RMO and Pre RMO Geometry Tutorial Worksheet 1: Based on Geometric Refresher

1) Show that quadrilateral ABCD can be inscribed in a circle iff \angle B and \angle D are supplementary.

2) Prove that a parallelogram having perpendicular diagonals is a rhombus.

3) Prove that a parallelogram with equal diagonals is a rectangle.

4) Show that the diagonals of an isosceles trapezoid are equal.

5) A straight line cuts two concentric circles in points A, B, C and D in that order. AE and BF are parallel chords, one in each circle. If CG is perpendicular to BF and DH is perpendicular to AE, prove that GF = HE.

6) Construct triangle ABC, given angle A, side AC and the radius r of the inscribed circle. Justify your construction.

7) Let a triangle ABC be right angled at C. The internal bisectors of angle A and angle B meet BC and CA at P and Q respectively. M and N are the feet of the perpendiculars from P and Q to AB. Find angle MCN.

8) Three circles C_{1}, C_{2},  C_{3} with radii r_{1}, r_{2}, r_{3}, with r_{1}<r_{2}<r_{3}. They are placed such that C_{2} lies to the right of C_{1} and touches it externally; C_{3} lies to the right of C_{2} and touches it externally. Further, there exist two straight lines each of which is a direct common tangent simultaneously to all the three circles. Find r_{2} in terms of r_{1} and r_{3}.

Cheers,

Nalin Pithwa

Basic Geometry: Refresher for pre-RMO, RMO and IITJEE foundation maths

1) Assume that in a \bigtriangleup ABC and \bigtriangleup RST, we know that AB=RS, AC=RT, BC=ST. Prove that \bigtriangleup ABC \cong \bigtriangleup RST without using the SSS congruence criterion.

2) Let \bigtriangleup ABC be isosceles with base BC. Then, \angle B = \angle C. Also, the median from vertex A, the bisector of \angle A, and the altitude from vertex A are all the same line. Prove this.

3) If two triangles have equal hypotenuses and an arm of one of the triangles equals an arm of the other, then the triangles are congruent. Prove.

4) An exterior angle of a triangle equals the sum of the two remote interior angles. Also, the sum of all three interior angles of a triangle is 180 degrees.

5) Find a formula for the interior angles of an n-gon.

6) Prove that the opposite sides of a parallelogram are equal.

7) In a quadrilateral ABCD, suppose that AB=CD and AD=BC. Then, prove that ABCD is a parallelogram.

8) In a quadrilateral ABCD, suppose that AB=CD and AB is parallel to CD. Then, prove that ABCD is a parallelogram.

9) Prove that a quadrilateral is a parallelogram iff its diagonals bisect each other.

10) Given a line segment BC, the locus of all points equidistant from B and C is the perpendicular bisector of the segment. Prove.

11) Corollary to problem 10 above: The diagonals of a rhombus are perpendicular. Prove.

12) Let AX be the bisector of \angle A in \triangle ABC. Then, prove \frac{BX}{XC} = \frac{AB}{AC}. In other words, X divides BC into pieces proportional to the lengths of the nearer sides of the triangle. Prove.

13) Suppose that in \triangle ABC, the median from vertex A and the bisector of \angle A are the same line. Show that AB=AC.

14) Prove that there is exactly one circle through any three given non collinear points.

15) An inscribed angle in a circle is equal in degrees to one half its subtended arc. Equivalently, the arc subtended by an inscribed angle is measured by twice the angle. Prove.

16) Corollary to above problem 15: Opposite angles of an inscribed quadrilateral are supplementary. Prove this.

17) Another corollary to above problem 15: The angle between two secants drawn to a circle from an exterior point is equal in degrees to half the difference of the two subtended arcs. Prove this.

18) A third corollary to above problem 15: The angle between two chords that intersect in the interior of a circle is equal in degrees to half the sum of the two subtended arcs. Prove this.

19) Theorem (Pythagoras): If a right triangle has arms of lengths a and b and its hypotenuse has length c, then a^{2}+b^{2}=c^{2}. Prove this.

