A fifth primer: plane geometry tutorial for preRMO and RMO: core stuff

  1. Show that three straight lines which join the middle points of the sides of a triangle, divide it into four triangles which are identically equal.
  2. Any straight line drawn from the vertex of a triangle to the base is bisected by the straight line which joins the middle points of the other sides of the triangle.
  3. ABCD is a parallelogram, and X, Y are the middle points of the opposite sides AD, BC: prove that BX and DY trisect the diagonal AC.
  4. If the middle points of adjacent sides of any quadrilateral are joined, the figure thus formed is a parallelogram. Prove this.
  5. Show that the straight lines which join the middle points of opposite sides of a quadrilateral bisect one another.
  6. From two points A and B, and from O the mid-point between them, perpendiculars AP, and BQ, OX are drawn to a straight line CD. If AP, BQ measure respectively 4.2 cm, and 5.8 cm, deduce the length of OX. Prove that OX is one half the sum of AP and BQ. or \frac{1}{2}(AP-BQ) or \frac{1}{2}(BQ-AP) according as A and B are on the same side or on opposite sides of CD.
  7. When three parallels cut off equal intercepts from two transversals, prove that of three parallel lengths between the two transversals the middle one is the Arithmetic Mean of the other two.
  8. The parallel sides of a trapezium are a cm and b cm respectively. Prove that the line joining the middle points of the oblique sides is parallel to the parallel sides, and that its length is \frac{1}{2}(a+b) cm.
  9. OX and OY are two straight lines, and along OX five points 1,2,3,4,5 are marked at equal distances. Through these points parallels are drawn in any direction to meet OY. Measure the lengths of these parallels : take their average and compare it with the lengths of the third parallel. Prove geometrically that the third parallel is the mean of all five.
  10. From the angular points of a parallelogram perpendiculars are drawn to any straight line which is outside the parallelogram : prove that the sum of the perpendiculars drawn from one pair of opposite angles is equal to the sum of those drawn from the other pair.  (Draw the diagonals,and from their point of intersection suppose a perpendicular drawn to the given straight line.)
  11. The sum of the perpendiculars drawn from any point in the base of an isosceles triangle to the equal to the equal sides is equal to the perpendicular drawn from either extremity of the base to the opposite side. It follows that the sum of the distances of any point in the base of an isosceles triangle from the equal sides is constant, that is, the same whatever point in the base is taken).
  12. The sum of the perpendiculars drawn from any point within the an equilateral triangle to the three sides is equal to the perpendicular drawn from any one of the angular points to the opposite side, and is therefore, constant. Prove this.
  13. Equal and parallel lines have equal projections on any other straight line. Prove this.

More later,

Cheers,

Nalin Pithwa.

A fourth primer: plane geometry question set including core theorems, preRMO and RMO

Hard core definitions of various special quadrilaterals:

  1. A quadrilateral is a plane figure bounded by four straight lines.
  2. A parallelogram is a quadrilateral whose opposite sides are parallel.
  3. A rectangle is a parallelogram which has one of its angles a right angle.
  4. A square is a rectangle which has two adjacent sides equal.
  5. A rhombus is a quadrilateral which has all its sides equal, but its angles are not right angles.
  6. A trapezium if a quadrilateral which has one pair of parallel sides.

Problem 1:

The straight lines which join the extremities of two equal and parallel straight lines towards the same parts are themselves equal and parallel. Prove this.

Problem 2:

The opposite sides and angles of a parallelogram are equal to one another, and each diagonal bisects the parallelogram.

Problem 3:

Corollary 1 of problem 2 above: If one angle of a parallelogram is a right angle, all its angles are right angles.

Corollary 2 of problem 2 above: All the sides of a square are equal; and all its angles are right angles.

Corollary 3 of problem 3 above: The diagonals of a parallelogram bisect one another.

Problem 4:

If the opposite sides of a quadrilateral are equal, then the figure is a parallelogram.

Problem 5:

If the opposite angles of a quadrilateral are equal, then the figure is a parallelogram.

Problem 6:

If the diagonals of a quadrilateral bisect each other, then the figure is a parallelogram.

Problem 7:

The diagonals of a rhombus bisect each other at right angles.

Problem 8:

If the diagonals of a parallelogram are equal, all its angles are right angles.

Problem 9:

In a parallelogram which is not rectangular, the diagonals are not equal.

