## Saturday, February 20, 2021

### 2021/010) Simplify $\cot 70^\circ +4\cos70^\circ=$

$\cot 70^\circ +4\cos70^\circ=$

$=\frac{\cos 70^\circ}{\sin 70^\circ} +4\cos70^\circ=$

$=\frac{\cos 70^\circ + 4\sin 70^\circ \cos 70^\circ}{\sin 70^\circ }$

$=\frac{\cos 70^\circ +2(2\sin 70^\circ \cos 70^\circ)}{\sin 70^\circ }$

$=\frac{\cos 70^\circ +2\sin 140^\circ}{\sin 70^\circ }$

$=\frac{\sin 20^\circ +2\sin 40^\circ}{\sin 70^\circ }$

$=\frac{(\sin 20^\circ +\sin 40^\circ) + \sin 40^\circ}{\sin 70^\circ }$

$=\frac{(2 \sin 30^\circ \cos 10^\circ) + \sin 40^\circ}{\sin 70^\circ }$ using sin A + sin B formula

$=\frac{\cos 10^\circ + \sin 40^\circ}{\sin 70^\circ }$

$=\frac{\sin 80^\circ + \sin 40^\circ}{\sin 70^\circ }$

$=\frac{2 * \sin 60^\circ \cos 20^\circ}{\sin 70^\circ }$

$=\frac{\sqrt{3} \sin 70^\circ}{\sin 70^\circ }$

$=\sqrt{3}$

## Thursday, February 18, 2021

### 2021/009) Solve for integers $(1+\frac{1}{a})(1+\frac{1}{b})(1+\frac{1}{c}) = 2$

Because of symmetry we have if $a,b,c$ is a solution then any permutation is a solution

Without loss of generality we choose $a \le b \le c$

So $(1+\frac{1}{a})(1+\frac{1}{b})(1+\frac{1}{c}) \le (1+\frac{1}{a})(1+\frac{1}{a})(1+\frac{1}{a})$

Or  $(1+\frac{1}{a})(1+\frac{1}{b})(1+\frac{1}{c}) \le (1+\frac{1}{a})^3$

So from given condition $(1+\frac{1}{a})^3 \ge 2$ or $(\frac{a+1}{a})^2 \ge 2$

The LHS is decreasing function of a and we have to see ratio of 2successsive cubes >2 as equal to 2 is not possibe

we have $\frac{64}{27} > 2$ and  $\frac{125}{64} <2$ giving $a < 4$

So we can have upto 3 values(as for some value we may not have a solution) for a . that is 1,2,3

putting a = 1 we get  $2(1+\frac{1}{b})(1+\frac{1}{c}) = 2$

or  $(1+\frac{1}{b})(1+\frac{1}{c})= 1$

as for b and c $\ge 1$ LHS is less than 1 this does not have a solution

putting a = 2 we get  $\frac{3}{2} (1+\frac{1}{b})(1+\frac{1}{c}) = 2$

or  $(1+\frac{1}{b})(1+\frac{1}{c}) = \frac{4}{3}$

or $3(b+1)(c+1) = 4bc$

or $3bc + 3b + 3c + 3 = 4bc$

or $bc -3b -3c -3 = 0$

or $b(c-3) -3 (c-3) = 12$

or $(b-3)(c-3) = 12 = 1 * 12 = 2 * 6 = 3 * 4$

by taking one set a a time(as we have $b \le c$  we get (b,c) = (4,15) one solution (5,9) another solution and (6,7) 3rd solution

putting a = 2 we get  $\frac{4}{3} (1+\frac{1}{b})(1+\frac{1}{c}) = 2$

or  $4 (1+\frac{1}{b})(1+\frac{1}{c}) = 2 * 3$

or $2 (1+\frac{1}{b})(1+\frac{1}{c}) = 3$

or $2(b+1)(c+1) = 3bc$

or $2bc + 2b + 2c + 2 = 3bc$

or $(bc-2b -2c = 2$

or $(b-2)(c-2) = 6 = 1 * 6 = 2 * 3$

giving 2 sets of solutions (b,c) as (3,8) or (4,5)

So we have full solution in (a,b,c) as (2,4,15), (2,5,9),(2,6,7), (3,3,8), (3,4,5) or any permutation of any of the solutions

## Saturday, February 13, 2021

### 2021/008) Given $\frac{x}{y+z} + \frac{y}{z+ x} + \frac{z}{x+y} = 1$ Find $\frac{x^2}{y+z} + \frac{y^2}{z+ x} + \frac{z^2}{x+y}$

Because 1st term of the LHS in the expected result has $x^2$ we need to multiply 1st term of the given expression by y , 2nd term by y and 3rd tert by z so we can multiply by x+y + z to get

$\frac{x*(x+y+z)}{y+z} + \frac{y*(x+y+z)}{z+ x} + \frac{z*(x+y+z)}{x+y} = 1$

Or

$\frac{x^2}{y+z} + x + \frac{y^2}{z+ x} + y + \frac{z^2}{x+y} + z = x+ y + z$

Or

$\frac{x^2}{y+z} + \frac{y^2}{z+ x} + \frac{z^2}{x+y} + x + y + z = x+ y + z$

Or

$\frac{x^2}{y+z} + \frac{y^2}{z+ x} + \frac{z^2}{x+y} = 0$

## Sunday, February 7, 2021

### 2021/007) Three real numbers are given, Fractional part of product of every 2 of them is $\frac{1}{2}$. Prove that these numbers are irrational.