20) Corollary to above theorem: Given a triangle ABC, the angle at vertex C is a right angle iff side AB is a diameter of the circumcircle. Prove this.

21) Theorem: The angle between a chord and the tangent at one of its endpoints is equal in degrees to half the subtended arc. Prove.

22) Corollary to problem 21: The angle between a secant and a tangent meeting at a point outside a circle is equal in degrees to half the difference of the subtended arcs.

23) Fix an integer, n \geq 3. Given a circle, how should n points on this circle be chosen so as to maximize the area of the corresponding n-gon?

24) Theorem: Given \bigtriangleup ABC and \bigtriangleup XYZ, suppose that \angle A = \angle X and \angle B= \angle Y. Then, prove that \angle C = \angle Z and so \bigtriangleup ABC \sim \bigtriangleup XYZ. Prove this theorem.

25) Theorem: If \bigtriangleup ABC \sim \bigtriangleup XYZ, then the lengths of the corresponding sides of these two triangles are proportional. Prove.

26) The following lemma is important to prove the above theorem: Let U and V be points on sides AB and AC of \bigtriangleup ABC. Then, UV is parallel to BC if and only if \frac{AU}{AB} = \frac{AV}{AC}. You will have to prove this lemma as a part of the above proof.

27) Special case of above lemma: Let U and V be the midpoints of sides AB and AC, respectively in \bigtriangleup ABC. Then, UV is parallel to BC and UV = \frac{1}{2}BC.

28) Suppose that the sides of \bigtriangleup ABC are proportional to the corresponding sides of \bigtriangleup XYZ. Then, \bigtriangleup ABC \sim \bigtriangleup XYZ.

29) Given \bigtriangleup ABC and \bigtriangleup XYZ, assume that \angle X = \angle A and that \frac{XY}{AB} = \frac{XZ}{AC}. Then, \bigtriangleup ABC \sim \bigtriangleup XYZ.

30) Consider a non-trivial plane geometry question now: Let P be a point outside of parallelogram ABCD and \angle PAB = \angle PCB. Prove that \angle APD = \angle CPB.

31) Given a circle and a point P not on the circle, choose an arbitrary line through P, meeting the circle at points X and Y. Then, the quantity PX.PY depends only on the point P and is independent of the choice of the line through P.

32) You can given an alternative proof of Pythagoras’s theorem based on the following lemma: Suppose \bigtriangleup ABC is a right triangle with hypotenuse AB and let CP be the altitude drawn to the hypotenuse. Then, \bigtriangleup ACP \sim \bigtriangleup ABC \sim \bigtriangleup CBP. Prove both the lemma and based on it produce an alternative proof of Pythagorean theorem.

33) Prove the following: The three perpendicular bisectors of the sides of a triangle are concurrent at the circumcenter of the triangle.

34) Prove the law of sines.

35) Let R and K denote the circumradius and area of \bigtriangleup ABC, respectively and let a, b and c denote the side lengths, as usual. Then, 4KR = abc.

36) Theorem: The three medians of an arbitrary triangle are concurrent at a point that lies two thirds of the way along each median from the vertex of the triangle toward the midpoint of the opposite side.

37) Time to ponder: Prove: Suppose that in \bigtriangleup ABC, medians BY and CZ have equal lengths. Prove that AB=AC.

38) If the circumcenter and the centroid of a triangle coincide, then the triangle must be equilateral. Prove this fact.

39) Assume that \bigtriangleup ABC is not equilateral and let G and O be its centroid and circumcentre respectively. Let H be the point on the Euler line GO that lies on the opposite side of G from O and such that HG = 2GO. Then, prove that all the three altitudes of \bigtriangleup ABC pass through H.

40) Prove the following basic fact about pedal triangles: The pedal triangles of each of the four triangles determined by an orthic quadruple are all the same.

41) Prove the following theorem: Given any triangle, all of the following points lie on a common circle: the three feet of the altitudes, the three midpoints of the sides, and the three Euler points. Furthermore, each of the line segments joining an Euler point to the midpoint of the opposite side is a diameter of this circle.