Problem 10:

Any straight line drawn through the middle point of a diagonal of a parallelogram and terminated by a pair of opposite sides is bisected at that point.

Problem 11:

In a parallelogram, the perpendiculars drawn from one pair of opposite angles to the diagonal which joins the other pair are equal. Prove this.

Problem 12:

If ABCD is a parallelogram, and X, Y respectively the middle points of the sides AD, BC, show that the figure AYCX is a parallelogram.

Problem 13:

ABC and DEF are two triangles such that AB, BC are respectively equal to and parallel to DE, EF; show that AC is equal and parallel to DF.

Problem 14:

ABCD is a quadrilateral in which AB is parallel to DC, and AD equal but not parallel to BC; show that (i) angle A + angle C = 180 degrees = angle B + angle D; (ii) diagonal AC = diagonal BD (iii) the quadrilateral is symmetrical about the straight line joining the middle points of AB and DC.

Problem 15:

AP, BQ are straight rods of equal length, turning at equal rates (both clockwise) about two fixed pivots A and B respectively. If the rods start parallel but pointing in opposite senses, prove that (i) they will always be parallel (ii) the line joining PQ will always pass through a certain fixed point.

Problem 16:

A and B are two fixed points, and two straight lines AP, BQ, unlimited towards P and Q, are pivoted at A and B. AP, starting from the direction AB, turns about A clockwise at the uniform rate of 7.5 degrees a second; and BQ, starting simultaneously from the direction BA, turns about B counter-clockwise at the rate of 3.75 degrees a second. (i) How many seconds will elapse before AP and BQ are parallel? (ii) Find graphically and by calculation the angle between AP and BQ twelve seconds from the start. (iii) At what rate does this angle decrease?

Problem 17 (Intercept theorem or Basic Proportionality Theorem):

If there are three or more parallel straight lines, and the intercepts made by them on any transversal are equal, then the corresponding intercepts on any other transversal are also equal.

Prove the corollary: In a triangle ABC, if a set of lines Pp, Qq, Rr, \ldots, drawn parallel to the base, divide one side AB into equal parts they also divide the other side AC into equal parts.

Definition:

If from the extremities of a straight line AB perpendiculars AX, BY are drawn parallel to a straight line PQ of indefinite length, then XY is said to be the orthogonal projection of AB on PQ.

Problem 18:

Prove: The straight line drawn through the middle point of a side of a triangle parallel to the base bisects the remaining side.

Problem 19:

The straight line which joins the middle points of two sides of a triangle is equal to half the third side.

More later,

Nalin Pithwa.

 

 

 

 

A third primer: plane geometry question set for preRMO and RMO

Problem 1:

Prove: Two right angled triangles which have their hypotenuses equal, and one side of one equal to one side of the other, are equal in all respects. (PS: Please do not use “short-cut or the magic formula — Pythagoras’s theorem; if you do, you are requested to prove Pythagoras’s theorem using plane geometry!) (PS:2: Remember Euclid’s geometry builds up from “scratch”…a small step at a time….axioms, definitions, lemmas, …; you can only use the theorems used and proved so far in “class”).

Problem 2:

If two triangles have two sides of the one equal to two sides of the other, each to each, but the angle included by the two sides of one greater than the angle included by the corresponding sides of the other; then the base of that which has the greater angle is greater than the base of the other.

Problem 3:

Prove converse of the above: If two triangles have two sides of the one equal to two sides of the other, each to each, but the base of one greater than the base of the other, then the angle contained by the sides of that which has the greater base is greater than the angle contained by the corresponding sides of the other.

Problem 4:

(i) The perpendicular is the shortest line that can be drawn to a given straight line from a given point. (ii) Obliques which make equal angles with the perpendicular are equal. (iii) Of two obliques the less is that which makes the smaller angle with the perpendicular. Prove all these.

Problem 5:

If two triangles have two sides of the one equal to two sides of the other, each to each, and have likewise the angles opposite to one pair of equal sides equal, then the angles opposite to the other pair of equal sides are either equal or supplementary, and in the former case, the triangles are equal in all respects.

More later,

Cheers,

Nalin Pithwa

A primer for preRMO and RMO plane geometry with basic exercises

Plane geometry is axiomatic deductive logic. I present a quick mention/review of “proofs” which can be “derived” in sequence….building up the elementary theorems …so for example, if there is a question like: prove that the three medians of a triangle are concurrent, please do not use black magic complicated machinery like Ceva’s theorem,etc; or even if say, the question asks you to prove Ceva’s theorem only, you have to prove it using elementary theorems like the ones presented below:

For the present purposes, I am skipping axioms and basic definitions and hypothetical constructions. I am using straight away the reference (v v v old text) : A School Geometry, Metric Edition by Hall and Stevens. (available almost everywhere in India):

Theorem 1:

The adjacent angles which one straight line makes with another straight line on one side of it are together equal to two right angles.

