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THE CHARM OF GEOMETRY
This collection of challenging problems is designed to improve visual spatial
abilities, creativity and the most important “the mathematical dream” without
which no discovery can be done.
Chapter 1
The goal of this chapter is to make you confident in your understanding, in your
mathematical dream, and in your intuition. One of the most famous mathematicians in the
history was Srinivasa Ramanujan. He let at his death few thousands formula describing a
new and extraordinary reality, but without any proofs. Now an entire army of
mathematicians works to find proofs at these formulas. He always said that an Indian
goddess inspired him in his work. The difference between proof and direct understanding
is the interference of a specific language. Another extraordinary mathematician,
Alexander Grothiendck Transformed his mathematical dream in very accurate proofs
going to postulates. Both these mathematicians spoke about their mathematical dream,
and many others lived this dream. The purpose of this chapter is to open a window for
this dream
The science of measuring, and the science of generating new objects using a set of
rules are very old, but in the same time very modern. The most ancient traces of geometry
are many millenniums old and are now archeology. The most modern deals with the laws
of the universe, from cellular automata theory to superstrings theory. This course will
start with the classical Euclidean geometry, and will continue with elements of projective
geometry.
Before Euclid, geometry was a big collection on technical prescriptions, mainly used
in building large temples. After Euclid it become a science that developed new branches
giving new mathematics. How the Euclidean geometry looked like? We have several
examples given by Euclid in his book “The Elements” You can find more about on:
http://www.obkb.com/dcljr/euclid.html
Examples
Problem 1
Take three points A,B,C in this order on a circle, and M the middle point of the arch
ABC. Be N the projection of M on the segment AB or BC that has the same ends as the
arch containing M. Prove that N is the middle point of the broken line ABC. (Figure 1)
Proof made before Euclid.
Take an isosceles triangle MAC with MA=MC, and B on AC, and fold it under the
line MB. You will obtain an inscriptible quadrilateral with vertexes A,B,C,M, because
two angles MAB and MBC are equal as angles in the isosceles triangle. (Figure 2)
If we project the vertex M on ABC it hit the middle point of the opposite side. After
we fold the triangle, this middle point will become the middle point of the broken line. So
we obtain our problem.
1
Note
For the ancient Greeks this demonstration practically proving the property was
enough. Euclid changed everything by introducing a methodology.
Euclid’s proof
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If we take two triangles that are in the case angle-side- side, we find two possibilities
(Figure 3)
As we can see angles 1 and 2 are 180 degrees together. So we can have three cases
with triangles in angle-side-side case. In two of them we have equal triangles, in the third
we have unequal triangles. In the last one these unequal triangle have two pairs of equal
sizes, one pair of equal angles opposite to one pair of sizes, and another pair of
supplementary angles opposite to the other pair of sizes.
Considering now figure 1 we may notice that triangles MAB and MBC are in the last
case, that means that two angles are supplementary. We can de-fold the figure now and
obtain the isosceles triangle we started that proving the problem.
This kind of precision in thinking introduced by Euclid was the greatest acquisition
of his axiomatic system. If you want to see more about ancient Greeks knowledge go
to “ Important lines in triangles, and associated circles”. Some of these properties were
well known by the ancient Greeks, some others were discovered later.
This unusual case of angle-size-size has applications in different other problems and
properties Many of these properties come from an extension of this case to similarity. If
one of these two unequal triangles that are in the case angle-side-side is magnified
preserving proportions, than the sizes of the not-magnified and of the magnified triangle
that oppose to the equal angles, or supplementary angles are proportional (Fig 4)
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This property can give a very easy solution for the bisector theorem (the bisector of
an angle cut the opposite side in two-segment proportional with the other two sizes), and
can be useful for other problems.
Problem 2** (orange belt)
Take a circle and two equal segments AB=AC with A, B, C on the circle. Take a
point P such as P see AB and BC from equal angles (^APB=^BPC). Find the geometric
place of P. (clue see in which case you will find your triangles, and consider any
hypothesis)
Geometry was a passion for many scientists, from the ancient times to the present
time. Many of laws discovered by Newton are geometrically proved. Newton’s interest in
geometry was very big he considering that universe obeys to geometric rules. Here is a
sample of Newton’s perception in geometry:
Problem 3 (attributed toto Newton)***(green belt)
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Take a quadrilateral with sizes tangent to a circle. Prove that the opposite tangent
points are joined by lines that are concurrent with the diagonals of the quadrilateral.
(Clue; consider two triangles determined by a diagonal and a line joining two
opposite middle points. For example AQT and TSC have two equal angles, and two
complementary angles. The rapport between sides opposing to equal angles equals the
rapport of sizes opposing to supplementary angles. Use this property for APT and TCR,
and try to arrange this information in your advantage).
Problem 4***(green belt)
Take ABCD in this order on a circle. AB and CD intersect each other in M , AC and
BD intersect each other in N. Take the bisectors of AMD and DNC. These bisectors
intersect the sizes of ABCD in P on BC, Q on CD, R on DA and S on AB. Prove that
SRQP is a rhomb. Prove that the sizes of this rhomb are parallel to the diagonals of
ABCD (Figure 6).
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(Clue consider relationships between angles and arches, bisector theorem and the case
ULL)
For solving these problems you need to know also the relationships between arches
and angles in the circle. Let’s remember: (Fig 7)
-1) the size of an angle with the vertex on a circle equals half of the arch delimited by
its lines.
-2) the size of an angle formed by a tangent and a chord equals half of the arch
delimited by the chord
-3) the size of an angle with vertex in the interior of a circle equals half of the sum of
arches delimited by its lines
-4) the size of an angle with vertex in the exterior of a circle equals half of the module
of difference between arches delimited by its lines.
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You need intelligence and patience to find solutions to these problems but not too
much knowledge. You are like ancient Greeks only with your mind in front of the
unknown, but this is the challenge of geometry. If you need more knowledge about
circles go to the section “Circles”
Ancient Greeks
What kinds of problems can be developed using Tales or Pythagorean theorems?
The complexity and beauty of these problems will amaze us. Some are very old,
some are more recent, but each gave the same happiness of understanding a little part of
universal rules. Ancient Greeks considered that Gods loved a human being that
discovered a new geometric property, or physics, engineering, astronomic or medicine
property. In their vision only such a human creature could open their minds to understand
a little bit of the universal rules. Let’s start with Tales
Problem 1* (yellow belt)
Take two lines and on each of them take three points: A, B, C on the first line and D,
E, F on the second line such as BD and CE are parallel, also AE and BF are Parallel.
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Prove that AD and CF are parallel.
(Clue take the two support lines, and intersect them in p. Consider Tales)
Many properties can extend in a very interesting way to new properties. The previous
one has some very interesting extensions:
Problem 2***(green belt)
Take a triangle ABC and a line d. From A, B, C take three parallel lines that intersect
d in A’, B’, C’. From A’ bring a line parallel to BC, From B’ a line parallel to AC, from
C’ a line parallel to AB. Prove that these lines intersect each other forming a triangle
equal to ABC.
(Clue; use the previous property, and the second clue translate the line until will pass
through a vertex of ABC. Nothing is changed in the problem because of this translation,
but you will find a new and better perspective to work
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with)
Extension in space*****(brown belt): Instead of triangle ABC take tetrahedron
ABCD and a line d. You will obtain similar conclusions if from A, B, C, D you will take
four planes that will intersect d in A’, B', C’, and D’. If from A’, B’, C’ D’ you will bring
planes parallel to the opposite face of the tetrahedron respectively BCD, ACD, ABD, and
ABC, these planes will intersect each other forming a tetrahedron equal with ABCD. If
instead of a line d you will take a plan alpha and four parallel lines coming from A, B, C,
D and intersecting alpha in A’, B’, C’, D’, and if from A’, B’, C’, D’ you will take planes
parallel to the opposite faces of ABCD you will obtain another tetrahedron twice bigger
that ABCD. Both these two extensions are particularly difficult to be visualized. This
impediment can be canceled by practicing three-dimensional problems. (See the section
Three-dimensional problems) (Try to find a metric relationship into the plane, and
extend it in the space)
Another interesting application of Tales reveals a different perspective about plane
and three-dimensional space. Newton didn’t consider this property when he described his
laws, but Einstein did it, describing a curved space (hyperbolic space). The property is
the following one:
Problem 3****(Blue belt)
Take a quadrilateral ABCD, and share the opposite sizes in the same number of equal
segments. Join points from the opposite sizes that count the same number of segments
counted from left to right or from up to down (Figure 8).
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Theoretical extension from natural numbers to rational numbers and fractions for
these proportions will lead us later to a higher level of geometry understanding describing
a double riglated surface.
To prove this property take first three equally distanced points on each size and than
generalize the procedure. Remember that the line that joins the middle points of two sizes
of a triangle is parallel to the third line.
Let us consider some particular cases for this problem
Case 1**(orange belt)
Take ABCD a quadrilateral, M and N on AB such as AM=MN=NB; O and P on BC
such as BO=OP=PC; R and S on CD such as CR=RS=SD; and T and U on DA such as
DT=TU=UA. Prove that UO is intersected by MS and NR in J and I; TL is intersected by
MS and NR in L and K and we obtain MI=IL=LS; NJ=JK=KR; UI=IJ=JO; and
TL=LK=KP. Prove also that the area of IJKL equal 1/9 from the area of ABCD (Figure
9)
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Figure 9 will help you to find the connection between Tales and this problem. For the
second point, show first that the area of IJKL is 1/3 from the area of ABCD. In order to
do this, work with triangle UTL, ILK, KJP, and than with triangles UIL, IJK and JOP.
