[G.Polya]
	        PART III. SHORT DICTIONARY OF HEURISTIC
		-------------------------------------------
 Analogy is sort of similarity. Similar objects agree with 
each-other in some respect, analogous objects  agree in certain 
relations of their perspective parts.
1.     A rectangular parallelogram is analogous to a rectangular
angular parallelepiped.  In fact, the relations between 
the sides of the parallelogram are similar to those between the faces 
of the parallelepiped;
       Each side of the parallelogram is parallel to just one
other side, and the perpendicular to the remaining sides.
Each face of the parallelepiped is parallel to just one
other her face and is perpendicular to the remaining faces.
Let us agree to call a side a "bounding element": of the parallelogram 
and face a "bounding element" of the parallelepiped. Then, we may 
contract the two foregoing statements into one that applies equally to 
both figures;
	Each bounding element is parallel to just one other
bounding elements and is perpendicular to the remaining 
bounding elements.
	Thus, we have expressed certain relations which are
common to the two systems of objects we compared, sides
of the rectangle and feces of the rectangular parallelepiped.  The 
analogy of these systems consists in this community of relations.
2. 	Analogy pervades all our thinking, our everyday 
speech and our trivial conclusions as well as artistic ways 
of expression and the highest scientific
achievements. Analogy is used on very different levels. People
often use vague, ambiguous, incomplete or incompletely
clarified analogies, but analogy may reach the level of 
mathematical precision. All sorts of analogy may play a
role the discovery of the solution and so we should
not neglect any sort.
3.    We may consider ourselves lucky when trying to 
solve a problem. In section 15, our original problem was 
concerned with the diagonal of a rectangular parallelepiped;
the consider of a simpler analogous problem, concerned with
the diagonal of a rectangle, led us to the solution 
of the original problem. We are going to discuss one more
 case of the same sort. We have to solve the following problem:
Find the centre of gravity of a homogeneous tetrahedron.
      Without knowledge of the integral calculus, and with
little knowledge of physics, this problem is not easy at all;
it was a serious scientific problem in the he days of 
Archimedes or Galileo. Thus if we wish to solve it with
as little preliminary knowledge as possible, we should
look around for a simpler analogous problem.  The corresponding
problem in the plane occurs here naturally;
Find the centre of gravity of homogeneous triangle.
Now, we have two questions instead of one. But two 
questions may be easier to answer than just one question
provided that the two questions are intelligently connected.
4.    Laying aside, for the moment, our original problem
concurring the tetrahedron, we concentrate upon the 
simpler analogous problem concerning the triangle. to
solve this problem, we have to know something about 
centres of gravity. The following principle is plausible and presents 
itself naturally.
       If a system of masses S consists of parts, each of which has its 
centre of gravity in the same plane, then this plane contains also the 
centre of gravity of the whole system S. 
      This principle yields all that we need in the case of the 
triangle. first, it implies that the centre of gravity of the triangle 
lies in the plane of the triangle. Then ,we may consider the triangle as 
consisting of fibres,(*thin strips, "infinitely narrow" 
parallelograms) parallel to a certain side of the triangle (the side AB 
in Fig 7). The centre of gravity of each fibre (of any parallelogram )
is obviously its midpoint, an and these midpoints lie on the 
line joining the vertex C opposite to the side AB to the 
midpoint M of AB (see fig 7.)
           C
          |||
         |_*_|
        |__|__|
       |___*___| 
      |____|____|
     |_____*_____|
    |______|______|
   |_______*_______|
  |________|________|
 |_________*_________|
A          M         B

Any plane passing through the median CM of the triangle
contains the centres of gravity of all parallel fibres 
which constitute the triangle. Thus, we are led to the 
conclusion that the centre of gravity of the whole triangle
angle lies on the same median.  Yet it must lie on the
other two medians just as well, it must be the common point of 
intersection of all three medians. 
It is desirable to verify know by pure geometry, independently
of any mechanical assumptions, that the three 
medians meet in the same point.

5. After the case of the triangle, the case of the tetrahedron is fairly 
easy. We have now solved a problem analogous to our proposed problem 
and, having solved it, we have a model to follow.
In solving the analogous problem which we use now as a model, we 
conceived the triangle ABC as consisting of 
fibres parallel to one of its sides, AB. Now we conceive 
the tetrahedron ABCD as consisting of filers parallel to 
one of its edges, AB.
The midpoints of the fibres which constitute the
triangle lie all on the same straight line, a median of the
triangle, joining the midpoint M of the side AB to the 
opposite vertex C. The midpoint of the fibres which con
constitute the tetrahedron lie all in the same plane, joining ng the 
midpoint M of the edge BV to the opposite edge CD
(see fig 8.) ; we may call this plane MCD in a median plane of 
the tetrahedron.