Corollary 1 of Theorem 1:

if two straight lines cut another, the four angles so formed are together equal to four right angles.

Corollary 2 of Theorem 1:

When any number of straight lines meet at a point, the sum of the consecutive angles so formed is equal to four right angles.

Corollary 3 of Theorem 1:

(a) Supplements of the same angle are equal. (ii) Complements of the same angle are equal.

Theorem 2 (converse of theorem 1):

If, at a point in a straight line, two other straight lines, on opposite sides of it, make the adjacent angles together equal to two right angles, then these two straight lines are in one and the same straight line.

Remark: this theorem can be used to prove stuff like three points are in a straight line.

Theorem 3:

If two straight lines cut one another, the vertically opposite angles are equal.

Theorem 4: SAS Test of Congruence of Two Triangles:

If two triangles have two sides of the one equal to two sides of the other, each to each, and the angles included by those sides equal, then the triangles are equal in all respects.

Theorem 5:

The angles at the base of an isosceles triangle are equal.

Corollary 1 of Theorem 5:

If the equal sides AB, AC of an isosceles triangle are produced, the exterior angles EBC, FCB are equal, for they are the supplements of the equal angles at the base.

Corollary 2 of Theorem 5:

If a triangle is equilateral, it is also equiangular.

Theorem 6:

If two angles of a triangle are equal to one another, then the sides which are opposite to the equal angles are equal to one another.

Corollary of Theorem 6:

Hence, if a triangle is equiangular, it is also equilateral.

Theorem 7 (SSS Test of Congruence of Two Triangles):

If two triangles have the three sides of the one equal to the three sides of the others, each to each, they are equal in all respects.

Theorem 8: 

If one side of a triangle is produced then the exterior angle is greater than either of the interior opposite angles.

Corollary 1 to Theorem 8:

Any two angles of a triangle are together less than two right angles.

Corollary 2 to Theorem 8:

Every triangle must have at least two acute angles.

Corollary 3 to Theorem 8:

Only one perpendicular can be drawn to a straight line from a given point outside it.

Theorem 9 :

If one side of a triangle is greater than another, then the angle opposite of the greater side is greater than the angle opposite to the less.

Theorem 10:

If one angle of a triangle is greater than another, then the side opposite to the greater angle is greater than the side opposite to the less.

Theorem 11: Triangle Inequality:

Any two sides of a triangle are together greater than the third side.

Theorem 12: Another inequality sort of theorem:

Of all straight lines drawn from a given point to a given straight line the perpendicular is the least.

Corollary 1 to Theorem 12:

Hence, conversely, since there can be only one perpendicular and one shortest line from O to AB: if OC is the shortest straight line from O to AB, then OC is perpendicular to AB.

Corollary 2 to Theorem 12:

Two obliques OP, OQ which cut AB at equal distance from C, the foot of the perpendicular are equal.

Corollary 3 to Theorem 12:

Of two obliques OQ, OR, if OR cuts AB at the greater distance from C. the foot of the perpendicular, then OR is greater than OQ.

Theorem 13 :

If a straight line cuts two other straight lines so as to make: (i) the alternate angles equal or (ii) an exterior angle equal to the interior opposite angle on the same side of the cutting line or (iii) the interior angles on the same side equal to two right angles, then in each case, the two straight lines are parallel.

Theorem 14:

If a straight line cuts two parallel lines, it makes : (i) the alternate angles equal to one another; (ii) the exterior angle equal to the interior opposite angle on the same side of the cutting line (iii) the two interior angles on the same side together equal to two right angles.

Theorem 15:

Straight lines which are parallel to the same straight line are parallel to one another.

Theorem 16:

Sum of three interior angles of a triangle is 180 degrees.

Also, if a side of a triangle is produced the exterior angle is equal to the sum of the two interior opposite angles.

Corollary 1:

All the interior angles of one rectilinear figure, together with four right angles are equal to twice as many right angles as the figure has sides.

Corollary 2:

If the sides of a rectilinear figure, which has no reflex angle, are produced in order, then all the exterior angles so formed are together equal to four right angles.

