Tag Archives: RMO Maths

Solutions of Triangles — a tricky IITJEE problem

December 11, 2014 – 12:55 pm

Question (IITJEE 1978). Suppose p_{1},p_{2},p_{3} are the altitudes through vertices A, B, C of a triangle ABC with area \Delta.

Prove that: 

\frac {1}{p_{1}} + \frac {1}{p_{2}} - \frac {1}{p_{3}}=\frac {2ab}{(a+b+c) \Delta} \cos^{2} (C/2)

Proof:

The RHS looks daunting. But, if we bring the factor \Delta in its denominator to the LHS, then the problem unfolds itself. Since, \Delta =(1/2)ap_{1}=(1/2)bp_{2}=(1/2)cp_{3}, the problem is equivalent to the following:

prove: a+b-c=\frac {4ab}{a+b+c}\cos^{2}(C/2)

If we write 2\cos^{2}(C/2) as 1+\cos C, and multiply both the sides of by a+b+c, then the problem is reduced to  proving that

(a+b)^{2}-c^{2}=2ab+2ab \cos C.

which is the same as the cosine formula 🙂 🙂 🙂

More later…

Nalin Pithwa

By Nalin Pithwa | Posted in IITJEE Foundation Math IITJEE Main and Advanced Math and RMO/INMO of (TIFR and Homibhabha) | Also tagged , , , , | Comments (0)

Congruency Tests of two triangles — for PreRMO, RMO, IITJEE Foundation Math

November 22, 2014 – 9:32 pm

Uptil now, we have discussed quite a bit of algebra, number theory and calculus for RMO, IITJEE Main and Advanced Mathematics. Let us switch back to plane geometry for a while.

In school, we all learn about tests of congruency of two triangles. But, have you wondered what is the exact meaning of “two plane figures are congruent”? Try to Google and find the answer for yourself. You can also post your answers as comments to this blog article.

The reasons I discuss these are two fold: (1) do you know the proofs of these congruency tests? In schools, students are just taught the meaning of these and how to apply them. Well, I will give you proofs of these in my blog at some later time. (2) There are some nuances to  these tests; when they cannot be applied, or rather, which *tests* fail to apply.

Here we go…

Two triangles are equal in all respects when the following three parts in each are severally equal:

1)Two sides, and the included angle. (the SAS Test for Congruency of 2 triangles).

2) The three sides. (the SSS Test for congruency of two triangles.

3) Two angles and one side, the side given in one triangle CORRESPONDING to that given in the other (ASA and AAS Test for Congruency of two triangles).

The two triangles are not, however, necessarily equal in all respects, when any three parts of one are equal to the corresponding parts of the other.

For example:

1) When the three angles of one are are equal to  the three angles of the  other, each to each, the fig 1 (attached, please download) shows  that the triangles need not be equal in all respects. (If you recall, such triangles are called similar, not  congruent because congruent means exactly same; two twins are similar, but not same).

2) When two sides and one angle in one are equal to two sides and one angle of the other, the given angles being opposite to equal sides, the  fig 2 (attached, please download) shows that the triangles need not be equal in all respects.

For, if $AB=DE$ and $AC=DF$ and the $\angle ABC = \angle DEF$, it will be seen that the shorter of the given sides in the triangle DEF may lie in either of  the positions DF or DF’.

Important Note: From these data, it may be shown that the angle opposite to the equal sides AB, DE are either equal (as for instance, the $\angle ACB$ and

$\angle DF’E$) or supplementary as (the $\angle ACB and \angle DFE$); and, that in the former case the triangles are equal in all respects. This is called the ambiguous case in the congruence of triangles. 

Note: ambiguous means that which has some uncertainty. In other words, there could be more than one possibilities.

If the given angles at B and E are right angles, the ambiguity disappears.

In  the next few blog articles, we will discuss the proofs of congruency tests of two triangles.

More later,

Nalin PIthwablogfigcongruencetests

 

By Nalin Pithwa | Posted in IITJEE Foundation Math IITJEE Main and Advanced Math and RMO/INMO of (TIFR and Homibhabha) | Also tagged , , , , , , | Comments (0)

Reciprocal equation for IITJEE and RMO/INMO

September 24, 2014 – 2:27 pm

In this set of little exercises, you will get a grip on reciprocal equations. 