42) Prove the following theorem and its corollary: Let R be the circumradius of triangle ABC. Then, the distance from each Euler point of \bigtriangleup ABC to the midpoint of the opposite side is R, and the radius of the nine-point circle of \bigtriangleup ABC is R/2. The corollary says: Suppose \bigtriangleup ABC is not a right angled triangle and let H be its orthocentre. Then, \bigtriangleup ABC, \bigtriangleup HBC, \bigtriangleup AHC, and \bigtriangleup ABH have equal circumradii.

43) Prove the law of cosines.

44) Prove Heron’s formula.

45) Express the circumradius R of \bigtriangleup ABC in terms of the lengths of the sides.

46) Prove that the three angle bisectors of a triangle are concurrent at a point I, equidistant from the sides of the triangle. If we denote the by r the distance from I to each of the sides, then the circle of radius r centered at I is the unique circle inscribed in the given triangle. Note that in order to prove this, the following elementary lemma is required to be proved: The bisector of angle ABC is the locus of points P in the interior of the angle that are equidistant from the sides of the triangle.

47) Given a triangle with area K, semiperimeter s, and inradius r, prove that rs=K. Use this to express r in terms of the lengths of the sides of the triangle.

Please be aware that the above set of questions is almost like almost like a necessary set of pre-requisites for RMO geometry. You have to master the basics first.

Regards,

Nalin Pithwa.

High School Geometry: Chapter I: Basic Theorem

If two chords of a circle intersect anywhere. at any angle, what can be said about the segments of each cut off by the other? The data seem to be too for any conclusion, yet an important and far reaching theorem can be formulated from only this meager amount of “given”:

Theorem 2. 

If two chords intersect, the product of the segments of the one equals the product of the segments of the other.

This theorem says that in Fig. 3A (attached as a jpeg picture), PA.PB=PC.PD (where the dot is the symbol for algebraic multiplication). What do we mean when we talk about “the product of two line segments”? (There is a way to multiply two lines with a ruler and a compass, but we won’t discuss it here.) What the theorem means is that the product of the respective lengths of the segments are equal. Whenever we put a line segment like PA into an equation, we shall mean the length of PA. (Although we shall soon have to come to grips with the idea of a “negative length” for the present we make no distinction between the length of PA and the length of AP; they are the same positive number).

The Greek geometers took great care to enumerate the different cases of the above theorem. Today, we prefer, when possible, to treat all variants together in one compact theorem. In Fig 3A, the chords intersect inside the circle, in 3B outside the circle, and in 3C one of the chords has become a tangent. Theorem 2 holds in all three cases, and the proofs are so much alike that one proof virtually goes for all.

You may object, and say that in Fig. 3B the chords don’t intersect. But, they do when extended, and in these blogs, we shall say that one line intersects a second when it in fact only intersects the extension of the second line. This is just part of a terminology, in quite general use today, that may be slightly more free-wheeling than what you have been accustomed to. In the same vein, P divided AB internally in Fig 3A, whereas in Fig 3B, P is said to divide the chord AB externally, and we still talk about the two segments PA and PB.

To prove theorem 2, we need to construct two lines, indicated in Fig 3A. Perhaps, you should draw them also in figures 3B and 3C, and, following the proof letter for letter, see for yourself how few changes are required to complete the proofs of these diagrams. In Fig 3A, \angle {1}= \angle {2} because they are inscribed in the same circular arc. And, \angle {5} = \angle {4}. Therefore, triangles PCA and PBD are similar, and hence have proportional sides:

\frac {PA}{PD} = \frac {PC}{PB} or PA. PB = PC.PD

In the next blog, we will discuss means and another basic theorem. theorem2Fig3

Till then, aufwiedersehen,

Nalin Pithwa

The Division Algorithm

One theorem, the Division Algorithm, acts as the foundation stone upon which our whole development of Number Theory rests. The result is familiar to most of us; roughly, it asserts that an integer a can be “divided” by a positive integer b in such a way that the remainder is smaller than b. The exact statement is given below:

Division Algorithm.