Case 2***(green belt)
ABCD is cut in an integer number of equal segments, opposite sides being cut in the
same number of segment (Figure 10). Conclusion is the same, but instead of equal
segment will be proportional
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segments.
Extension 1**(orange belt)
Take a similar situation in the tri-dimensional space considering segments AB and
CD not in the same plan. In order to proof this case project the entire figure on a plan
perpendicular on AD than on a plan perpendicular on BC and see what is happening.
(Do you see Einstein’s perspective now?)
From Tales to similarity
Problem 1**(orange belt)
Take two segments AB and CD. Find a point O in their plane such as triangles OAB
and OCD are similar (From O, AB and CD look the same) (Figure 11)
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It is a simple exercise to show that angles noted with the same figure are equal. Point
O is called center of similitude. In fact there are two centers of similitude. The other one
can be obtained taking R the intersection of BC with AD, and the circles determined by
RCD and RAB. The other point of intersection of these circles is also a center of
similitude.
Let’s connect facts: if instead of two segments we have two similar figure containing
these segments we can see them the same from the same centers of similarity from which
we see the two segments.
Problem 1****(blue belt)
Take two directly similar triangles ABC and A’B’C’. On AA’, BB’, CC’ draw other
three direct similar triangles AA’A”; BB’B”, and CC’C”. Prove that A”B”C” is similar to
ABC (Figure 12)
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This figure describes very well these amassing phenomena, but another figure will
help you more to see the solution. (Figure 13)
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If you look from the perspective of O the center of similitude you will find a lot of
similar figures that will lead you to the solution. After you will find it, try to do the same
thing, using figure 12.
Observation 1
If all triangles ABC, A’B’C’, A”B”C”, AA’A”,BB’B”, and CC’C” are reduced to
segments with one vertex as interior point, you have a proof for problem 3 of the double
riglated problem where interior points cut their segment in rapport that are real numbers.
Observation 2
If we take points P, P, P” such as to form similar triangles with the vertexes of ABC,
A’B’C’, and A”B”C” respectively, than PP’P” will be similar to AA’A”.
Consequence 1**(orange belt)
Take ABC a triangle A’B’C’ three points on its sizes such as ABC and A’B’C’ are
directly similar. Prove that A”, B”, C” points on AA’, BB’, CC’ respectively dividing
their segments in the same rapport form a similar triangle with ABC. Prove that their
centers of gravity are collinear, and the center of gravity of A”B”C” divide the segment
determined by centers of gravity of ABC and A’B’C’ in the same rapport. (Clue try to
degenerate the main problem)
Search for new thinks
There are several questions that can be searched regarding centers of similarity and
similarity. One of them regards the fact that we have two centers of similarity. If the
network respects one center of similarity will respect also the other one Can we find a
relationship between these two centers and the double network of similar triangles
(figures) developed in the plane that respect these two centers? This kind of approach was
developed in the twentieth century geometry: algebraic geometry, differential geometry,
algebraic varieties, and differential varieties. These centers and these families of triangles
are typical examples of geometry regarded from the structural point of view, not only
from the synthetic point of view. This new point of view is used in the modern geometry
to describe numerous phenomena including physical space, cosmology, atomic and
subatomic particles.
Let’s remember
We have a double riglated surface in the plane, and also in the tri-dimensional space.
In the plane this double riglated surface is determined as a degenerated case for a
complex structure covering the plane formed by two classes of directly similar triangles.
Do we have something similar in the three-dimensional space? In fact this is a real
problem of geometry in which we need to discover new phenomena. This kind of
problem challenges our curiosity and skills and helps us to develop creative thinking and
to discover new phenomena. We don’t know how much about double riglated surfaces
ancient Greeks knew. What we know is that Archimede used some quadratic surfaces
(some double riglated) for creating strong focalizing mirrors for defending Syracuse
against the Roman Army. One example of such a surface is the following one:
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Three-dimensional space
Problem 1***(green belt)
Take an object with six faces formed by plane quadrilaterals. Let’s note it
ABCDA’B’C’D’, where AA’, BB, CC’ and DD’ are segments belonging to two faces.
Prove that A’, B’, C, and D’ are not in the same plane. Prove that if you will divide each
side in the same number of equal segments and join them as to form a double riglated
surface, knots from double riglated surfaces corresponding to ABCD, A’B’C’D’, and
A”B”C”D” that have similar positions, will be collinear describing two new sets of
double riglated surfaces. Make an extension of this property. More about double riglated
surfaces you will find in the section “Perpendicularity in the three-dimensional space”
In fact we find what is the space correspondent of a plane double riglated surface. We
will call it a lattice. This kind of geometric structure, which is able to give a division of
the three-dimensional space in a natural way, can give us a referential for new and more
complex properties. Take a look to the degenerated case when the double riglated surface
and lattice have right angles. We obtain in this case Cartesian plan and space where we
make all our calculus. Is it possible to find a different way for calculus extending it on
double riglated surfaces and three-dimensional lattices?
You just found a field of creative thinking and discovery. Maybe you will rediscover
known facts, maybe new facts. Anyway you will enjoy very much to do it. Good luck on
finding new ideas.
Right angles and Pythagora
If Thales of Milet used to sell olive oil to the Greek army in order to live, Pithagora
choused to create a school in which to introduce all revolutionary ideas of his time. One
of them was his theorem. Let’s see some applications:
Problem 1
Take two segments AB and CD in the plane or in space, that makes a right angle to
each other. Prove that AD at power 2 minus AC at power 2 equal BD at power 2 minus
BC at power 2.
Problem 2**(orange belt)
Take two triangles ABC and A’B’C’ with the property that projections from A, B, C
on B’C’, A’C’, A’B’ are concurrent. Prove that projections from A’, B’, C’ on B’C’,
A’C’, and A’B’ are also concurrent (These triangles are called orthologic triangles).
Prove the same property for tetrahedrons. (Indication: Prove first the metric relationship
BA’ at power 2 minus A’C at power 2 plus CB’ at power 2 minus B’A at power 2 plus
AC’ at power 2 minus C’B at power 2 equals zero.) (See the Figure)
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General advice like parallelism problems, perpendicularity problems are also many
times indifferent to translations. This fact is due to the preservation of angles between a
line and a translated other line. This property can be frequently used in perpendicularity
problems.
Problem 3 ****(blue belt)
Take ABC and A’B’C’ two orthologic triangles. Take on AA’, BB’, and CC’ the
points A”,B”, and C” such as AA”/A”A’=BB”/B”B’=CC”/C”C’= k. Prove that A”B”C”
is orthologic with ABC and A’B’C’ and with any other triangle like it with another “k”.
We found an interesting property in the plane regarding orthology that is specific for
the plan. In the tri-dimensional space we can’t find fascicles of orthologic figures with
vertexes on the same lines. Isn’t it due to the fact that perpendicularity in the three
dimensional space is a different phenomena than in plane?
Problem 4 ***(green belt)
Take a tetrahedron ABCD. Prove that the planes passing through the middle points of
any side and perpendicular on the other sides are concurrent. (To see more go to the
section Tetrahedrons and Parallelepipeds)
Problem 5*(yellow belt)
Take a triangle ABC and a line “d”. Take three parallel lines passing through A, B,
and C and intersecting “d” in A’, B’ and C’. Prove that perpendicular lines from A’, B’,
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and C’ on the opposite sides of the triangle BC, CA and AB are concurrent. (See the
figure)
Extend this property in the three dimensional space. (Clue, see the line as a
degenerated triangle, and parallel lines as concurrent in a point at infinity)
Problem 6 *****(brown belt)
Take two tetrahedrons ABCD and A’B’C’D’ such as the plans passing through the
middle point of each side of ABCD and perpendicular on the opposite side on A’B’C’D’
are concurrent. Prove that the plans passing through middle points of A’B’C’D’ and
perpendicular on the opposite sides of ABCD are also concurrent.
This problem shoes that the right angle is only simpler to use in calculus, but space is
not structured in a natural way conforming to plan right angles. We will understand this
fact more after understanding new phenomena (clue translating one tetrahedron or
magnifying it you can arrange these two tetrahedrons in a better position without
modifying the problem).
Perpendicularity concept has an interesting history and evolution. From perpendicular
lines, to perpendicular curves, planes, spaces, and even orthogonal, matrices, and
structures, this concept become more and more rich and complex.
Symmetry
Symmetry is another fundamental concept. In fact it is maybe the most fundamental
of all concepts, deriving every other concept. We can recognize this concept in
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everywhere in nature, in many ways, and mathematics considered it from various
perspectives. One of these is the following one:
Take a segment AB and an interior point P. Let it be R the symmetrical point of P
with regard to A and Q the symmetrical point of P with regard to B. Take M the middle
point of RQ. Prove that MB=AP. Now make an extension considering an angle AOB
instead of a segment, A line OP instead of a point P and symmetrical lines with regard to
OA and OB. Instead on a middle point Q take the bisector of ROQ named OM. Prove
that angles POA and MOB are equal (isogonal lines for the angle AOB).