A reciprocal polynomial has the form

ax^{n}+bx^{n-1}+cx^{n-2}+...+cx^{2}+bx+a

in which a \neq 0 and the coefficients are symmetric about the middle one. A reciprocal equation is of the form p(t)=0 with p(t) a reciprocal polynomial.

1(a) Verify that each of the following polynomials is a reciprocal polynomial:

x^{3}+4x^{2}+4x+1

3x^{6}-7x^{5}+5x^{4}+2x^{3}+5x^{2}-7x+3

1(b) Show that 0 is not a zero of any reciprocal polynomial.

1(c) Show that -1 is a zero of any reciprocal polynomial of odd degree, and deduce that any reciprocal polynomial of odd degree can be written in the form (x+1)q(x), with q(x) a reciprocal polynomial of even degree.

1(d) Show that, if r is a root of a reciprocal equation, then so also is 1/r.

2(a) Let ax^{2k}+bx^{2k-1}+...+rx^{k}+...+bx+a be a reciprocal equation of even degree 2k. Show that this equation can be rewritten as

a(x^{k}+x^{-k})+b(x^{k-1}+x^{-k+1})+...+r=0

2(b) Let t=x+x^{-1}. Verify that x^{2}+x^{-2}=t^{2}-2 and that x^{3}+x^{-3}=t^{3}-3t. Prove that, in general, x^{m}+x^{-m} is a polynomial of degree m in t.

2(c) Use the substitution in 2b to show that the reciprocal equation in 2a can be rewritten as an equation of degree k in the variable t. Deduce that the solution of a reciprocal equation of degree 2k can in general be reduced to solving one polynomial equation of degree k as well as at most k quadratic equations.

3(a) Show that a product of reciprocal polynomials is a reciprocal polynomial.

3(b) Show that, if f, g,  h are polynomials with f=gh and f and h are both reciprocal polynomials, then g is also a reciprocal polynomial.

Please do post your comments, solutions…you can take a snapshot in your smartphone, and just post it 🙂

More later…

Nalin Pithwa

 

By Nalin Pithwa | Posted in IITJEE Foundation Math IITJEE Main and Advanced Math and RMO/INMO of (TIFR and Homibhabha) | Also tagged , , , , , | Comments (0)

Pythagorean Triples

September 12, 2014 – 5:51 pm

The Pythagorean Theorem, that “beloved” formula of all high school geometry students, says that the sum of the squares of the sides of a right triangle equals the square of the hypotenuse. In symbols,

a^{2}+b^{2}=c^{2} where a and b are the sides and c is the hypotenuse.

Since we are interested in number theory, that is, the theory of the natural numbers, we will ask whether there are any Pythagorean triangles all of whose sides are natural numbers. There are many such triangles. The most famous has side 3,4 and 5. Here are the first few examples:

3^{2}+4^{2}=5^{2}

5^{2}+12^{2}=13^{2}

8^{2}+15^{2}=17^{2}

28^{2}+45^{2}=53^{2}

Our first naive question is whether there are infinitely many Pythagorean triples, that is, triples of natural numbers (a,b,c) satisfying the equation a^{2}+b^{2}=c^{2}

The answer is “YES” for a silly reason. If we take a Pythagorean triple (a,b,c) and multiply it by some other number d, then we obtain a new Pythagorean triple

(ad,bd,cd). This is true because

(da)^{2}+(db)^{2}=d^{2}(a^{2}+b^{2})=d^{2}c^{2}=(dc)^{2}

Clearly, these new Pythagorean triples are not very interesting. So we will concentrate our attention on triples with no common factors. We will even give them a name:

A primitive Pythagorean triple (or PPT for short) is a triple of numbers (a,b,c) so  that a,b, and c have no common factors and satisfy

a^{2}+b^{2}=c^{2}.

To investigate whether are infinitely many PPT’s is the same as asking whether there is a formula to find as many PPT’s as we want — the formula would contain relationship(s) between a, b and c. 

As explained in the previous blog, the first step is to accumulate some data. I used a computer to substitute in values for a and b and checked if a^{2}+b^{2} is a square. Here are some PPT’s that I found:

(3,4,5); (5,12,13); (8,15,17); (7,24,25);

(20,21,29); (9,40,41); (12,35,37); (11,60,6);

(28,45,53); (33,56,65); (16,63,65.

A few conclusions can be easily drawn even from such a short list. For example, it certainly looks like one of a and b is odd and the other is even. It also seems that c is always odd.

