We want to show that :
Part I: TPT:
Proof of Part I: let
Then, 
Hence, 
That is, 
Hence, 
Hence, 
Hence, 
Hence, 
Therefore, 
Consider 
Therefore, 
Therefore, 
Therefore, 
Therefore, 
Hence, 
That is, 
From I and II, 
Part II: TPT: 
Quite simply, reversing the above steps we prove part II.
QED.
Cheers,
Nalin Pithwa
Reference: A Beautiful Mind by Sylvia Nasar.
Comment: This is approach is quite similar to what Prof. Joseph Silverman explains in his text, “A Friendly Introduction to Number Theory.”
Peter Sarnak, a brash thirty-five-year-old number theorist whose primary interest is the Riemann Hypothesis, joined the Princeton faculty in the fall of 1990. He had just given a seminar. The tall, thin, white-haired man who had been sitting in the back asked for a copy of Sarnak’s paper after the crowd had dispersed.
Sarnak, who had been a student of Paul Cohen’s at Stanford, knew Nash by reputation as well as by sight, naturally. Having been told many times Nash was completely mad, he wanted to be kind. He promised to send Nash the paper. A few days later, at tea-time, Nash approached him again. He had a few questions, he said, avoiding looking Sarnak in the face. At first, Sarnak just listened politely. But within a few minutes, Sarnak found himself having to concentrate quite hard. Later, as he turned the conversation over in his mind, he felt rather astonished. Nash had spotted a real problem in one of Sarnak’s arguments. What’s more, he also suggested a way around it. “The way he views things is very different from other people,” Sarnak said later. ‘He comes up with instant insights I don’t know I would ever get to. Very, very outstanding insights. Very unusual insights.”
They talked from time to time. After each conversation, Nash would disappear for a few days and then return with a sheaf of computer printouts. Nash was obviously very, very good with the computer. He would think up some miniature problem, usually very ingeniously, and then play with it. If something worked on a small scale, in his head, Sarnak realized, Nash would go to the computer to try to find out if it was “also true the next few hundred thousand times.”
{What really bowled Sarnak over, though, was that Nash seemed perfectly rational, a far cry from the supposedly demented man he had heard other mathematicians describe. Sarnak was more than a little outraged. Here was this giant and he had been all but forgotten by the mathematics profession. And the justification for the neglect was obviously no longer valid, if it had ever been.}
Cheers,
Nalin Pithwa
PS: For RMO and INMO (of Homi Bhabha Science Foundation/TIFR), it helps a lot to use the following: (it can be used with the above mentioned text of Joseph Silverman also): TI nSpire CAS CX graphing calculator.
or 
according as A and B are on the same side or on opposite sides of CD.
cm.More later,
Cheers,
Nalin Pithwa.
Hard core definitions of various special quadrilaterals:
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, 
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.
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
What is the next number in sequence?
A) 15, 20, 20, 6, 6, 19, 19, 5, 14, 20, 5, ?
Solution to A:
Ans is 20. The sequence is the position in the letter of the alphabet of the first letter in the numbers 1 to 12, when given in full. e.g. ONE: O=15.
B) 1, 8, 11, 18, 80, ?
Ans is 81. The sequence comprises whole numbers beginning with a vowel.
C) 1, 2, 4, 14, 21, 22, 24, 31, ?
Ans is 32, The sequence comprises whole numbers containing the letter O.
D) 4, 1, 3, 1, 2, 4, 3, ?
Ans. is 2. The sequence is as follows: there is one number between the two I’s, two numbers between the two 2’s, three numbers between the two 3’s and four numbers between the two 4’s.
E) 1, 2, 4, 7, 28, 33, 198, ?
Ans is 205.
F) 17, 8, 16, 23, 28, 38, 49, 62, ?
Answer is 70. Sum of digits in all previous numbers in the sequence.
G) 27, 216, 279, 300, ?
Ans is 307. Difference divided by 3 and added to the last number.
H) 9,7,17,79,545, ?
Answer is 4895. Each number is multiplied by its rank in the sequence, and the next number is subtracted.
I) 2,3,10,12,13, 20,?
Answer is 21. They all begin with the letter T.
J) 34, 58, 56, 60, 42, ?
Answer is 52. The numbers are the totals of the letters in the words ONE, TWO, THREE, FOUR, FIVE, SIX when A=1, B=2, C=3, etc.
Regards,
Nalin Pithwa.
Age old questions:
Dave is younger than Fred and older than George.
Alan is younger than Ian and older than Colin.
Ian is younger than George and older than John.
John is younger than Colin and older than Edward.
Fred is younger than Barry and older than Harry.
Harry is older than Dave.
Who is the youngest?
Answer: You have to try to create an “ascending” or “a descending” sequence and try to “fill in the gaps” :— answer is Edward.
Regards,
Nalin Pithwa.
Some blogs away I had posted several interesting, non-trivial, yet do-able-with-some-effort problems for preRMO and RMO.
A student of mine, RI has submitted the following beautiful solution to the chess problem. I am reproducing the question for convenience of the readers:
Question:
The sixty four squares of a chess board are filled with positive integers one on each in such a way that each integer is the average of the integers on the neighbouring squares. (Two squares are neighbours if they share a common edge on vertex. Thus, a square can have 8, 5 or 3 neighbours depending on its position.) Show that all the sixty four entries are in fact equal.
Answer (by RI):
Let us denote the set of all integers on the chess board by S (assume they are distinct). [Now, we can use the Well-ordering principle: every non-empty set of non-negative integers contains a least element. That is, every non-empty set S of non-negative integers contains an element a in S such that 
QED.
Three cheers for RI 🙂 🙂 🙂
Regards,
Nalin Pithwa