Problem 1**(orange belt)
Take a triangle ABC and An interior point P. Note angles PAB=1, PPBC=2, PCA=3.
Take a point Q such as angles QAC=1, and QCB=3 Prove that QBA=2 (We call P and Q
isogonal points) (clue take A’, B’, C’ the symmetrical points of P to the sides of ABC.
Prove that the perpendicular bisectors of A’B’C’ are QA, QB, and QC).
Problem 2**(orange belt)
Take P, and Q two isogonal points, and take the projections of P and Q on the sides of
the triangle ABC. Prove that:
-ABC and the triangles determined by projection point of P or Q on ABC are
ortologic triangles.
-All six projections of P and Q on the sides of ABC are on the same circle.
Problem 3***(green belt)
Do the same thing for a tetrahedron with P interior point. Instead of angles with sides
consider dihedral angles between planes determined by P and a side and a face containing
the same side. Prove that P has an isogonal point Q. Prove that the tetrahedron
determined by the projection points of P or Q on the faces of the tetrahedron ABCD and
the tetrahedron ABCD are orthological tetrahedrons. Prove that projection points of P and
Q on the faces of ABCD are not on the same sphere, but Projection points of P and Q on
the sides of the tetrahedron are on the same sphere. (Clue, consider symmetrical points of
P considering the planes of the tetrahedron. Take the perpendicular lines on the planes of
triangles generated by any three points of this set, and passing through the centers of
circumscribed circles. Prove that these lines are concurrent in the center of the sphere
determined by the four symmetrical points. Prove that these perpendiculars are the
isogonal lines we need)
Problem 4* (yellow belt)
Take a triangle ABC. Prove that the isogonal lines for a set of parallel lines passing
through its vertexes intersect on the circle ABC (see the figure)
(Clue, consider the inscriptible quadrilaterals)
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Problem 5**(orange belt)
Take a triangle ABC. Prove that the isogonal transformation of points of a circle
passing through two vertexes of the triangle is also a circle passing through two vertexes
of the triangle. (Clue consider angles in circle)
Problem 6*****(brown belt)
Take a quadrilateral ABCD and two points isogonal for any angle. Prove that the
segment, which joins these isogonal points, is divided in two equal segments by the line
that joins the middle points of the two diagonals of the quadrilateral. (Clue consider
projections of these isogonal points on the sides of quadrilateral. Prove that these
projections are on the same circle. Find the relationships between areas of triangles
formed by the projections, isogonal points. Connect these relationships to the relationship
between areas described by a point on the line joining the middle points of the diagonals.
Extend the property to a complete quadrilateral, see complete quadrilateral)
Problem 4
Instead of equal angles PAB=QAB take segments on a side of the triangle, like M,
and N on AB such as AM=NB and CM contains P, CN contains Q. If you obtain two
equal segments on AC you will obtain equal segments on AB. These points will be called
isothomic points. If you will take the same problem on a tetrahedron ABCD, a point P
and all planes passing from P and a side of the tetrahedron, and the intersection points of
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these planes with the sides of the tetrahedron we will obtain a number of six new point on
the sides of the tetrahedron. Taking their isothomics and joining with the opposite side
we will obtain six new planes. Prove that all these planes are concurrent in the same point
called Q. We say that P and Q are isothomic points.
Problem 5
Take ABCD a tetrahedron, cut by a plan alpha. Prove that isotomic points on
intersection points with each size of the tetrahedron are on the same plan
You can solve problems 4 and 5 if you know Ceva and Menelaos theorems. To see
more go to “ Section Ceva and Menelaos Theorems”
Comment 1
Isogonality and isothomy are transformations of the n dimensional space that respect
several characteristics: Both need a referential that is a simplex (a triangle in the plane,
and a tetrahedron in the three-dimensional space). Both transformations respect the
interior of the simplex transforming interior points or figures in interior points and
figures, and similar for the exterior points. There are some differences. Isogonality is
excepted for the points that are sited on the sides; vertexes or “n-1” border simplexes of
the “n” dimensional simplex. Isothomy is not excepted. Isogonality is described first in
the two-dimensional space isothomy is described firs in the one-dimensional space.
Can we find a new transformation of their kind specific for the three dimensional
space?
Comment 2
These problems prove us that perpendicularity and symmetry are closely related. This
fact is not very surprising, because symmetry’s definition involves perpendicularity. We
see here one interesting fact: If we cultivate perpendicularity we obtain perpendicularity.
If we cultivate equality we obtain equality. This observation encourages us to find
helping lines or points able to open the problem. In the same time these problems develop
visual abilities. Let’s test it.
Problems using similarity are also related to symmetry, rotations and homotety
(Dilations). In fact we considered only some aspects regarding symmetry in the plane. If
you want to have a better image about symmetry visit:
http://www.ca/~mathed/Geometry/Transformations/Symmetry.html
http://www.ucs.mun.ca/~mathed/Geometry/Transformations/Transformations.html
http://www.geom.umn.edu/docs/reference/CRC-formulas/
You will have the opportunity to see many other kinds of symmetries that can
develop interesting properties you can obtain.
Problem for testing visual abilities
Take a room and two balls one red and one blue. Find the way in which you can hit
the blue ball such as it to hit consecutively each wall and than the red ball. (Clue:
consider all walls made by mirrors and try to see the rout of the moving ball in each wall)
Section Circle
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Circles are interesting figures. To understand their behavior we need to know the
basic rules that lead everything in this area.
Rule #1
The angle with a vertex on the center of the circle equals the arch limited by its side.
Every other rule derives from this one. Let’s make a small list:
Rule #2
An angle with a vertex on a circle equals half of the arch delimited by its side
Rule #3
Two parallel lines delimit equal arches on the same circle.
These basic rules permit us to transfer information from angles to arches, and from
arches or angles to parallel lines. We have now a game with few rules, but that is able to
describe extremely complex characteristics. Let’s see some applications:
Problem 1
Take a circle and a pair of points A, and B. From these points take two parallel lines
that will cut the circle again in A'’ and B'’ Continue the procedure several times. Finally
you will obtain a pair of points that joined with A and B such as not to cross, will
determine a pair of parallel lines.
Problem 2 The shell*(yellow belt) (see the figure)
Take a circle and a number of points that form a regular polygon. Lets take two
consecutive points A and B on the circle. Join A and B with all the other vertexes of the
regular polygon. These lines will intersect each other many times, but if you chose
intersection points such as to have an order you will obtain a shell. Prove that each curve
of this shell is a circle, and each point of intersection on these circles belongs to a regular
polygon with the same number of sides like the first one.
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(Clue Consider relationships among angles, chords and arches)
Problem 3
Take a broken line with n sides with the property that any line that passes through the
middle points of these sides and are perpendicular on these sides, are concurrent to each
other in the same point. Show that all vertexes of this broken line are on the same circle.
A particular object related to circles is inscriptible quadrilateral. This one has the
following properties:
-1) Two opposite angles form together 180 degrees (are supplementary)
-2) the angle between a diagonal and a side equals the angle between the other
diagonal and the opposite side.
Other properties are derived from these two.
Problem 3
Take two circles that intersect to each other in P and Q. Through P take a line that
cuts the two circles in M and N. Take the tangents on each circle from these two points M
and N. These tangents will intersect in R. Prove that the quadrilateral RMQN is
inscriptible.
Problem 4**(orange belt)
Take three circles with equal radiuses that have a common point P. Prove that the
circle passing through the other three intersection points of these circles taking two by
two has the same radius like the first three. (Clue take the radiuses involved in this figure,
and parallel lines from centers and points of intersection of circles)
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Case Study
Take a circle and ABC a triangle with vertexes on the circle. Take the altitudes of the
circle AA’, BB’ CC’ and take the points of intersection A”, B”, C” with the circle. If H is
the intersection point of these altitudes prove that:
-HA’=A’A”; HB’=B’B”; HC’=C’C”
-The circles determined by HAB; HBC and HCA are equal with the circle ABC
-Prove that the angle between the altitude passing through A and the bisector passing
through A equals the angle between the bisector passing through A and the diameter of
the circle ABC passing through A.
-Using the last property proof Ptolemeu formula: If ABCD is an inscriptible
quadrilateral with diagonal AC and BD show that ABCD+ADBC=ACBD
Problem5***(green belt)
Take an inscriptible quadrilateral ABCD, and take the circles that are the symmetrical
circles to the initial one considering the sides of ABCD. Prove that these circles intersect
each other in four other points that are the vertexes of an inscriptible quadrilateral equal
with ABCD. (Clue take the lines joining the centers of the involved circles. These lines
and the common chords are diagonals of rhombs)
Now a strong problem
Problem6*****(Brown belt)
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Take a regular polygon with an odd number of sides. Prove that joining one vertex
with the others taken two by two by diagonals, you will obtain a number of segments.
Taking the sum of lengths of these segments will obtain a number that will equal the sum
of the other diagonals of the polygon. (Very difficult but very interesting problem. You
need a lot of skills or some knowledge of complex numbers or analytic geometry to prove
it). Good luck in solving it. Try first the case of an equilateral triangle and extend the
procedure (see the figure).
Now you have the minimum knowledge to appreciate a strategic problem that is
considered to be discovered by Napoleon Bonaparte. This famous person had a very high
strategic thinking proved by this problem. Why strategy is what you will see? Just follow
the reasoning.
Problem 4 partially attributed to Napoleon****(blue belt)
Take three equilateral triangles OAB, OCD, and OEF with a common vertex O.