Theorem 17: AAS test of congruence of two triangles:

If two triangles have two angles of one equal to two angles of the other, each to each, and any side of the first equal to the corresponding side of the other, the triangles are equal in all respects.

Theorem 18:

Two right angled triangles which have their hypotenuses equal, and one side of one equal to one side of the other are equal in all respects.

Theorem 19:

If two triangles have two sides of the one equal to two sides of the other, each to each, but the angle included by the two sides of one greater than the angle included by the two corresponding sides of the other, then the base of that which has the greater angle is greater than the base of the other.

Conversely, 

if two triangles have two sides of the one equal to two sides of the other, each to each, but the base of one greater than the base of the other, then the angle contained by the sides of that which has the greater base is greater than the angle contained by the corresponding sides of the other.

Theorem 20:

The straight lines which join the extremities of two equal and parallel straight lines towards the same parts are themselves equal and parallel.

Theorem 21:

The opposite sides and angles of a parallelogram are equal to one another, and each diagonal bisects the parallelogram.

Corollary 1 to Theorem 21:

If one angle of a parallelogram to a right angle, all its angles are equal.

Corollary 2 to Theorem 21:

All the sides of a square are equal and all its angles are right angles.

Corollary 3 to Theorem 21:

The diagonals of a parallelogram bisect each other.

Theorem 22:

If there are three or more parallel straight lines, and the intercepts made by them on any transversal are equal, then the corresponding intercepts on any other transversal are also equal.

Tutorial exercises based on the above:

Problem 1: In the triangle ABC, the angles ABC, ACB are given equal. If the side BC is produced both ways, show that the exterior angles so formed are equal.

Problem 2: In the triangle ABC, the angles ABC, ACB are given equal. If AB and AC are produced beyond the base, show that the exterior angles so formed are equal.

Problem 3: Prove that the bisectors of the adjacent angles which one straight line makes with another contain a right angle. That is to say, the internal and external bisectors of an angle are at right angles to one another.

Problem 4: If from O a point in AB two straight lines OC, OD are drawn on opposite sides of AB so as to make the angle COB equal to the angle AOD, show that OC and OD are in the same straight line.

Problem 5: Two straight lines AB, CD cross at O. If OX is the bisector of the angle BOD, prove that XO produced bisects the angle AOC.

Problem 6: Two straight lines AB, CD cross at O. If the angle BOD is bisected by OX, and AOC by OY, prove that OX, OY are in the same straight line.

Problem 7: Show that the bisector of the vertical angle of an isosceles triangle (i) bisects the base (ii) is perpendicular to the base.

Problem 8: Let O be the middle point of a straight line AB, and let OC be perpendicular to it. Then, if P is any point in OC, prove that PA=PB.

Problem 9: Assuming that the four sides of a square are equal, and that its angles are all right angles, prove that in the square ABCD, the diagonals AC, BD are equal.

Problem 10: Let ABC be an isosceles triangle: from the equal sides AB, AC two equal parts AX, AY are cut off, and BY and CX are joined. Prove that BY=CX.

Problem 11: ABCD is a four-sided figure whose sides are all equal, and the diagonal BD is drawn : show that (i) the angle ABD = the angle ADB (ii) the angle CBD = the angle CDB (iii) the angle ABC = the angle ADC.

Problem 12: ABC, DBC are two isosceles triangles drawn on the same base BC, but on opposite sides of it: prove that the angle ABD = the angle ACD.

Problem 13: ABC, DBC are two isosceles triangles drawn on the same base BC, but on the same side of it: prove that the angle ABD = the angle ACD.

Problem 14: AB, AC are the equal sides of an isosceles triangle ABC, and L, M, N are the middle points of AB, BC and CA respectively; prove that (i) LM = NM (ii) BN = CL (iii) the angle ALM = the angle ANM.

Problem 15: Show that the straight line which joins the vertex of an isosceles triangle to the middle points of the base (i) bisects the vertical angle (ii) is perpendicular to the base.

Problem 16: If ABCD is a rhombus, that is, an equilateral four sided figure, show by drawing the diagonal AC that (i) the angle ABC = the angle ADC (ii) AC bisects each of the angles BAD and BCD.

Problem 17: If in a quadrilateral ABCD the opposite sides are equal, namely, AB = CD and AD=CB, prove that the angle ADC = the angle ABC.

