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

Adding above 3 we get

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

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 $(x+y)^2 = 0$