Notation of these triangles is made in the same sense of rotation. Take I the middle point
of BC, J the middle point of DE, and K the middle point of FA. Prove that IJK is also an
equilateral triangle (See Figure Napoleon’s problem)
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Proof
Strategic thinking means that we have not only strategy in constructing additional
lines but also many steps of demonstration.
-Step1) AC=BD, CE=DF, EA=FB. Focus your attention on two equilateral triangles
-Step2) Triangles ACE and BDF are equal rotated to each other with 60 degrees
-Step3) if we note with L, M, and N the middle points of CD, EF, and AB, than NI is
middle line for the triangle ABC and IL is a middle line for the triangle BCD.
-Step4) NI=IL, and the angle NIL is 120 degrees.
-Step5) we discover three isosceles similar triangles with one angle of 120 degrees:
NIL, LJM, and MKN. If we complete them so that I, J, K to become the centers of some
26
equilateral triangles we obtain a new figure (Napoleon 2)
We continue our demonstration following a new line given by the new figure
-Step 6) PM=RN and between these lines is a 60 degrees angle
-Step 7) PLTN is an inscriptible quadrilateral, so is QTMR, as a result
-Step 9) TMSN is also inscriptible.
-Step 10) PM, RN, and LS are concurrent in T, and between them are 60 degrees
angles.
-Step 11) Noting with U the intersection between IJ and LT, V the intersection
between JK and TM, and W the intersection between KI and NT we see that IJ is
perpendicular on LT, JK is perpendicular on Tm and NT is perpendicular on KI (the line
that join the centers of two secant circles is perpendicular on the common chord).
-Step 12) Quadrilaterals IUTV, JVTU, and VKWT are inscriptible with a 120 degrees
angle (formed by PM, RN and SL to each other), and two 90 degrees angles each. In
conclusion the last angles are 60 degrees each. These angles are the angles of IJK that is
proved now to be equilateral.
What do you say now about Napoleon?
We can see that inscriptible quadrilaterals were just instruments in solving problems
made with points and lines. Is fascinating how circles interfere with polygons in
structured properties, we will see later why this is happening. Now let’s see a problem
that characterizes some properties specific to inscriptible quadrilaterals.
Problem 5
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Take three inscriptible quadrilaterals ABCD, CDEF, and EFAB noted in the same
sense of rotation. Prove that all points A, B, C, D, E, and F are on the same circle (See the
figure).
We use absurd reduction method to prove this property. First we take six points on
the same circle We keep fixed the quadrilateral ABCD, and make a variation of CDEF
such as to remain inscriptible but E’ and F’ to be out of the initial circle. Bringing a
parallel line to EF that will cut CF in F ‘and DE in E’ did this variation. We have now
two inscriptible quadrilaterals and we can see what will happen to the third one ABE’F’.
This one ABE’F’ will be no more inscriptible because the sum of two opposite angles
will be no more 180 degrees. Prove this fact, and prove the case when E’ and F’ are
interior to the circle. Now is a simple matter of logic to eliminate the impossible cases
and to remain with the lonely possible case that becomes certitude.
This problem is a very useful instrument in solving many other problems. You
already know one; Euler’s circle.
Do you want to know more about circles? Japanese tradition accorded a high interest
in circles geometry. You can find more visiting the following site:
http://www.sciam.com/1998/0598issue/0598rothman.html
Important lines in triangles and associated circles
Ancient Greeks starting with Euclid considered as important lines the lines
considering perpendicularity and middle points. We have the following list:
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-Perpendicular bisector of a segment is a line perpendicular on the segment passing
through the middle point of the segment. The image is of a belt formed by parallel lines,
intersected by a perpendicular one, and the parallel line that can be obtained by folding
this belt, and working as a middle of the belt. You can fold a triangle with respect to a
middle point and the line where this middle point exists on a side, and you obtain a
perpendicular bisector of a triangle.
-Bisector of an angle is a similar concept but instead of belt we have an angle formed
by concurrent lines. The “middle “ line obtained by folding the angle is the bisector. You
can fold a triangle considering only one angle and you obtain the bisector of an angle in a
triangle.
-Median of a triangle is a line that joins a vertex with the middle point of the opposite
side. It separates the triangle into two triangles of equal areas.
-Middle line of a triangle is the segment that joins the middle points of two sizes. This
line can be obtained by folding a triangle with respect to two middle points of two sizes
of the triangle.
-Altitude is the line passing through a vertex perpendicular on the opposite side.
These lines can be obtained by folding a triangle such as to respect a vertex and the
opposite line.
Try alone and see if you can fold a triangle in a significant way obtaining other lines.
Ancient Greeks used these folding procedures to demonstrate some interesting
properties, for example that the sum of all three angles in a triangle is 180 degrees.
You can try also making the following steps:
- Take a triangle ABC, and the middle points M of AB and N of AC. Fold the
triangle after the middle line MN.
- Fold the triangle such as M and BC line, are considered, N and BC line are
considered. You will obtain a rectangle where all the angles of the triangle form
the angles in the half of plane on a side of a line. This means 180 degrees.
Ancient Greeks used to use a scissors too for proving properties. So they knew that
the sum of angles of a quadrilateral is 180 degrees, following the following steps:
- Take a quadrilateral ABCD, and take M on AB, N on BC, P on CD, and R on DA
the middle points of these sides.
- Join the opposite middle points, and cut these two lines with a scissors
- Rotate each quadrilateral in the same plan such as the points A, B, C, and D to be
in the same place, and the points M, N, O, and P separated by the scissors in
different quadrilaterals to be also two by two in the same place (P1 and P2 for
example to be the same point). You will obtain a parallelogram with angles A, B,
C, and D in the interior completing each other to 360 degrees, being angles
around a point.
Problems, for some of them you need to read first Ceva and Menelaos theorems.
Problems
Take a triangle and a point P. Take AP, BP, and CP three lines.
Problem 1
If you take the symmetrical lines considering the bisectors of the triangle for AP, BP,
and CP, you must prove that these symmetrical lines will intersect in the same point
Q
Problem 2
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If AP, BP, and CP, intersect the opposite sides in three points, and you take the
symmetrical points of these considering the middle points of their own side, and join
with the opposite vertexes, prove that you obtain concurrent lines.
Problem 3
Take the projections of P on each of the sides of the triangle, take the symmetrical
points considering the middle points of each side, and take the perpendicular lines
from these symmetrical points on their sides. Prove that these lines are concurrent.
Isn’t it interesting that starting from the concept of symmetry in describing important
lines these properties can be extended in a similar way for all derived lines? Try to
discover other applications using the same methodology.
Connections among important lines
Problem 1
Prove that medians are concurrent. Prove the same think for bisectors, perpendicular
bisectors, and altitudes. Prove that the intersection of bisectors is the center of the
circle tangent to all sizes of the triangle, intersection of perpendicular bisectors is the
center of the circle passing through all vertexes of the triangle, intersection of
altitudes determine with triangle vertexes and with its altitudes inscriptible
quadrilaterals, and intersection of medians is the center of gravity of the triangle.
Problem 2
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Take a triangle ABC, bisectors of interior angles and bisectors of exterior angles (An
exterior angle is delimited by a side of the triangle and the prolonging line of another
side of the triangle). Prove that interior bisector of an angle and exterior bisectors of
the others two angles intersect in centers of circles that are tangents to a side and to
prolonging other two sides of the triangle. If we call these points A’, B’, C’, show that
highs of A”B”C” are bisectors for ABC.
Problem 3**** (Blue belt)
Take an altitude AA’ of a triangle ABC and a point P on this altitude. Join B and C
with this point P and note B’ and C’ the points of intersection with the opposite sides
of BP and CP. Prove that the angles B’A’ A and C’A’A are equal. (Difficult but
interesting problem requiring strategic thinking. You can use Tales to create
additional constructions).
The richness of classical geometry is not yet considered in its integrity by the modern
geometry. Only some parts of the classical geometry were starting points for the modern
geometry. In fact humans make mathematics from a short historical period, and even if
the quantity of mathematical properties discovered until now is huge, it covers only a
small part of the universal rules. Mathematics is a new field for every generation, and
everybody can discover new things. In fact we are the product of our efforts and of own
understanding.
Solving techniques
To solve a problem is frequently a very difficult task because it has a hidden solution.
If we know how to make a good interpretation of the initial data this could be very
helpful to find a solution. Usually the problem considers a property of the space that
keeps us inside of phenomena. If we find the hidden phenomena behind the problem
we find more than a solution. Let’s consider as example a problem discovered in the
XVI century, the butterfly (see the figure)
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Problem 1 the butterfly***(green belt)
Take a circle, AC and BD two chords that intersect in P. Take a line perpendicular on
OP where O is the center of the circle. This line intersects AB in U and CD in V.
Prove that PU=PV.
Comment
The most unusual thing in this problem is the forced introduction of a right angle
OPV. This “seed” of a new concept is the key for the solution. If we take OM
perpendicular on AB and ON perpendicular on CD, we obtain two new inscriptible
quadrilaterals MUPO and NVPO. Studying the angles of these quadrilaterals we find
that OU=OV that means OUV is isosceles, and we obtain easily the solution.
Sometimes we have a middle point introduced in the problem. We probably need to
consider other middle points that can configure a new structure able to give a
direction in finding the solution. The same can be done with parallel lines that require
other parallel lines as additional constructions.
