion by the
same letter as the circle itself. We assume that all the
quaternions, obtained from the given one by the multiplication by
a real number, correspond to the same circle. Then, for example,
the quaternion $AB-BA$ corresponds to the line passing through the
centers of the circles $A$ and $B$, and the quaternion
$ABC-ACB+BCA-BAC+CAB-CBA$ corresponds to the circle orthogonal to
three circles $A$, $B$ and $C$. And the commutator of a quadruple
of quaternions $ABCD-ABDC+...$ always equals to zero. This
identity implies, for example, the following geometrical theorem:

\smallskip
\noindent{\bf Proposition.} Let $D$ be the center of the
circumscribed circle of the triangle $ABC$, and let $A'$, $B'$ and
$C'$ be the centers of circumscribed circles of the triangles
$BCD$, $CAD$ and $ABD$ respectively. Then the lines $AA'$, $BB'$
and $CC'$ have a common point.
\smallskip

\noindent{\bf 33.} Assume that $\vec A-\vec B+\vec C-\vec D=0$.
Consider the vector $\vec P=\frac{[\vec A,\vec B]+[\vec B,\vec
C]+[\vec C,\vec A]}{(\vec A,\vec B,\vec C)}$. Assume that it
corresponds to some circle $P$. Then by Problem 32 the circle $P$
is the common perpendicular to $A$, $B$ and $C$. Notice that
$$
(\vec P,\vec D)=(\vec P,\vec A-\vec B+\vec C)=1-1+1=1,
$$
hence by Problem 31 the circle $P$ is orthogonal to the circle
$D$. In other words, the four circles $A$, $B$, $C$ and $D$ have a
common perpendicular.

\noindent{\bf 34.} First let us prove a Lemma:

\smallskip
\noindent{\bf Lemma.} If there are nonzero numbers $x$, $y$ and
$z$, such that $x+y+z=0$ and $x\vec A+y\vec B+z\vec C=0$, then the
circles $A$, $B$ and $C$ have a common point.
\smallskip

\noindent{\it Proof of the lemma.} The assumption of the lemma
means that the ends of the vectors $\vec A$, $\vec B$ and $\vec C$
lie on one plane. Draw a plane $\pi$ passing through this line and
the point $O$. Let $\omega$ be the circle, which is the
intersection of the sphere and the plane $\pi$. Let $H$ be the
pole of our line in the plane $\pi$ with respect to the circle
$\gamma$. Draw a line $h$ passing through the point $H$ and
orthogonal to the plane $\pi$. Then one can check that all the
circles $A$, $B$ and $C$ pass through the intersection point of
the sphere and the line $h$.

\smallskip
\noindent Now let us solve our problem. Consider a circle $C$
corresponding to the vector $\vec C=\frac{d_B}{d_B-d_A}\vec
A-\frac{d_A}{d_B-d_A}\vec B$. Then by our lemma the circle $C$
passes through both intersection points of the circles $A$ and
$B$. Denote by $\alpha$, $\beta$ and $\gamma$ the angles between
the pairs of circles $B$ and $C$, $C$ and $A$, $A$ and $B$
respectively. Then by the formula from the solution of the problem
31 we have
$$
\cos \beta = \frac{(\vec A,\vec C)-1}{d_A
d_C}=\frac{\frac{d_B}{d_B-d_A}(\vec A,\vec
A)-\frac{d_A}{d_B-d_A}(\vec A,\vec B)-1}{d_A d_C}=
\frac{d_Ad_B(1-\cos\gamma)}{d_C(d_B-d_A)}.
$$
Writing down the analogous formula for $\cos \alpha$, we get $\cos
\alpha=\cos \beta$. In other words, $C$ is a circle passing
through both intersection points of $A$ and $B$ and dividing the
angle between this two circles into two equal parts ('{\it
bisector}').

\smallskip

{\bf Acknowledgements.} The authors are grateful to V. Arnold for
important remarks, to V. Dremov, whose valueable ideas have been
realized in this project, and also to O. Karpenkov, I. Losev, and
to the jury of the XVIII Summer conference of the Tournament of
towns for useful discussions.

\small

\bigskip

\noindent[1] V.I. Arnold, Theorem on altitudes of a triangle in
the Lobachevsky geometry as the Jacobi identity in the Lie algebra
of quadratic forms in the simplectic plane, Math. Prosveschenie,
Third Series 9(2005), pages 93--99 (in Russian).

\flushpar[2] O.Ya. Viro, Yu.V. Dobdotukhina, Interlacements of
crossing lines, Quant 3(1988), pages 12--19 (in Russian).


\flushpar[3] Yu.P. Solov'ev, A.B. Sossinsky, Geometry of sliding
vectors, Quant 8(1985), pages 9--17.

\flushpar[4] J. Conant, R. Schneiderman, P. Teichner, Jacobi
identities in low-dimensional topology, preprint (2006).

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