Problem 18: If ABC and DBC are two isosceles triangles drawn on the same base BC, prove that the angle ABD = the angle ACD, taking (i) the case where the triangles are on the same side of BC (ii) the case where they are on the opposite sides of BC.

Problem 19: If ABC, DBC are two isosceles triangles drawn on opposite sides of the same base BC, and if AD be joined, prove that each of the angles BAC, BDC will be divided into two equal parts.

Problem 20: Show that the straight lines which join the extremities of the base of an isosceles triangle to the middle points of the opposite sides are equal to one another.

Problem 21: Two given points in the base of an isosceles triangle are equidistant from the extremities of the base: show that they are also equidistant from the vertex.

Problem 22: Show that the triangle formed by joining the middle points of the sides of an equilateral triangle is also equilateral.

Problem 23: ABC is an isosceles triangle having AB equal to AC, and the angles at B and C are bisected by BC and CO: prove that (i) BO = CO (ii) AO bisects the angle BAC.

Problem 24: Show that the diagonals of a rhombus bisect one another at right angles.

Problem 25: The equal sides BA, CA of an isosceles triangle BAC are produced beyond the vertex A to the points E and F, so that AE is equal to AF and FB, EC are joined: prove that FB is equal to EC.

Problem 26: ABC is a triangle and D any point within it. If BD and CD are joined, the angle BDC is greater than the angle BAC. Prove this (i) by producing BD to meet AC (ii) by joining AD, and producing it towards the base.

Problem 27: If any side of a triangle is produced both ways, the exterior angles so formed are together greater than two right angles.

Problem 28: To a given straight line, there cannot be drawn from a point outside it more than two straight lines of the same given length.

Problem 29: If the equal sides of an isosceles triangle are produced, the exterior angles must be obtuse.

Note: The problems 30 to 43 are based on triangle inequalities:

Problem 30: The hypotenuse is the greatest side of a right angled triangle.

Problem 31: The greatest side of any triangle makes acute angles with each of the other sides.

Problem 32: If from the ends of a side of a triangle, two straight lines are drawn to a point within the triangle, then those straight lines are together less than the other two sides of the triangle.

Problem 33: BC, the base of an isosceles triangle ABC is produced to any point D; prove that AD is greater than either of the equal sides.

Problem 34: If in a quadrilateral the greatest and least sides are opposite to one another, then each of the angles adjacent to the least side is greater than its opposite angle.

Problem 35: In a triangle, in which OB, OC bisect the angles ABC, ACB respectively: prove that if AB is greater than AC, then OB is greater than OC.

Problem 36: The difference of any two sides of a triangle is less than the third side.

Problem 37: The sum of the distances of any point from the three angular points of a triangle is greater than half its perimeter.

Problem 38: The perimeter of a quadrilateral is greater than the sum of its diagonals.

Problem 39: ABC is a triangle, and the vertical angle BAC is bisected by a line which meets BC in X, show that BA is greater than BX, and CA greater than CX. Obtain a proof of the following theorem : Any two sides of a triangle are together greater than the third side.

Problem 40: The sum of the distance of any point within a triangle from its angular points is less than the perimeter of the triangle.

Problem 41: The sum of the diagonals of a quadrilateral is less than the sum of the four straight lines drawn from the angular points to any given point. Prove this, and point out the exceptional case.

Problem 42: In a triangle any two sides are together greater than twice the median which bisects the remaining side.

Problem 43: In any triangle, the sum of the medians is less than the perimeter.

Problem 44: Straight lines which are perpendicular to the same straight line are parallel to one another.

Problem 45: If a straight line meets two or more parallel straight lines, and is perpendicular to one of them, it is also perpendicular to all the others.

Problem 46: Angles of which the arms are parallel each to each are either equal or supplementary.

Problem 47: Two straight lines AB, CD bisect one another at O. Show that the straight line joining AC and BD are parallel.

Problem 48: Any straight line drawn parallel to the base of an isosceles trianlge makes equal angles with the sides.

More later. Get cracking. This perhaps the simplest introduction, step by step, to axiomatic deductive logic…discovered by Euclid about 2500 years before ! Hail Euclid !