                          D

                        _/T\
                     _-~/=| \
                  _-~  /===. \
               _-~    /====|  \
            _-~      /=====.   \ 
         _-~        /=====_| _  \
      _-~___-----~~/__--~~~   \. \ 
     -_~__________/=____________\_\
 
  A                M               B

In the case of the triangle, we had three medians like MC, each of which 
has to contain the centre of gravity of the triangle Therefor, these 
three medians must meet
in one point which is precisely the centre of gravity.  In the case of 
the tetrahedron, we have six median planes 
like MCD, joining the midpoint of some edge to the opposite edge,
each of which has to contain the centre of 
gravity of the tetrahedron.  Therefore, these six median
planes must meet in one point which is precisely the 
centre of gravity.
6.	Thus, we have solved the problem of the centre of
gravity of the homogeneous tetrahedron.  To complete 
our solution, it is desirable to verify now by pure geometry,
independently of mechanical considerations, that the
six median planes mentioned pass through the same
point.
	When we have solved the problem of the centre of gravity
of the homogeneous triangle, we found it desirable
to verify, in order to complete our solution, that the three
medians of the triangle pass through the same point.
This problem is analogous to the foregoing but visibly
simpler.
	Again we may use, in solving the problem concerning
the tetrahedron, the simpler analogous problem concerning the triangle 
(which we may suppose here as solved).
In fact, consider the three median planes, passing through the three 
edges DA, DB, DC issued from the 
vertex D; each passes also through the midpoint of the 
opposite edge(the median plane through DC passes through M, see Fig. 8).  
Now, these three median planes
intersect the plane of /\ABC in the three medians of this 
triangle.  These three medians pass through the same point (this is the 
result of the simpler analogous problem) and this point, just as D, is 
a common point of the three median planes. The straight line, joining 
the two 
common points, is common to all three median planes.
	We proved that those 3 among the 6 median planes which pass 
through the vertex D have a common straight 
line. The same must be true of those 3 median planes
which pass through A; and also of the 3 median planes
through B; and also of the 3 through C. Connecting these facts suitably, 
we may prove that the 6 median 
planes have a common point. (The 3 median planes
passing through the sides of /\ABC determine a common-point,
and 3 lines of intersection which meet in the 
common point.  Now, by what we have just proved,
through each line of intersection one more median plane
must pass.)
	7. Both under 5 and under 6 we used a simpler analogous
problem, concerning the triangle, to solve a problem
about the tetrahedron.  Yet the two cases are different in an 
important respect.  Under 5, we used the method of the simpler 
analogous problem whose solution we imitated
 point by point.  Under 6, we used the result of the 
simpler analogous problem, and we did not care how
this result had been obtained. Sometimes yes, we may be
able to use both the method and the result of the simpler
analogous problem. Even our foregoing example shows
this if we regard the considerations under 5 and 6 as 
different parts of the solution of the same problem.
	Our example is typical. In solving a proposed problem.
we can often sure the solution of a simpler analogous
problem; we may be able to use its methods, or its result,
or both. Of course, in more difficult cases, complications
may arise which are not yet shown by our example. 
Especially, it can happen that the solution of the analogous
 problem cannot be immediately used for our 
original problem. Then, it may be worth while to reconsider
the solution, to vary and to modify it till, after having 
tried various forms of the solution, we find eventually
one that can be extended to our original problem.
	8. It is desirable to foresee the result, or at least, some 
features of the result, with some degree of plausibility. 
Such plausible forecasts are often based on analogy. 
	Thus we may know that the centre of gravity of a 
homogeneous triangle coincides with the centre of gravity
of its three vertices (that is, of three material points with 
equal masses, placed in the vertices of the triangle).
Knowing this, we may conjecture that the centre of 
gravity of a homogeneous tetrahedron coincides with the 
centre of gravity of its four vertices.
	This conjecture is an "inference by analogy." Knowing that 
the triangle and the tetrahedron are alike in many
respects, we conjecture, that they are alike in one more 
respect. It would be foolish to regard the plausibility of
such conjectures as certainty, but it would be just as 
foolish, or even more foolish , to disregard such plausible 
conjectures. 
	Inference by analogy appears to be the most common 
kind of conclusion, and it is possibly the most essential 
kind. It yields more or less plausible conjectures which
may or may not be confirmed by experience and stricter
reasoning. The chemist, experimenting  on animals  in 
order to foresee the influence of his drugs on humans,
draws conclusions by analogy. But so did a small boy I 
knew. His pet dog had to be taken to the veterinary, and 
he inquired;
"Who is the veterinary?":
"The animal doctor."
"Which animal is the animal doctor?"
9.	An analogical conclusion from many parallel cases
is stronger than one from fewer cases. Yet quality is still 
more important here than quantity. Clear-cut analogies 
weigh more heavily than ague similarities, systematically
arranged instances count for more than random collections 
of cases.
	In the foregoing (under 8) we put forwards a conjecture
about the centre of gravity of the tetrahedron. This
conjecture was supported by analogy; the case of the
tetrahedron is analogous to that of the triangle. We may 
strengthen the conjecture by examining one more analogous
case, the case of a homogeneous rod (that is, a 
straight line-segment of uniform destiny).
	The analogy between
	segment 	triangle	tetrahedron

