## Tag Archives: IITJEE maths

### Solutions of Triangles — a tricky IITJEE problem

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

### Implicit differentiation example — Helga von Koch’s snowflake curve (1904)

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Let us continue further our exploration of IITJEE Calculus. Especially, implicit differentiation.

Start with an equilateral triangle, calling it curve 1. On the middle third of each side, build an equilateral triangle pointing outward. Then, erase the interiors of the old middle thirds. Call the expanded curve curve 2. Now, put equilateral triangles, again pointing outward, on the middle thirds of  the sides of curve 2. Erase the interiors of the old middle thirds to  make curve 3. Repeat the process, as shown, to define an infinite sequence of plane curves. The limit curve of the sequence is Koch’s snowflake curve.

The snowflake curve is too rough to  have a tangent at any point. In other words, the equation $F(x,y)=0$ defining the curve does not define y as a differentiable function of x or x as a differentiable function of y at any point. We will encounter snowflake again when we study length.

More later…

Nalin Pithwa

### Reciprocal equation for IITJEE and RMO/INMO

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

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.

More later…

Nalin Pithwa

### Pythagorean Triples

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