Cheers,

Nalin Pithwa

Why do we need proofs? In other words, difference between a mathematician, physicist and a layman

Yes, I think it is a very nice question, which kids ask me. Why do we need proofs? Well, here is a detailed explanation (I am not mentioning the reference I use here lest it may intimidate my young enthusiastic, hard working students or readers. In other words, the explanation is not my own; I do not claim credit for this…In other words, I am just sharing what I am reading in the book…)

Here it goes:

What exactly is the difference between a mathematician, a physicist, and a layman? Let us suppose that they all start measuring the angles of hundreds of triangles of various shapes, find the sum in each case and keep a record. Suppose the layman finds that with one or two exceptions, the sum in each case comes out to be 180 degrees. He will ignore the exceptions and say “the sum of the three angles in a triangle  is 180 degrees.” A physicist will be more cautious in dealing with the exceptional cases. He will examine them more carefully. If he finds that the sum in them is somewhere between 179 degrees to 180 degrees, say, then he will attribute the deviation to experimental errors. He will then state a law: The sum of three angles of any triangle is 180 degrees. He will then watch happily as the rest of the world puts his law to test and finds that it holds good in thousands of different cases, until somebody comes up with a triangle in which the law fails miserably. The physicist now has to withdraw his law altogether or else to replace it by some other law which holds good in all cases tried. Even this new law may have to be modified at a later date. And, this will continue without end.

A mathematician will be the fussiest of all. If there is even a single exception he will refrain from saying anything. Even when millions of triangles are tried without a single exception, he will not state it as a theorem that the sum of the three angles in ANY triangle is 180 degrees. The reason is that there are infinitely many different types of triangles. To generalize from a million to infinity is as baseless to a mathematician as to generalize from one to a million. He will at the most make a conjecture and say that there is a strong evidence suggesting that the conjecture is true. But that is not the same thing as a proving a theorem. The only proof acceptable to a mathematician is the one which follows from earlier theorems by sheer logical implications (that is, statements of the form : If P, then Q). For example, such a proof follows easily from the theorem that an external angle of a triangle is the sum of the other two internal angles.

The approach taken by the layman or the physicist is known as the inductive approach whereas the mathematician’s approach is called the deductive approach. In the former, we make a few observations and generalize. In the latter, we deduce from something which is already proven. Of course, a question can be raised as to on what basis this supporting theorem is proved. The answer will be some other theorem. But then the same question can be asked about the other theorem. Eventually, a stage is reached where a certain statement cannot be proved from any other earlier proved statement(s) and must, therefore, be taken for granted to be true. Such a statement is known as an axiom or a postulate. Each branch of math has its own axioms or postulates. For examples, one of the axioms of geometry is that through two distinct points, there passes exactly one line. The whole beautiful structure of geometry is based on 5 or 6 axioms such as this one. Every theorem in plane geometry or Euclid’s Geometry can be ultimately deduced from these axioms.

PS: One of the most famous American presidents, Abraham Lincoln had read, understood and solved all of Euclid’s books (The Elements) by burning mid-night oil, night after night, to “sharpen his mental faculties”. And, of course, there is another famous story (true story) of how Albert Einstein as a very young boy got completely “addicted” to math by reading Euclid’s proof of why three medians of a triangle are concurrent…(you can Google up, of course).

Regards,

Nalin Pithwa

 

 

Axiomatic Method : A little explanation

I) Take an English-into-English dictionary (any other language will also do). Start with any word and note down any word occurring in its definition, as given in the dictionary. Take this new word and note down any word appearing in it until a vicious circle results. Prove that a vicious circle is unavoidable no matter which word one starts with , (Caution: the vicious circle may not always involve the original word).

For example, in geometry the word “point” is undefined. For example, in set theory, when we write or say : a \in A ; the element “a” ‘belongs to’ “set A” —- the word “belong to” is not defined.

So, in all branches of math or physics especially, there are such “atomic” or “undefined” terms that one starts with.

After such terms come the “axioms” — statements which are assumed to be true; that is, statements whose proof is not sought.

The following are the axioms based on which equations are solved in algebra:

  1. If to equals we add equals, we get equals.
  2. If from equals we take equals, the remainders are equal.
  3. If equals are multiplied by equals, the products are equal.
  4. If equals are divided by equals (not zero), the quotients are equal.

More later,

Nalin Pithwa.

Pre-RMO training; a statement and its converse; logic and plane geometry

I hope the following explanation is illuminating to my readers/students:

How to prove that two lines are parallel ? (Note that we talk of parallel lines only when they lie in the same plane; on the other hand: consider the following scenario — your study table and the floor on which it stands. Let us say you draw a straight line AB on your study table and another line PQ on the floor on which the study table is standing; then, even though lines AB and PQ never meet, we do not say that they are parallel because they lie in different planes. Such lines are called skew lines. They are dealt with in solid geometry or 3D geometry or vector spaces).