Given integers a and b, with b>0, there exist unique integers q and r satisfying

a=qb+r, 0 \leq r < b.

The integers q and r, are called, respectively, the quotient and remainder in the division of a by b.

Example.

We propose to show that the expression a(a^{2}+2)/3 is an integer for all a \geq 1. According to the Division Algorithm, every a is of the form 3q, 3q+1, or 3q+2. Assume the first of these cases. Then,

\frac{a(a^{2}+2)}{3}=q(9q^{2}+2), which clearly is an integer. Similarly, if a=3q+1, then

\frac{(3q+1)((3q+1)^{2}+2)}{3}=(3q+1)(3q^{2}+2q+1)

and a(a^{2}+2)/3 is an integer in this instance also. Finally, for a=3q+2, we obtain

\frac{(3q+2)((3q+2)^{2}+2)}{3}=(3q+2)(3q^{2}+4q+2)

an integer once more. Consequently, our result is established in all cases.

More later,

Nalin Pithwa

 

 

Early Number Theory

Inegral numbers are the fountainhead of all mathematics. —- H. MINKOWSKI.

Before becoming weighed down with detail, we should say a few words about the origin of number theory. The theory of numbers is one of the oldest branches of mathematics; an enthusiast, by stretching a point here and there, could extend its roots back to a surprisingly remote date. Although it seems probable that the Greeks were largely indebted to Babylonians and ancient Egyptians for a core of information about the properties of natural numbers, the first rudiments of an actual theory are generally credited to Pythagoras and his disciples.

Our knowledge of the life of Pythagoras is very scanty, and little can be said with any certainty. According to the best estimates, he was born between 580 and 562BC on the Aegean island of Samos. It seems that he studied not only in Egypt, but may even have extended his journeys as far east as Babylonia. When Pythagoras reappeared after years of wandering, he sought out a favorable place for a school and finally settled upon Croton, a prosperous Greek settlement on the heel of the Italian boot. The school concentrated on four mathemata, or subjects of study: arithmetica (arithmetic in the sense of number theory rather than the art of calculating), harmonia (music), geometria (geometry), and astrologia (astronomy). This four field division of knowledge became known in the Middle Ages as the quadrivium, to which was added the trivium of logic, grammar and rhetoric. These seven liberal arts came to be looked upon as the necessary course of study for an educated person.

Pythagoras divided those who attended his lectures into two groups — the Probationers (or listeners) and the Pythagoreans. After three years in the first class, a listener could be initiated in the second class, to whom were confided the main discoveries of the school. The Pythagoreans were a closely knit brotherhood, holding all worldly goods in common and bound by an oath not to reveal the founder’s secrets. Legend has it that a talkative Pythagorean was drowned in a shipwreck as the gods’ punishment for publicly boasting that he added the dodecahedron to the number of regular solids enumerated by Pythagoras. For a time, the autocratic Pythagoreans succeeded in dominating the local government in Croton, but a popular revolt in 501BC led to the murder of many of its prominent members, and Pythagoras himself was killed shortly thereafter. Although the political influence of the Pythagoreans was thus destroyed, they continued to exist for at least two centuries more as a philosophical and mathematical society. To the end, they remained a secret order, publishing nothing and, with noble self-denial, ascribing all their discoveries to the Master.

The Pythagoreans believed that the key for an explanation of the universe lay in number and form, their general thesis being that “Everything is Number.” By number, they meant of course a positive integer. For a rational understanding of nature, they considered it sufficient to analyze the properties of certain numbers. Pythagoras himself, we are told “seemed to have attached supreme importance to the study of arithmetic, which he advanced and took out of the realm of commercial utility.”