Problem 2**(orange belt)
Take a triangle ABC. From B and C take two interior lines such as their angles with AB
and Ac to be equal. These lines intersect in P. Project the point P on AB in I and on AC
in J. Prove that IM =M where M is the middle point of BC. (Clue take the middle point of
BP and CP and use them making additional lines.
Problem 3**(orange belt)
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Take a hexagon ABCDEF that has opposite sides, parallel to each other. Prove that the
triangles ACE and BDF have equal areas (see the figure).
(Clue study this figure, and consider the triangle ACD too).
Sometimes studying such a construction involved in the solution of a problem you
can find interesting phenomena that can be generalized or a theorem. Let’s see some
examples:
Section Ceva and Menelaos Theorems
Menelaos lived in ancient Greeks period, Ceva in a more modern period, but they
discovered both across age new phenomena related to the same mathematical
perspective.
Menelaos Theorem
Take a triangle ABCD and a line d that intersect the sides of the triangle in A’, B’,
and C’ on the opposite faces of A, B, and C respectively. In these conditions
AC’/C’B BA’/A’CCB’/B’A=1 (See Figure Ceva and Menelaos)
Ceva Theorem
Take a triangle ABC and a point P that is not contained by any side of this triangle.
Take A’, B’, C’ the intersection points of AP, BP, CP (cevian lines or cevians) with
BC, AC, and AB respectively. Prove the same relationship like in Menelaos theorem.
From now on we will call the lines passing through a vertex Cevian Lines, and the
lines intersecting all sides of a triangle transversal lines. If you note the angle formed
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by cevian lines with A1,A2,B1,B2,C1,C2 following the same sense of rotation prove
that sinA1/sinA2sinB1/sinB2sinB1/sinC2=1
An interesting application is Triliniar Polar
Application
Take a triangle ABC, a point P the cevians AA’, BB’ CC’, and the transversals A’B’,
B’C’, and C’A’. Prove that the points of intersection of AB with A’B’ (M); BC with
B’C’ (N); and AC with A’C’ (R) are on the same line.
This property gives us a transformation of the plan in itself. For each point (P) we
find a unique line (MNR), and for each line a unique point P that generates this line. We
will see in the section Transformations how important are these transformations in
developing mathematics and solving problem abilities.
A historical application due to Blaise Pascal
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Take a circle and six points on the circle forming a hexagon. The points obtained by
intersecting the opposite sides of the hexagon are on the same line. (See the Figure)
Another very important application is due to Desargues
Desargues Theorem
Take two triangles ABC and A’B’C’ such as AA’, BB’, and CC’ to intersect in the
same point P. Prove that if AB and A’B’ intersect in M, BC and B’C’ intersect in N, and
CA and C’A’ intersect in R, than M, N, and R are on the same line. Triangles in these
conditions are called homological triangles, P is called center of homology, and the line
QRS is called axes of homology.
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Do you want a difficult to prove but very interesting application?
Problem *****(brown belt)
Take a circle and six points on this circle. Join these points three by three forming
two triangles. Prove that these triangles intersect each other forming hexagons with
concurrent diagonals. (See the Figure Concurrent Diagonals)
36
Remark
We can see that in these problems we can choose randomly our points and we will
obtain the same property. In Menelaos theorem the line d can intersect the triangle
outside of its sides. In Ceva the point P can be everywhere. In Pascal problem we can
choose any close broken line with vertexes in A,B,C,D,E,F and if we count opposite sides
as in a convex hexagon we obtain the same property. Even more, we can degenerate these
properties. Let’s take as example of degeneration, Pascal problem
Degeneration
Watch the following figures and find new properties. Develop new problems.
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Remark
Prove all these properties in the space by transferring everything on cones. The
procedure is simple: Take the figure in the plan and a point P in the space not belonging
to the plane. Than join the point P with every other points in the plane. You will obtain
cones or pyramids, planes passing through P instead of lines, and lines passing through P
instead of points. If you intersect all this structure with a plan you will obtain conics at
intersection instead of circles, and lines and points similar with the plan figure and in the
same relationship. Using this trick you can extend many properties from circles to conics.
A very useful metric relationship
Take two omological triangles ABC and A’B’C’. Their sides intersect each other in
I,J,K,L,M, and N (see the figure omology). In these conditions we have the following
metric relationship that extends Ceva theorem, and Desargues theorem:
38
This metric relationship is the same if L, K, L, M, N, P, and R are points on the sides of a
triangle belonging in the same time to a conic. Can you find the passage from two
homologic triangles and conics intersecting a triangle using con’s method?
Application
Show that if three cevians in the previous figure intersect each other in the same
point, than the other three will also intersect each other in a point.
Search for new things
As we can notice there is a relationship between omology and orthology. Both of
these phenomena need two related triangles. The metric relationships regarding these
phenomena are very similar, just replacing substation of square sums with produces of
raports; and both these phenomena develop classes of triangles with the same center of
omology and axe of homology or with the same center of orthology and cocenter of
orthology. Can we find any relationship between the centers of homology and orthology
if two triangles are in the same time omologic and orthologic?
Do you like spiders?
Unless you are entomologists probably not, but certainly you admired their net. Is so
regular and adapted to different situations. We wonder if these nets don’t have anything
mathematical in them. The answer is positive, and is incredible how an insect with few
thousands of neurons know this mathematics. Look at the following figure and try to
solve spider’s problem.
39
The most interesting fact is that spiders create their nets making inscriptible
quadrilaterals. So A’ABB’ is inscriptible, and all the others are the same. Prove that if
ABCDEFGHIJ is a polygon with all its vertexes on the same circle, than
A’B’C’D’E’F’G’H’I’J’ has all vertexes also on the same circle. The trick is simple, you
need first to prove that OAOA’=OBPB’=OCOC’ etc
In fact we have here another transformation, “the inversion”. In this case this
transformation transform a circle into a different circle. The inversion can transform a
circle into a line if the point O (the center of inversion) is on the circle (see the
40
figure)
In this case the circle determined by OABCDE is transformed in a line A’B’C’D’E’F’
It seems that this transformation creates another interesting instrument for geometry.
First we have to notice that OAB and OA’B’ are similar triangle, but with inverse
similarity. This means that we can ‘t obtain one from the second one by magnifying and
rotating it. So the angle OAB equals the angle OB’A’.
Problem1
Take a quadrilateral ABCD, and note with E the intersection between AB and CD
and with F the intersection between AD and BC. This figure is called a complete
quadrilateral. Its sides form four triangles. The circles passing through the vertexes of
these triangles are concurrent in the same point I (point of Miguel). Show that an
inversion with center I of the entire figure formed by lines and circles will give a similar
figure.
Problem2
Prove that taking a line that pass through O and intersect two curves one being the
inversion of the other one, than the angles between this line and the tangent lines on the
intersection points with these curves are equal.
Problem3
Take two circles one interior to the next one and a number of circles two by two
tangents and tangent to the first two circles. Prove that if you start making these tangent
41
circles starting from any point between the first two circles, the chain of tangent circles
will close. (Clue: transform the entire figure by an inversion such as the first two circles
to become concentric, see remarkable property)
Problem 4******(black belt)
Attack this problem when you will be confident on your strength. Is a very interesting
property, but need a lot of skills to prove it. Take a pentagon ABCDE, and join its sides
until you will obtain a star formed by a closed broken line. Through each triangle forming
a corner of the star pass one circle. These circles intersect two by two in five new points.
Prove first that these points are on the same circle. Prove than inverting the entire figure
containing all lines and circles determined by triangle you obtain a similar figure.
An important application of the inversion is the power of a point for a circle, and the
radical axe.
Take a circle and an exterior point P. Through P take two lines that intersect the circle
in A, B and A’, B’. Than OA.OA’=OB.OB’. If a point A describe the circle, the point A’
will describe the circle in the same time but in the other sense. The radical axe of two
circles is the geometric place of all points with equal powers to two circles. If these
circles will intersect each other the radical axe is the common cord. If they are tangent,
the radical axe will be the common tangent in the tangent point.
Let us see some properties generated by this transformation and by the radical axe.
Problem 5 Brianchon Theorem
Take a hexagon that has all sides tangent to the same circle (conic). Prove that the
diagonals that join the opposite vertexes of the hexagon are concurrent (see the figure
42
Brianchon).
See the picture and study its properties. You have some clues: The first clue is
HR=GM (tangents at two circles). The second clue is ER=ES (tangents to the same circle
from a point). The third clue is to choose the exterior circles such as the diagonals of the
hexagon are radical axes of them taken two by two. After you will finish to prove this
problem try to degenerate it at pentagons, quadrilaterals, and triangle. Can you find an
extension in the three-dimensional space, using cones instead of tangent lines, and radical
plans instead of radical axis? Try to visualize the curious shape of the body that will
replace the initial hexagon. You can have an image about how geometry can extend to
other figures or bodies that the usual ones. Consider two cases: of three external spheres,
and of four external spheres, knowing that any three spheres have the same radical axes,
and any four spheres have the same radical center.
Problem***(green belt)
Take a cone that intersects a sphere. If one of the intersection curves is a circle prove
that the second one is also a circle.
Let’s see more attentively the radical axes. We can find a very remarkable property.