Coming back to the question — when can we say that two lines are parallel?

Answer:

Suppose that a transversal crosses two other lines.

1) If the corresponding angles are equal, then the lines are parallel.
2) If the alternate angles are equal, then the lines are parallel.
3) If the co-interior angles are supplementary, then the lines are parallel.

A STATEMENT AND ITS CONVERSE

Let us first consider the following statements:

A transversal is a line that crosses two other lines. If the lines crossed by a transversal are parallel, then the corresponding angles are equal; if the lines crossed by a transversal are parallel, then the alternate interior angles are equal; if the lines crossed by a transversal are parallel, then the co-interior angles are supplementary.

The statements given below are the converses of the statement given in the above paragraph; meaning that they are formed from the former statements by reversing the logic. For example:

STATEMENT: If the lines are parallel then the corresponding angles are equal.

CONVERSE: If the corresponding angles are equal, then the lines are parallel.

Pairs such as these, a statement and its converse, occur routinely through out mathematics, and are particularly prominent in geometry. In this case, both the statement and its converse are true. It is important to realize that a statement and its converse are, in general, quite different. NEVER ASSUME THAT BECAUSE A STATEMENT IS TRUE, SO ITS CONVERSE IS ALSO TRUE. For example, consider the following:

STATEMENT: If a number is a multiple of 4, then it is even.
CONVERSE: If a number is even, then it is a multiple of 4.

The first statement is clearly true. But, let us consider the number 18. It is even. But 18 is not a multiple of 4. So, the converse is not true always.

\it Here \hspace{0.1in} is \hspace{0.1in}an \hspace {0.1in}example \hspace{0.1in}from \hspace{0.1in}surfing

STATEMENT: If you catch a wave, then you will be happy.
CONVERSE: If you are happy, then you will catch a wave.

Many people would agree with the first statement, but everyone knows that its converse is plain silly — you need skill to catch waves.

Thus, the truth of a statement has little to do with its converse. Separate justifications (proofs) are required for the converse and its statements.

Regards,
Nalin Pithwa.

Reference: (I found the above beautiful, simple, lucid explanation in the following text): ICE-EM, year 7, book 1; The University of Melbourne, Australian Curriculum, Garth Gaudry et al.

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.

Pure plane geometry problems for RMO training or practice

Similar Figures:

Definition:

Polygons which are equiangular and have their corresponding sides proportional are called similar.

If, in addition, their corresponding sides are parallel, they are said to be similarly situated or homothetic.

Theorem 1:

If O is any fixed point and ABCD…X any polygon, and if points A^{'}, B^{'}, C^{'}, \ldots X^{'} are taken on OA, OB, OC, …OX (or those lines produced either way) such that \frac{OA^{'}}{OA} = \frac{OB^{'}}{OB} = \ldots = \frac{OX^{'}}{OX}, then the polygons ABCD...X and A^{'}B^{'}C^{'} \ldots X^{'} are homothetic.

Before we state the next theorem, some background is necessary.

If O is a fixed point and P is a variable point on a fixed curve S, and if P^{'} is a point on OP such that \frac{OP^{'}}{OP} = k, a constant, then the locus of P^{'} is a curve S^{'}, which is said to be homothetic to S; and P, P^{'} are corresponding points.

O is called the centre of similitude of the two figures.

If P and P^{'} lie on the same side of O, the figures are said to be directly homothetic w.r.t.O, and O is called the external centre of similitude.

If P and P^{'} lie on the opposite sides of O, the figures are said to be inversely homothetic w.r.t. O, and O is called the internal centre of similitude.

If we join A and B in the first case, we say that the parallel lines AB, A^{'}B^{'} are drawn in the same sense, and in the second case, in opposite senses.

Theorem 2:

Let A, B be the centres of any two circles of radii a, b; AB is divided externally at O and internally at O_{1} in the ratio of the radii, that is, \frac{AO}{BO} = \frac{AO_{1}}{O_{1}B} = \frac{a}{b} then the circles are directly homothetic w.r.t. O and inversely homothetic w.r.t. O_{2}, and corresponding points lie on the extremities of parallel radii.

Now, prove the above two hard core basic geometry facts. ! 🙂 🙂 🙂 

Cheers,

Nalin Pithwa.