THE CHARM OF GEOMETRY
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:
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.
[pic]
Note
For the ancient Greeks this demonstration practically proving the property was enough. Euclid changed everything by introducing a methodology.
Euclid’s proof
If we take two triangles that are in the case angle-side- side, we find two possibilities (Figure 3)
[pic]
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)
[pic]
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)
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. [pic]
(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).
[pic]
(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.
[pic]
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. Prove that AD and CF are parallel. [pic]
(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 with)[pic]
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).
[pic]
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)
[pic]
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 segments.[pic]
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)
[pic]
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)
[pic]
This figure describes very well these amassing phenomena, but another figure will help you more to see the solution. (Figure 13)
[pic]
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:
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)
[pic]
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’, and C’ on the opposite sides of the triangle BC, CA and AB are concurrent. (See the figure)[pic]
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 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)
[pic]
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 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:
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
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.
[pic]
(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)
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
[pic]
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)
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).
[pic]
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)
[pic]
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 equilateral triangles we obtain a new figure (Napoleon 2) [pic]
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
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).
[pic]
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:
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:
-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
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
[pic]
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
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)
[pic]
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)
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).
[pic]
(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 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
[pic]
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
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) [pic]
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.
[pic]
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)
[pic]
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.
[pic]
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:
[pic]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.
[pic]
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 figure)[pic]
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 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 Brianchon).[pic]
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
[pic]
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.
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:
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.
Universe is geometry, even life is geometry, and do you want to see more? Than visit:
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.
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 multi-dimensional 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 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:
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:
Geometry has many directions of development. If you want to know more, visit:
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).
[pic]
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.
The most interesting application of this transformation is the transformation of the line RQ in the point P. (see the figure)
[pic]
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?
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)
[pic]
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.
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:
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 three-dimensional 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’; 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.
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:
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.
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).
[pic]
(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 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:
If you are interested to share your ideas you can visit
annie@forum.swarthmore.edu
Do you want to know ancient Chinese perspective about Cosmos, humanity and geometry, visit:
Do you want to know more about Projective geometry, and its specific transformations, visit:
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:
For the national and international competitions you can train using the following sites:
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