Remarkable property
43
Let’s study these phenomena. We take firs a set of circles with a common cord as
radical ax. On this common cord we take a point O and we get the tangent lines at all
circles already described. We have a surprise. All the segments obtained are equal
(OA=OB=OC=...). This means that O is the center of a circle that cut all the set of circles
with radical axe PR. The second surprise is that this circle is orthogonal with all the
circles of the set. That means that if we the tangent in A to the circle with center O, and if
we take the tangent in A (OA) from the circle of the set containing A, and, these two
tangents are perpendicular. The third surprise. If we change the point O with O’ we
obtain a different circle orthogonal to the set. If we vary O we obtain a second set of
circles that have all the same radical axe generated by the centers line of the first set.
These circles are not tangent nor secant to each other, they form a sequence of circles
with limit the point P. And the last surprise, if we make an inversion of the entire
structure with the center of inversion in P we will obtain a set of concentric circles and a
set of diameters for all of them.
Problem
Take two perpendicular lines “d” and “g”. On “d” take a point P, and two equal
segment PR=PS=a. Take RT perpendicular on PR, with T on “g” Note the size of RT
with b. Take SU perpendicular on PS, with U on “g”, and note its size with c. Take a
point V on “d” such as VT=VR=b. Prove that VU=US=c.
Note
This problem is a competition like problem. You need to find out the phenomena that
better describes your problem, and to use it.
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Properties in the space
We can make inversions in the space. These inversions will transform a sphere into
another sphere or in a plan if the center of inversion ion the sphere.
Problem*****(brown belt)
Take a tetrahedron ABCD, and on each side take a point. Take the spheres generated
by a vertex and by point on sides that pass through this vertex. Prove that all these four
spheres have a common point.
To find a solution to this problem is difficult because you can’t draw on a plan very
well figures from the three dimensional plane. We need to see in our mind. Let’s try.
Take one of these spheres, and its intersections with the three plans of the tetrahedron
containing its generators. You will find three circles. Take the intersection of this sphere
with the other three spheres. You will find other three circles. Let us see more attentively
these circles. First if we consider three spheres passing through the three vertexes of a
face of the tetrahedron and we intersect them with this face, we obtain three circles that
intersect each other in point on the sides of the triangle. The second observation based on
the previous one is about the sphere containing the six circles. These circles intersect
each other three by three in points. We don’t know if the circles determined by the
intersection of the last three spheres with the first have the same property. If yes, our
problem is solved, because the last point will belong to all spheres. Now we desperately
need a miracle. This miracle is brought by the inversion. If we invert all the structure
formed by circles on the sphere taking as center of inversion a vertex of the tetrahedron
that is on the sphere, the sphere will be transformed in a plan. The three circles passing
through the vertex will become lines and the other three will become circles. We obtain a
very peaceful and easy problem. We have a peaceful triangle with a point on each side,
and three circles generated by a vertex and two points on the sides passing through this
vertex. All these three circles have a common point.
We used in this proof the fact that inversion preserve incidence. In fact inversion
preserves also the angles between the lines forming a figure when this figure is inverted.
The new one will have the same angles for the inverted components.
Do you like inversion? A mathematician, (Steiner) who was illiterate until 16 years,
being shepherd in mountains developed the geometry of inversion. He learned to read and
write after 16 and at 23 years he was professor in a German university. This geometry is
his understanding about the Universe rules that he made in his childhood looking to the
stars and to mountains and to his animals. His courses were interesting but without
figures, because he used to use the internal vision that can make us understand in the
N-dimensional spaces. If you don’t believe me try to extend the property described in the
previous problem in n-dimensional space. You can do the same thing with isogonal and
isotomic point in n-dimensions, and with other problems. Not any property can be
extended. Most properties are characteristic for one kind of n-dimensional space, but not
for a different space. Still among most of properties you can find relationships across
dimensions. The common properties are frequently related to the internal transformations
of the n-dimensional space that are similar to different dimensions.
Do you want to know more about inversions, than visit:
http://www.maths.gla.ac.uk/~wws/cabripages/inversive/inversive0.html
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Translations
The most common activity needs rotations and translations. Moving objects from a place
to another is usual experience that doesn’t destroy objects’ shapes. We can understand
now why.
Problem 1
Take a triangle ABC on a fixed position. Take another triangle A’B’C’ on a mobile
position obtained by translation. Show that the triangle A”B”C” obtained by joining the
middle points of AA’, BB’, and CC’ doesn’t change the shape when A’B’C’ translates?
Problem 2
Show that if ABC and A’B’C’ are directly similar to each other, than the triangle
A”,B”,C” obtained by joining the middle points of AA’, BB’, and CC’ is also similar to
ABC and A’B’C’ (clue translate one of these two triangles until A and A’ will be the
same point). Is an extension in three-dimensional space possible? Is similarity in the
plane and in the three-dimensional space the same concept?
Translation is useful in area or volume problems.
Problem 3
Take two lines “d” and “g”, and AB on d, CD on g two mobile segments containing the
intersection point between d and g and preserving their measures. Prove that ACBD will
preserve its area when AB and CD will translate on their lines d and g. (clue, take
through A,B,C,D lines parallel to d and g. Compare the rapport between the areas of
these two figures.)
Extend this property in the space taking three concurrent fixed lines, “d”, “g”, and “h”,
and on each a segment with fixed length but mobile on its line and containing the
intersection point of these lines. Prove that the polyhedron obtained will preserve its
volume.
Sometimes we don’t need to translate something, it is useful to construct parallel lines
instead of translating figures.
Problem5*****(brown belt)
Take a square and divide it in a number of triangles with equal areas. Prove that you will
obtain an even number of triangles.
Comment
Usually translation is associated with another transformation “omothety” This
transformation has a center, and changes the size of the figure proportionally. In fact it is
a generalization from Tales. The common characteristics of translation and omothety are
the respect of direction, the respect of angles and of proportions. For translation
proportion is 1. Both transformations are frequently used together in perpendicularity
problems, where you can translate and change the proportions of a figure until you will
arrange it in a good perspective. If you want to see some examples try orthology section.
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Universe is geometry, even life is geometry, and do you want to see more? Than visit:
http://classes.yale.edu/99-00/math190a/TransfGeom.html
Cavalieri Principle
This principle was probably at the basis of the calculus derivatives and antiderivatives.
Let’s see if calculus procedures can’t be completed using the same principle:
Problem1***(green belt)
Take a pyramid and a line passing through its vertex P On this line take two segments PR
and UV, R being out of the pyramid in up direction, U being on the base of the pyramid
and V on the line but in down direction. Take for each face of the pyramid a prism that
has the face like a base, with the property that the other parallel base pass through R if the
face is lateral, and through V if the face is the base. Prove that the sum of volumes of the
prisms constructed on the lateral faces equal the volume of the prism developed on the
base.
Problem2***(green belt
Take a tetrahedron ABCD, and a line “d” that intersect the opposite sides AB and CD in
P and R. Take PQ=RS with Q and S on d and outside of the tetrahedron. Now take the
prisms that have the as bases the faces of the tetrahedron and that have another property.
The other basis pass through Q or S. Prove that the volumes of prisms that have faces
passing through Q equal the volume of prisms that has faces passing through S
Problem3***(green belt)
Take a tetrahedron ABCD and the line d that join the middle points of AB and CD. Prove
that any plan containing this line d divide the tetrahedron in two parts with equal
volumes.
Problem4****(blue belt)
Take two tubes with equal radiuses, perpendicular to each other and with concurrent axis.
Find the volume of the intersection of these two tubes. (Clue, use planes parallel to the
two axes that cut the body in slices. Take in the same time the entire structure and find
the invariants.
Interesting perspective, isn’t it? It seems that Archimedes understood this perspective,
and even developed the nucleus of differential and integral calculus. Leibniz and Newton
had rediscovered almost two millenniums after his death, this perspective. If Archimedes
would have enough time to finish the description of this perspective the history should be
maybe different.
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Rotations
Do you remember direct similarity and centers of similarity? If instead of two unequal
segments you take two equal segments you will obtain a rotation instead of a similarity
and a center of rotation instead of a center of similarity.
Problem 1
Take two triangles ABC and AB’C’ equilateral with A common vertex. Prove that BAB’
and CAC’ are equal triangles rotated to each other with 60 degrees. Considering this
property, take the following problem: Take ABC an equilateral triangle, P an interior
point, and PAA’, PBB’, and PCC’ three equilateral triangles such as A’, B’ C’ are on the
same side of PA, PB, and PC. Prove that A’B’C’ is also equilateral, equal and with
parallel sides to ABC.
Problem 3**(orange belt)
Take a quadrilateral ABCD and build in its exterior four squares: ABIJ, BCKL, CDMN,
and DAPR. Prove that the two lines joining the centers of the opposite squares are equal
and perpendicular on each other (clue, consider the middle point of a diagonal as a
potential center of rotation.)
Problem 4***(orange belt)
Take a triangle ABC, and build of its sides on the exterior the triangles ABIJ, BCKL, and
CANM. Show that the line joining a vertex of the triangle with the center of the opposite
square is equal to and perpendicular on the line that joins the centers of the other two
squares.
As we can notice direct similarity is a composition of two elementary transformations:
rotation, and omothety. Inverse similarity is a composition of symmetry with respect to a
line, rotation and omothety.