The Pythagorean doctrine is a curious mixture of cosmic philosophy and number mysticism, a sort of supernumerology that assigned to everything material or spiritual a definite integer. Among their writings, we find that 1 represented reason, for reason could produce only one consistent body of truth; 2 stood for man and 3 for woman; 4 was the Pythagorean symbol for justice, being the first number that is the product of equals; 5 was identified with marriage, because it is formed by the union of 2 and 3; and so forth. All the even numbers, after the first one, were capable of separation into other numbers, hence, they were prolific and were considered as feminine and earthy — and somewhat less highly regarded in general. Being a predominantly male society, the Pythagoreans classified the odd numbers, after the first two, as masculine and divine.

Although these speculations about numbers as models of “things” appear frivolous today, it must be borne in mind that the intellectuals of the classical Greek period were largely absorbed in philosophy and that these same men, because they had such intellectual interests, were the very ones who were engaged in laying the foundations for mathematics as a system of thought. To Pythagoras and his followers, mathematics was largely a means to an end, the end being philosophy. Only with the founding of the School of Alexandria do we enter a new phase in which the cultivation of mathematics was pursued for its own sake.

It was at Alexandria, not Athens, that a science of numbers divorced from mystic philosophy first began to develop. For nearly a thousand years, until its destruction by the Arabs in 641AD, Alexandria stood at the cultural and commercial center of the Hellenistic world. (After the fall of Alexandria, most of its scholars migrated to Constantinople. During the next 800 years, while formal learning in the West all but disappeared, this enclave at Constantinople preserved for us the mathematical works of the various Greek schools.) The so-called Alexandrian Museum, a forerunner of the modern university, brought together the leading poets and scholars of the day; adjacent to it there was established an enormous library, reputed to hold over 700,000 vokumes — hand-copied at its height. Of all the distinguished names connected with the Museum, that of Euclid (circa 300BC), founder of the School of Mathematics, is in a special class. Posterity has come to know him as the author of the Elements, the oldest Greek treatise on mathematics to reach us in its entirety. The Elements is a compilation of much of the mathematical knowledge available at that time, organized into thirteen parts or Books, as they are called. The name of Euclid is so often associated with geometry that one tends to three of the Books were devoted to Number Theory.

Euclid’s Elements constitutes one of the great success stories of world literature. Scarcely, any other book save the Bible has been more widely circulated or studied. Over a thousand editions of it have appeared since the first printed version in 1482 and before its printing, manuscript copies dominated much of the teaching of mathematics in Western Europe. Unfortunately, no copy of the work has been found that actually dates from Euclid’s own time, the modern editions are descendants of a television prepared by Theon of Alexandria, a commentator of the 4th century AD.

More later,

Nalin Pithwa

 

 

High School Geometry: Chapter 1: Introduction: part 1

IMAG0267A practical problem:

The owner of a drive-in theatre has been professionally informed on the optimum angle \theta that the screen AB should present to the viewer (Figure 1). But, only one customer can sit in the preferred spot V, directly in front of the screen. The owner is interested in locating other points, U, from which the screen subtends the same angle \theta.

The answer is the circle passing thru the three points A, B and V and the reason is:

Theorem 1:

An angle inscribed in a circle is measured by half the intercepted arc.

Inasmuch as \angle{ABC} is measured by the same arc as \angle{AVB}, the angle at U is the same as the angle at V. (If you want a proof of the theorem, just let me know).

A typical example of a calculus problem is this: Of all triangles on the same base and with the same vertex angle, which has the greatest area? We are able to steal the show briefly from calculus and solve this problem almost at a glance, with the aid of Theorem 1. If AB is the base and the given vertex angle is \theta, then all the triangles have their vertices lying on the circle of Figure 1. But, area= (1/2) base \times altitude.

One half the base is a constant, and the altitude (and hence, the area) is greatest for AVB, the isosceles triangle.

Even to set up the problem for the calculus approach is awkward (try it and see!) and several lines of careful calculations are required to solve it.

From Theorem 1, we also have the useful corollary that any angle inscribed in a semicircle is a right angle (measured by half the 180 \deg arc of the semicircle). If by any chance we can show that some angle ACB (Fig 2) is a right angle, then we know that a semicircle on AB as a diameter must pass through C. This is the converse of the corollary.

More later,

Nalin Pithwa