It is not hard to prove that these conjectures are correct. First, if a and b are both even, then c would also be even. This  means that a,b and c would have a common factor of 2, so the triple would not be primitive. Next, suppose that a and b are both odd, which means that c would have to be even. This means that there are numbers x,y and z so  that

a=2x+1 and b=2y+1 and c=2z

We can substitute this in the equation a^{2}+b^{2}=c^{2} to get

(2x+1)^{2}+(2y+1)^{2}=(2z)^{2}

Hence, 2x^{2}+2x+2y^{2}+2y+1=2z^{2}

This last equation says that an odd number is equal to an even number, which is impossible, so a and b cannot both be odd. Since, we have just checked that they cannot both be even and cannot both be odd, it must be true that one is even and the other is odd. It’s then obvious from the equation a^{2}+b^{2}=c^{2} that c is also odd. 

We can always switch a and b, so our problem now is to find all solutions in natural numbers to the equation

a^{2}+b^{2}=c^{2} with a odd, b even and a,b,c, having no common factors.

The tools we will use are factorization and divisibility.

Our first observation is that if (a,b,c) is a primitive PPT, then we can factor

a^{2}=c^{2}-b^{2}=(c-b)(c+b)

Here are a few examples from the list given earlier, where note that we always take n to be odd and b to be even:

3^{2}=5^{2}-4^{2}=(5-4)(5+4)=1.9

15^{2}=17^{2}-8^{2}=(17-8)(17+8)=9.25

35^{2}=37^{2}-12^{2}=(37-12)(37+12)=25.49

33^{2}=65^{2}-56^{2}=(65-56)(65+56)=9.121

It looks like c-b and c+b are themselves always squares. We check this observation with a couple more examples:

21^{2}=29^{2}-20^{2}=(29-20)(29+20)=9.49

63^{2}=65^{2}-16^{2}=(65-16)(65+16)=49.81

How can we prove that c-b and c+b are squares? Another observation apparent from our list of examples is that c-b and c+b seem to have no common factors. We can prove this last assertion as follows. Suppose that d is a common factor of c-b and c+b, that is, d divides both c-b and c+b. Then, d also divides

(c+b)+(c-b)=2c and (c+b)-(c-b)=2b

Thus, d divides 2b and 2c. But, b and c have no common factor because we are assuming that (a,b,c) is a primitive Pythagorean triple. So, d must equal 1 or 2. But, d also divides (c-b)(c+b)=a^{2} and a is odd, so d must be 1. In other words, the only number dividing both c-b and

c+b is 1, so c-b and c+b have no common factor.

We now know that c-b and c+b have no common factor, and that their product is a square since (c-b)(c+b)=a^{2}. The only way that this can happen is if c-b and c+b are themselves squares. So, we can write

c+b=s^{2} and c-b=t^{2} where s>t \geq 1 are odd integers with no common factors. Solving these two equations for b and c yields

c=(s^{2}+t^{2})/2 and b=(s^{2}-t^{2})/2 and then

a=\sqrt{(c-b)(c+b)}=st

We have finished our first proof of elementary number theory! The following theorem records our accomplishment:

Theorem (Pythagorean Triple Theorem) You will get every primitive Pythagorean triple (a,b,c) with a odd and b even by using the formulas

a=st and b=(s^{2}-t^{2})/2 and c=(s^{2}+t^{2})/2 where

s>t \geq 1 are chosen to be any odd integers with no common factors.

More later…

Nalin Pithwa

By Nalin Pithwa | Posted in IITJEE Foundation Math IITJEE Main and Advanced Math and RMO/INMO of (TIFR and Homibhabha) | Also tagged , , , , , | Comments (3)

From scratch — the 3 medians of a triangle are concurrent

July 16, 2014 – 8:59 am

page1medianspage2mediansT

Please download the attachments, pages 1 and 2 respectively of the proof that the 3 medians of a triangle are concurrent. That’s called the power of axioms!! This is the example which mesmerized Albert Einstein as a child to Mathematics and later to  Physics.

More later…

Nalin

By Nalin Pithwa | Posted in IITJEE Foundation Math IITJEE Main and Advanced Math and RMO/INMO of (TIFR and Homibhabha) | Also tagged , , | Comments (1)

Math from scratch

June 27, 2014 – 4:22 am

Math from scratch!