has many aspects. A segment is contained in a straight line
a triangle in a  plane, a tetrahedron in space. Straight
line-segments are the simplest one-dimensional bounded
figures, triangles the simplest polygons, tetrahedrons the 
simplest polyhedrons.
	The segment has 2 zero-dimensional bounding elements (2 
end-points) and its interior is one-dimensional.
bounding elements (3 vertices, 3 sides) and its
interior is two-dimensional
	The tetrahedron has 4 zero-dimensional, 6 one-dimensional, 
and 4 two-dimensional bounding elements (4 vertices, 6 edges, 4 faces) 
and its interior is three dimensional.
	These numbers can be assembled into a table. The successive-
 columns contain the numbers for the zero, one, 
two, and three dimensional elements, the successive rows
the numbers for the segment, triangle, and tetrahedron;

2	1
3	3	1
4	6	4	1

Very little familiarity with the powers of a binomial is
needed to recognise in these numbers a section of Pascal's
triangle. We found a remarkable regularity in segment,
triangle, and tetrahedron.
10.	If we have experienced that the objects we 
compare are closely connected, "inference by analogy," as 
the following, may have a certain weight with us.
	The centre of gravity of a homogeneous rod coincides 
with the centre of gravity of its 2 end-points. The centre
of gravity of a homogeneous triangle coincides with the 
centre of gravity of its 3 vertices. Should we not suspect
that the centre of gravity of a homogeneous tetrahedron
coincides with the centre of gravity of its 4 vertices?
	Again the centre of gravity of a homogeneous rod
divides the distance between its end-points in the proportion
 1:1. The centre of gravity of a triangle divides the distance 
between any vertex and the midpoint of the opposite side in the 
proportion 2:1. Should we not suspect that the centre of gravity of a 
homogeneous tetrahedron divides the distance between any vertex and the 
centre of gravity of the opposite face in the proportion 3:1?
	It appears extremely unlikely that the conjectures suggested by
these questions should be wrong, that such a beautiful regularity 
should be spoiled. The feeling that  harmonious simple order cannot
be deceitful guides the discoverer both in the mathematical and in the other
sciences, and is expressed by the Latin saying :  simplex sigillum 
veri  (simplicity is the seal of truth).)

[The preceding suggests an extension to n dimensions. It appears 
unlikely that what is true in the first three dimensions, for 
n=1,2,3, should cease to be true for higher values of n., 
This conjecture is an "inference by induction?"; it illustrates that 
induction is naturally based on analogy. See INDUCTION AND 
MATHEMATICAL INDUCTION.]
[	11.	We finish the present section by considering briefly
the most important cases in which analogy attains the precision of 
mathematical ideas.
(I)	Two systems of mathematical objects, say S and S' are so 
connected that certain relations between the objects of S are governed 
by the same laws as those between the objects of S'
This kind of analogy between S and S' is exemplified 
by what we have discussed under 1; take as S the sides of 
a rectangle, as S' the faces of a rectangular parallelepiped.
(II)	 There is a one-one correspondence between the 
objects of the two-systems S and S', preserving certain 
relations. That is, if such a relation holds between the 
objects of one system , the same relation holds between
the corresponding objects of the other system. Such a 
connection between two systems is very precise sort of 
analogy; it is called isomorphism (or holohoedral isomorphism).
(III) There is a one-many correspondence between the objects of the two 
systems S and S' preserving certain
relations. Such a connection (which is important in the various
 branches of advanced mathematical study, especially 
in the Theory of Groups, and need not be discussed here
in detail) is called merohedral isomorphism (or homo-
morphism; homoiomorphism would be, perhaps, a better term). 
Meroherdral isomorphism may be considered as 
another very precise sort of analogy.]