**Problem:**

Find the value of .

**Problem:**

Find the value of

**Problem:**

Find the value of

**Problem:**

Prove that , if n is even, and is zero, if n is odd.

More later,

Nalin Pithwa

Mathematics demystified

October 17, 2016 – 5:18 am

**Problem:**

Find the value of .

**Problem:**

Find the value of

**Problem:**

Find the value of

**Problem:**

Prove that , if n is even, and is zero, if n is odd.

More later,

Nalin Pithwa

June 9, 2016 – 10:31 pm

**Problem 1:**

Given , , , prove that the triangle is ambiguous and find the other angles, using five figure tables (log/trig).

*Kindly send your answers, comments, etc.*

Nalin Pithwa

PS: Can you check/prove/verify the ambiguous case of solutions of triangles using plane geometry of your high school days?

June 8, 2016 – 6:13 pm

The three sides a, b, c and the three angles A, B, C are called the elements of the triangle ABC. When any three of these six elements (except all the three angles) of a triangle are given, the triangle is known completely; that is, the other three elements can be expressed in terms of the given elements and can be evaluated. This process is called the solution of triangles.

- If the three sides a, b, and c are given, angle A is obtained from or . B and C can be obtained in a similar way.
- If two sides b and c and the included angle A are given, then gives . Also, , so that B and C can be evaluated. The third side is given by , or, .
- If two sides b and c and the angle B (opposite to side b) are given, then . And, , and give the remaining elements.

By applying the cosine rule, we have:

, or if we manipulate this, we get

or,

This equation leads to the following cases:

**Case I:**

If , no such triangle is possible.

**Case II:**

Let . There are further following two cases:

**Sub-case II a:**

B is an obtuse angle, that is, is negative. There exists no such triangle.

**Sub-case II b:**

B is an acute angle, that is, is positive. There exists only one such triangle.

**Case III:**

Let . There are following two cases further here also:

**Sub-case IIIa:**

B is an acute angle, that is, is positive. In this case, two values of a will exist if and only if or, , which means two such triangles are possible. If , only one such triangle is possible.

**Sub-case IIIb:**

B is an obtuse angle, that is, is negative. In this case, triangle will exist if and only if . So, in this case, only one such triangle is possible. If , there exists no such triangle.

**Note:**

If one side a and angles B and C are given, then , and and .

If the three angles A, B and C are given, we can only find the ratios of the three sides a, b, and c by using the sine rule(since there are infinite number of similar triangles possible).

More theory later,

Nalin Pithwa

May 1, 2016 – 7:47 pm

**Problem: (China 2003)**

Let n be a fixed positive integer. Determine the smallest positive real number such that for any , , , , in the interval , if

, then

.

**Solution:**

**(by Yumin Huang)**

The answer is:

, if

, if

, if

The case is trivial.

If ,we claim that with equality if and only if . It suffices to show that

or,

Because , we get

.

By setting,

, the last inequality becomes

, or

Squaring both sides and clearing denominators, we get , that is,

. This establishes our claim.

Now, assume that . We claim that . Note that ; by setting and letting , we find that , and so the left hand side of the desired inequality approaches . It suffices to show that

.

Without loss of generality, assume that . Then,

It suffices to show that **relation ***

But, because ,

Consequently, by the **arithmetic geometric mean inequality, **

Because

we have

or $latex $

April 9, 2016 – 10:22 am

**Ref: Titu Andreescu and Zumin Feng.**

**Problem:**

Let x, y, z be positive real numbers such that . Determine the minimum value of

**Solution:**

An application of **Cauchy**–**Schwarz inequality **makes this as a one step problem. Nevertheless, we present a proof which involves only the easier inequality for real numbers x and y by setting first and and second and .

Clearly, z is a real number in the interval . Hence, there is an angle a such that . Then, , or . For an angle b, we have . Hence, we can set , and for some b, it suffices to find the minimum value of

or

Expanding the right hand side gives

Equality holds when and , which implies that and . Because , equality holds when and , that is, , , .

More later.

Nalin Pithwa

January 2, 2016 – 6:48 pm

**Exercise XXVII. Problem #32. The sides of a triangle are in AP and the greatest and least angles are ** **and** . **Prove that**

**Proof:**

Let a, b, c be in AP. Hence, , which in turn implies c is greatest and a is the least.

Hence, and .

**Want: **

**Given: **

. Hence,

Now, we have and this is equal to the following:

.

Also, similarly, we have the following:

but it is given that , hence, , . So the above expression changes to

.

**QED.**

December 30, 2015 – 4:26 pm

**Exercise XXVII. Problem 30.**

If a, b, c are in AP, prove that , , are in AP.

**Proof:**

Given that

TPT: **. —— Equation 1**

Let us try to utilize the following formulae:

which implies the following:

and

Our strategy will be reduce LHS and RHS of Equation I to a common expression/value.

which is equal to

which is equal to

which is equal to

which in turn equals

From the above, consider only the expression, given below. We will see what it simplifies to:

—- **Equation II.**

Now, consider RHS of Equation I. Let us see if it also boils down to the above expression after simplification.

From equation II and above, what we want is given below:

that is, want to prove that

but, it is given that and hence, , which means and

that is, want to prove that

i.e., want:

i.e., want:

i.e., want:

Now, in the above,

.

Hence, .

**QED.**

October 22, 2015 – 8:58 pm

Let us a discuss yet one more complex number based IITJEE mains problem.

**Problem:**

Let and be nth roots of unity, which subtend a right angle at the origin. Then, integer n must be of the form __________________. (fill in the blank).

**Solution:**

nth roots of unity are given by

where .

Let

Let where ,

As the join of and subtends a right angle at the origin, we deduce that is purely imaginary.

, for some real k

. Thus, n must be of the form 4k.

More later,

Nalin Pithwa

July 28, 2015 – 11:23 pm

**Question:**

If , then

equals

(a) -1

(b) 1

(c) -i

(d) i

**Solution.**

Using De Moivre’s theorem,

which in turn equals

Hence, .

More complex stuff to be continued in next blog (pun intended) 🙂

Nalin Pithwa

July 23, 2015 – 11:06 am

**Problem.**

Let ABC be a triangle and be the altitude through A. Prove that

.

(As usual, a, b, c denote the sides BC, CA, AB respectively.)

**Proof.**

The given inequality is equivalent to .

where is the area of triangle ABC. Using the identity

, we see that the inequality to be proved is

(here we use ), which is true. Observe that equality holds iff . QED.

More later,

Nalin Pithwa