Rotation is a plane phenomenon. Even if you can rotate an object in the space in a very
random way you don't find the same characteristics with plane rotations. In the space you
can obtain a complex rotation preserving the center of rotation by composing as many
plane rotations you want with the condition that all these rotations will have the same
center. If you have an axis of rotation in the space you can reduce your problem to a
plane rotation.
We find as you see transformations that are specific for an “n” dimensional space but not
for another, and transformations that are general in any dimension. From this perspective
translation is a transformation on the one-dimensional space, and applied in a multidimensional space will describe vectors and a linear perspective about the space.
Rotations are two-dimensional transformations. They give circles as invariants and these
circles can be extended to conics by composing with a different transformation. Rotations
will give us a quadratic vision about the space that can be in the three- dimensional space
correlated with double riglated surfaces and quadric equations.
Can you find a specific transformation for the three dimensional space? Try with an
accurate vision about one and two dimensional space transformation. Tranlation refers to
points and can be extended to segments. Rotation refers to lines and can be extended to
angles. Invariant for translation is direction. Invariant for rotation is the family of circles
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described by any point. Which object will replace in the three-dimensional space the
segment and the angle from the one and two-dimensional spaces? Which structure of
invariants will describe, and what transformation?
You need a new perspective? Than visit:
http://www.intl-light.com/handbook/ch07.html
These questions are part of a mathematical philosophy able to create new perspectives in
understanding. Such a philosophy was first expose by Felix Klein at the famous congress
of Erlangen.
If you want to see more about this perspective, visit:
http://plato.stanford.edu/entries/geometry-19th/
Geometry has many directions of development. If you want to know more, visit:
http://medialab.di.unipi.it/web/IUM/Waterloo/node34.html
The vision about a mathematics that can be done using transformations due to Felix Klein
was a very revolutionary idea in mathematics. Considering transformations we can
discover a new universe of mathematical properties that has its own potential and gives
new perspectives. In fact this is one of the most interesting conclusions about
mathematics and science development. We own most of these developments to new
perspectives about the same subject. We depend on these perspectives that are able to
enlighten old facts giving them a meaning.
One of these new (old) perspectives was introduces by a French scientist “Poncelet”.
Poncelet was a soldier in Napoleon’s army and was captured by the Russian army, and
imprisoned in Siberia in a salt mine. In this period he created a new chapter in
mathematics, the projective geometry, and a new transformation “the polar
transformation” This new chapter of mathematics changed radically the perspective in
mathematics and sciences.
The Polar Transformation
What we can find searching for general phenomena?
Do you remember about Poncelet? In his prison in the salt mine where he was forced
to work for 18 years, he found a hope in searching the beauties of the universe transferred
in geometry problems. One of these problems is extremely difficult, but was discovered
as a consequence of a very deep understanding of very complex phenomena. This
problem is an example of a trained mind search in the treasure room of the universe.
Poncelet problem******(black belt)
Take two circles one interior to the other one. Show that if you can draw a close
broken line with vertexes on the big circle and with tangent segments to the small circle
and starting from a point of the big circle, than you can do this starting from any point of
the big circle (see the figure Poncelet).
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Try to solve this problem with conventional tool starting from a close broken line
forming a triangle. Try than to find the phenomena that lead Poncelet to this problem. If
you succeed you are able to discover too. If not you need more training, but finally with
the same result: The Treasure.
Birapport and anarmonic rapport
Take the bisector theorem but now consider the two bisectors coming through the vertex
A, the interior and the exterior one. The interior intersects BC in P and the exterior
intersects BC in R. In these conditions RC/RB=PC/PB. We can rewrite this as
RC/RBPB/PC=1 and this produce we call birraport. If the produce equals a number
different from 1 will be called anarmonic rapport.
Important properties
-If a fascicle of four concurrent lines is crossed by two secant lines, on each of these
secant lines the points of intersection will give the same anarmonic rapport.
-If we take four points on a line and the geometric place of any other points that see the
four points under the same anarmonic rapport; this geometric place will be a conic. If
instead of anarmonic rapport we take the birraport, we obtain a circle.
-If we take a circle, an exterior point P, a line passing through P that intersect this circle
in A and B, and Q and R the tangent points of the circle of the mobile line, than P, A, B
and S the intersection point between AB and QR are in the same birapport.
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The most interesting application of this transformation is the transformation of the line
RQ in the point P. (see the figure)
In the case when the point P is on the circle it will be transformed in the tangent line
coming through P. In the space a tangent plane will transform in the tangency point.
Because of this property problems about concurrence (point) will transform in problems
about coliniarity (line). So Brianchon and Pascal problems will be the same one
transformed through the polar transformation.
Application
Show that the points N, A, C’, and B from the triliniar polar are in a birapport.
Problem****(blue belt)
Take a polyhedron that has eight triangular faces concurrent four by four. Prove that if
seven of them are tangent to the same sphere, that the eighth one is also tangent to the
same sphere. (Clue, Transforming everything using the polar transformation, this
polyhedron will be transformed in another one containing six vertexes instead of six
planes. Instead of four planes concurrent in a point we will find four points on the same
plane, this means that our new polyhedron will have six quadrilateral faces. Instead of
five faces tangent to the same circle we will have five faces that have all vertexes on the
same circle. The conclusion is also transferred: instead of the six faces tangent to the
circle we will have the six faces have vertexes on the same circle. The last sentence we
can prove transferring this information to inversion problems. In fact we already met it
under a different form. Do you recognize it?
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Another interesting problem is due to an ancient Greek mathematician “Pappus”. It can
be extended from two lines to a conic and solved using polar transformation and
birraport. (See the figure)
Problem*****(brown belt)
If A, B, C and D, E, F are on two different lines than the point of intersection of the
following pairs of lines AE and BD; BF and CE; AF with DC are also on a line.
(Clue You need to be creative put more lines such as to form a fascicle cut by a line and
giving the same anarmonic rapport. It is not easy but Papus found it with very elementary
instruments long time ago. Isn’t it interesting?
The connection between polar transformation and conics is very close. We need to know
that through any five points pass a unique conic. Through four points pass an infinity of
conics, but through six point pass only one conic if taken two points and the fascicles
with these points as vertexes and connecting the other four points we find two fascicles
with the same anarmonic rapport. In fact if we keep four fixed points and we vary the
sixth one such as the fascicle preserves the same anarmonic rapport we obtain a conic as
geometric place of the fifth point.
We can use this property in many problems connecting our knowledge across geometry.
Problem****(blue belt)
Prove that the geometric place of the isogonal points of a line considering a triangle is a
conic.
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Problem****(blue belt)
Prove that the intersection between a double riglated surface and a plan is a conic.
Projective geometry gives a perspective about perspectives. Renaissance painters were
first preoccupied by perspective. Than architects become also preoccupied. Perspective
rules developed after this new mathematical perspective, and now the projective
geometry is a fundamental instrument in understanding the universe. From quantum
mechanics physics to neural networks different characteristics and objects are described
in the projective space.
Do you want to know more about the projective geometry, than visit:
http://www.anth.org.uk/NCT/basics.htm
Perpendicularity in the three dimensional space
-Take a trihedral angle Oa, Ob, and Oc, and planes passing through each of these lines
and perpendicular on the planes generated by the other two lines. Prove that these plans
intersect each other after a line (the first line with the role of an altitude).
-Take a tetrahedron and take the line passing through the intersection point of the
altitudes of a face, and perpendicular on this face. Prove that these lines generate a double
riglated surface, the other riglature being given by the perpendicular lines from vertexes
to the opposite faces.
-Take a tetrahedron and the lines perpendicular on each pair of opposite sides. (Common
perpendiculars). Prove that if the opposite sides of the tetrahedron are perpendicular, that
these lines are concurrent.
-Take a tetrahedron, and the middle points of each side, and bring planes perpendicular
on the opposite sides. Prove that these planes are concurrent.
-Take a tetrahedron and than the perpendicular lines from each vertex to the opposite
side. (The real altitudes). Prove that these altitudes are concurrent if and only if the
opposite sides are also perpendicular.
What we can notice is the great variety of lines and planes involved in the
perpendicularity concept in the space. We will find even more properties. Because this
diversity is already a little bit confusing we need to find the relationship between all these
lines and planes to have a good image. It is possible to find that the relationship doesn’t
correspond with our image about perpendicularity extended from plane to the threedimensional plane. In this case we need to find the truth; do you have any idea?
Let’s consider first the easiest tetrahedron to study. This tetrahedron has the opposite
sides perpendicular two by two, and is called orthogonal tetrahedron.
Problem 1
Take the altitudes and the common perpendiculars of an ortogonal tetrahedron. Proof
that:
-The altitudes and the common perpendicular are concurrent
-If A’, B’. C’, D’ are the intersection points of the altitudes with the faces of the
tetrahedron, than if you bring perpendicular lines from A,B,C,D on B’C’D’; A’C’D’;
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A’B’D’; and A’B’C’ respectively, than these perpendiculars are concurrent in the same
point. Noting the point of intersection with P prove that AP.PA’= BP.PB’=
CP.PC’=DP.DP’=MP.NP= RP=.SP=TP.UP where MN , RS and TU are common
perpendiculars
-Taking the plans generated by points, which are on the sides of a trihedron and common
perpendicular you obtain another tetrahedron. Prove that this new tetrahedron and ABCD
have perpendicular sides to each other. Prove also that the point of intersection of
altitudes in ABCD, is the center of the sphere tangent to the surfaces of the new
tetrahedron. Prove that the new tetrahedron and A’B’C’D’ have parallel sides and faces.