Math differs from all other pure sciences in the sense that it can be developed from *scratch*. In math jargon, it is called “to develop from first principles”. You might see such questions in Calculus and also Physics.

This is called Axiomatic-Deductive Logic and was first seen in the works of Euclid’s Elements (Plane Geometry) about 2500 years ago. The ability to think from “first principles” can be developed in high-school with proper understanding and practise of Euclid’s Geometry.

For example, as a child Albert Einstein was captivated to see a proof from *scratch* in Euclid’s Geometry that “the three medians of a triangle are concurrent”. (This, of course, does not need anyone to *verify* by drawing thousands of triangles and their mediansJ) That hooked the child Albert Einstein to Math, and later on to Physics.

An axiom is a statement which means “self-evident truth”. We accept axioms at face-value. So, there are axioms, definitions, propositions, lemmas, theorems and corollaries.

Euclid’s geometry rests on the following fundamental axioms:

1)      There can be one and only one straight line joining two given points.

2)      (a) If O is a point in a straight line AB, then a line OC, which turns about O from the position OA to the  position OB must pass through one position, and only one,, in which it is perpendicular to AB.

(b) All right angles are equal.

3) (a) If a point O moves from A to B along the straight line AB, it must pass through one         position in which it divides AB into two equal parts.

3)(b) If a line  OP, revolving about O, turns from OA to OB, it must pass through one position in which it divides the angle AOB into two equal parts.

4) Magnitudes which can be made to coincide with one another are equal.

5) Playfair’s Axiom: Through a given point, there can be only one straight line parallel to a given straight line.

Note that these are the only basic assumptions to be used in geometric constructions also with ruler and compass.

PS: article reblogged and slightly modified:

Reference: To start geometry from scratch, you can start working from the first page of a classic text ” A School Geometry” by Hall and Stevens, Metric Edition; For example, Amazon India link is:

https://www.amazon.in/School-Geometry-H-S-Hall/dp/9385923331/ref=sr_1_4?crid=6QK90ZAHHFAU&keywords=a+school+geometry+hall+and+stevens&qid=1561783333&s=books&sprefix=A+School%2Caps%2C253&sr=1-4

Or Infibeam: https://www.infibeam.com/Books/school-geometry-part-1-6-pb-hall-h-s/9788183552806.html

I would like to add a few more details as there are some students/readers who want to pursue this further. {By the way, I have used the above reference only. One more thing,…Dover publications still prints/publishes/sells the original volumes of Euclids books}:

\textbf{Hypothetical Constructions}

From the above axioms, it follows that we may suppose:

i) A straight line can be drawn perpendicular to a given straight line from any point in it.

ii) A finite straight line (that is, a segment) can be bisected.

iii) Any angle can be bisected by a line (we call such a line its angle bisector).

\textbf{Superposition and Equality}

AXIOM: Magnitudes which can be made to coincide with one another are equal.

This axiom implies that any line, angle, or figure may be taken up from its position, and without change in size or form, laid down upon a second line, angle, or figure, for the purpose of comparison, and ti states that two such magnitudes are equal when one can be exactly placed over the other without overlapping.

This process is called superposition, and the first magnitude is said to be applied to the other. (Note: this is the essence of “congruency” relation in geometry).

\textbf{Postulates}

In order to draw geometric figures, certain instruments are required. These are — a straight ruler, and a pair of compasses. The following postulates (or requests) claim the use of these instruments, and assume that with their help the processes mentioned below may be duly performed:

Let it be granted:

  1. That a straight line may be drawn from any one point to any other point.
  2.  That a finite (or terminated) straight line may be produced (that is, prolonged) to any length in that straight line.
  3. That a circle may be drawn with any point as centre and with a radius of any length.
  4. some notes: Postulate 3 above implies that we may adjust the compasses to the length of any straight line PQ, and with a radius of this length draw a circle with any point O as centre. That is to say, the compasses may be used to transfer distances from one part of a diagram to another.
  5. Hence, from AB, (a given terminated line), the greater of two straight lines, we may cut off a part equal to PQ the less. Because, if with centre A, and radius equal to PQ, we draw an arc of a circle cutting AB at X, it is obvious that AX is equal to PQ.

More later,

Regards,

Nalin Pithwa.

By Nalin Pithwa | Posted in geometry, Pre-RMO, pure mathematics | Also tagged , , , , , , | Comments (0)