Let the three numbers be a,b,c as product of every 2 numbers has fractional part $\frac{1}{2}$ so each of the product is an integer plus $\frac{1}{2}$

So there exists integers p,q r such that $ab = \frac{2p+1}{2}\cdots(1)$

$bc = \frac{2q+1}{2}\cdots(2)$

$ca = \frac{2r+1}{2}\cdots(3)$

multiplying all 3 we get $(abc)^2 = \frac{(2p+1)(2q+1)(2r+1)}{8}$

or $(abc) = \frac{\sqrt{(2p+1)(2q+1)(2r+1)}}{2\sqrt{2}}$

Now numerator is square root of odd number and denominator is product of 2 and$\sqrt{2}$ so the number is irrational and as product of all 3 is irrational and product of 2 number is rational so 3rd number is irrational.

this is true for each pair and hence all 3 numbers are irrational. Hence proved

## Saturday, January 30, 2021

### 2021/006) Let a,b,c be real with $a,b,c > 1$ and $\frac{1}{a^2-1} + \frac{1}{b^2-1} + \frac{1}{c^2-1} = 1$ Show that $\frac{1}{a+1} + \frac{1}{b+1} + \frac{1}{c+1} <= 1$

We have $\frac{1}{a^2-1} + \frac{1}{b^2-1} + \frac{1}{c^2-1} = 1\cdots(1)$

Now

$\frac{2}{a^2-1} = \frac{1}{a+1} + \frac{1}{a-1}$

As $a > 1$ so we have $a-1 > 0$ and hence

$\frac{1}{a+1} < \frac{1}{a-1}$

adding  $\frac{1}{a+1}$ on both sides

$\frac{2}{a+1} < \frac{1}{a-1} + \frac{1}{a+1}$

Or $\frac{2}{a+1} > \frac{2}{a^2-1}$

Or $\frac{1}{a+1} < \frac{1}{a^2-1}\cdots(2)$

Similarly

Or $\frac{1}{b+1} < \frac{1}{a^2-1}\cdots(3)$

Or $\frac{1}{c+1} < \frac{1}{a^2-1}\cdots(4)$

$\frac{1}{a+1} + \frac{1}{b+1} + \frac{1}{c+1} < \frac{1}{a^2-1} + \frac{1}{b^2-1} + \frac{1}{c^2-1}$

or $\frac{1}{a+1} + \frac{1}{b+1} + \frac{1}{c+1} < 1$

## Tuesday, January 26, 2021

### 2021/005) Solve $|z - i | = |z + i |$

Method 1

Let $z = x + iy$
Then we have $|x + iy - i | = |x + iy + i |$
Or $|x + i(y -1) i | = |x + i(y + 1) |$
Knowing that $|x + iy| = \sqrt{x^2+y^2}$ we get
$\sqrt{x^2 + (y-1)^2} = x^2 + (y+1)^2$
Or $x^2 + (y-1)^2 = x^2 + (y+1)^2$
Or $(y-1)^2 = (y+1)^2$
or $y^2 - 2y + 1 = y^2 + 2y +1$ or $4y=0$ or $y = 0$
so the solution set is x + 0i

Another method that is Method 2

We have z equidistant from the points (0,1), (0, -1) in the complex plane
So the point z lies on the perpendicular bisector of the segment joining (0,1) and (0,-1) .
The midpoint of the segment is (0,0) and as this is the y axis the line that is perpendicular
to this is the x axis or any point (x,0) equivalent to the number x+0i is the solution

## Sunday, January 24, 2021

### 2021/004) Given $(\sqrt{x+\sqrt{x^2+1}})(\sqrt{y+\sqrt{y^2+1}}) =1$ find $(x+y)^2$

Let $x=\tan\theta$ $\theta$ in the range $0 ..\frac{\pi}{2}$ so we have $sec\theta$ positive

Then we have $\sqrt{x+\sqrt{x^2+1}} = \sqrt{\tan\theta + \sec\theta}$

From the given condition we have

$\sqrt{y+\sqrt{y^2+1}} = \frac{1}{\sqrt{x+\sqrt{x^2+1}}}$

Or

$\sqrt{y+\sqrt{y^2+1}} = \frac{1}{\tan \theta + \sec\theta}$

$=\frac{\sec^\theta - tan^2\theta }{\tan \theta + \sec\theta}$

or $y+\sqrt{y^2+1} =\sec\theta - \tan \theta\cdots(1)$

if we take $y=\tan\alpha$ $\alpha$ in the range $-\frac{\pi}{2} ..\frac{\pi}{2}$ so we have $sec,\theta$ positive

then so $y + \sqrt{y^2+1}= \sec\alpha \pm \tan \alpha\cdots(2)$

comparing (1) and (2) we have $y =-\tan \theta$

so $x+y=0$ or $(x+y)^2 = 0$