-Take the common perpendicular of the tetrahedron ABCD considering the sides AB and
CD. Take a point P on it, and from P bring perpendicular lines PI on AC, PJ on AD, PK
on BC, and PL on BD. Prove that IJKL are on the same plane, and IJKL are on the same
circle. Prove that ABIJKL and CDIJKL belongs to two spheres centered in the middle on
AB respectively CD.
As we can see in these properties, perpendicularity is associated with circles and
spheres. The more general case when we have a common tetrahedron we will associate
perpendicularity with double riglated surfaces and quadrics.
Spheres
Spheres are interesting objects that have their own geometry. There are not simple
extensions of circles and most of circle properties don’t fir with sphere’s properties. For
example in the circle if you fix two points A, and B on the circle and take a mobile point
C on the same side of AB on the circle, that the bisector of ACB pass through a fixed
point. This property doesn’t extend to sphere if we take a trihedral angle. Instead of these
extensions we can find new properties:
Problem 1***(green belt)
Take a quadrilateral pyramid with equal lateral sides. Prove that the sum of two
opposite dihedral angles equal the sum of the other two dihedral angles (dihedral angle is
the angle between two planes. It can be obtained as a plan angle considering the
intersection between the two planes and a plan perpendicular on their common side)
Problem 2*(yellow belt)
Take sphere and a quadrilateral that has its sides part of big circles on the sphere
(circles that have the same center as the sphere), and vertexes on the same plane. Prove
that the sum of the opposite angles is the same (the angle between two curves is the angle
between the tangents on these curves in the point of intersection)
Problem 3****(blue belt)
Take two intersecting spheres, their common circle, A, B, C and D on this circle.
Take P and Q on one sphere R, and S on the other sphere such as PA and QD intersect in
R, PB and QC intersect in S, RA and SB intersect in P, RD and SC intersect in Q. Prove
that PQ and RS are perpendicular to each other.
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Problem4***(green belt)
Take a number of equal spheres in the tri-dimensional space, and suppose that each of
them enlightens the others. Prove that taking together the shadowed areas of all spheres
you will obtain the area of one sphere.
Properties regarding double riglated surfaces
You can find new perspectives regarding double riglated surfaces studying:
http://www.acm.org/sigs/sigchi/chi95/proceedings/papers/jl_bdy.htm
Newton Gauss lines and plans
The concept of the middle seemed to be universal. It became “center” for various
transformations, and for different structures generated by points and lines that are
connected to transformations or not. Do you remember Newton Gauss line? This line
passes through the middle points of the three diagonals of a complete quadrilateral. It also
can be considered as an extension for the “ middle” concept from the structural
perspective, but can be also associated with polar transformation and anarmonic rapport.
Starting from Newton-Gauss line you can see now how this concept develops:
-Taking any three lines there is only one line intersecting any of these three lines and
passing through a point of one of them. Is easy to proof but is an important characteristic
Taking different points from one of these three lines and constructing the set of lines
that pass through these points you obtain a double riglated surface. The main property of
this is metric. If you take four lines from one of the two sets of lines, and intersect with
other four lines from the second set of lines you will obtain point on each line of a set that
have the same anarmonic rapport.
In fact considering all the perpendicularity characteristics in a tetrahedron we can
notice that the point of intersection is a degenerated case of a double riglated surface. So
you can fix your attention no the double riglated surfaces generated by common
perpendiculars, altitudes, and others.
A special case is the following one:
Problem***(green belt)
Take a tetrahedron ABCD, take the middle points of any side and take the planes
passing through these points and perpendicular on the opposite sides. Show that these
plans are concurrent.
The problem looks very difficult and it is if we don’t find its roots.
Prof. Take through any side of the tetrahedron a plan that is parallel to the opposite
side. You will obtain a parallelipiped in which the sides of the tetrahedron are diagonals.
There is another tetrahedron that can be obtained in a similar way taking the remained
vertexes of the parralelipiped. The planes that pass through the middle points of each side
and are perpendicular on the other sides of the first tetrahedron have a function for the
second tetrahedron that you can discover. Continue to search these properties and find out
some more characteristics.
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This problem is interesting because it associated perpendicularity with a system of
reperation that looks more Cartesian. In fact discerning among facts we can see that in the
three-dimensional space the natural way to look is through double riglated surfaces if we
consider lines perpendicular on other lines or planes, and through a cartezian reperation
system if we consider planes perpendicular on other plans. This new way of thinking
requires more instruments related to the projective geometry. These instruments will be
developed in a different section. Now we need to focus on a different aspect of the space
geometry: coplanarity.
The problem starts from a plan property; the Newton Gauss line.
Problem****(blue belt problem)
Take a quadrilateral and join its opposite sides forming a complete quadrilateral.
Prove that the middle points of the three diagonal of the quadrilateral are on the same
line. There are more than a hundred of demonstrations for this property, no one easy. In
the space it looks even more difficult: (See the figure).
(Clue: take a point on T on MN and prove that the sums of areas
ABT+CDT=BCT+ADT. Extent this property for the entire line MNP and take the point T
as a geometric place)
Problem*****(brown belt)
Take five lines in a plane that intersects each other. Show that Newton-Gauss lines of
the complete quadrilaterals generated by any four of these lines are concurrent in the
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same point (In order to solve this problem see the following connected problems that will
introduce you in this phenomenon:
Additional problem1
Take three lines “a”, “b”, and “c” parallel with the line “d”, other three lines “a’”,
“b’”, and “c’” parallel with another line “d’”. Be A1, B1, C1 A2, B2, C2, respective the
intersections of the following pairs of lines in this order: (b, c’), (c, a’), (a, b’), (c, b’),
(a, c’), (b, a’). Prove that the lines A1A2, B1B2, and C1C2 are concurrent.
Additional problem 2 for the additional problem 1
Take two sets of three concurrent lines Ia, Ib, Ic, and I’a’, I’b’, I’c’. Be A1, B1, C1
and A2, B2, C2 the points of intersection of pairs of lines in the following order: (Ib,
I’c’), (Ic, I’a’), (Ia, I’b’), (Ic, I’b’), (Ia, I’c’), and (Ib, I’a’). Prove that A1A2, B1B2, and
C1C2 are concurrent.
Additional problem 3 for the additional problem 2 See Pappus problem
Papus problem will help you to solve additional problem 2 that will help you to solver
additional problem 1 that will help you to solve the main problem. This is a usual event in
mathematics where you need a long chain of partial conclusions that will lead you to the
main theorem. From the main theorem you can develop another chain of properties.
Problem*****(Brown belt)
Take five plans and the Newton Gauss lines of the complete quadrilaterals generated
by the intersection of a plane with the other four. Prove that all these Newton Gauss lines
are parallel with the same plane.
Problem ******(Black belt)
Take a tetrahedron ABCD and a line “d” that intersects the faces of the tetrahedron in
A’, B’, C’, and D’, each of these points belonging to the opposite face of A,B,C,and D.
Prove that the middle points of AA’, BB’, CC’, and DD’ are on the same plan.
The number of properties regarding planes can continue by varying the line, by
taking a line in a plane and considering the five planes, and by taking fascicles of lines
and planes and searching how the coplanar points will behave. If you are able to deal
with these problems you are prepared to be a researcher in mathematics even if you need
to learn much more. Your mind is prepared to discover new things. Good hunting.
If you want to see where this geometry will lead you visit:
http://archives.math.utk.edu/topics/geometry.html
If you are interested to share your ideas you can visit
[email protected]
Do you want to know ancient Chinese perspective about Cosmos, humanity and
geometry, visit:
http://m759.freeservers.com/PHiching.html
Do you want to know more about Projective geometry, and its specific
transformations, visit:
http://www.cs.elte.hu/geometry/csikos/proj/proj.html
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Chapter 2
Probably most of you did not solved all problems found in chapter 1, but you still
have a better vision about geometrical phenomena. For a researcher in mathematics, both
these characteristics are absolutely necessary. We need to train our mathematical dream
and our intuition as well as our solving capacities. Chapter 1 is designed to open your
minds and to create a base for your intuition. Chapter 2 is designed to improve these
solving capacities. Chapter 2 contains also problems, but these problems will contain
complete solutions. They will also contain the following links:
- What knowledge do I need to understand the statement
- Which theorems are necessary to figure out the solution
- Where and how can be applied these properties
- Which phenomena express better these properties
- What kinds of relationships are developed by these properties, and what
techniques does they suggest.
- What is the main mathematical philosophy that includes or extends these
properties
Each link can lead to a new problem that is also contain the same links
Problems and solutions
Other additional materials you can find on:
http://forum.swarthmore.edu/geometry/geom.puzzles.html
http://www.ics.uci.edu/~eppstein/junkyard/polyomino.html
http://www.math.binghamton.edu/zaslav/Matroids/matroidprobs.html
http://www.ics.uci.edu/~eppstein/gina/geom.html
http://www.c3.lanl.gov/mega-math/welcome.html
http://forum.swarthmore.edu/geometry/geom.units.html
http://modelingnts.la.asu.edu/html/UAFCG.html
For the national and international competitions you can train using the following
sites:
http://www.unl.edu/amc/a-activities/a4-for-students/reading-list.html
http://imo.wolfram.com/problemset/index.html.
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