Doubts about elementary geometric proof in Arnolds' "Lectures for young mathematicians"? I'm reading Arnolds' "Lectures for young mathematicians", there is this proof:


I am a bit confused about two things:

*

*In the first line of the centralized equations, they write: $area(OACB)=area(OACB)+ area(OA_1 A)+ area(BC_1 C)$ - Is this a typo? How can the area of something be equals to itself plus something else?


*What is the meaning of $area(OA_1 A C C_1 B)$? To me, it seems they are describing a solid, I don't know what area they are pointing with this.
 A: I originally interpreted and wrote this up as a 3D problem.  But a commenter points out that it is more likely a 2D problem, so I've added that.  But surprisingly, the statement in 3D can be shown to be true as well.
$ \newcommand{\area}{\operatorname{area}} $
2D
There is indeed a typo.  The first two lines should read
$$
\begin{aligned}
\area(OACB)&=\area(OACB)+ \area(OA_1 A)- \area(BC_1 C) \\
&=\area(OA_1ACC_1B)
\end{aligned}
$$
Here, by adding and subtracting two congruent triangles, parallelogram $OACB$ is converted into a hexagon $OA_1ACC_1B$ with the same area.
3D
The 3D version is essentially about directed areas (in 2D it would be signed areas, but we're working in 3D here, so we need to talk about spatial direction, not just positive and negative).  It's a little ambiguous whether the vertices in the diagram are points or vectors, but let's think of them as points.  Choose any 3D point $O^*$ as an origin.  Define a pseudo cross product ($*$) between points $X,Y$ in terms of the vector cross product ($\times$):
$$
X*Y = (X-O^*)\times(Y-O^*)/2.
$$
Define the directed area of a polygon $V_1V_2V_3\dots V_n$ as a sum of pseudo cross products:
$$
\area(V_1V_2V_3\dots V_n)= V_1*V_2+V_2*V_3+\dots +V_n*V_1 
$$
For readability and conciseness, and if there is no ambiguity, we will write $X*Y$ as $XY.$  Then
$$
XY = (X-O^*)\times(Y-O^*)/2.
$$
and
$$
\area(V_1V_2V_3\dots V_n)= V_1V_2+V_2V_3+\dots +V_nV_1 \tag{1}
$$
Convince yourself that

*

*the value of $\area(V_1\dots V_n)$ is independent of choice of $O^*$

*for a polygon in the $xy$ plane, $\area(V_1\dots V_n)$ is a vector whose magnitude is the area of the polygon and whose sign is determined by the winding direction of the vertices.

*for a general 3D polygon $\area(V_1\dots V_n)$ is a vector whose magnitude is the area of the polygon and whose direction is determined by the right hand rule applied to the vertex sequence.

$\area(V_1\dots V_n)$ gives a value for any non-degenerate sequence of $V_i$, so it is defined even for non-planar polygons.  For example, the polygon $OA_1 A C C_1 B$ in the OP.  This may sound a bit weird at first, but there is a simple physical concept that may be helpful.  Consider a container filled with pressurized gas escaping through a hole with boundary $V_1\dots V_n$.  The thrust generated will be proportional to $\area(V_1\dots V_n)$ and this will be true whether the polygon is planar or not.
OK, we're ready to get to the OP.
Convince yourself that

*

*for points $X,Y$, $XY=-YX$ (hint: for cross products, $X\times Y=-Y\times X$)

*for a vector $V$ and translation $T:X\to X+V$, $\area(XYZ)=\area(T(X)T(Y)T(Z))$ (hint: either expand both sides and simplify, or recall that $\area()$ is independent of choice of $O^*$)

*$\area(OA_1 A) = \area(BC_1 C)$

*$\area(OA_1 A) = -\area(BC C_1)$

*$\area(OA_1 A) + \area(BC C_1)=0$
By the latter point, the first line of the centered equations should be either
$$
\area(OACB)=\area(OACB)+ \area(OA_1 A)- \area(BC_1 C)
$$
or
$$
\area(OACB)=\area(OACB)+ \area(OA_1 A)+ \area(BC C_1)
$$
For either way of writing it, $0$ area is being added to $\area(OACB)$.
As for the second question, $OA_1 A C C_1 B$ is a non-planar hexagon and $\area(OA_1 A C C_1 B)$ is as defined above.
As for the rest of the equations, one can verify or correct them by expanding $\area()$ in terms of its definition and then simplifying.
